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Plus Two Computer Science Chapter Wise Questions and Answers Chapter 5 Web Designing Using HTML

April 13, 2021February 1, 2020 by Prasanna

Students can Download Chapter 5 Web Designing Using HTML Questions and Answers, Plus Two Computer Science Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus Two Computer Science Chapter Wise Questions and Answers Chapter 5 Web Designing Using HTML

Plus Two Computer Science Web Designing Using HTML One Mark Questions and Answers

Question 1.
Name the tag to which the attribute frame is associated.
Answer:
<Table>
Frame attribute specifies the border lines around the table. Possible values are void, box, above, below, hsides, Hsides, Lhs, Rhs Eg. <Table Frame = “below”>

Question 2.
The tag used to create combo in HTML is _______
Answer:
<Select>

Question 3.
The option attribute is associated with _______ <tag>
Answer:
<Select>

Question 4.
Consider the following.
<FRAMESET Cols= “50%, *”>
What will be the output of the above HTML statement?
Answer:
It divides browser window with frames of equal width in column-wise.

Question 5.
Consider the following.
<FRAMESET Rows= “50%,*”>
What will be the output of the above HTML statement?
Answer:
It divides browser Window into two frames in row-wise equally.
Plus Two Computer Science Chapter Wise Questions and Answers Chapter 5 Web Designing Using HTML img2

Question 6.
<Select> Tag create ______ in HTML.
Answer:
Combo box

Question 7.
The default align value for a table is ______
(a) Left
(b) Right
(c) Center
(d) Justify
Answer:
(a) Left

Question 8.
A submit button can be created by____ tag
(a) <SUBMIT>
(b) <INPUT>
(c) <SELECT>
(d) <ACTION>
Answer:
(b) <INPUT>

Question 9.
_____ Tag enclosed the heading cells in a table.
(a) <TABLE>
(b) <TR>
(c) <TH>
(d) <TD>
Answer:
(c) <TH>

Question 10.
______ is an empty tag.
(a) <FRAME>
(b) <FORM>
(c) < FRAMESET
(d) <TABLE>
Answer:
(a) <FRAME>

Question 11.
Pick the odd one out. Justify your answer
(a) NOFRAME
(b) Body
(c) SRC
(d) HEAD
Answer:
(c) SRC. It is an attribute. All other are tags.

Question 12.
Name the tag which is used as an attribute to another tag?
Answer:
<Center>or<Frame>

Question 13.
The borderlines around a table is specified by _____ Attribute.
(a) Frame
(b) cellborder
(c) Background
(d) Border
Answer:
(d) Border

Question 14.
To change the background colour of a table, which attribute of <table>tag is used for this purpose.
Answer:
BGCOLOR – This attribute specifies the background colour of the table.
<TABLE B0RDER=2 BGCOLOR= “Blue”>

Question 15.
Which tag is used to divide the window more than one?
Answer:
<Frameset>

Question 16.
_____ tag is used to pass information from web viewers to web server.
Answer:
<Form>

Question 17.
_______ tag provides a label for the form control.
Answer:
<Label>

Question 18.
A_____ has no <body> section.
Answer:
<Frameset> tag

Question 19.
A <frameset> tag no _____ tag.
Answer:
<Body> tag

Question 20.
Name the attribute used to merge two or more rows of a table in an HTML document.
Answer:
Rowspan

Question 21.
In HTML ______ tag is used to create a row in a table
Answer:
<TR>

Question 22.
The space between the bonder of the cell and its content can be adjusted by an attribute of the < TABLE > tag. Identify this attribute.
Answer:
Cell padding

Question 23.
Give the value of the frame attribute of <TABLE> tag to get the outer border only.
Answer:
box.

Question 24.
Baiju wants to place a picture in a table cell. Which attribute of the <TD> tag will be used for this.
Answer:
Background

Question 25.
_____ tag forms the definition term in a definition
(a) <DD>
(b) <DT>
(c) <DL>
(d) <DR>
Answer:
(b) <DT>

Question 26.
Name the possible values of type attribute of UnOrdered list.
Answer:
tag <UL> can take values square, circle or disc

Question 27.
To create a list using Upper case letters use ______?
Answer:
<OLType=”A”>

Question 28.
To create a list using Lower case letters use ______?
Answer:
<OLType=”a”>

Question 29.
To start a list from the count of 3 using _____?
Answer:
a <OLstart=”3”>

Question 30.
Mr. Suresh wants to prepare a list of students with register number. But he wants to start numbering from 5? How can it be done using HTML?
Answer:
Ordered list is used for this <OL start = “5”>

Question 31.
Pick the wrong one from the statements given below:
(a) <OL>and <UL> have Type attribute
(b) Default numbering scheme in <OL> is 1,2, 3…
(c) In Definition List, <DD> tag is used to give definition of terms
(d) Start attribute, of ordered list, should always be set to 1
Answer:
(d) Start attribute, of ordered list, should always be set to 1

Question 32.
Which of the following is the correct way to create an email link?
(a) <A href= “abc@xyz”>
(b) <mail href= “abc@xyz”>
(c) <mail> “abc@xyz”>
(d) <A href= “mailto: abc@xyz”>
Answer:
(d) <A href= “mailto: abc@xyz”>

Question 33.
There are two web pages in the class project created by Mathew. The second page should appear in the browser when clicked at a particular text in the first page. What do you call this feature? Name the tag and attribute needed for creating such a feautre
Answer:
This feature is called link
Tag used is <A> and attribute is HREF

Question 34.
Observe the table with two rows. Which of the following is used with TD tag to merge the cells C and D?
Plus Two Computer Science Chapter Wise Questions and Answers Chapter 5 Web Designing Using HTML img3
(a) Merge=colspan 2
(b) Rowspan= “2”
(c) Colspan= “2″
(d) Merge=raw2
Answer:
(c) Colspan= “2″

Question 35.
Why do we use <NOFRAME> tag?
Answer:
<NOFRAME> tag is used to give content when some browser that does not support frameset.

Question 36.
Which of the following tag is used to create a list box in a html Form?
(a) <SUBMIT>
(b) <INPUT>
(c) <SELECT>
(d) <ACTION>
Answer:
(c) <SELECT>

Question 37.
The tag used for creating a drop down list in HTML is ______
Answer:
<select> tag

Plus Two Computer Science Web Designing Using HTML Two Mark Questions and Answers

Question 1.
Manju wants to display 3 web pages on the same screen horizontally. Which HTML statements can be used for this?
Answer:
<Frameset Rows=“33%, *, * ” >
<Frame SRC = Page1. html>
<Frame SRC = Page2. html>
<Frame SRC – Page3. html>
< / Frameset >

Question 2.
Adithya College of Engineering wants to Create its web site, in which the home page is to be designed as a combination of two Vertical panes.
Answer:

Suggest suitable tags used for this.
Write the HTML statements to get this type of page

Answer:
1. < Frame set > and < Frame > tags

2. < Frameset cols = “ 50%, * ” >
< Frame SRC = “ Page1. HTML” >
< Frame SRC = “ Page2. HTML ” >
< / Frameset >

Question 3.
‘Kerala Communication channel’ conducts a TV program based on Education policies of Kerala Govt. They want to take the feedback from the viewers through their website.

While creating this site, which type of tag is used to accept multiple line of text from the viewers,
Name any two attributes of this tag.

Answer:

<TEXTAREA>
Cols, rows, name

Question 4.
Distinguish between cell spacing and cell padding attribute of <Table> tag.
Answer:

Cellspacing: It specifies the space between two table cells
Cellpadding: It specifies the space between cell border and content

Question 5.
Match the following.
Plus Two Computer Science Chapter Wise Questions and Answers Chapter 5 Web Designing Using HTML img4
Answer:

Question 6.
Differentiate <frame> and <frameset> tags.
Answer:
The <frameset> tag defines the frame sections and the <frame> tag specifies the pages to be loaded in each frame
<FRAMESET> tag And <FRAME>tag Attributes of <FRAMESET>

Rows: Used to divide screen in row wise
Cols: Used to divide screen in column wise Attributes of <FRAME>
SRC: Specifies name of web page to be loaded in Frame
Scrolling: Enables the web page displayed to be scrolled

Eg: <FRAMESET Cols=“50%,*”>
< FRAME SRC=“page1.html”>
< FRAME SRC=“page2.html”>
</FRAMESET>

Question 7.
How can you merge cells in a table?
Answer:
By using attributes Colspan or Rowspan
1. colspan: It Is used to span or to stretch a cell, over a number of columns.
Eg: <TD Colspan=3> spans the cell over three columns

2. rowspan: It is used to span or to stretch a cell over a number of rows.
Eg: <TD Rowspan=3> spans the cell over three rows.

Question 8.
Raju created a web page as follows:
<HTML>
<HEAD>
<TITLE>SampieTable</TITLE>
</HEAD>
<BODY>
<TABLE><TR><TH>Roll.No.</TH><TH>Name
</TH><TR>
<TR><TD>1</TD><TD>Raju</TD></TR>
<TR><TD>2</TD><TD>Ramu</TD></TR>
</TABLE>
</BODY>
</HTML>
But he is unable to view any tabular format in the . web page, when it is displayed in the browser.
Find the reason for it and correct the same.
Answer:
Without the “Border” attribute it never shows tabular form, Border attribute is missing.
<TABLE BORDER=3>

Question 9.
Name the possible values of type attribute of Ordered list.
Answer:
Values as follows

type = 1 for 1, 2, 3,….
type = i for i, ii, iii ……..
type = I for I, II, III …….
type = a for a, b, c,…
type = A for A, B, C ………

Question 10.
Predict the output of the following HTML segment.
<OLType = “1” start = “5”>
<Li> Chocolate</Li>
<Li> Milk </Li>
<Li> Coffee</Li>
</OL>
Answer:
5. Chocolate
6. Milk
7. Coffee

Question 11.
Compare the use of Type attribute in Ordered and Unordered list in HTML?
Answer:
1. Unordered List (<UL>) – Items are displayed with square, circle or disc in front.
Eg: <UL TYPE=”circle”>

2. Ordered List (<OL>) – Items are displayed with the following type values.
Type = 1 for 1, 2, 3,
Type = i for i, ii, iii
Type = I for I, II, III
Type = a for a, b, c,
Type = A for A, B, C
Eg: <OLTYPE=”A”>

Question 12.
Differentiate internal linking and external linking with examples.
Answer:
1. External Linking -: Used to connect two different webpages
Eg:<A href = “School. html”> School</A>

2. Internal linking: Internal links are given to a section in the same document.
<A href =“#top”>Goto Top </A>
<A href = “#bottom>Goto Bottom </A>

Question 13.
While moving the mouse pointer over a web page, the mouse pointem changes its shape to hand icon symbol.

Give reason for this change in mouse pointer.
Name the tag and attributes used for it.

Answer:

It is a hyper link
<A> tag, href attribute

Question 14.
HTML has facility to provide external and internal hyper links.

Which tag is used to include a hyperlink?
Explain two attributes needed for creating internal

Answer:

<A>
name, href

Question 15.
Match the following.

EMBED	href
OL	loop
A	start
BGSOUND	hidden

Answer:
EMBED- hidden, OL – start, A – href, BGSOUND – loop

Question 16.
Categorize the following tags into container tags and empty tags.
<A>, <FRAME>, <FRAMESET>, <INPUT>
Answer:
1. Empty tags
<FRAME> & <INPUT>

2. Container tags
<FRAMESET> & <A>

Question 17.
The <FORM> tag is used to accept data and communicate with a server program.

Name any two attributes of FORM tag.
How will you create a “SUBMIT” button and a “RESET” button within the FORM tag?

Answer:

Action, Method
<INPUT Type = “submit”>
<INPUT Type = “reset”>

Question 18.
Aliya wants to display three webpages (A.htm, B.htm, C.htm) on the same screen horizontally at the ratio 20%, 40%, 40%. Write the HTML code for the same.
Answer:
<FRAMESET ROWS=”20%, 40%, 40%”>
<FRAME Src= “A.html”>
<FRAME Src= “B.html”>
<FRAME Src= “C.html”>
</FRAMESET >

Question 19.
Distinguish Cellspacing and Cellpadding attributes of <TABLE> tag.
Answer:
Cell spacing: Specifies space between table cells Cell padding- Specifies space between cell border and content.

Plus Two Computer Science Web Designing Using HTML Three Mark Questions and Answers

Question 1.
Point out the difference between relative and absolute URL.
Answer:
URL means uniform Resource Locator.
Two type of URL
1. Relative URL – Here we explicitly give the web site address
Eg: <A href=http://www.hscap.kerala.gov.in>

2. Absolute URL – Here we implicitly give the web site address. The path is not specified here.
Eg: Consider the web pages index.html and school.html saved in folder C:\BVM.The file indexs.html contains the following.
<A href=”school.html”>. Here we did not specify the full path of the file school.html. But this implicitly points to the file stored in C:\BVM

Question 2.
Name the tag which is used to play the music in background while the webpage is being viewed.
Answer:
<BGSOUND>
<HTML>
<HEAD>
<Title> Table </Title>
</HEAD>
<BODY>
<Table Border = “1”>
<TR align = “Center”>
<TH COLSPAN = “2” > ANIMALS</TH>
</TR>
<TR align-‘center”>
<TD> WILD <BR>BEAR, TIGER</TD> <TD>DOMESTIC <BR>GOAT, DOG</TD>
</TR>
</TABLE>
</BODY>
</HTML>

Question 3.
Differentiate <FRAME>, <FRAMESET> and <NOFRAME> tags.
Answer:

<FRAMESET> tag is used to divide the window more than one pane. It has no body section.
<FRAME> It specifies the pages within a frameset.
<NOFRAME> <NOFRAME> tag is used to give content when some browser that does not support frameset.

Question 4.
In a web page, user needs to enter the address of persons. Name the tag used for this. List and explain any two main attributes of it.
Answer:
<TEXTAREA>. This is used to enter multiple lines in a Text Box of a web page. Main attributes are

Rows: Specifies the height of text area control, ie. The number of Lines the Text Area should have
Cols: Specifies width ie number of characters per line.
Name: Gives a variable name to the Text Area control. Eg: <TEXTAREA Name = ”Address” Cols = 20 Rows = 5>

Question 5.
Consider the following table.
Plus Two Computer Science Chapter Wise Questions and Answers Chapter 5 Web Designing Using HTML img6
Write the HTML code for the above.
Answer:
<HTML>
<HEAD>
<TITLE>TABLE</TITLE>
</HEAD>
<BODY>
<CENTER><B>TABLE</CENTER>
<TABLE BORDER = 1>
<TR>
<TH>Batch</TH>
<TH>Boys</TH>
<TH>Girls</TH>
</TR>
<TR>
<TH>Science</th>
<TD>25</TD>
<TD>26</TD>
</TR>
<TR>
<TH>COMMERCE</TH>
<TD>20</TD>
<TD>30</TD>
</TR>
</TABLE>
</BODY>
</HTML>

Question 6.
Name the tag which permits a user to add more than one web page in a single window. List any attributes of it with explanation.
Answer:
<FRAMESET> tag and <FRAME>tag Attributes of <FRAMESET>

Rows: Used to divide screen in row wise
Cols: Used to divide screen in column wise Attributes of <FRAME>
SRC: Specifies name of web page to be loaded in Frame
NAME: Gives a name for the frame.

Eg. <FRAMESET Rows= “50%,*”>
<FRAME SRC= “page1.html”> <FRAMESRC=“page2.htm”>
</FRAMESET>

Question 7.
In VB, a Programmer can use Option Button, Text Box, Combo box, etc. to accept inputs from the user. But in HTML <INPUT> tag is used for creating ail the above controls. Which attributes of the <INPUT> tag is used for this. List and explain the possible values of it.
Answer:
The ‘type’ attribute of <INPUT> tag is used to create different control. The main values of type attribute is given below.

Text: Creates a single line Text Box. Eg. <INPUT Type= “Text”>
Password: creates a password box in which characters are displayed by symbols like asterisk(*) Eg. <INPUTType= “Password”>
Check Box: Creates a check box. Eg. INPUT Type = “Check Box”>
Radio: Creates option button (Radio Button) Eg. <INPUT Type= “Radio” Name=“sex” Value= “M”>Male
Reset: Creates reset button. It is used to clear all the data entered Eg. <INPUT Type=“Reset”>
Submit: Creates a submit Button. When click on it data Entered in the form will sent to web server. Eg. INPUT Type = “Submit”>

Question 8.
Write the HTML code for the following.
Plus Two Computer Science Chapter Wise Questions and Answers Chapter 5 Web Designing Using HTML img7
Answer:
<HTML>
<HEAD>
<TITLE>TABLE</TITLE>
</HEAD>
<BODY>
<TABLE BORDER = ”1”>
<TR>
<TH> Computer </TH>
<TH> Acer </TH>
<TH>HP</TH>
</TR>
<TR>
<TH>PC</TH>
<TD>30,000</TD>
<TD>32,000</TD>
</TR>
<TR>
<TH>Laptop</TH>
<TD colspan = 2>40,000</TD>
</TR>
</Table>
</Body>
</HTML>

Question 9.
Write the HTML Code for the following.
Plus Two Computer Science Chapter Wise Questions and Answers Chapter 5 Web Designing Using HTML img8
Answer:
<HTML>
<HEAD>
<TITLE>TABLE EXAMPLE</TITLE>
</HEAD>
<BODY>
<TABLE BORDER = ‘1’>
<TR>
<TH>Subject</TH>
<TH>Type of<BR>Exam</TH>
<TH>Mark</TH>
</TR>
<TR>
<TH Rowspan=2>Computer<BR>Application</TH> <TH>Theory</TH>
<TD>42</TD>
</TR>
<TR>
<TH>Practical</TH>
<TD>26</TD>
</TR>
</Table>
</Body>
</HTML>

Question 10.
Name the attributes in HTML, which can present in more than one tag.
Answer:
The attributes are

Border: lt can act in <Table> and in <Frameset>Tag.
Bgcolor: It can act in <Table> and in <Body>tag.
Type: It can act in <OL> and in <lnput>tag.

Question 11.
What are the main attributes of the <Form> tag?
Answer:
1. Method:
It determines the method of submission of form data to the server. Get and Post are the two form submission methods. Post method is used to pass large volume of data. Also, post method is more secure as data entered is not visible during submission. The get method is faster and is used to send lesser volume of data and it is not secure.

2. Action:
The URL of the server side program to process the form data is specified by the Action attribute.
Eg: <Form Action=”http://www.scert.com/asp/. process. asp”>

Question 12.
What are the difference between get method and post method?
Answer:

Post Method	Get Method
It is used to pass large volume of data The data is not visible during submission It is slower. It is secure.	It is used to pass lesser volume of date The data is visible, during submission It is faster It is not secure

Question 13.
Lena wants to create a web page, to select the district name from a combo. By default the combo box contain the district Trichur. Help her to do so.
Answer:
<HTML>
<HEAD>
<TITLE>COMBOBOX</TITLE>
</HEAD>
<BODY>
DISTRICT
<SELECT>
<OPTION SELECTED>TRICHUR <OPTION>ERNAKULAM <OPTION>PALAKKAD </SELECT>
</BODY>
</HTML>

Question 14.
Ram creates a web page to record the sex of a student. But he has made a mistake that the student can select both male and female choices at a time. What is the reason for the same? Help him to correct the mistake.
Answer:
To record sex of student radio buttons are used. Usu-ally radio buttons are provided as a group from which exactly one can be selected at a time by giving the same name for both radio buttons. But here Ram did not give same narfie for both buttons. Therefore the mistake. The correct code is as follows,
<HTML>
<HEAD>
<TITLE>RADIO BUTTON</TITLE>
</HEAD>
<BODY>
sex<input type=”radio” name=’’sex” value= “Male”> Male
<input type=”radio” name=”sex” value= “Female”> Female
</BODY>
</HTML>

Question 15.
Explain the attributes of <Frameset> Tag.
Answer:

Cols: It determines the dimension of vertical frames(Columns) in the frameset page. Eg: <Frameset Cols=”50%,*”> creates two vertical frames with equal width
Rows: It determines the dimension of horizontal frames(Rows) in the frameset page. Eg: <Frameset Rows=”50%,*”> creates two horizontal frames with equal height.
Border: It specifies the thickness of the border for the frames.
Border color: It specifies the colour for the frame border.

Question 16.
Explain the attributes of <Frame tag>
Answer:

Src: It specifie the URL of the document to be loaded in the frame.
Scrolling: It indicates scroll bar is to be shown in the frame or not, values are yes, no or auto
No resize: It stops the resizing of the frame, no value is to be assigned.
Margin width and Marginheight: It sets the hori¬zontal and vertical margins, values are in pixels.
Name: It gives a name for the frame.
Target: It specifies the target frame.

Question 17.
Mr. Sonet visited a website that contains two frames. He tries to resize the first frame by mouse.But he failed to do so. What is the reason behind. Explain?
Answer:
This is because the web designer used Noresize attribute of frame tag while he design the page. No resize attribute stops the resizing of the frame, no value is to be assigned.
Eg: <frame src=”page1.html” noresize>

Question 18.
We know that an HTML document contains two sec¬tions head and body section. While designing a web page as follows what will happen?
Answer:
<html>
<head>
</head>
<body>
<frameset cols=”50%,*”>
<frame src=”page1 .html” noresize> <frame src=”page2.html”>
</frameset>
</body>
</html>
There is no output because a <frameset> tag has no body tag. It is very important. So the correct code is as follows,
<html>
<head>
</head>
<frameset cols=”50%,*”>
<frame src=”page1 .html” noresize>
<frame src=”page2.html”>
</frameset>
</html>

Question 19.
Create a web page as follows to display a list contains items.
Plus Two Computer Science Chapter Wise Questions and Answers Chapter 5 Web Designing Using HTML img9
Answer:
T0 create a list box set the size property of <Select> tag to more than 1.
<html>
<head>
</head>
<body>
House Hold ltems<Br>
<select size=3>
<option>TV
<option>Fridge
<option selected>Washing Machine </select>
</body>
</html>

Question 20.
In VB there are separate controls to create List Box and Combo Box. But in HTML these controls can be created by using a single tag.

Name the tag used for this?
Which attribute is used for this and how?

Answer:
1. < SELECT >

2. Size attribute
< SELECT Size = 1> gives combo box
< SELECT Size = 3> gives a list box

Question 21.
Write HTML code for creating the following webpage using tag.
Plus Two Computer Science Chapter Wise Questions and Answers Chapter 5 Web Designing Using HTML img10
Answer:
<HTML>
<HEAD>
<TITLE>List Tag</TITLE.
</HEAD>
<BODY>
<H2 Align=”centre” ><B>Net Link Ltd></ B><H2><BR>
<H2 Align=”centre”><B>Bangalore></B></H2><BR>
<HR>
<B>Our products</B>
<BR>
<OLtype=1>
<LI>Television <LI>Washing Machine <UL>
<LI><Model 2005XP Series <LI> Model 2006 ST Series </UL>
</OL>
</Body>
</HTML>

Question 22.
Write the HTML code for creating the following web page using List tag.
Plus Two Computer Science Chapter Wise Questions and Answers Chapter 5 Web Designing Using HTML img11
Answer:
HTML>
<HEAD>
<TITLE>Definition List Tag</TITLE>
</HEAD>
<BODY>
<H1 Align= “center“><B>COMPUTER TERMS</B>< H1><BR>
<DL>
<DT>CPU
<DD>Central Processing Unit <DT>ALU
<DD>Arithmetic and Logic Unit <DT>WWW <DD>World Wide Web </DL>
</BODY>
</HTML>

Question 23.
What will be the output of the following?
<html>
<head>
</head>
<body>
<OL type=”A” start=5>
<LI>COMPUTER
<LI>BIOLOGY
</OL>
</body>
</html>
Answer:
The output is as follows
E.COMPUTER, F.BIOLOGY

Question 24.
Find the errors from the following and correct it.

<ULtype=”A” start=5>
<IMG src=”picture.jpg”size=100>
<HTML>
<HEAD><TITLE></HEAD></TITLE> <BODY>This is a sample web page</BODY>

Answer:

UL has only specified types and has no start attribute
IMG has no size attribute, use height or width attribute
<HTML>
<HEAD><TITLE></TITLE></HEAD> <BODY>This is a sample web page</BODY> </HTML>

Question 25.
Your brother requested you to prepare a list of best friends, from your class using HTML.

Which type of list you will prefer?
Write HTML code to create such a list of 4 students.

Answer:
1. Ordered List<OL> Can be used

2. <HTML>
<HEAD>
<Title>list<Title>
</HEAD>
<BODY>< center><B><U>List of best studentsc/ B></U></Center>
<OL type=1>
<LI> Christeena
<LI>Vijitha
<LI>Archa
<U>Tija
</OL>
</BODY>
</HTML>

Question 26.
Ravi wants to display the name of 6 subjects as a list by using Upper case Roman Numerals. Help him to do so.
Answer:
<OL type= “I”>
<LI>English
<LI> Sanskrit
<LI>Business studies
<LI>Accountancy
<LI>Economics
<LI>Computer Applications
</OL>

Question 27.
Your class selected you as the group leader of Group V. In your group there are 5 students with Roll No. 20 to 24. Prepare a list using appropriate tag in HTML.
Answer:
<html>
<head>
<title>
Ordered list </title>
</head>
<body>
<B><U>Group V</U></B>
<OL start = 20>
<LI>Manisha
<LI>Anisha
<LI>Nisha
<LI>lsha
<LI>Sha
</OL>
</body>
</html>

Question 28.
Write HTML code to get the following output.
Plus Two Computer Science Chapter Wise Questions and Answers Chapter 5 Web Designing Using HTML img12
Answer:
<html>
<head>
<title>
Unordered list </title>
</head>
<body>
<OL type=”a”> <LI><B>Computer</B> <UL >
<LI>Laptop
<LI>desktop
</UL>
</OL>
</body>
</html>

Question 29.
Write HTML code to get the following table as output.
Plus Two Computer Science Chapter Wise Questions and Answers Chapter 5 Web Designing Using HTML img13
Answer:
<html>
<head>
<title>
Table
</title>
</head>
<body>
<table border=1> <tr>
<th>Name
<th>Score
</tr>
<tr>
<td>Raju
<td>90
</tr>
<tr>
<td>Reena
<td>84
</tr>
</table>
</body>
</html>

Question 30.
Write the tags to define the following:

Text Box
Subnnit Button
Reset Button
Radio Button

Answer:

.<input type=”text” name=”txtname”>
.<input type=”Submit” value=”Send”>
.<input type=”Reset” value=”Clear”>
.<input type=”Radio” name=”Sex” value= “Male”>Male <input type=”Radioi’ name=”Sex’’ value=” Female”>Female

Plus Two Computer Science Web Designing Using HTML Five Mark Questions and Answers

Question 1.
Write an HTML code to create a web page with 3 frames as shown below:
Plus Two Computer Science Chapter Wise Questions and Answers Chapter 5 Web Designing Using HTML img14
Answer:
Step 1: Take a notepad, type the following and save it as main.html on Desktop.
<html>
<head>
<title>
Main page </title>
</head>
<body bgcolor = “red”>
<center> main page.
</center>
</body>
</html>

Step 2: Take another notepad, type the following and save it as page1.html on Desktop.
<html>
<head>
<title>
pagel
</title>
</head>
<body bgcolor = “yellow”>
<center> page 1 </center>
</body>
</html>

Step 3: Take another notepad, type the following and save it as page 2.html on Desktop.
<html>
<head>
<title> page 2 </title>
</head>
<body bgcolor = “blue”>
<center> page 2 </center>
</body>
</html>

step 4: Take another notepad, type the following and save it as frame.html on Desktop.
<html>
<head>
<title> frame </title>
</head>
<frameset rows = “30%, *”>
<frame src = “main.html ”>
<frameset cols = “50%, *”>
<frame src = “page1.html”>
<frame src= “page2.htmr>
</frameset>
</frameset>
</html>

Step 5. Finally run the frame.html file.

Question 2.
Explain any three attributes of <FORM>tag.
Answer:
1. Action – Here we give the name of program (including the path) stored in the Webserver.

2. Method – There are 2 types of methods get and post.

Get method	Post method
1. Faster	1. Slower
2. To send small volume of data	2. To send large volume of data
3. Less secure	3. More secure
4. Data visible during submission	4. Data not visible during submission

3. Target – Specifies the target window for displaying the result. Values are given below.

_blank-Opens in a new window
_self-Opens in the same frame
_parent – Opens in the parent frameset
_top-Opens in the main browser window
name – Opens in the window with the specified name.

Question 3.
Explain the attributes of <Table> tag?
Name any six attributes of <table> tag that determine the general layout of table.
Answer:

Border: It specifies the thickness of the border lines around the table
Bordercolor: It specifies the colour for border lines
Align: It specifies the table alignment, the values can b,e left, right or center
Bgcolor: It specifies the back ground colour for the table.
Cellspacing: It specifies the space between two table cells
Cellpadding: It specifies the space between cell border and content
Cols: It specifies the number of columns
Width: It determines the table width
Frame: It specifies the border lines around the table, values are void, border, box, above, below,…

Question 4.
Explain the attributes of <TH> and <TD>?
Answer:

Align: It specifies the horizontal alignment of the content, the values can be left, right, center and justify.
Valign: It specifies the vertical alignment of the content, the values can be top, middle, Bottom, and baseline.
Bgcolor: It specifies the background colour for the cell.
Colspan: It is used to span or to stretch a cell over a number of columns. Eg: <TD Colspan=3> spans the cell over three columns
Rowspan: It is used to span or to stretch a cell over a number of rows. Eg: <TD Rowspan=3> spans the cell over three rows.

Question 5.
Create an HTML page as shown below using lists.
The recipe for preparation
1. The ingredients
•100g flour
• 10g sugar
• 1 cup water
• 2 egg
• Salt and pepper

2. The procedure
A. Mix dry ingredients thoroughly
B. Pour in wet ingradients
C. Mix for 10 mts
D. Bake for 1 hr at 100 degree C temperature <HTML>
Answer:
<HTML>
<HEAD>
<TITLE>
List Demo </TITLE>
</HEAD>
<BODY Bgcolor = “Green”>
The recipe for preparation <OL>
<LI> The ingredients </LI>
<ULtype=”disc”>
<LI> 100 g flour </LI>
<LI> 10 g Sugar </LI>
<LI> 1 cup water </LI>
<LI>2egg </LI>
<LI> Salt and pepper </LI>
</UL>
<LI> The procedure </LI>
<OL TYPE =”A”>
<LI> Mix dry ingredients thoroughly </LI>
<LI> Pour in wet ingredients </LI>
<LI> Mix for 10 mts </LI>
<LI> Bake for 1 hr at 100 degree C temperature </LI>
</OL>
</OL>
</BODY>
</HTML>

Question 6.
Create HTML code for the following output.
1. Flowers
• Jasmine
• Rose
• Lily
2. Vegetables
• Beetroot
• Cabbage
• Cucumber
3. Fruits
i. Apple
ii. Orange
iii. Pineapple
Answer:
<HTML>
<HEAD>
<TITLE>
Nested List </TITLE>
</HEAD>
<BODY Bgcolor = “Blue”> <OL>
<LI> Flowers </LI>
<UL TYPE=”disc”>
<LI> Jasmine </LI>
<LI> Rose </LI> <LI>Lily</LI>
</UL>
<LI> Vegetables </LI>
<ULTYPE=”disc”>
<LI>Beetroot</LI>
<LI>Cabbage</LI>
<LI>Cucumber</LI>
</UL>
<LI> Fruits</LI>
<OL type=T> <LI>Apple</LI> <LI>Orange</LI> <LI>Pineapple</LI>
</OL> </OL>
</BODY>
</HTML>

Plus Two Maths Chapter Wise Questions and Answers Chapter 2 Inverse Trigonometric Functions

April 13, 2021January 31, 2020 by Prasanna

Students can Download Chapter 2 Inverse Trigonometric Functions Questions and Answers, Plus Two Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 2 Inverse Trigonometric Functions

Plus Two Maths Inverse Trigonometric Functions Three Mark Questions and Answers

Question 1.
Prove the following
Plus Two Maths Inverse Trigonometric Functions 3 Mark Questions and Answers 1
Answer:

Plus Two Maths Inverse Trigonometric Functions 3 Mark Questions and Answers 3

Question 2.
Find the value of
Plus Two Maths Inverse Trigonometric Functions 3 Mark Questions and Answers 5
Answer:

Question 3.
If tan^-1x + tan^-1y + tan^-1z = π, show that x + y + z = xyz
Answer:
Given;
tan^-1x + tan^-1y + tan^-1z = π
⇒ tan^-1x + tan^-1y = π – tan^-1z
Plus Two Maths Inverse Trigonometric Functions 3 Mark Questions and Answers 7
⇒ x + y + z = xyz.

Question 4.
Match the following
Plus Two Maths Inverse Trigonometric Functions 3 Mark Questions and Answers 8
Answer:

Question 5.
Solve 2 tan^-1(cos x) = tan^-1(2 cos x)
Answer:
2 tan^-1(cosx) = tan^-1(2cosx)
⇒ $\frac{2 \cos x}{1-\cos ^{2} x}$ = 2cosx
⇒ 1 = 1 – cos² x ⇒ 1 = sin²x
⇒ x = ±$\frac{\pi}{2}$.

Question 6.
Solve the following

2tan^-1(cosx) = tan^-1(2cosecx)
tan^-12x + tan^-13x = $\frac{\pi}{4}$

Answer:
1. 2tan^-1(cosx) = tan^-1(2cosecx)
Plus Two Maths Inverse Trigonometric Functions 3 Mark Questions and Answers 10

2. tan^-12x + tan^-13x = $\frac{\pi}{4}$
Plus Two Maths Inverse Trigonometric Functions 3 Mark Questions and Answers 11
⇒ (6x – 1)(x + 1) = 0
⇒ x = $\frac{1}{6}$, x = – 1
Since x = – 1 does not satisfy the equation, as the LHS becomes negative. So x = $\frac{1}{6}$.

Question 7.
Solve 2 tan^-1(cos x) = tan^-1(2 cos x)
Answer:
2 tan^-1(cos x) = tan^-1(2 cos x)
⇒ $\frac{2 \cos x}{1-\cos ^{2} x}$ = 2cosx
⇒ 1 = 1 – cos² x
⇒ 1 = sin² x ⇒ x = ±$\frac{\pi}{2}$

Plus Two Maths Inverse Trigonometric Functions Four Mark Questions and Answers

Question 1.
Prove that $\sin ^{-1} \frac{12}{13}+\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{63}{16}=\pi$
Answer:
Plus Two Maths Inverse Trigonometric Functions 4 Mark Questions and Answers 12

Plus Two Maths Inverse Trigonometric Functions 4 Mark Questions and Answers 13

Question 2.

Find the principal value of sec^-1$\left(-\frac{2}{\sqrt{3}}\right)$ (1)
if sin$\left(\sin ^{-1}\left(\frac{1}{5}\right)+\cos ^{-1}(x)\right)=1$, then find the value of x. (3)

Answer:
1. principal value of:
Plus Two Maths Inverse Trigonometric Functions 4 Mark Questions and Answers 14

2. find the value of x:
Plus Two Maths Inverse Trigonometric Functions 4 Mark Questions and Answers 15

Question 3.
Solve the following
Plus Two Maths Inverse Trigonometric Functions 4 Mark Questions and Answers 16
Answer:

Plus Two Maths Inverse Trigonometric Functions 4 Mark Questions and Answers 18

The value x = –$\frac{\sqrt{3}}{\sqrt{28}}$ makes the LHS negative, so rejected.

Question 4.
(i) Choose the correct answer from the bracket. cos(tan^-1 x), |x| < 1 is equal to (1)
Plus Two Maths Inverse Trigonometric Functions 4 Mark Questions and Answers 21

Answer:

(draw a right triangle to convert ‘tan’ to ‘sin’).

Question 5.
(i) In which quadrants are the graph of cos^-1 (x) lies, x ∈ [-1,1 ] (1)
(ii) If cos^-1x + cos^-1y = $\frac{\pi}{3}$, then
sin^-1x + sin^-1y = ……… (3)
(a) $\frac{2 \pi}{3}$
(b) $\frac{\pi}{3}$
(c) $\frac{\pi}{6}$
(d) $\frac{\pi}$
(iii) If tan^-1x + tan^-1y = $\frac{\pi}{4}$ then prove that x + y + xy = 1 (2)
Answer:
(i) First and Second quadrant
Plus Two Maths Inverse Trigonometric Functions 4 Mark Questions and Answers 27

⇒ x + y = 1 – xy ⇒ x + y + xy = 1.

Question 6.
(i) sin(tan^-1(1)) is equal to
Plus Two Maths Inverse Trigonometric Functions 4 Mark Questions and Answers 29

Answer:

Plus Two Maths Inverse Trigonometric Functions Six Mark Questions and Answers

Question 1.
Show that sin^-1$\frac{3}{5}$ – sin^-1$\frac{8}{17}$ = cos^-1$\frac{84}{85}$.
Answer:
Plus Two Maths Inverse Trigonometric Functions 6 Mark Questions and Answers 34

(draw a right triangle to convert ‘tan’ to ‘cos’).

Question 2.
(i) Choose the correct answer from the Bracket.
If cos^-1x = y, then y is equal to (1)
(a) π ≤ y ≤ π
(b) 0 ≤ y ≤ π
(c) $-\frac{\pi}{2}$ ≤ y ≤ $\frac{\pi}{2}$
(d) 0 ≤ y ≤ π
(ii) Find the value of cos^-1 cos$\left(\frac{7 \pi}{3}\right)$ (3)
(iii) Solve for x if, tan^-1$\left(\frac{1+x}{1-x}\right)$ = 2 tan^-1x (2)
Answer:
(i) Range of cos^-1x is [0, π] ⇒ 0 ≤ y ≤ π

(ii) Here $\left(\frac{7 \pi}{3}\right)$ lie outside the interval [0, π]. TO make it in the interval proceed as follows.
Plus Two Maths Inverse Trigonometric Functions 6 Mark Questions and Answers 36

Plus Two Computer Science Chapter Wise Questions and Answers Chapter 4 Web Technology

April 13, 2021January 31, 2020 by Prasanna

Students can Download Chapter 4 Web Technology Questions and Answers, Plus Two Computer Science Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus Two Computer Science Chapter Wise Questions and Answers Chapter 4 Web Technology

Plus Two Computer Science Web Technology One Mark Questions and Answers

Question 1.
Specify an attribute of the HTML tag.
Answer:
DIR: Indicates documents direction. It can take values rtl or ltr.
Eg. <HTML DIR= “rtl”> This specify that document is to read from right to left.

Question 2.
Mention the default value of size attribute of <BASEFONT> tag.
Answer:
3.

Question 3.
Name the tag which has ‘Noshade’ attribute.
Answer:
<HR>Tag

Question 4.
Maximum possible value of the size attribute of the <BASEFONT>tag?
Answer:
7.

Question 5.
Salim developed a personal website. In which he has to create an e-mail link. Can you suggest the protocol used to achieve this task?
Answer:
Mailto.
Eg. <A href= “Mailto:[email protected]” >Mail to me</A>.

Question 6.
You want to connect your webpage to the web portal www.yahoo.com. Mention the tag that can be used for this.
Answer:
<A> Anchor Tag
Eg. <Ahref=”www. yahoo.com”>Yahoo</A)

Question 7.
Specify the main attribute of <IMG> tag used to include an image file in web page.
Answer:
SRC. It specify the name of the image file to be included in the page.
Eg. <IMG SRC= “C:/home.Jpg”>

Question 8.
Select the attribute associated with <IMG> tag from the following: (Name, size, Align, value)
Answer:
Align

Question 9.
Write HTML code forgiving hyperlink in webpage.
Answer:
<A HREF= “page2.htm”>Page2</A>

Question 10.
The default colour of Vlink is ________
(a) Blue
(b) Green
(c) Red
(d) Yellow
Answer:
(c) Red

Question 11.
ThedefauKcolour of A link is ________
(a) Blue
(b) Green
(c) Red
(d) Yellow Green
Answer:
(b) Green

Question 12.
Saritha is assigned the task of writing explanatory notes in an HTML code. Which tag she can utilize.
Answer:
Explanatory notes can be given using <comment> tag. Using two mark up elements. as<!— and →

Question 13.
Pick the odd man out.
(a) BODY
(b) HTML
(c) CENTER
(d) ALIGN
Answer:
(d) ALIGN, all others are tag

Question 14.
Your school has a web site, www.myschool.otg. You want to create a link to this site from your website. Write the code for implementing this.
Answer:
<A href=”www.myschool.org”>My School</A>

Question 15.
HTML was developed by ________
Answer:
Tim Berners-Lee

Question 16.
What are the two major sections of an HTML document?
Answer:
Head section and Body section

Question 17.
An HTML file is saved with _____ extension.
(a) .vbp
(b) .mdb
(c) .htm
(d) .frm
Answer:
(c) .htm or .html

Question 18.
The software used to view web page is _____
(a) Notepad
(b) web browser
(c) Webserver
(d) Web Editor
Answer:
(b) web browser

Question 19.
The default alignment of image in HTML is _____
(a) Left
(b) Right
(c) Center
(d) Inline with content
Answer:
(a) Left

Question 20.
_______ is an alternative for centralizing a paragraph other than using <P> tag
(a) <AUGN>
(b) <C>
(c) <CENTER>
(d) <CENTRE
Answer:
(c) <CENTER>

Question 21.
______ attribute Of <A> tag specifies the URL of the hyper linked document.
(a) Name
(b) Target
(c) HREF
(d) SRC
Answer:
(c) HREF

Question 22.
What does HTML stands for?
Answer:
Hyper Text Markup Language

Question 23.
Communication on web can be classified into _____ and ______ .
Answer:
Client to Server and Server to Server

Question 24.
The protocol which is responsible for splitting the data into smaller packets is _____
Answer:
TCP.

Question 25.
The protocol which is responsible for the routing of data packets through the correct destination is _____
Answer:
IP (Internet Protocol)

Question 26.
TCP/IP stands for
Answer:
Transmission Control Protocol/lntenet Protocal

Question 27.
Identify the protocol responsible for e-mail communication.
Answer:
(a) DNS
(b) HTTP
(c) TCP/IP
(d) SMTP
Answer:
(d) SMTP

Question 28.
In server to server communication, authentication is done with help of _____
(a) HTTP
(b) Digital certificate
(c) Client
(d) DNS
Answer:
(b) Digital certificate

Question 29.
Which server acts between merchant server and bank server for transferring data in encrypted format?
Answer:
Payment Gateway

Question 30.
Identify the name of a place where servers and networking systems are placed with high security.
(a) Head office
(b) DNS
(c) Data centre
(d) IIS
Answer:
(c) Data centre

Question 31.
Identify the port number which requests for the service of sending e-mail communication.
(a) 22
(b) 25
(c) 53
(d) 80
Answer:
(b) 25

Question 32.
The IP address corresponding to a domain name is ______ presenting server.
Answer:
DNS

Question 33.
Programs embedded in HTML documents are termed as ______
Answer:
Scripts

Question 34.
Running of ____ scripts can be blocked by the user
(a) Client side
(b) Server side .
(c) Both client side and server side.
(d) None of these
Answer:
(a) Client side

Question 35.
A platform-independent serverside scripting language is ______
Answer:
PHP

Question 36.
Which among the following tools is used for easy formatting and defining style of a document written in HTML?
(a) Ajax
(b) CSS
(c) JSP
(d) JavaScript
Answer:
(b) CSS

Question 37.
Pick the Odd one from the following list and give reason. (IMG, FONT, BR, ALIGN, PRE)
Answer:
ALIGN which is an attribute, all others are tags.

Question 38.
Choose the correct HTML statement to display an image with file name “kerala.jpg” as the background of the web page.
(a) <IMG src=”kerala.jpg”>
(b) <BODYbgcolor=”kerala.jpg”>
(c) <BODY src=”kerala.jpg”>
(d) <BODY background=”kerala.jpg”>
Answer:
(d) <BODYbackground=”kerala.jpg”>

Question 39.
Two of the following HTML tags have same attribute ‘Align’. Identify them.
(<IMG>, <MARQUEE>, <B>, <P>, <BODY>)
Answer:
<IMG> AND <P>

Question 40.
Identify the correct HTML statement to draw a horizontal line with half the width of the screen.
(a) <HR width=“50%” size= “3″>
(b) <HR size= “50%” width= “3″>
(c) <HR length=“50%” size= “3″>
(d) <HR width= “50%” length= “3″>
Answer:
(a) <HR width=“50%” size= “3″>

Question 41.
A student wants to display a poem in a web page just like as he entered in the text editor. Which tag in HTML will help him?
Answer:
<PRE>tag

Question 42.
A student created a web page about his school. The school name is displayed in the page. He wanted to change the style, colour, and size of the school name. Identify the most appropriate tag in HTML needed for that.
Answer:
<FONT> tag

Plus Two Computer Science Web Technology Two Mark Questions and Answers

Question 1.
A student developed a web page about India. He wanted to display a scrolling text moving from right side to left side with a background colour blue.
The text is “I Love My Country”.

Identify the tag needed for it.
Write the HTML statement to do the task.

Answer:

<MARQUEE>
<MARQUEE direction=”left” bgcolor=”blue” > I Love My country </MARQUEE>

Question 2.
Write HTML statement for displaying the following text items:

A₂B³
A>B

Answer:

A <SUB>2</SUB> B <SUP>3</SUP>
A>B

Question 3.
Two HTML tags are given. They are <BODY> and <FONT>. Identify and write the attribute of each from the following list. (Size, Text, Link, Bgcolor, Color)
Answer:
The attributes of <BODY> tag are Text, Link and Bgcolor The attributes of <FONT> tag are size and color.

Question 4.
Write the use of Border and Alt attribute of <IMG> tag.
Answer:
Border: This attribute is used forgiving border to an image
Alt: This attribute is sued for giving an alternate text. When there is no image in the specified location or the browser doesn’t support the image them this text will be displayed.

Question 5.
When a client send request to a server, the server must know which service is demanded by the client.

How does the server identify the type of service requested?
Write the name of any one of the services in the Webserver.

Answer:

Port number
Any service like FTP, SMTP, HTTP etc

Question 6.
Following are steps for searching the IP address of a domain name by a browser. Rearrange them in proper order.

Look in the local memory of ISP
Look in the DNS servers starting from the root server .
Look in the local memory of brower
Look in the local memory of Operating System

Answer:
Correct order is 3, 4, 1, 2

Question 7.
Categorise the following tags into Container tags and empty tags.
<B>, <BR>, <A>,<FRAME>, <FRAMESET>, <LI>, <HR>
Answer:
1. Container tags:
<B>, <A>,<FRAMESET>

2. Empty tags:
<BR>, <HR>, <LI>, <FRAME>

Question 8.
Write the HTML statement to get the following output.

Commerce
Humanities

Answer:

<B>Commerce</B>
<l>Humanities</l>

Question 9.
Write the HTML statement to get the following output.

H₂SO₄
a² + b²
Computer

Answer:

H<sub>2</sub>SO<sub>4</sub>
a<sup>2</sup> + b<sup>2</sup>
<b>Computer</b>

Question 10.
Write HTML code to display as follows.
The <IMG> tag is used for placing images
Answer:
The &lt IMG &gt tag is used for placing images.

Question 11.
Name some browsers.
Answer:

Internet Explorer
Netscape Navigator
Opera
Eudora

Question 12.
What is a website?
Answer:
A website is a collection of webpages. A webpage is created by using HTML tags.

Question 13.
What is the role of attributes in an HTML tag?
Answer:
Attributes are parameters for providing additional information within a tag. Attribute values specify certain characteristics of the tag.
Eg: <P align=”right”>

Question 14.
What is HTML?
Answer:
HTML stands for HyperText Markup Language. It is used to create web pages. It has two types of tags empty and container. The important thing we have to remember in container tag is first opened tag must be closed last.

Question 15.
Name the main attributes of the <HR> tag.
Answer:

Size: It specifies the line thickness
Width: It specifies the length
Noshade: It specifies no shade is given for the line.
Color: It specifies the color.

Question 16.
Mary wants to display her name in various head-ings. Name the heading tags available in HTML.
Answer:
Heading can be given in six levels from <H1> to <H6>.The tag <H1> produces big heading. The tag <H6> produces small heading. The heading size re¬duce from <H1> to <H6> sequentially.

Question 17.
Differentiate between <FONT> and <BASEFONT>tags
Answer:
<BASEFONT>tag sets the normal font for the entire document text. The font specified will be taken as the default font for the entire document. The main attributes are size, face, and color. The <FONT> tag defines the font characteristics of the text enclosed. <Font> tag change the font property of the text enclosed within <Font> and </Font> whereas <BaseFont> tag specifies the default font characteristics.

Question 18.
Explain the different types of hyperlinks.
Answer:
The two types of Hyper Links are

External HyperLink: This is used two connect the locations of two different web pages
Internal HyperLink: This is used to connect the different locations of the same web page

Question 19.
Write the HTML code to display the following list:

Form
TextBox
Label
Command Button

Answer:
<html>
<head>
<title>
list
</title>
</head>
<body bgcolor=”vbcyan”>
<ul type=”circle”>
<li>Form
<li>TextBox
<li>Label
<li>CommandButton</ul>
</body>
</html>

Question 20.
The body section forms the content displayed in the browser window. Briefly explain any four attributes in the BODY tag.
Answer:

Bgcolor – It is used to set background colour
Background – It is used to set a background picture
Text – it is used to set the foreground colour
Left margin – It is used to set the left margin

Question 21.
HTML has the facility to provide External Jinks as well as Internal links.

Which tag is used to include an External link?
How will you construct an Internal link?

Answer:
1. Anchor Tag i.e. <a href=”filename”>

2. Internal link is used to link two places of the same web page
<html>
<head>
<title>
Internal link
</title>
</head>
<body bgcolor=”vbcyan”>
<a name=top href=#bottom>Goto Bottom</a>
<ul type=”circle”>
<li>Form
<li>TextBox
<li>Label
<li>CommandButton
</ul>
<a name=bottom href=#top>Goto top</a>
</body>
</html>

Question 22.
Categorise the following tags in HTML and write the criterian for the categorisation.
<BR>, <P>, <BODY>, <B>, <HR>, <IMG>
Answer:
1. Empty tag:
<BR>
<HR>
<IMG>
2. Container tag
<P>
<BODY>
<B>

Question 23.
Differentiate empty tags and container tag with example.
Answer:
1. Empty tags: It has opening tag only, no closing tag
Eg; <hr>, <br> etc
2. Container tag: It as both opening and closing tag.
Eg: <html> </html>
<body></body>etc.

Question 24.
Write True or False

Text is an attribute of <BODY> tag to insert a text matter in the web page.
<EM> tag functions similar to <l> tag.

Answer:

False
True

Plus Two Computer Science Web Technology Three Mark Questions and Answers

Question 1.
Write and explain any four text formatting tags in HTML.
Answer:
Text formatting tags are given below.

<B>: This tag is based to make the text Bold Eg: <B> Computer application </B>
<l>: This tag is used to make the text in italics eg: <l> computer aplication</l>
<U>: This tag is used to underline the text eg: <U> computer aplication</U>
<S>: This tag is used for striking out the text eg: <S> computeraplication</S>

Question 2.
Match the following.
Plus Two Computer Science Chapter Wise Questions and Answers Chapter 4 Web Technology img1
Answer:
Correct match as given below

<H2> – Heading-Align
<MARQUEE> – Scrolling text – Bgcolor
<IMG> – Inserting picture-Src

Question 3.
Briefly explain the use of tags <Q>, <PRE> and <ADDRESS> tags.
Answer:

<Q>: It is used to give text within double quotes
<PRE>: This tag is used to display the content as we entered in the text editor.
<ADDRESS>: This tag is used to provide information of the author or owner

Question 4.
Write a HTML code to develop a web page about Kerala state as shown below:
Plus Two Computer Science Chapter Wise Questions and Answers Chapter 4 Web Technology img2
The specifications for the page are:

The main heading must be of bigger in size, centralised and bold.
Subheadings must be lesser size than main heading and in italics.
There should be a picture at the center of the page with filename“tree.jpg”.
The background colour of the page must me blue.

Answer:
<HTML>
<HEAD>
<TITLE>Kerala State</TITLE>
</HEAD>
<BODY bgcolor = “blue”>
<center> <H1> kerala state </H1 >
<I><H2> <br> God’s own country </H2></l> </center>
<img src = “tree.jpg” align = “middle”>
<br>
Capital: Thimvanthapuram </BODY>
</HTML>

Question 5.
PHP is a popular scripting language.

Write whether it is client side or server side.
Write a brief note on PHP.

Answer:

Server side
PHP (PHP Hypertext Preprocessor)
- It is an open-source, general-purpose scripting language.
- It is a server-side scripting language
- Introduced by Rasmus Lendorf
- A PHP file with extension .php
- It support database programming the default * DBMS is MySQL
- It is platform-independent
- PHP interpreter in Linux is LAMP(Linux, Apache, MySQL, PHP)

Question 6.
How client side scripting differs from server side scripting?
Answer:
Following are the differences

Client-Side Scripting	Server Side Scripting
Script is copied to client browser	To the webserver.
Executed by the client.	Executed by the server and result gets back to the browser window
Used for Client level validation	Connect to the database in the server
It is possible to block by the user	Cannot possible
Client-side scripts depend on the type and version of the browser	It does not depend on the type and version of the browser

Question 7.
Briefly explain the two types of communication on the web.
Answer:
The two types of communication on the web are given below.
1. Client to Web server communication:
This communication is carried out. between client to the webserver (shopping site). The technology used to protect data that are transferred from client to web server is https.

2. Web server to web server communication:
This communication is usually carried out between web sever (seller) to another web server (normally bank). For the safe transaction. Digital certificate issued by third-party web sites are used.

Question 8.
Compare static and dynamic webpages.

Static web pages	Dynamic web pages
Content and layout is fixed	Content and layout is changed frequently
Never use database	Database is used
Run by browser	It runs on the server and result get back to the client(browser)
Easy to develop	Not at all easy

Question 9.
Differentiate Empty tag and Container tag.
Answer:
There are two types of tags, opening tag, and closing tag
1. Empty tag: It has only opening tag and no closing tag
Eg: <br>, <hr>,..

2. Container tag: It has both opening and closing tags. This tag contains some text data
Eg:<html>, <head>, <body>,….

Question 10.
Create a webpage using HTML to display the following message.
Answer:
‘The symbol H₂O represents water’.
<HTML>
<HEAD>
<TITLE>Subscript</TITLE>
</HEAD>
<Body>
<B>’The symbol H<SUB>2</SUB>O represents water’<B> .
</Body>
</HTML>

Question 11.
While designing a webpage Neena wants to explain the meaning of each step. Can you help her?
Answer:
Neena can use Comments while writing the code. It is a good programming practice. Comments improve readability. It is not the part of a program. Comments <!—and —>
Eg.
<!—This is used to display an information—> <HTML>
<!—head section—>
<HEAD>
<TITLE>Subscript</TITLE>
</HEAD>
<!—BODY section—>
<Body>
<B> The symbol H <SUB>2</SUB>O represents water’</B>
</Body>
</HTML>

Question 12.
Complete the following table.
Plus Two Computer Science Chapter Wise Questions and Answers Chapter 4 Web Technology img3
Answer:

Question 13.
Give HTML tag to display the sentence “WELCOME TO HTML” as centralized heading, having red colour.
Answer:
<HTML>
<HEAD><TITLE>My page</TITLE></HEAD> <BODY>
<FONTCOLOR=“RED”>
<H1 ALIGN= “CENTER”>WELCOME TO HTML</ H1>
</FONT>
</BODY>
<HTML>

Question 14.
Explain the Main attributes of font tag.
Answer:
<FONT>tag defines the font properties of text en-closed
The main attributes are:

Face – This specifies the type of font.
Color – This specifies the colour of the text enclosed
Size – This specifies the font size
Eg. <FONT Face= “Arial” size = “3” color= “magenta”>

Question 15.
John visited a website, it is found that when clicking on a particular text the browser open a new web page. Name the feature and Identify the tag used for this purpose. Write the HTML code to link to a file name “main.html”.
Answer:

Hyper Linking. By clicking on hypertext we can see or go to other webpages or to other section of same document.
<A> tag is used
<A href=”main.htmr>Main</A>

Question 16.
Antony visited his school website, he could not see the picture of his school instead of that there is a text message “Your browser could not support images”. Why it is happened and write the html code for this.
Answer:
Because the browser he used cannot load the image. He can use Alt attribute of IMG Tag.
<IMG SRC =“D:\school.jpg” Alt-Your browser could not support images”>

Question 17.
Suppose you want to display a picture named school.jpg located in the ‘Photos’ subdirectory of the directory ‘My documents’ of C drive in your web page.

Name the tag used for this purpose.
Write HTML code for the web page.

Answer:
1. <IMG>tag. It is used to display images in webpage.

2. code for the web page.
<HEAD><TITLE>
</HEAD>
<Body>
<IMG SRC= ‘C:/My documents/Photos/School.jpg’></Body>
<html>

Plus Two Computer Science Web Technology Five Mark Questions and Answers

Question 1.
Explain the main list tags in HTML?
Answer:
HTML provides three basic types of lists- unordered, ordered and definition list.
1. Unordered list:
Unordered list arranges the list items with bullet symbols in front. <UL> and </UL> tag encloses an Unordered list. List items are specified by <LI> tag. The tag <UL> can take values square, circle or disc. The default type is disc.
Eg: <UL>
<LI>COMPUTER
<LI>BIOLOGY
</UL>

2. Ordered list:
In Ordered list, the list items are numbered in sequence. <OL> and </OL> tag encloses an Ordered list. List items are specified by <LI> tag. The tag <OL> can take values as follows

type = 1 for 1,2,3
type = i for i, ii, iii
type = I fori, II, III,….
type = afora,b,c,…
type = AforA,B,C

Eg:<OL>
<LI>COMPUTER
<LI>BIOLOGY
</OL>

3. Definition List:
It is formed by a group of definitions and their descriptions. No bullet symbol or number is provided for the list items. The <DL> and </DL> tags enclose the definition list. The <DT> tag contains the definition term and <DD> tag specifies the description.
Eg: <DL>
<DT>Echeque
<DD>Electronic cheque
</DL>

Question 2.
Explain the use of <BODY> tag and list any four of its attributes
Answer:
Web page contents are given int the body section. Attributes of body tag are:
1. BGCOLOR – Specifies background color for the document Body
Eg. <BODY BGCOLOR= “RED”>

2. BACKGROUND – Sets the image as background for the document body
Eg. <BODY BACKGROUNG= “C:\result.jpg”>

3. TEXT – Specifies the color of the text content of the page
Eg. <BODYTEXT= “Red”>

4. LINK – Specifies colour of the hyperlinks that are not visited by the user

5. ALINK – Specifies the colour of hyperlinks

6. VLINK – Specifies the color of hyperlinks which are already visited by the viewer.
Eg. < BODY ALINK= “Cyan” LINK=” Magenta” VLINK= “Orange”>

7. Left margin and Right margin-Sets margin from left and top of the document window.

Plus Two Maths Chapter Wise Questions and Answers Chapter 1 Relations and Functions

April 13, 2021January 31, 2020 by Prasanna

Students can Download Chapter 1 Relations and Functions Questions and Answers, Plus Two Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 1 Relations and Functions

Plus Two Maths Relations and Functions Three Mark Questions and Answers

Question 1.
Show that the relation R in the set is given by A = {x ∈ Z, 0 = x = 12}
R = {(a,b): a – b is a multiple of 4}
Answer:
Given; R = {(a,b): a – b is a multiple of 4}
Clearly a – a = 0 is a multiple of 4. So
(a, a) ∈ R, hence Reflexive
(a,b) ∈ R ⇒ a -b is a multiple of 4, imply that b – a is a multiple of 4 ⇒ (b, a) ∈ R So Symmetric.
(a, b), (b, c) ∈ R ⇒ a – b and b – c are multiple of 4.
a – b + b – c is a multiple of 4
a – c is a multiple of 4 ⇒ (a, c) ∈ R So Transitive.
Hence R is an equivalence Relation.

Question 2.
Let R be a Relation in the set A = {1, 2, 3, 4, 5, 6} defines as R = {(x,y): y = 2x – 1}

Write R in roster form and find it’s domain and range
Is R an equivalence relation ? Justify.

Answer:
1. R ={(1, 1), (2, 3), (3, 5)}
Domain = {1, 2, 3} ; Range ={1, 3, 5}

2. Since (2, 2) ∉ R, R is not reflexive
(2, 3) ∈ R but (3, 2) ∉ R
R is not symmetric.
(2, 3) ∈ R, (3, 5) ∉ R but, (2, 5) ∉ R
R is not transitive
∴ R is not an equivalence relation.

Question 3.
Plus Two Maths Relations and Functions 3 Mark Questions and Answers 1
Answer:
Given;

Question 4.
Let ‘ * ’ is a binary operation on the set Q of rational numbers as follows;

a * b = a – b
a * b = a + ab
a * b = a² + b²
a * b = (a – b)²

Find which of the binary operation are commutative and associative. Also find the identity element if exists.
Answer:
1. Given; a * b = a – b
b * a = b – a ≠ a – b = a * b
Hence not commutative.
Now; a * (b * c) = a * (b – c)
= a – (b – c) – a – b + c
(a * b) * c = (a – b) * c = a – b – c
a * (b * c) ≠ (a * b) * c
Hence not associative.
Now; a * e = a ⇒ a – e = a ⇒ e = 0
e * a = a ⇒ e – a = a ⇒ e = 2a
a * e ≠ e * a.
So identity element does not exist.

2. Given; a * b = a + ab
b * a = b + ab ≠ a + ab = a * b
Hence not commutative.
Now; a * (b * c) = a * (b + bc)
= a + a(b + be) = a + ab + abc
(a * b) * c = (a + ab) * c
= a + ab + c(a + ab)
= a + ab + ac + abc
a * (b * c) * (a * b) * c
Hence not associative.
Now; a * e = a ⇒ a + ae = a ⇒ e = 0
e * a = a ⇒ e + ea = a ⇒ e = $\frac{a}{1+a}$
a * e ≠ e * a.
So identity element does not exist.

3. Given; a * b = a² + b²
b * a = b² + a² = a² + b² = a * b
Hence commutative.
Now; a * (b * c) = a* (b² + c²) = a² + (b² + c²)²
(a * b) * c = (a² + b²) * c
= (a² + b²)² + c²
a * (b * c) ≠ (a * b) * c
Hence not associative.
Now; a * e – a ⇒ a² + e² = a
⇒ e = $\sqrt{a-a^{2}}$ ∉ Q
So identity element does not exist.

4. Given; a * b = (a – b)²
b * a = (b – a)² = (a – b)² = a * b
Hence commutative.
Now; a * (b * c) = a * (b – c)² (a – (b-c)²)²
(a * b) * c – (a – b)² * c = ((a – b)² – c)²
a * (b * c) ≠ (a * b) * c
Hence not associative.
Now; a * e = $\sqrt{a}$ ⇒ e = a – $\sqrt{a}$ ∉ Q
So identity element does not exist.

Question 5.
Show that the relation R on the set of natural numbers defined as R: { (x, y): y – x is a multiple of 2} is an equivalance relation
Answer:
Since x – x = 0 is multiple of 2, (x, x)∈ R
Therefore reflexive.
If y – x is a multiple of 2 then x – y is also a multiple of 2. Therefore (x, y) ∈ R ⇒ (y, x) ∈ R. Hence symmetric.
If y – x is a multiple of 2 and z-y is a multiple of 2, then their sum y – x + z – y = z – x is a multiple of 2. Therefore (x, y), (y, z) ∈ R ⇒ (x, z) ∈ R
Hence transitive.
Therefore R is an equivalance relation.

Question 6.
Consider f : [3, ∞) → [1, ∞) given by f(x) = x² – 6x + 10. Find f^-1
Answer:
f(x) = (x – 3)² + 1
y = (x – 3)² + 1
⇒ (x – 3)² = y – 1
⇒ x = $\sqrt{y-1}$ + 3
∴ f^-1(x) = $\sqrt{x-1}$ + 3

Question 7.
‘*’ be a binary operation on N × N defined as (a, b) * (c, d) – (ac, bd)

Show that * is commutative.
Find the identity element of * if any
Write an element of N × N which has an inverse.

Answer:
1. (a, b) * (c, d) = (ac, bd) = (ca, db)
= (c, d) * (a, b)
Hence * is commutative.

2. (a, b) * (e, e) = (a, b)
⇒ (ae, be) = (a, b)
⇒ ae = a ⇒ e = 1
⇒ be = b ⇒ e = 1
Therefore (1,1) is the identity element.

3. Invertible element is (1,1).

Question 8.
Show that the relation R in the set R of Real numbers defined asR = {(a, b): a ≤ b²} is neither reflexive, nor symmetric nor transitive.
Answer:
a ≤ a² is not true for all a. So (a, a) ∉ R hence not Reflexive.
(a, b) ∈ R ⇒ a ≤ b² does not imply that b ≤ a². So (b, a) ∉ R, hence not Symmetric. (a, b),(b, c) ∈ R ⇒ a ≤ b² and b ≤ c² does not imply that a ≤ c². So (a, c) ∉ R, hence not Transitive.

Question 9.
Consider f :R⁺ → [-5, ∞) given by f(x) = 9x² + 6x – 5. Show that fis invertible with f^-1(y) = $\frac{\sqrt{y+6}-1}{3}$
Answer:
Plus Two Maths Relations and Functions 3 Mark Questions and Answers 3
f is invertible and f^-1(y) = $\frac{\sqrt{y+6}-1}{3}$.

Question 10.
Let A = R – {3} and B = R – {1} consider the function f: A → B defined by f(x) = $\frac{x-2}{x-3}$ is f one-one and onto? Justify your answer.
Answer:
Given; f: A → B, A = R – {3} and B = R – {1}
f(x₁) = f(x₂) ⇒ $\frac{x_{1}-2}{x_{1}-3}=\frac{x_{2}-2}{x_{2}-3}$
⇒ (x₁ – 2)(x₂ – 3) = (x₂ – 2)(x₁ – 3)
⇒ x₁x₂ – 2x₂ – 3x₁ + 6 = x₂x₁ – 2x₁ – 3x₂ + 6
⇒ -2x₂ – 3x₁ = – 2x₁ – 3x₂ ⇒ x₁ = x₂
Hence one-one
Let y ∈ B ⇒ f(x) = y ⇒ $\frac{x-2}{x-3}$ = y
⇒ x – 2 = xy – 3y
⇒ x – xy = 2 – 3y
⇒ x(1 – y) = 2 – 3y ⇒ x = $\frac{2-3 y}{1-y}$ ∈ A=
Hence onto. Therefore bijective.

Question 11.
Plus Two Maths Relations and Functions 3 Mark Questions and Answers 4
Answer:

Plus Two Maths Relations and Functions Four Mark Questions and Answers

Question 1.
The relation R defined in the set A = {-1, 0, 1} as R = {(a,b):a = b²}

Check whether R is reflexive, symmetric and transitive.
Is R an equivalence relation

Answer:
1. (-1, -1) ∉ R, R is not reflexive
(-1,1) ∈ R but (1, -1) ∉ R, R is not symmetric
(-1,1) ∈ R, (1, 1) ∈ R and (-1, 1) ∈ R,
R is transitive.

2. R is not reflexive, not symmetric and not transitive. So R is not an equivalence relation.

Question 2.
Let A = {1, 2, 3}. Give an example of a relation on A which is

Symmetric but neither reflexive nor transitive.
Transitive but neither reflexive nor symmetric.

Answer:
1. R = {(1, 2),(2, 1)}
(1, 1) ∉ R ⇒ R is not reflexive
(1, 2) ∈ R ⇒ (2, 1) ∈ R, R is symmetric
(1, 2) ∈ R, (2, 1) ∈ R but (1,1) ∉ R,
R is not transitive

2. R = {(1, 2),(1, 3),(2, 3)}
(1, 1) ∉ R ⇒ R is not reflexive
(1, 2) ∈ R ⇒ but (2,1) ∉ R, R is not symmetric
(1, 2) ∈ R, (2, 3) ∈R ⇒ (1, 3) ∈ R,
R is transitive.

Question 3.
Find fog and gof if

f(x) – |x| and g(x) – |3x + 4|
f(x) = 16x⁴ and g(x) = x ^{$\frac{1}{4}$}

Answer:
1. Given; f(x) = |x| and g(x) = |3x + 4|
fog(x) = f(g(x))
= f(|3x + 4|) = ||3x + 4|| = |3x + 4|
gof(x) = g(f(x)) = g(|x|) = |3|x| + 4|

2. fog(x) = f(g(x)) = g(x^{$\frac{1}{4}$}) = 16(x^{$\frac{1}{4}$})⁴ = x
gof (x) = g(f(x)) = g(16x⁴) = (16x⁴)^{$\frac{1}{4}$} = 4x.

Question 4.
Consider the binary operation* : Q → Q where Q is the set of rational numbers is defined as a * b = a + b – ab

Find 2 * 3 (1)
Is identity for * exist? If yes, find the identity element. (2)
Are elements of Q invertible? Is yes, find the inverse of an element in Q (1)

Answer:
1. 2 * 3 = 2 + 3 – 6 = -1.

2. a * e = a ⇒ a + e – ae = a
⇒ e(1 – a) – 0 ⇒ e = 0
e * a = a ⇒ e + a – ea = a
⇒ e(1 – a) = 0 ⇒ e = 0 is the identity element.

3. a * b = 0 ⇒ a + b – ab = 0 ⇒ b = $\frac{a}{a-1}$

Question 5.
‘*’ is a binary operation on R defined as a * b = 2ab

Determine whether * is commutative and associative
Find the identity element if exists.
Find the inverse element, if exists

Answer:
1. a * b = 2ab = 2ba = b * a
Therefore commutative
a * (b * c) = a * (2bc) = 4abc
(a * b) * c = (2ab) * c = 4 abc
Therefore is associative.

2. a * e = a ⇒ 2ae = a ⇒ e = $\frac{1}{2}$
e * a = a ⇒ 2ea = a ⇒ e = $\frac{1}{2}$
Therefore identity element is $\frac{1}{2}$.

3. a * b = $\frac{1}{2}$ ⇒ 2ab = $\frac{1}{2}$ ⇒ b = $\frac{1}{4a}$, a ≠ 0.

Question 6.

If f: R – {2} → R – {2}, defined by f(x) = $\frac{2 x-3}{x-2}$ then find fof
Which of the following satisfies the condition f^-1 ≠ f.

(a) f : R – {0} → R – {0}, f(x) = $\frac{1}{x}$
(b) f :R → R, f(x) = -x
(c) f : R – {-1} → R – {-1}, f(x) = $\frac{x}{x+1}$
(d) f: R – {2} → R – {2}, f(x) = $\frac{2 x-3}{x-2}$

Answer:
1. fof(x) = f(f(x)) = f$\left(\frac{2 x-3}{x-2}\right)$
Plus Two Maths Relations and Functions 4 Mark Questions and Answers 7

2. Following satisfies the condition f^-1 ≠ f:
(a) f : R – {0} → R – {0}, f(x) = $\frac{1}{x}$
(b) f :R → R, f(x) = -x
The graph of functions in (a) and (b) symmetric with respect to the line y = x. The function in (d) we have already shown that fof (x) = x. So the answer is (c)
f : R – {-1} → R – {-1}, f(x) = $\frac{x}{x+1}$.

Question 7.
If f :R → R is a function defined by f(x) = 3x – 2

Show that f is one-one.
Find fof(x).
Find the inverse of f if exists.

Answer:
1. f(x₁) = f(x₂)
⇒ 2x₁ – 3 = 2x₂ – 3 ⇒ x₁ = x₂
Therefore one-one.

2. fof(x) = f(f(x)) = f(3x – 2)
f (x) = 3(3x – 2) – 2 = 9x – 6 – 2 = 9x – 8.

3. Let g(x) = $\frac{x+2}{3}$
fog(x) = f(g(x))
= f$\left(\frac{x+2}{3}\right)$ = 3$\left(\frac{x+2}{3}\right)$ – 2 = x
Similarly we can show that
gof(x) = g(f(x)) = x
Therefore; f^-1 = $\frac{x+2}{3}$.

Question 8.
Let A = N × N and be the binary operation. On A defined by (a, b) * (c, d) = (a+c, b+d). Show that ‘ * ’ is commutative and associative. Find the identity for ‘ * ’ on A if any.
Answer:
Given; A = N × N and
(a, b) * (c, d) = (a + c, b + d)
(c, d) * (a, b) = (c + a, d + b)
= (a + c, b + d) = (a, b) * (c, d))
Hence commutative.
Now; (a, b) * [(c, d) * (e, f)] = (a, b) * [c + e, d+f]
= (a + c + e, b + d + f)
[(a, b) * (c, d)] * (e, f) = [a + c, b + d] * (e, f)
= (a + c + e, b + d + f)
Hence associative.
(a, b) * (e, e) = (a, b) ⇒ (a + e, b + e) = (a, b)
⇒ a + e = a, b + e = b ⇒ e=0, e = 0 ⇒ (0,0) ∉ A
So identity element does not exist.

Question 9.
If f(x) = $\left(\frac{4x+3}{6x-4}\right)$, x ≠ $\frac{2}{3}$

Show that fof{x)=x, for all x ≠ $\frac{2}{3}$.
What is the inverse of ‘f?

Answer:
1.
Plus Two Maths Relations and Functions 4 Mark Questions and Answers 10
2. Since; fof(x) = x
f is the inverse of it self.

Question 10.
1. Let f be a function defined by f(x) = $\sqrt{x}$ is a function if it defined from (1)
(a) f : N → N
(b) f : R → R
(c) f : R → R⁺
(d) f : R⁺ → R⁺
2. Check the injectivity and surjectivity of the following functions (3)
(a) f : N → N defined by f(x) = x³
(b) f : R → R given by f(x) = [x]
Answer:
1.
(d) f : R⁺ → R⁺

2.
(a) For x, y ∈ N,
f(x) = f(y) ⇒ x³ = y³ ⇒ x = y
Therefore, f is injective
For 2 ∈ N, there does not exist x in the domain N such that f(x) = x³ = 2.
∴ f is not surjective.

(b) f : R → R given by f(x) = [x]
It seen that f(1.1) = 1 and f(1.8) = 1;
But 1.1 ≠ 1.8;
∴ f is not injective
There does not exist any element x ∈ R
such that f(x) = 0.7
∴ f is not surjective.

Plus Two Maths Relations and Functions Six Mark Questions and Answers

Question 1.
Consider the function f(x) $=\frac{x+1}{x-1}$; x ≠ 1
1. fof(2) = ____
(a) 1
(b) 2
(c) 3
(d) 4
2. What is the inverse of f?
3. f(3) + f^-1(3)
Answer:
1. fof(2):
(b) 2
Plus Two Maths Relations and Functions 6 Mark Questions and Answers 11

2. Let g : range of f → R- {1} inverse of f Let ‘y be any arbitrary element in the range of f
then y = f(x) = $\frac{x+1}{x-1}$
⇒ xy – y = x + 1 ⇒ x(y – 1) = y +1
Plus Two Maths Relations and Functions 6 Mark Questions and Answers 12
Let us defi ne g : range of f → R- {1} as

3. f(3) = 2, f^-1(3) = 2
f(3) + f^-1(3) = 2 + 2 = 4.

Question 2.
1. Show that the function f: R → R defined by f(x) = 2x – 3 is one-one and onto. Find f^-1
2. Which of the following figure represents the graph of a function on R which is onto but not one-one.
Plus Two Maths Relations and Functions 6 Mark Questions and Answers 14

3. Write a function on R which is onto but not one-one
Answer:
1. f(x) = 2x – 3
f(x₁) = f(x₂)
⇒ 2x₁ – 3 = 2x₂ – 3 ⇒ x₁ = x₂
Therefore f is one-one.
Let f(x) = yeR, then
y = 2x – 3 ⇒ 2x = y + 3 ⇒ x = $\frac{y+3}{2}$ ∈ R
Therefore f is onto. Then f^-1(x) = $\frac{x+3}{2}$.

2. Option (c)
[(b),(c),(d) are not one-one since for different values of x, we have same value of y. ie; horizontal line meets at more than one point, (b) and (d) are not onto since range and codomain are different].

3. f(x) = x², f : R → [0, ∞) (Any other function).

Question 3.
A = {1, 2, 3, 4, 6}, * is a binary operation on A is defined as a * b = HCF of a and b.

Represent * with the help of an operation table.
Find the identity element.
Write a commutative binary operation on A with 3 as the identity element. (Hint: Operation table may be used.

Answer:
1.
Plus Two Maths Relations and Functions 6 Mark Questions and Answers 16

2. From the table it is clear that the identity element is 6.

3. We can write any operation table which is commutative with 3 as the identity element.
Plus Two Maths Relations and Functions 6 Mark Questions and Answers 17

Question 4.
Let * be a binary operation on the set of all real numbers R defined by a * b = a + b + a²b for a, b R.

Find 2 * 6 and 6 * 2. (2)
Prove that * is neither commutative nor associative. (2)
Find the identity elements in R if exists. (2)

Answer:
1. a * b = a + b + a²b
2 * 6 = 2 + 6 + 2² × 6 = 8 + 24 = 32
6 * 2 = 6 + 2 + 6² × 2 = 8 + 72 = 80.

2. a * b = a + b + a²b a * b ≠ b * a
b * a = b + a + b²a
∴ is not commutative.
(ab)c = (a + b + a²b) c
= a + b + a²b + c + (a + b + a²b)²c
a * (b * c) = a * (b + c + b²c)
= a + b + c + b²c + a²(b + c + b²c)
∴ is not associative.

3. Let e is the identity element.
∴ a * e = a ⇒ a + e + a²e = a ⇒ e(1 + a²) = 0
∴ e = 0
∴ e * a = a ⇒ e + a + e²a = a ⇒ e + e²a = 0
⇒ e(1 + ae) = 0
⇒ e = 0; (1 + ae) = 0
ea = -1 ⇒ e = – $\frac{1}{a}$
Identity element does not exists.

Question 5.

Consider f: {3, 4, 5, 6} → {8, 10, 12, 13, 14} And f = { (3, 8), (4, 10), (5, 12), (6, 14)}. State whether f has inverse ? Give reason. (2)
Consider f : R → R given by f(x) = 3x + 2 Show that f is invertible. Find the inverse of f

Answer:
1. Distinct elements in set {3, 4, 5, 6} has distinct images, under f.
∴ f is one- one
But 14 in the codomain has no pre image.
∴ f is not onto.
∴ f has no inverse.

2. f(x) = 3x + 2; then
f(x₁) = f(x₂) = 3x₂ + 2 ⇒ x₁= x₂
Hence F is one – one
For y ∈ R, let y = 3x + 2 ⇒ x = $\frac{y-2}{3}$ ∈ R
Plus Two Maths Relations and Functions 6 Mark Questions and Answers 18

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 3 Electrochemistry

April 13, 2021January 31, 2020 by Prasanna

Students can Download Chapter 3 Electrochemistry Questions and Answers, Plus Two Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 3 Electrochemistry

Plus Two Chemistry Electrochemistry One Mark Questions and Answers

Question 1.
The metal which acts as the anode in Daniell cell is _______
Answer:
Zinc

Question 2.
Electrolysis of dilute H₂SO₄ liberates ________ gas at cathode.
Answer:
Hydrogen

Question 3.
The charge required to reduce 1 mole of I₂ to I_– is
(a) 96500 C
(b) 2 × 96500 C
(c) $\frac{1}{2}$ × 96500 C
(d) 4 × 96500 C
Answer:
(b) 2 × 96500 C

Question 4.
Of the following metals that cannot be obtained by electrolysis of the aqueous solution of their salts are
(a) Ag and Mg
(b) Ag and Al
(c) Mg and Al
(d) Cu and Cr
Answer:
(c) Mg and Al

Question 5.
Chemically, rust is _______
Answer:
Hydrated ferric oxide (Fe₂O₃.xH₂O)

Question 6.
One of your friends argues that in the laboratory CuS04 is not stored in Aluminium bottles.

Do you agree?
Justify.

Answer:
1. Yes.

2. Aluminium displaces copper from copper sulphate solution. Because, in electrochemical series Al stands before copper. The reduction potential of Cu is greater than that of Al. Greater the value of reduction potential, more easily is the substance reduced.
E°_(Al⁺³/Al) = – 1.66 V,
E°_(Cu⁺²/Cu)= + 0.34V

Question 7.
Arrange the following metals in the order in which they displace each other from the solution of their salts:
Al, Cu, Fe, Mg and Zn
Answer:
The correct order is Mg, Al, Zn, Fe, Cu

Question 8.
The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm^-1. Calculate the molar conductivity.
Answer:
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 3 Electrochemistry one mark q8 img 1

Question 9.
The limiting molar conductivity of HCl, CH₃COONa and NaCl are respectively 425, 90 and 125 mho cm²mol^-1 at 25°C. The molar conductivity of 0.01 m. CH₃COOH solution is 7.5 mho cm² mol^-1 at the same temperature. The degree of dissociation of 0.1 as acetic acid solution at the same temperature is _________
Answer:
0.02

Question 10.
The ion of least limiting molar conductivity among the following is
(a) SO^2-₄
(b) H⁺
(c) Ca²⁺
(d) CH₃COO^–
Answer:
(d) CH₃COO^–

Question 11.
Standard electrode potential of three metals XY and Z are 0.52V, 2.87 V and -0.44 V respectively. The reducing power of these metals are
Answer:
Y > Z > X

Question 12.
A current of strength of 9.65 ampere is passed through excess fused AlCl₃ for 5 hours. How many liters of chlorine will be liberated at STP.
Answer:
20.16

Question 13.
Consider the following four electrodes
P = Cu²⁺(.0001 M/Cu(s) Q = Cu²⁺(0.1 M)/Cu(s)
R = Cu²⁺(0.01 M)/Cu(s) S = Cu²⁺(0.001 M)Cu(s)
If the standard reduction potential of Cu²⁺/Cu is +.34V the reduction potential in volts of the above electrode follows the order.
Answer:
Q > R > P

Plus Two Chemistry Electrochemistry Two Mark Questions and Answers

Question 1.
Resistance of lead is greater than that of gold.

Do you agree?
What do you mean by resistance?
What do you mean by resistivity?

Answer:

Yes.
Resistance is the hindrance offered by a substance to the flow of electricity through it.
Resistivity is the resistance of a conductor of length 1 cm and area of cross section 1 cm².

R ∝ $\frac{\ell}{\mathrm{A}}$ Or R = ρ$\frac{\ell}{\mathrm{A}}$ where R → Resistance
and ρ → resistivity.
When l = 1 cm and A = 1 cm²
then, R = ?

Question 2.
In the chemistry lab, a student immersed two platinum electrodes in a solution of CuSO₄ and electric current was passed through the solution. After sometime he noticed that the colour of CuSO₄ disappeares with the evolution of a gas at the electrode. In his Practical Log, he recorded that the colourless solution contains H₂SO₄.

Is it true?
Comment on his findings and justify your answer.

Answer:
1. Yes.

2. When aqueous CuSO₄ solution is electrolysed using Pt electrodes, Cu is deposited at the cathode and O₂ is liberated at the anode.
CuSO₄(aq) → Cu²⁺(aq) + SO₄^2-(aq)
Cu²⁺(aq) + 2$\bar { e }$ → Cu(s) (at cathode)
H₂O(l) → 2H+(aq) + 1/2 O₂(g) + 2$\bar { e }$ (at anode)
Thus, H₂SO₄ is formed as the secondary product.

Question 3.
Consider the argument, “On the basis of position in the electrochemical series, Al does not displace hydrogen from water and acids

Do you agree with the argument?
Substantiate your answer.

Answer:

No.
Metals which are above hydrogen in the electro chemical series can displace H₂ from acids, i.e., elements having lower reduction potential than hydrogen, can liberate H₂ from acids and water. Al displaces hydrogen from water and acids.

Question 4.
How much electricity in Faraday is required to produce

20.0 g of Ca from molten CaCl₂?
45.0 g of Al from molten Al₂O₃?

(Hint: Atomic masses – Ca: 40 u, Al: 27 u)
Answer:
1. The electrode reaction is Ca²⁺(aq) + 2e^– → Ca(s)
Electricity required to produce 40 g Ca = 2 F
Electricity required to produce 20 g Ca = 1 F

2. The electrode reaction is.
Al³⁺(aq) + 3e^– → Al(s)
Electricity required to produce 27 g of Al = 3 F
Electricity required to produce 45 g Al = 5 F

Question 5.
In a debate, a student said “Electrolytes dissolved in water dissociate into constituent ions”. His friend supported this statement and said “The degree of dissociation of an electrolyte increase with the increasing concentration of the electrolyte.”

Do you agree with these statements?
If you agree or not, give the reason.

Answer:

The first statement is correct, but the second statement is wrong.
Degree of dissociation of an electrolyte increases with dilution and not by increasing the concentration.

Question 6.
The standard electrode potential of the elements A, B and C are 0.68, – 2.50 and – 0.50 volts respectively.

Arrange them in the order of their reducing power.
Justify your answer.

Answer:

B < C < A
Greater the value of reduction potential, greater will be the reducing power.

Question 7.
In a standard hydrogen electrode, is it possible to replace the Platinum wire in the glass tube with Aluminium wire? Justify your answer.
Answer:
No. Platinum is an inert electrode and moreover the coefficient of cubical expansion of platinum is equal to that of glass. But, aluminium being an active metal will interfere in the cell reaction.

Question 8.
Listen to the following statements.
“Zinc displaces silver from AgNO₃ solution, because standard oxidation potential of zinc is greaterthan that of silver”.

Is it true?
What happens when we use copper instead of zinc?
Justify your answer.

Answer:

Yes.
Copper also displaces Ag from AgNO₃ solution.
In electrochemical series, Cu lies above Ag. The reduction potential of copper is +0.34 V. But in the case of silver, the reduction potential is +0.80 V.

Question 9.
“Metal having -ve reduction potential when dipped in the solution of its own ion has a tendency to go into the solution”.

Analyse the above statement.
Compare the reduction potential of H⁺ ion with that of other metal ions.

Answer:

The statement is correct.
The reduction electrode potentials of elements lying above hydrogen in the electrochemical series are negative while those of elements lying below hydrogen are positive. The standard potential for the reaction 2H⁺ + 2e^– → H_2(g) is taken as 0.00 V.

Question 10.
Consider the statement ‘Specific conductance increases on dilution’.

Is it true?
What is your opinion about it? Justify.
Define specific conductance and give its equation.

Answer:

No.
This is because, the number of ions per unit volume that carry the current in a solution decreases on dilution.
Specific conductance/Conductivity of a conductor is the conductance offered by the conductor of unit length and unit area of cross section.

k = $\frac{l}{R A}=\frac{1}{R} \times \frac{l}{A}$ where $\frac{l}{A}$ is the cell constant.

Question 11.
Write Nernst equation and calculate the emf of the following cell at 298 K.
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 3 Electrochemistry two mark q11 img 2
Answer:

Question 12.
The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001 M KCl solution at 298 K is 0.146 × 10^-3 S cm^-1?
Answer:
Cell constant = $\frac{\text { Conductivity }}{\text { Conductance }}$
i.e., Cell constant = Conductivity × Resistance
= 0.146 × 10^-3 S cm^-1 × 1500 Ω
= 0.219 cm^-1

Question 13.

Give the name for the iron coated with the Zinc.
Can we store Cl₂ gas in copper vessel? Give reason.
(E°_{(Cu²⁺/ Cu)} = + 0.34 V E°_{(Cl²/Cl^–)} = +1 -36 V)(2)
The equilibrium constant of a reaction which is difficult to calculate can be calculated by knowing the standard potential of the cell in which reaction takes place. Give the mathematical expression relating equilibrium constant and cell potential (2)

Answer:

Galvanised iron
No. Since the reduction potential of chlorine is greater than that of copper, chlorine reacts with copper to form cuprous chloride. E°_{(Cl²/Cl^–)} > (E°_{(Cu²⁺/ Cu)})
E°_cell = $\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log \mathrm{K}_{\mathrm{c}}$

where E°_cell is the standard cell potential (E°_cell = E°_cathode – E°_anode and K_c is the equilibrium constant of the cell reaction.

Plus Two Chemistry Electrochemistry Three Mark Questions and Answers

Question 1.
Corrosion causes huge loss of metals.

Can you suggest two examples of corrosion from everyday life?
Suggest any two factors which enhance the rate of corrosion.
What is the chemical name of rust?

Answer:

Rusting of iron, Tarnishing of silver
two factors which enhance the rate of corrosion.
- Presence of impurities
- Air and moisture
Hydrated ferric oxide (Fe₂O₃.xH₂O)

Question 2.
Write a note on primary cells, secondary cells and fuel cells.
Answer:

Primary cell	Secondary cell
Electrode reaction cannot be reversed.	Electrode reaction can be reversed by an external electric energy source.
Reaction occurs only once & after use they become dead; not chargeable.	Reaction can occur many times in both directions; Rechargeable
Can’t be reused	Can be reused
Cannot be recharged	Can be recharged

Question 3.
Various factors influence the conductance of solutions.

What do you mean by conductance?
Write its formula.
Write the factors influencing it.

Answer:
1. It is also defined as the reciprocal of resistance. It is denoted by C.

2. Conductance, C = $\frac{1}{\mathrm{R}}$
But, R = ρ$\frac{l}{A}$
∴ C = $\frac{1}{\rho} \frac{\mathrm{A}}{\ell}$

3. The factors influencing it.

The nature of the electrolyte added.
Nature of solvent and its viscosity.
Concentration of the solution.
Temperature.

Question 4.
The position of some metals in the electrochemical
series in the decreasing order of electropositive
character is given:
Mg > Al > Zn > Cu > Ag

What would happen if a copper spoon is used to stir a solution of aluminium?
Is there any reaction?
Comment on your response and justify.

Answer:

Nothing will happen. The Al solution can be safely stirred using a Cu spoon.
No.
The Cu spoon will not react with the solution of Al. Greater the value of reduction potential of a metal, more easily it is reduced. The element having low reduction potential is more reactive and is oxidised. Cu has greater reduction potential than Al and is more active metal than Cu.

Question 5.
In the lab a student added a zinc piece to CuSO₄ solution,

What may be his observation?
What did that observation indicate?
Give reason for the above observation.

[Hint: Process related to electrolysis]
Answer:

Zinc pieces will dissolve in CuSO₄ solution and the intensity of the blue colour of CuSO₄ solution decreases.
Zinc displaces copper from copper sulphate solution.
Lower the value of reduction potential more easily is the substance oxidised and greater the value of reduction potential, more easily is the substance reduced. Zinc is more reactive. Hence Zinc displaces copper from copper sulphate solution.

Question 6.
Observe the given figure.

Identify the process.
Identify A and B.
Illustrate the mechanism of this process.

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 3 Electrochemistry three mark q6 img 4

Answer:

Electrolysis.
A: Anode B: Cathode
During electrolysis, initially the electrolyte gets ionised. On passing electricity, the ions migrate to oppositely charged electrodes. Cations move towards cathode and anions move towards anode. Cations gain electrons from cathode while anions give out electrons to anode during discharge. The movement of ions in solution is responsible for conduction of electricity.

Question 7.
In the Chemistry lab the teacher told the students to label some electrolytes suitably. The labelling given by one student is as given:
Strong electrolyte – H₂SO₄, NH₄OH, KOH, Ca(OH)₂
Weak electrolytes – NaOH, HCN, NaCl, HCl

Is it correct? Justify.
Correct it.
What do you mean by electrolysis?

Answer:

No.
Strong electrolytes – H₂SO₄, KOH, HCl, NaOH, NaCl
Weak electrolytes – NH₄OH, HCN, Ca(OH)₂
The process of chemical decomposition of electrolyte by the passage of electricity through its molten or dissolved state is called electrolysis.

Question 8.
This figure shows the corrosion of iron.
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 3 Electrochemistry three mark q8 img 5

What happens to a metal during corrosion – oxidation or reduction?
Explain the electrochemical theory of rusting of iron.
Write two examples for metals which undergo corrosion.

Answer:
1. Oxidation.

2. The impure surface of iron behaves like as electrochemical cell. Pure iron functions as the anode while the impure surface acts as cathode. The electrolyte is moisture containing dissolved O₂ or CO₂.
At anode: 2Fe(s) → 2Fe²⁺(aq) + 4e^–
At cathode: 4H⁺(eq) + O₂(g) + 4e^– → 2H₂O(l)
The overall reaction is
2Fe(s) + O₂(g) + 4H⁺(aq) → 2Fe²⁺(aq) + 2H₂O(l)
Ferrous ions so formed are oxidised by atmospheric oxygen to ferric ions. The rust formed is chemically hydrated ferric oxide (Fe₂O₃.xH₂O)

2Fe²⁺(aq) + 2H₂O + 1/2 O₂ → Fe₂O₃ + 4H⁺
Fe₂O₃ + xH₂O → Fe₂O₃.xH₂O (rust)

3. Iron and Copper.

Question 9.
In an experimental set-up for the measurement of potential of a half-cell using the reference electrode and a salt bridge, the salt bridge was suddenly removed.

What happens to the voltage when the salt bridge is removed?
Justify your answer.
Write any two examples of inert electrolytes used for the construction of salt bridge.

Answer:

Voltage ceases.
When the salt bridge is removed the migration of ions between the half cells is hindered which results in the accumulation of charge near the electrodes. This decreases the cell voltage.
KCl, NH₄NO₃

Question 10.
In the class, teacher wrote some ions on the black board. It is given below.
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 3 Electrochemistry three mark q10 img 6

Which among the ions, gain electron more easily?
Is it true that the reduction potential of hydrogen is less than that of Na and Mg?
Justify your answer.

Answer:

Mg²⁺
No. The reduction potential of hydrogen is greater than that of both Na and Mg.
In electrochemical series, sodium and magnesium lie above hydrogen. Standard reduction potential of hydrogen electrode is taken as zero volt.

Question 11.
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 3 Electrochemistry three mark q11 img 7

What does this figure represent?
What is its potential?
How can we construct this?

Answer:

SHE (Standard Hydrogen Electrode) Or NHE (Normal Hydrogen Electrode).
Its potential is taken as zero.
It consists of a platinum wire sealed into a glass tube carrying a platinum foil, coated with finely divided Pt, at one end. The electrode is dipped in an aqueous acid solution of 1M concentration. Pure hydrogen gas at 1 bar pressure is bubbled through the solution at 298 K.

Question 12.
Given the standard electrode potential values
E° _(Al³⁺/Al) = -1 -66 v E°_(Zn²⁺/Zn) = -0.764 V.

Draw a neat diagram of galvanic cell using the above electrodes and write the cell reaction.
Calculate the emf of the above cell.

Answer:
1.
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 3 Electrochemistry three mark q12 img 8

2. Cell emf, E_cell = E°_(Zn²⁺/Zn) – E°_(Al³⁺/Al)
= – 0.764 – (-1.66) = + 0.896 V

Question 13.
Depict the galvanic cell in which the reaction Zn(s) + 2Ag⁺(aq) → Zn²⁺ + 2Ag(s) takes place. Further show:

Which of the electrodes is negatively charged?
The carriers of current in the cell.
Individual reaction at each electrode.

Answer:
The electrochemical cell which will involve the given cell reaction is depicted as Zn(s) | Zn²⁺(aq) || Ag⁺(aq) | Ag(s)

In this cell, the electrode Zn(s) | Zn²⁺(aq) will act as negative terminal and Ag | Ag⁺(aq) electrode will act as positive terminal.
The conventional current will flow from silver to zinc electrode in the external circuit.
Individual reaction at each electrode.
- Cathodic reaction: 2Ag⁺(aq) + 2e^– → 2Ag(s)
- Anodic reaction: Zn(s) → Zn²⁺(aq) + 2e^–

Question 14.
How much electricity in terms of Faraday is required to produce

20.0 g of Ca from molten CaCl₂?
40.0 g of Al from molten Al₂O₃?

Answer:
1. The electrode reaction is
Ca²⁺+ 2e^– → Ca
Electricity required to produce 40 g of Ca = 2 F
Electricity required to produce 20 g of Ca = 1 F

2. The electrode reaction is Al³⁺ + 3e^– → Al
Electricity required to produce 27 g of Al = 3 F
Electricity required to produce 40 g of Al = $\frac{3 \mathrm{F} \times 40}{27}$ = 4.44 F

Question 15.
We can calculate the molar conductivity of NH₄OH at infinite dilution if Λ° of Ba(OH)₂, BaCl₂ and NH₄Cl are known.

Can you agree with this statement? Explain the law behind it.
Give the equation for Λ° of NH₄OH from the above data.

Answer:
1. Yes. According to Kohlrausch’s law the molar conductivity of an electrolyte at infinite dilution is equal to the sum of the molar ionic conductivities of the cations and anions at infinite dilution.

Question 16.

Represent the cell in which the following reaction takes place.
Mg(s) + 2Ag²⁺(0.0001M) → Mg²⁺(0.130M) + 2 Ag(s). Calculate its E_cell if E°_cell = + 3.17 V.
Discuss the variation of molar conductivity with concentration for acetic acid and KCI.

Answer:
1. Mg(s) | Mg²⁺(0.130M) || Ag⁺(0.0001M) | Ag(s)

2. Conductivity decreases with dilution because the number of ions per unit volume that carry the current in a solution decreases on dilution. For strong electrolytes molar conductivity increases steadily with increase in dilution due to decrease in interionic attraction. For weak electrolytes molar conductivity increases with dilution due to increase in degree of dissociation.
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 3 Electrochemistry three mark q16 img 10

Question 17.

Write the relation between molar conductance and molarity of an electrolyte solution.
λ_m^∞ of NaCl, HCl and CH³COONa are 126.4, 425.9 and 91 S cm² mol^-1 respectively. Calculate the λ_m^∞ of acetic acid.

Answer:
1. Λ_m = $\frac{1000 \mathrm{k}}{\mathrm{M}}$

2. Λ°_CH₃COOH = Λ°_CH₃COONa + Λ°_HCl – Λ°_NaCl
= 91 + 425.9 – 126.4
= 390.5 Ω^-1 cm² mol^-1

Question 18.
Certain cells are given below.

Classify them into primary, secondary and fuel cells. (Mercury cell, Nickel-Cadmium cell, Lead storage battery, H_2- O₂ cell, Dry cell)
Represent the cell in which the overall reaction is,
Cd + 2Ag⁺ → Cd²⁺ + 2Ag

Answer:
1. Primary cells: Mercury cell, Dry cell
Secondary cells: Nickel-Cadmium cell, Lead storage battery.
Fuel cell : H₂ – O₂ cell

2. Cd → Cd²⁺ + 2e^– (Oxidation half cell)
2Ag⁺ + 2e^– → 2Ag (Reduction half cell)
Therefore the cell is represented as Cd | Cd²⁺ || Ag⁺ | Ag

Plus Two Chemistry Electrochemistry Four Mark Questions and Answers

Question 1.
Every conductivity cell has a definite value for its cell constant.

What is cell constant?
How is it related with specific conductance?
Calculate the cell constant of a conductivity cell containing 0.05M KCl solution, whose resistance is 100 ohm and conductivity is 0.0075 ohm^-1 cm^-1.

Answer:
1. The quantity $\ell / \mathrm{A}$ is called cell constant.
A = Area of cross section of the electrodes.
$\ell$ = distance of separation of two electrodes.

2. Specific conductance = Cell constant × Conductance

3. Cell constant = Specific conductance × Resistance
= 0.0075 ohm 1 cm^-1 × 100 ohm = 0.75 cm^-1

Question 2.

Explain the working of a H₂ – O₂ fuel cell.
What are the advantages of fuel cells over other cells?

Answer:
1. In H₂ – O₂ fuel cell, hydrogen and oxygen are bubbled through porous carbon electrodes into concentrated aqueous NaOH solution, which acts as the electrolyte. The electrode reactions are,
Anode: 2H₂(g) + 4 OH^–(aq) → 4 H₂O(l) + 4 $\bar { e }$
Cathode: O₂(g) + 2 H₂O(l) + 4 $\bar { e }$ → 4 OH^–(aq)
Overall reaction: 2H₂(g) + O₂(g) → 2 H₂O(l)

2. The advantages of fuel cells over other cells

They are pollution free.
More efficient than conventional methods.
Runs continuously as long as the reactants are supplied.
Electrodes are not affected.

Question 3.
Consider the equation, λ_m^∞HCl = λ_m^∞H⁺ + λ_m^∞Cl^–

Is the above equation correct?
Explain the law.
Write the application of this law.

Answer:
1. The above equation is correct.

2. Kohlrausch’s law. Molar conductance at infinite dilution of an electrolyte is the sum of the molar ionic conductivities at infinite dilution of the cations and anions present in the solution.

3. The application of this law:

To find the molar conductivity at infinite dilution (Λ°m) of weak electrolytes.
To find the degree of dissociation (α) of weak electrolytes.
α = $\frac{\Lambda_{m}}{\Lambda_{m}^{0}}$
To determine the dissociation constant of weak electrolytes.
Ka = $\frac{\mathrm{c}_{\alpha}^{2}}{1-\alpha}$

Question 4.
1. Depict the galvanic cell in which the reaction Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s) takes place.
2. Further show

Which of the electrodes is negatively charged?
The carriers of current in the cell.
Individual reaction at each electrode.

Answer:
1. Zn(s) | Zn²⁺(aq) || Ag⁺(aq) | Ag(s)

2. Further show

Zn electrode is negatively charged (Anode).
Ions
Individual reaction at each electrode.
- At anode : Zn(s) → Zn²⁺(aq) + 2e^–
- At cathode : 2Ag⁺(aq) + 2e^– → 2Ag(s)

Question 5.

It is not possible to determine the molar conductivity of weak electrolytes at infinite dilution graphically. Do you agree with this? Justify your answer.
Calculate the molar conductivity of NH₄OH at infinite dilution if molar conductivity at infinite dilution of Ba(OH)₂, BaCl₂, and NH₄Cl are 523.4, 280 and 130 ohm^-1 cm² mol^-1 respectively.

Answer:
1. Yes. I agree with this statement because for weak electrolytes such as acetic acid, molar conductivity increases rapidly with dilution.
Hence, the plot of Λ_m against $\sqrt{\mathrm{C}}$ (C = concentration of electrolyte) is not linear. Therefore the value of Λ°_m for weak electrolytes cannot be obtained by extraploting the graph.
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 3 Electrochemistry four mark q5 img 11

Question 6.
Standard reduction potentials of some elements are given below.
Al = – 1.66 V, Cd = – 0.40 V, Ag = + 0.80 V & Au = + 1.36 V

Which of these can displace hydrogen from hydrochloric acid. Why?
Construct a galvanic cell using any two electrodes. Write the cell reaction and calculate the standard emf.

Answer:
1. Al and Cd can displace hydrogen from HCl. Metals with low reduction potential than H (i.e., with -ve value) can liberate H₂ from acids.

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 3 Electrochemistry four mark q6 img 12

Question 7.
Molar conductance at infinite dilution of acetic acid cannot be determined experimentally.

How can it be determined?
State the law.
Give two sets of electrolytes which can be used to find out the molar conductance of acetic acid at infinite dilution.

Answer:

It can be determined using Kohlrausch’s law of independent migration of ions.
Molar conductivity at infinite dilution of an electrolyte is the sum of molar ionic conductivities of the cation and anion at infinite dilution.
CH₃COONa, HCl, NaCl & CH₃COOK, HCl, KCl.

Question 8.
An electrochemical cell is a combination of two half cells consisting of a metal with its own ions in solution.

Represent the Daniell cell.
E°_cathode and E°_anode for the above cell are + 0.34 V and – 0.76 V respectively. Calculate the E°_cell?
Give the graphical interpretation of the following graph.

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 3 Electrochemistry four mark q8 img 13
Answer:

Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)
E°_cell = E°_cathode – E°_anode = (-0.34V) – (-0.76V) = +1.1V
The graph indicates the variation of conductance with dilution for strong and weak electrolytes. For both of them Λ_m increases with dilution. For strong electrolytes like KCl, the graph is a straight line since the variation is linear. But for weak electrolytes like CH₃COOH the graph is a curve due to step increase in the value of molar conductance near infinite dilution.

Question 9.
The standard electrode potential of some electrodes are given below.

Ag⁺/Ag = + 0.80 V
Fe²⁺/Fe = – 0.44 V
Cu²⁺/Cu = + 0.34 V
Al³⁺/Al = – 1.66 V

Construct 4 possible electrochemical cells and calculate the cell potential of the corresponding cells?
Answer:
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 3 Electrochemistry four mark q9 img 14

Question 10.

What do you mean by cells?
What is the difference between primary cells and secondary cells?
Fill in the following table:

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 3 Electrochemistry four mark q10 img 15
Answer:
1. Cells are the devices in which chemical energy is converted into electrical energy.

Primary cell	Secondary cell
Electrode reaction cannot be reversed.	Electrode reaction can be reversed by an external electric energy source.
Reaction occurs only once & after use they become dead; not chargeable.	Reaction can occur many times in both directions; Rechargeable
Can’t be reused	Can be reused
Cannot be recharged	Can be recharged

a – Zinc
b – Graphite
c – Zinc-Amalgam
d – HgO and Carbon
e – Paste of NH₄Cl and ZnCl₂
f – Paste of KOH and ZnO

Question 11.
Iron gets destroyed due to continuous contact with the environment.

Name the phenomenon.
Suggest some prevention methods.
Explain any one prevention method

Answer:
1. Corrosion

2. Some prevention methods

By using non-metallic coating
By using metallic coating
Cladding
Cathodic protection
Alloy formation

3. Metallic coating or electroplating iron is coated with zinc. Iron thus obtained is called galvanised iron. Coating of an electropositive metal on the metal surface.

Question 12.
Corrosion causes enormous damage to various articles made of metal.

What do you mean by corrosion?
What is barrier protection?
Name any two metals which can be used as sacrificial electrodes in the electrochemical method of protecting iron.

Answer:
1. Destruction of a metal.

2. In this method, a barrier film is introduced between metal and atmospheric oxygen and moisture. Painting or with a thin film of oil or grease electroplaing.

3. Two metals which can be used as sacrificial electrodes in the electrochemical method of protecting iron

Question 13.
Scientists suggest fuel cells as an alternative for conventional methods of power generation.

What are fuel cells?
Give the overall cell reaction of the H₂ – O₂ fuel cell. Also draw a neat labelled diagram of this cell.
What are the advantages of the fuel cells?

Answer:
1. Fuel cells are Galvanic cells designed to convert the energy of combustion of fuels directly into electrical energy.

2. The overall reaction taking place in an H₂ – O₂ fuel cell is 2H₂(g) + O₂(g) → 2H₂O(l)
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 3 Electrochemistry four mark q13 img 16

3. Advantages:

High efficiency
Steady voltage
Pollution free

Question 14.
Certain cells are given below.

In Daniel cell, if Zn electrode is replaced by Mg what happens? Why?
Zinc displaces copper from copper sulphate solution while copper doesn’t replace zinc from zinc sulphate solution. Justify your answer.

Answer:
1. If Zinc electrode is replaced by Mg, a greater voltage (> 1.1 V) will be produced by the cell. Because Mg lies above Zn in electrochemical series.
E°_Mg²⁺Mg) = -2.36 V
E°_Zn²⁺Zn) = -0.76 V
i. e., Mg is a stronger reducing agent than Zn.

2. Zinc lying above hydrogen in the electrochemical series has relatively greater tendency to lose electrons. Hence Zn is capable of displacing hydrogen from acids. Moreover, a more active metal Zn, displaces, a less reactive copper from the salt solution. But Cu, being less active than Zn cannot displace Zn from its salt, ZnSO₄.

Plus Two Chemistry Electrochemistry NCERT Questions and Answers

Question 1.
Depict the galvanic cell in which the reaction Zn(s) + 2Ag⁺(aq) → Zn²⁺ + 2Ag(s) takes place. Further show:

Which of the electrodes is negatively charged?
The carriers of current in the cell.
Individual reaction at each electrode.

Answer:
The electrochemical cell which will involve the given cell reaction is depicted as Zn(s) | Zn²⁺(aq) || Ag⁺(aq) | Ag(s)

1. In this cell, the electrode Zn(s) | Zn²⁺(aq) will act as negative terminal and Ag | Ag⁺(aq) electrode will act as positive terminal.

2. The conventional current will flow from silver to zinc electrode in the external circuit.

3. Individual reaction at each electrode:

Cathodic reaction: 2Ag⁺(aq) + 2e^– → 2Ag(s)
Anodic reaction: Zn(s) → Zn²⁺(aq) + 2e^–

Question 2.
The resistance of a conductivity cell containing 0.001
M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001 M KCl solution at 298 K is 0.146 × 10^-3 S cm^-1?
Answer:
Cell constant = $\frac{\text { Conductivity }}{\text { Conductance }}$
i.e., Cell constant = Conductivity × Resistance = 0.146 × 10^-3 S cm^-1 × 1500 Ω
= 0.219 cm^-1

Question 3.
How much electricity in Faraday is required to produce

20.0 g of Ca from molten CaCl₂?
40.0 g of Al from molten Al₂O₃?

Answer:
1. The electrode reaction is
Ca²⁺(aq) + 2e^– → Ca(s)
Electricity required to produce 40 g Ca = 2 F
Electricity required to produce 20 g Ca = 1 F

2. The electrode reaction is
Al³⁺(aq) + 3e^– → Al(s)
Electricity required to produce 27 g of Al = 3 F
Electricity required to produce 40 g Al = $\frac{3 F \times 40}{27}$ = 4.44 F

Question 4.
The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm^-1. Calculate the molar conductivity.
Answer:
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 3 Electrochemistry one mark q8 img 1

Plus Two Computer Science Chapter Wise Questions and Answers Chapter 3 Data Structures and Operations

April 13, 2021January 31, 2020 by Prasanna

Students can Download Chapter 3 Data Structures and Operations Questions and Answers, Plus Two Computer Science Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus Two Computer Science Chapter Wise Questions and Answers Chapter 3 Data Structures and Operations

Plus Two Computer Science Data Structures and Operations One Mark Questions and Answers

Question 1.
Write down the full form of FIFO.
Answer:
First In First Out

Question 2.
A queue is implemented in a ………… manner
(a) fifo
(b) filo
(c) life
(d) lilo
Answer:
(a) fifo

Question 3.
People waiting in a cinema theatre counter for taking tickets is an example for …………
(a) stack
(b) queue
(c) array
(d) none of these
Answer:
(b) queue

Question 4.
A linked list is a linear collection of data elements called ………….
Answer:
nodes

Question 5.
A stack can be grow or shrink, so it can be considered as ……………… data structure.
Answer:
dynamic

Question 6.
The data structure in which elements are arranged in non-sequence is named as type of data structure.
(a) Heterogeneous data structure
(b) Synthetic data structure
(c) Linear data structure
(d) Non-linear data structure
Answer:
(c) Linear data structure

Question 7.
Consider the following two statements.
(i). Memory of size dynamic data structures can be changed during execution.
(ii). Static data structures are associated with primary memory
(a) Statement (i) and statement (ii) are not true
(b) Statement (i) is true and statement (ii) is false
(c) Statement (i) is false and statement (ii) is true
(d) Statement (i) and statement (ii) are false
Answer:
(a) Statement (i) and statement (ii) are not true

Question 8.
……………… is a data structure in which items are added at one end and removed from the other,
(a) Stack
(b) Queue
(c) List
(d) None of the above
Answer:
(b) Queue

Question 9.
…………………. is very useful in situation when data have to be stored and then retrieved in reverse order,
(a) Stack
(b) Queue
(c) List
(d) Link list
Answer:
(a) Stack

Question 10.
Which of the following data structure is linear type?
(a) Graph
(b) Trees
(c) Binary tree
(d) Stack
Answer:
(d) Stack

Question 11.
Combining the elements of two sorted data structures to form a new one is referred as
(a) Merging
(b) Sorting
(c) Traversal
(d) Searching
Answer:
(a) Merging

Question 12.
Which term is applicable with stack data structure?
(a) LILO
(b) FILO
(c) LIFO
(d) FIFO
Answer:
(c) LIFO

Question 13.
Placing glasses one above another can be considered similar to ……….. data structure.
(a) Queue
(b) Stack
(c) Record
(d) Graph
Answer:
(b) Stack

Question 14.
Each node containing data and a pointerto the next node is applicable with …………. data structure.
(a) Array
(b) Linked List
(c) Stack
(d) Queue
Answer:
(b) Linked List

Question 15.
Identify and correct mistake in the following Stack – PUSH algorithm,
start
if stack is full
return null
endif
top = top -1
stack[top] = data
stop
Answer:
top = top – 1 is to be replaced by top = top + 1

Question 16.
People waiting in a cinema theatre counter for taking tickets is an example for ………………
(a) stack
(b) queue
(c) array
(d) none of these
Answer:
(b) queue

Plus Two Computer Science Concepts of Object-Oriented Programming Two Mark Questions and Answers

Question 1.
A word, say “computer” is stored in an array. Another array is to be created by storing the reverse of the word. How can stack support you to perform this task? Explain the algorithm.
Answer:
This can be performed by pushing each character of the word “computer” onto a stack as it is read. After the word is finished, then the characters are popped off the stack, so they will come in the reverse order such as “retupmoc” in the desired output.

Question 2.
Name the principle by which tickets are issued in a Cinema Ticket counter. Which data structure supports this principle?
Answer:
It is an example for Queue. Here the persons are added at the back end and tickets are issued to the front end. The FIFO method is used.

Question 3.
Define stack. Write down short notes about stack.
Answer:
A stack is a linear structure in which items can be added or removed only at one end called top. Add an item into the stack is called push and deleting an item from the stack is called pop.

Question 4.
Write short notes about queue.
Answer:
A queue is also a linear structure in which the items can be added at the back end called rear and the items are removed from the front end. There are two operations adding an item at the rear end and removing an item from the front end.

Question 5.
Write an algorithm for Push operation on a stack;
Answer:

Step 1: If top = N. Then print “OVERFLOW” and return
Step 2: Set top = top + 1
Step 3: Set Stack[top] = item
Step 4: stop

Question 6.
Write an algorithm for Pop operation on a stack;
Answer:

Step 1: If top = Null. Then print “UNDERFLOW” and return
Step 2: Set item=Stack[top]
Step 3: Set-top = top – 1
Step 4: stop

Question 7.
A linked list containing all the name of students in your class is to be created. Write its C++ structure to define the node.
Answer:
struct node
{
char name[15];
node *link;
}

Question 8.
Write a short note about circular queue.
Answer:
ClrcularQueue Is a linear data structure in which the operations are performed based on FIFO (First In First Out) principle and the last position is connected back to the first position to make a circle.

Question 9.
With suitable diagrams and notes, Explain the different classifications of data structures.
Plus Two Computer Science Chapter Wise Questions and Answers Chapter 2 Concepts of Object-Oriented Programming img1

Plus Two Computer Science Concepts of Object-Oriented Programming Three Mark Questions and Answers

Question 1.
Write an algorithm for inserting an element into a queue
Answer:

Step 1: If front = 1 and rear = N or front =rear + 1. Then print “OVERFLOW” and return
Step 2: If front = Null then Set front = 1 and rear = 1 Else if rear = N then set rear * 1 Else Set rear* rear + 1 End if
Step 3: Set Queue[rear] = item
Step 4: stop

Question 2.
Write an algorithm for deleting an element from a queue
Answer:

step 1: If front = Null then print “UNDERFLOW” and return
Step 2: Set item = Queue[front]
Step 3: If front = rearthen Set front = Null and rear = Null Else if front = N then set front = 1 Else Set front = front + 1 End if
Step 4; stop

Question 3.
Write the steps for inserting a new element to a queue.
Answer:

Step 1: If front = 1 and rear = N or front = rear +1 Then print “ OVERFLOW and return
Setp 2: If front = Null then Setfront = 1 and rear = 1 Else if near = N then Set rear * 1 Else Set rear* rear + 1 End if
Step 3: Set queue [rear] * item.
Step 4: Stop

Question 4.
Write the steps for deleting an element from a queue
Answer:
Deletion operation
It is. the process of deleting(removing) a data item. from the queue from the front. If the queue is empty ‘ v ‘.and we try to delete an item from the queue makes 1 the queue underflow. Algorithm is given below

Step 1: If front = Null then print “UNDERFLOW: and return
Step 2: Set item = Queue [front]
Step 3: If front = Null and rear = Null Else if front N then set front = 1 Else Set front = front + 1 End if
Step 4: stop

Question 5.
Write the steps for deleting a node from a linked list?
Answer:
Deletion from a linked list
It is the removal of a node from the data structure.

Step 1: Get the address of the previous node (POS – 1) and next node (POS + 1) in the pointers PreNode and PostNode respectively.
Step 2: Copy the contents of PostNode into the link part of node at position (POS – 1).
Step 3: Free the node at position POS.

Plus Two Computer Science Concepts of Object-Oriented Programming Five Mark Questions and Answers

Question 1.
Prepare a short note about all the operations associated with data structures:
Answer:
Operations on data structures

Traversing – Accessing or visiting or reading all elements of a data structure is called Traversal.
Searching – Searching is the process of finding elements in a data structure
Inserting – It is the process of adding new data at particular location is called insertion.
Deleting – It is the process of removing a particular element from the data structure.
Sorting – Arranging elements in a particular order(ascending or descending) is called sorting. Examples are bubble sort and selection sort.
Merging – This is the process of combining two sorted data structure and form a new one.

Question 2.
You are given a queue containing 50 students. Who will get the first chance and who will be irt the last position if the queue is in linear form.

In which position you add a new student?
In which problem you remove a student?
Which operations are performed for this purpose?
Write all steps for adding new students to a queue
Write all steps for removing a student from the queue?

Answer:
1. At the back or rear position
2. From the front position
3. Add a new student to the queue is insertion and removing a student from the queue is called deletion
4.

Step 1: If front =1 and rear=N or front =rear+1. Then print “OVERFLOW” and return
Step 2: If front = Null then Set front = 1 and rear =1 Else if rear 3 N then set rear = 1 Else Set rear = rear+1 End if
Step 3: Set Queue[rear]=item Step 4: Stop

Step 1: If front = Null then print “UNDERFLOW and return
Step 2: Set item = Queue[front]
Step 3: If front = rear then Set front Null and rear =Null Else if front N then set front = 1Else Set front = front + 1 End if
Step 4: stop

Question 3.
Class teacher said to the class leader to collecting their observation books and place them in a table one by one. Teacher should take books one by one from this group always from the top. Suppose that books be arranged in linear form from bottom to top

In which position a new book can be placed?
How can a book be taken from this group?
Which operations are performed for this purpose?
Write all steps for adding new books to this group or
Write all steps for removing books one by one from this group.

Answer:
1. At the top
2. From the top.
3. To place a new book it is push operation and taking books from the stack is pop operation
4.

Step 1: If top = Null. Then print “UNDERFLOW” and return
Step 2: Set item = Stack[top]
Step 3: Set-top = top – 1
Step 4: stop

Step 1: If front = 1 and rear=N or front =rear+1. Then print “OVERFLOW and return”
Step 2: If front = Null then Set front = 1 and rear =1 Else if rear = N then set rear = 1 Else Set rear = rear + 1
End if

Plus Two Computer Science Chapter Wise Questions and Answers Chapter 2 Concepts of Object-Oriented Programming

April 13, 2021January 30, 2020 by Prasanna

Students can Download Chapter 2 Concepts of Object-Oriented Programming Questions and Answers, Plus Two Computer Science Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus Two Computer Science Chapter Wise Questions and Answers Chapter 2 Concepts of Object-Oriented Programming

Plus Two Computer Science Concepts of Object-Oriented Programming One Mark Questions and Answers

Question 1.
Protecting data from access by unauthorized function is Data
(a) Polymorphism
(b) encapsulation
(c) data abstraction
(d) Inheritance
Answer:
(b) encapsulation

Question 2.
The act of partitioning a program into individual compoents is called,
(a) Polymorphism
(b) encapsulation
(c) data abstraction
(d) modularity
Answer:
(d) modularity

Question 3.
Which of the following is not an OOP concept?
(a) Overloading
(b) procedural programming
(c) data abstraction
(d) Inheritance
Answer:
(b) Procedural programming

Question 4.
The ability for a message or data to be processed in more than one form is called 6303 message or data.
(a) Polymorphism
(b) encapsulation
(c) data abstraction
(d) Inheritance
Answer:
(a) Polymorphism

Question 5.
C++ is a ……. language.
(a) Qbject based
(b) non-procedural
(c) Object-oriented
(d) procedural
Answer:
(c) Object-oriented

Question 6.
Which of the following is not a characteristic of OOP?
(a) It emphasises more on procedure rather than data.
(b) It models real world
(c) It wraps up related data items and associated functions in the unit.
(d) None of the above
Answer:
(a) It emphasises more on procedure rather than data.

Question 7.
Which among the following is true about OOPs.
(a) It supports data abstraction
(b) It supports polymorphism
(c) It supports structured programming.
(d) It supports all the above
Answer:
(d) It supports all the above

Question 8.
Identify the correct statements from the following.
(a) Procedural programming approach aims more at procedures.
(b) The object-oriented approach views a problem in terms of objects involved.
(c) Real life cannot be modeled in OOP.
(d) Procedures takes a secondary status in object-oriented approach.
Answer:
(a), (b), (d)

Question 9.
“One form many behaviours” represents
(a) modularity
(b) Abstraction
(c) polymorphism
(d) Inheritance
Answer:
(c) polymorphism

Question 10.
What do you understand by programming paradigm?
Answer:
Paradigm means organizing principle of a program. It is an approach to programming.

Question 11.
What are the characteristics of procedural paradigm? Procedural programming.
Answer:
In procedural paradigm, the emphasis is on doing things. Data is the reason for a program’s existence. Yet data is given second class status while programming.

Question 12.
Which of the following is not a programming paradigm?
Answer:
(a) Object oriented
(b) Procedure oriented
(c) Both a and b
(d) Standard
Answer:
(d) Standard. This is not a programming paradigm

Question 13.
In programming, modules are composed of
(a) data types
(b) functions
(c) statements
(d) programs
Answer:
(b) functions

Question 14.
The following function prototypes implement an OOP concept. Identify the concept and explain it.
Answer:
Polymorphism

Plus Two Computer Science Concepts of Object-Oriented Programming Two Mark Questions and Answers

Question 1.
State whether the following statements are true or false.

In procedural oriented programming, data is the reason for a program’s existence.
The real word concept gets simplified using concept of abstraction.

Answer:
1. False. In procedural programming, the emphasis is on doing actions or operations. Here data is not secure.

2. True. The real world concept gets simplified using concept of abstraction. In object-oriented programming the problems are viewed in terms of objects involved. Thus modeling real-world becomes easy in object oriented programming. Objects essential features alone are represented. This property is known as data abstraction.

Question 2.
How does inheritance support ‘reusability’?
Answer:
Inheritance allows the addition of additional features to an existing class without modifying it. One can derive a new class (subclass) from an existing one and add new features to it. There is reduction in amount of typing and efforts.

Question 3.
Define the termbase class, subclass and the relationship between a base class and subclass.
Answer:
Inheritance is the capability of a class to inherit properties from another class. That class that inherits from other class is subclass and the other class is base class. For eg: when we say that the class ‘student’ inherits from the class ‘person’, then ‘person’ is a base class of ‘student’ and ‘student’ is a subclass of ‘person’.

Question 4.
How the data is hidden and safe if encapsulation is implemented?
Answer:
Encapsulation is the way of combining both data and the functions that operate on that data. Encapsulation is most often achieved through information hiding, which is the process of hiding all the secrets of an object that do not contribute to its essential characteristics. The structure of an object is hidden.

Question 5.
Differentiate object and Class.
Answer:
An object is an identifiable entity with some characteristics and behavior. It represents an entity that can store data and its associated functions. A class is a group of objects that share common properties and relationships. It represents a group of similar objects

Question 6.
Distinguish class and object with the help of examples.
Answer:

Class – A class is a collection of objects with similar attributes.
Object – An object is an instance of the class. Furniture class is an example for class. Chair, Table, etc. are the instances ie. objects of the class furniture

Question 7.
When we switch on the computer, booting takes place. We do not know what all things are happening inside. Identify the OOP concept that resembles this scenario. Give another situation that resembles this concept.
Answer:
Data abstraction: It refers to the act of representing essential features without including the background details. Switchboard is an example for this.

Question 8.
Which of the following members of a C++ class data type are hidden from accessing?
(a) Private and protected members
(b) Private and public members
(c) Public and protected members
(d) Private, public and protected members
Answer:
(a) Private and protected members

Question 9.
Explain how operator overloading implements polymorphism.
Answer:
Polymorphism is the ability for a message or data to be processed in more than one form. This is achieved by function overloading, operator overloading and dynamic binding.

Plus Two Computer Science Concepts of Object-Oriented Programming Three Mark Questions and Answers

Question 1.
You have seen a flask. When you put some ice-cubes in it, it will stay there for hours without melting and when you pour some hot water it will stay hot.

Name the OOP feature that we can correlate with this situation.
Write a short note about that feature

Answer:
1. Polymorphism

2. Polymorphism is an essential feature of object oriented programming. Polymorphism means having many forms. It is the capability of an object to behave differently in response to a message or action. That is, the same operation is performed differently depending upon the type of data it is working with. Polymorphism is implemented in C++ in two ways, compile time polymorphism and runtime polymorphism

Question 2.
Suppose your school is selected as the venue for conducting District youth festival. For the smooth functioning of the programme different committees are formed and each committee is assigned separate duties.

Which OOP feature can be correlated to this situation?
Write a brief note on that feature

Answer:
1. Modularity

2. When programs become larger, the program is broken into functions or modules. A module is separate unit in itself. Each unit can be compiled separately. All modules work hand in hand in order to achieve the program’s goal. The act of partitioning a program into individual unit is called modularity. It reduces program’s complexity.

Question 3.
You have seen the enquiry machine in a railway station. The user just types in a few details like train number, station, and date. The details of train, seat availability, fare, etc. will be displayed. The user is not aware of its working. Which OOP feature can you correlate with this situation?
Answer:
Data Abstraction:
Data abstraction is an important feature of OOP. It refers to the act of representing essential features without including the background details or explanations. Abstraction supports data hiding so that only the relevant information is expressed to the user and the rest of the information remains hidden from the user.

Question 4.
The famous luxury ship ‘TITANIC’ has provided all facilities like food, entertainment and medical aid available at that time for its passengers. Can you correlate this with any of the OOP features you have left. Justify.
Answer:
Encapsulation:
Encapsulation is the most fundamental concept of OOP. Placing data and functions together is the central idea of object oriented programming. This is known as Encapsulation

Question 5.
Using a real world example explain the concept of inheritance.
Answer:
Parent child relationship is the best example of inheritance. Child derives properties of parents. C++ supports inheritance.

It is the capability of one class to inherit properties from another class. In other words, inheritance is the process of creating new classes called derived classes, from existing base classes. The derived class inherits all the properties of base class but can add features of its own. The base class is unchanged by this process.
Plus Two Computer Science Chapter Wise Questions and Answers Chapter 2 Concepts of Object-Oriented Programming img1

Question 6.
In an ATM machine the user will be given options to access their account details, withdraw money, deposit money, etc. only. The background details are not shown to the user.

Name the object oriented feature that we can correlate with this situation.
Briefly explain about the feature.

Answer:
1. Data abstraction

2. Abstraction refers to the act of representing essential features without including the background details or explanations. Abstraction supports data fiding so that only the relevant information is expressed to the user and the rest of information remains hidden from the user.

Question 7.
Consider the diagram. Can you correlate any OOP feature from this? Write a note on the feature.
Plus Two Computer Science Chapter Wise Questions and Answers Chapter 2 Concepts of Object-Oriented Programming img2
Answer:
Data Encapsulation. Placing data and functions together is the central idea of object oriented programming. This is encapsulation. It is a way to implement abstraction by wrapping up data and associated functions into a single unit.

Question 8.
Consider the following statements.
6 + 9 = 15 but ‘x’ + ‘yz’ = ‘xyz’ Which OOP feature can you infer from the above. Explain.
Answer:
Polymorphism. Polymorphism is the ability for a message or data to be processed in more than one form. The same operation is performed differently depending upon type of data it is working with.

Question 9.
Abstraction and encapsulation are complementary concepts of OOP. Justify.
Answer:
Abstraction and encapsulation are complementary concepts of OOP. Abstraction refers to the act of representing essential features without including the background details. Encapsulation is wrapping up of data and functions into a single unit. Abstraction focuses upon the observable behaviour of an object, whereas encapsulation focuses upon the implementation that gives rise to this behaviour

Question 10.
Write four reasons for the increasing complexity of procedural programming.
Answer:
Following are the main reasons of increasing the procedural language complexity.

Data is undervalued
Adding new data requires modifications to all/many functions.
Creating new data types is difficult
Provides poor real world modeling

Question 11.
List the advantages of Object Oriented Programming.
Answer:

OOP allows modularity (divide the large programs into smaller ones.)
it is good for defining abstract data types
It allows data abstraction
It allows code reusability
Real life entities can be easily created
It supports to create new data types.

Question 12.
Identify a suitable attribute from the 2nd column and an associated behaviour from 3^rd column for each object in the 1^st column.
Plus Two Computer Science Chapter Wise Questions and Answers Chapter 2 Concepts of Object-Oriented Programming img3
Answer:
i) → b → 4
ii) → d → 3
iii) → a → 2

Question 13.
“School kalolsavam” is organised with the help of various committees. Each committee is assigned with a specific task. Identify the OOP concept that is analogous to this situation. Give another real life case that resembles with this concept.
Answer:
The OOP concept used in this situation is modularity. Consider the situation to fill up the application forms of students in a class. Suppose one application form takes nearly 5 minutes of time.

So a class teachers will take 5 * 50 = 250 minutes to fill up the application forms of the entire class. Instead of this, the teacher distributes the application forms to the students, that takes only 5 minutes and teachers gives 5 minutes instructures to the students how to fill up the forms:

Next 5 minutes enough for the students to complete the work. After that the next 5 minutes to return back the filled up application forms. So the entire work will take only 5 + 5 + 5 + 5 = 20 minutes

Question 14.
Look at the following figure:
Plus Two Computer Science Chapter Wise Questions and Answers Chapter 2 Concepts of Object-Oriented Programming img4

Identify the OOP concept shown by the figure.
Write short note about this concept.

Answer:
1. Inheritance

2. Inheritance is the capability Of one class of things to inherit properties from another class. The class from which attributes are inherited is called the base class. The inherited class is called derived class.

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Kinetics

April 13, 2021January 30, 2020 by Prasanna

Students can Download Chapter 4 Chemical Kinetics Questions and Answers, Plus Two Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Kinetics

Plus Two Chemistry Chemical Kinetics One Mark Questions and Answers

Question 1.
The rate of a reaction is expressed as

The reaction is
(a) 4A + B → 2C + 3D
(b) B + 3D → 4A + 2C
(c) A + B → C + D
(d) B + D → A + C
Answer:
(b) B + 3D → 4A + 2C

Question 2.
Order of a reaction can
(a) rational
(b) zero
(c) integer
(d) all the above
Answer:
(d) all the above

Question 3.
The unit of rate constant for a zero order reaction is.
(a) L s^-1
(b) L mol^-1 s^-1
(c) mol L^-1 s^-1
(d) mol s^-1
Answer:
(c) mol L^-1 s^-1

Question 4.
The expression for half-life period of a first order is ___________
Answer:
t_1/2 = $\frac{0.693}{k}$

Question 5.
A plot of log k vs $\frac{1}{T}$ helps to calculate
(a) Energy of activation
(b) Rate constant of the reaction
(c) Order of the reaction
(d) Energy of activation as well as the frequency factor.
Answer:
(d) Energy of activation as well as the frequency factor

Question 6.
When the initial concentration of the reactant in a reaction is doubled the half life period of the reaction gets halved. The order of the reaction is _____
Answer:
2

Question 7.
In first order reaction, the concentration of the reactant decreases from 0.6 M to 0.3 M in 30 minutes. The time taken for the concentration to change from 0.1 M to 0.25 M is
Answer:
60 minute

Question 8.
The rate constant of a first order reaction is doubled when the temperature is increased from 20°C to 25°C. How many times to rate constant will increase if the temperature is raised from 20°C to 40°C
Answer:
16

Question 9.
The unit mol L^-1 s^-1 is meant for the rate constant of the reaction having the order _____
Answer:
Zero

Question 10.
If the activation energy for the forward reaction is 150 kJ/mol and that of the reverse reaction is 260 kJ/ mol. What is the enthalpy change?
Answer:
110 kJ/mol

Plus Two Chemistry Chemical Kinetics Two Mark Questions and Answers

Question 1.
Explain a graphical method for determination of activation energy.
Answer:
Activation energy can be determined graphically from the Ink vs 1/T graph. From the graph,
In k = ln(Ae^-E_a/RT)
In k = In A + in e^-E_a/RT
In k = In A + -E_a /RT
This is in the form of y = mx + c
When a graph is plotted between 1/T and ln k, a straight line is obtained.
Slope of the line = ^–E_a/R
Therefore, E_a = slope × R

Question 2.
Choose the correct answer. The equation for t_1/2 of first order reaction is

t_1/2 = 0.693/k
t_1/2 = [R]₀/2k

Answer:
1. t_1/2 = 0.693/k

Question 3.
In a Chemistry class, teacher asked to write the rate expressions for the reaction,
NO₂ + CO → NO + CO_2-
Student 1 wrote: Rate ∝ [NO_1/2] [CO]
Student 2 wrote: Rate ∝ [NO_1/2] [NO_1/2]

Which is correct?
Justify your answer.

Answer:

Student 2 is correct.
In fact the reaction takes place in 2 steps. They are,

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Kinetics two marks q3 img 1

According to the law of chemical kinetics, the rate of a reaction is determined by the slowest step. Here step 1 is the slowest step i.e., rate determining step. By the law of mass action, rate of a chemical reaction is directly proportional to the product of molar concentration of the reactants.
rate = k[NO₂][NO₂]= k[NO₂]²

Question 4.
What do you mean by rate of a reaction? Explain the method for the determination of the rate of reaction.
Answer:
Rate of a chemical reaction is the amount of product formed in unit time or the amount of reactant that reacted in unit time.
Rate = dx/dt. Here dx represents the amount of the product formed in a small interval of time dt.

Question 5.
1. Mention the order of the following reactions:

CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH
C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ + C₆H₁₂O₆

2. Unit of the rate constant of a reaction depends on the order of the reaction. If the concentration is expressed in mol/L and time in second. Find the unit of k for zero, first and second order reactions.
Answer:
1. Order of the reactions:

Pseudo 1^st order
Pseudo 1^st order

2. Zero order = L^-1 mol¹ s^-1
1^st order = s^-1
2^nd order = L mol^-1 s^-1

Question 6.
Match the following:

Type of reaction	Order
Photochemical	3
Nuclear reaction	2
2NO + O₂ → 2NO₂	1
N₂ + O₂ → 2NO	0

Answer:

Type of reaction	Order
Photochemical	0
Nuclear reaction	1
2NO + O₂ → 2NO₂	3
N₂ + O₂ → 2NO	2

Question 7.
Mention the factors that affect the rate of a chemical reaction.
Answer:
The various factors that affect the rate of a chemical reaction are:

Concentration of reactants
Temperature of reaction
Presence of catalyst
Nature of reactants
Surface area
Exposure to radiation

Plus Two Chemistry Chemical Kinetics Three Mark Questions and Answers

Question 1.
Derive the expression for the half-life of a chemical reaction of first order?
Answer:
For a first order reaction,
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Kinetics three marks q1 img 2
Thus, for a first order reaction, half-life period is constant, i.e., it is independent of initial concentration of the reacting species.

Question 2.
Say True or False.

The rate of the reaction NO₂ + CO → NO + CO₂ can be written as a rate a [NO₂]¹ [CO]¹.
The reaction NO₂ + CO → NO + CO₂ takes place through one step.
If a reaction takes place through more than two steps, the rate of the reaction is determined by the slowest step.

Answer:

False
False
True

Question 3.
The rate of a chemical reaction is influenced by many factors.

Write any five factors influencing rate of a reaction.
Write a short note about temperature dependence of rate of a reaction.

Answer:

Nature of the reactants, concentration of the reactants, temperature, pressure, catalyst, surface area, influence of radiation.
Usually the rate of chemical reaction increases when the temperature increases. The rate of a chemical reaction is found to be twice that of the original value for every 10° raise of temperature.

Question 4.
Consider the statement: “Sum of average kinetic energy and activation energy is equal to threshold energy”.

Is this correct?
Define activation energy.
Name the two methods used to determine the activation energy.

Answer:

Yes.
The excess amount of energy that is to be given to the reactant molecules in order to involve in a chemical reaction is called activation energy.
Theoretical method and graphical method.

Question 5.
“Nuclear reactions are first order reactions.”

Do you agree?
What do you mean by order of a reaction?
Write two differences between order and molecularity.

Answer:

Yes.
Order of a chemical reaction is the sum of the powers of the concentration terms which occur in the rate expression.
Two differences between order and molecularity:
- Order: Experimental, can be zero.
- Molecularity: Theoretical, cannot be zero.

Question 6.
Analyse the statements:
Statement 1: “Half life of every reactions depend upon the initial amount.’’
Statement 2: “Half life depends on the initial amount except in the case of first order reactions.”

Which statement is correct?
Give the expressions for half life of a first order reaction. State how it is related to the initial concentration.
Show that t_1/2 of zero order reaction depends on initial amount?

Answer:
1. Statement 2.

2. t_1/2 = $\frac{0.693}{k}$
t_1/2 of a first order reaction is independent of the initial concentration of the reacting species.

3. For a zero order reaction,
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Kinetics three marks q6 img 3
Where [R]₀ is the initial amount.
i.e., t_1/2 of zero order reaction is directly proportional to the initial amount.

Question 7.
A relation is given below. Answer the following questions.
Average kinetic energy + Activation energy = Threshold energy

Explain Average Kinetic Energy, Activation Energy, and Threshold Energy.
Give the equation showing the relation between rate constant and temperature and explain the terms.
Name the above equation.

Answer:
1. Average kinetic Energy – The energy possessed by the reactant molecules which are involved in a chemical reaction.
Threshold energy – The minimum amount of energy required for the reactant molecules in order to involve in a chemical reaction.
Activation Energy – The excess amount of energy that is to be given to the reactant molecules in order to involve in a chemical reaction.

2. k = A e^-E_a/RT

A → constant (frequency factor)
T → Temperature in Kelvin
E_a → Activation energy
K → Rate constant

3. Arrhenius equation.

Question 8.
Inversion of sucrose in presence of dilute acid is said to be pseudo first order reaction.

Why is it said to be so?
Give another example of pseudo order reaction.

Answer:
1.

The rate depends only on the concentration of sucrose and it is independent of the concentration of water. Thus, the reaction appears to be of second order but actually follows first order kinetics,

2. Hydrolysis of ester in presence of acid.

Question 9.
For a reaction A + B → C + D, the rate equation is given by, r = k[A] [B],

What is the order of the above reaction?
Find the unit of k in the above rate equation.
What happens to the order of the reaction if‘B’is taken in large excess?

Answer:

2^nd order
L mol^-1s^-1
If B is taken in large excess, the reaction behaves as 1^st order reaction. Such reactions are called pseudo 1^st order reactions.

Question 10.
For a reaction A → P, the rate of the reaction doubles when the concentration of A is increased by two times.

Determine the order of the reaction.
Derive an expression to determine the 14^th life period of above order reaction.

Answer:

First order reaction.
1/4^th life of a first order reaction can be determined as follows.

Question 11.
Fora reaction A + B → products, the rate equation is given as rate = k[A]^1/2 [B]²

What is the order of the above reaction?
Distinguish between order and molecularity.

Answer:
1. Order of the reaction = $2+\frac{1}{2}=\frac{5}{2}$
2.

Molecularity	Order
1. Experimental	1. Theoretical
2. Can be zero	2. Can not be zero
3. Can be a fraction	3. Cannot be a fraction
4. No.of reacting species taking part in an elementary reaction, which must collide simultaneously to bring about a chemical reaction	4. Sum of the powers of the concentration of the reactants in the rate law expression

Question 12.
For the reaction:
2A + B → A₂B
the rate = k[A][B]² with k = 2.0 × 10^-6 mol^-2 L² s^-1. Calculate the initial rate of the reaction when [A] = 0.1 mol L^-1 [B] = 0.2 mol L^-1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L^-1.
Answer:
Initial rate = k[A][B]²
But [A] = 0.1 M, [B] = 0.2 M and
k = 2.0 × 10^-6 mol^-2 L² s^-1
∴ Initial rate
= 2.0 × 10^-6 mol^-2 L² s^-1 × 0.1 M × (0.2 M)²
= 8 × 10^-9 M s^-1

From the equation 2A + B → A₂B it is clear that when 2 moles of A are used then 1 mol of B is used in the same time. Therefore, when A has been reduced to 0.06 M , 0.04 M of A and hence 0.02 M of B have reacted. Thus,
Concentration of A left = [A] = 0.06 M
Concentration of B left = [B] = (0.2 M – 0.02 M) = 0.18 M
Rate = k[A][B]²
= 2.0 × 10^-6 mol^-2 L² s^-1 × 0.06M × (0.18 M)²
= 3.89 × 10^-9M s^-1

Question 13.
The decomposition of NH₃ on platinum surface is zero order reaction. What are the rates of production of N₂ and H₂ if k = 2.5 × 10^-4 mol^-1 L s^-1?
Answer:
The decomposition of NH₃ is
2NH₃ → N₂ + 3H₂
Rate of reaction,

where k is the rate constant. Since, reaction is of zero order,
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Kinetics three marks q13 img 8

Question 14.
The rate constant of a first order reaction is 60 s^-1. How much time will it take to reduce the initial concentration of the reactant to its 1/16^th value?
Answer:
Rate constant of reaction, k = 60 s^-1
t_15/16 = ?
Rate constant of first order reaction is given as,
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Kinetics three marks q14 img 9

Question 15.
The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
Answer:
T₁ = 293K, T₂ = 313K
According to Arrhenius equation,
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Kinetics three marks q15 img 10
2.303 × 8.314 J K^-1 mol^-1 × 4585.45 K × 0.6021
= 52863.33 J mol^-1 = 52.86 kJ mol^-1

Plus Two Chemistry Chemical Kinetics Four Mark Questions and Answers

Question 1.

Which type of the reaction is this?
Write the order and molecularity of the reaction.
What is the unit of rate constant?
Write the 3 steps involved in this reaction?

Answer:

Photochemical reaction
Order = 0 Molecularity = 2
L^-1 mol¹ sec^-1
Initiation, propagation, termination

Question 2.
K = Ae^-E_a/RT

Write the name of this equation.
What do you mean by E_a?
Draw the graph between 1/T and Ink.
Write the significance of k.

Answer:
1. Arrhenius equation.

2. The excess amount of energy that is to be given to the reactant molecules in order to involve in a chemical reaction is called activation energy (E_a)

3.
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Kinetics four marks q2 img 11

4. By noting the value of k, we can predict the speed of a chemical reaction. In the case of a fast reaction, the value of k is high and in the case of a slow reaction value of k is small. By noting the unit of rate constant, we can predict the order of chemical reaction.

Question 3.
Match the following:

First order reaction	2NO + O₂ → 2NO₂
Second order reaction	CH₃CHO → CH₄+ CO
Third order reaction	Photochemical
1¹/₂ order reaction	N₂+ O₂ → 2NO
Zero order reaction	H₂O → 2H₂O + 1/2O₂

Answer:

First order reaction	H₂O₂ → H₂O + 1/2O₂
Second order reaction	N₂+ O₂ → 2NO
Third order reaction	2NO + O₂ → 2NO₂
1¹/₂ order reaction	CH₃CHO → CH₄+ CO
Zero order reaction	Photochemical

Question 4.
Consider the reaction

Answer the following questions

Write down 2 steps through which the reaction takes place.
Find out the slowest step.
Which is the rate-determining step?
Find out the order of the reaction.

Answer:
1. 2H₂O₂ + I^– → H₂O + IO^–
H₂O₂ + IO^– → H₂O + I^– + O₂

2. 2H₂O₂ + I^– → H₂O + IO^–

3. 2H₂O₂ + I^– → H₂O + IO^–

4. Order = 2

Question 5.
A chemical reaction is conducted within two hours. But after one hour half of the product is obtained.

Name the term used to designate this time?
Give the expression for this term if it is a first order reaction.
Calculate the rate constant of the above reaction.

Answer:

Half life of a chemical reaction.
t_1/2 = $\frac{0.693}{k}$
k = $=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{1 \mathrm{hr}}$ = 0.693 hr^-1

Question 6.

Define half-life period of a reaction.
A first order reaction is found to have a rate constant k = 5.5 × 10^-14 s^-1. Find the half-life period of the reaction.

Answer:
1. Time in which the concentration of the reactant is reduced to one half of its initial concentration.
2.

Question 7.
Order of a reaction is an experimental quantity.
(1) Calculate the overall order of a reaction which has the rate expression:
a) Rate = k[A]^1/2[B]^3/2
b) Rate = k[A]^3/2[B]^-1
(2) The initial concentration of N₂O₅ in the following first order reaction N₂O₆ (g) → 2NO₂(g) + 1/2 O₂(g) was 1.24 × 10^-2 mol/L at 318 K.

The concentration of N₂O₅, after 60 min was 0.20 × 10^-2 mol/L. Calculate the rate constant of the reaction at 318 K.
Answer:
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Kinetics four marks q7 img 14

Question 8.
1. Write the rate expression for the following reaction.
2N₂O₅(g) → 4NO₂(g) + O₂(g)
2. For a hypothetical reaction, aX + bY → Products, the rate law is given as $\frac{d x}{d t}$ = k[X]²[Y]^1/2. What happens to the rate of the reaction when

The concentration of ‘X’ is doubled keeping that of ‘Y’ constant.
The concentration of both ‘Y’ is doubled keeping that of ‘X’ constant.

Answer:
1. Rate expression:

2.

Increases by 4 times
Increases by $\sqrt{2}$ times.

Question 9.
1. Identify the reaction order.

k= 2.3 × 10^-5 L/mol/s
k= 3 × 10^-4 s^-1

2. Calculate the activation energy. The rate constants of a reaction are 1 × 10^-3 s^-1 and 2 × 10^-3 s^-1 at 30°C and 40°C respectively.
Answer:
1. The reaction order:

Second order
First order

2.
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Kinetics four marks q9 img 16

Question 10.
For a reaction A + B → Product, the rate law is given by r= k[A] [B],

What is the order of the reaction?
What will be the effect of temperature on rate constant?
Can activation energy for reactions be zero? Justify your answer.

Answer:
1. Order is the sum of the powers of the concentration terms in the rate law.
rate = K [A]^x [B]^y
∴ Order = x + y

2. According to Arrhenius equation, at a particular temperature for a definite value of activation energy, rate constant,
k ∝ e^-1/T

3. If average kinetic energy of reactant molecule is equal to threshold energy, then activation energy can be zero.

Plus Two Chemistry Chemical Kinetics NCERT Questions and Answers

Question 1.
For the reaction:
2A + B → A₂B
the rate = k[A][B]² with k = 2.0 × 10^-6 mol^-2 L² s^-1. Calculate the initial rate of the reaction when [A] = 0.1 mol L^-1 [B] = 0.2 mol L^-1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L^-1.
Answer:
Initial rate = k[A][B]²
But [A] = 0.1 M, [B] = 0.2 M and
k = 2.0 × 10^-6 mol^-2 L² s^-1
Initial rate
= 2.0 × 10^-6 mol^-2 L² s^-1 × 0.1 M × (0.2 M)²
= 8 × 10^-9 M s^-1

From the equation 2A + B → A₂B it is clear that when 2 moles of A are used then 1 mol of B is used in the same time. Therefore, when A has been reduced to 0.06 M , 0.04 M of A and hence 0.02 M of B have reacted. Thus,
Concentration of A left = [A] = 0.06 M.
Concentration of B left = [B] = (0.2 M – 0.02 M) = 0.18 M
Rate = k[A][B]²
= 2.0 × 10^-6 mol^-2 L² s^-1 × 0.06M × (0.18 M)²
= 3.89 × 10^-9M s^-1

Question 2.
Mention the factors that affect the rate of a chemical reaction.
Answer:
The various factors that affect the rate of a chemical reaction are:

Concentration of reactants
Temperature of reaction
Presence of catalyst
Nature of reactants
Surface area
Exposure to radiation

Question 3.
The decomposition of NH₃ on platinum surface is zero order reaction. What are the rates of production of N₂ and H₂ if k = 2.5 × 10^-4 mol^-1 L s^-1?
Answer:
The decomposition of NH₃ is
2NH₃ → N₂ + 3H₂
Rate of reaction,

where k is the rate constant. Since, reaction is of zero order,
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Kinetics three marks q13 img 8

Question 4.
The rate constant of a first order reaction is 60 s^-1. How much time will it take to reduce the initial concentration of the reactant to its 1/16^th value?
Answer:
Rate constant of reaction, k = 60 s^-1
t_15/16 = ?
Rate constant of first order reaction is given as,
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Kinetics three marks q14 img 9

Question 5.
The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
Answer:
T₁ = 293K, T₂ = 313K
According to Arrhenius equation,
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Kinetics three marks q15 img 10
2.303 × 8.314 J K^-1 mol^-1 × 4585.45 K × 0.6021
= 52863.33 J mol^-1 = 52.86 kJ mol^-1

Plus Two Computer Science Chapter Wise Questions and Answers Chapter 1 Structures and Pointers

April 13, 2021January 30, 2020 by Prasanna

Students can Download Chapter 1 Structures and Pointers Questions and Answers, Plus Two Computer Science Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus Two Computer Science Chapter Wise Questions and Answers Chapter 1 Structures and Pointers

Plus Two Computer Science Structures and Pointers One Mark Questions and Answers

Question 1.
Name a method used to put logically related elements in one unit.
Answer:
Structure or class

Question 2.
…………… keyword is used to specify a structure.
Answer:
struct.

Question 3.
State True or False.
Answer:
‘A structure can be assigned to another structure only if both the structures are of same structure type’.

Question 4.
A structure brings together a group of
(a) Items of the same data type
(b) Integers with user defined names.
(c) Related data items and functions, if needed.
(d) None of the above
Answer:
(c) Related data items and functions, if needed.

Question 5.
Pick the odd one out from the following.
(a) int
(b) struct
(c) char
(d) long
Answer:
(b) struct

Question 6.
Select the structure member operator. Structure member operator select
(a) *
(b) .
(c) ®
(d) [ ]
Answer:
(b) .

Question 7.
The address of a computer memory starts at ………
(a) 0
(b) 1
(c) 1024
(d) none of these
Answer:
(a) 0

Question 8.
The area used for dynamic memory allocation is called
(a) heap
(b) stack
(c) RAM
(d) Hard Disk
Answer:
(a) heap

Question 9.
Given int *P, x = 35; P = &x; what is the value of *P?
(a) 34
(b) 35
(c) 0
(d) none of these
Answer:
(b) 35

Question 10.
int *P; p is having the address 2340. What will be the value of P after the execution of the statements P++; P = P – 3;
(a) 2340
(b) 2342
(c) 2338
(d) 2332
Answer:
(d) 2332

Question 11.
Dynamic allocation means memory allocation at the time of ……….
(a) compilation
(b) debugging
(c) execution
(d) programming
Answer:
(c) execution

Question 12.
The ‘new’ operator should be matched with a ……….
(a) pointer
(b) reference
(c) delete
(d) address
Answer:
(c) delete

Question 13.
In the declaration void ff (int &, float &); the parameters are passed by …….
(a) reference
(b) value
(c) pointers
(d) default
Answer:
(a) reference

Question 14.
Using ……… the memory location of a variable can be directly accessed and manipulated.
Answer:
Pointers.

Question 15.
Anil argued that memory can be allocated as and when required during runtime. Do you agree with this opinion? Justify.
Answer:
Yes. Memory can be allocated during run time using the operator new.

Question 16.
Name the unary operator that returns the address of its operand.
Answer:
& operator. Operator (Address operator)

Question 17.
Which among the following operators cannot be used with pointers?
(a) +
(b) –
(c) /
(d) None of these
Answer:
(c) /

Question 18.
The address of the first byte is known as
(a) start
(b) Base address
(c) int address
(d) none of these
Answer:
(b) Base address

Question 19.
What are the free store operators?
Answer:
‘new’ and ‘delete’ operators.

Question 20.
Give the syntax of new operator.
Answer:
= new ;
eg: int *P = new int;

Question 21.
What does the following stmt mean?
int *P = new int [15];
Answer:
Allocating memory for an integer array of 15 elements.

Question 22.
The ……. operator reverses the action of the ‘new’ operator.
Answer:
delete.

Question 23.
Which operator is used to access individual elements of a structure ora class while using pointer variables?
Answer:
(a) *
(b) .
(c) ?
(d) None of these
Answer:
(b) .

Question 24.
A pointer is
(a) the address of a variable
(b) an indication of the variable to be accessed next
(c) a variable for storing addresses
(d) the datatype of an address variable
Answer:
(c) a variable for storing addresses

Question 25.
The expression * test can be said to ‘test
(a) be a ponterto test
(b) refer to the contents of test
(c) dereference test
(d) refer to the value of the variable pointed to by test
Answer:
(d) refer to the value of the variable pointed to by test

Question 26.
The expression double * can be said to double*
(a) be a pointer pointing to variable double
(b) refer to contents of double
(c) be a pointer to a double type value
(d) refer to the value of the variable pointed to by double.
Answer:
(c) be a pointer to a double type value

Question 27.
The new operator –
(a) returns a pointer to a variable
(b) creates a variable called new
(c) obtains memory for a new variable
(d) tells how much memory is available
Answer:
(c) obtains memory for a new variable

Question 28.
Write a declaration for an array of 8 pointers to float float values
(a) float a []
(b) float*[*a] [8]
(c) float *a[8]
(d) float [*a][8]
Answer:
(b) float*[*a] [8]

Question 29.
A pointer to void can hold pointers to ………….
(a) void
(b) int
(c) float
(d) any data type
Answer:
(d) any data type

Question 30.
The contents of two pointers that point to adjacent variable of type float differ by
(a) 4
(b) 2
(c) 1
(d) 8
Answer:
(a) 4

Question 31.
In the …….. memory allocation, the amount of memory to be allocated is not known beforehand
(a) Static
(b) binding
(c) object
(d) dynamic
Answer:
(d) dynamic

Question 32.
Improper use of ‘new’ and ‘delete’ may lead to ….. ‘new’, ‘delete’ operator
(a) increase of memory
(b) memory leaks
(c) destruction of memory
(d) syntax errors
Answer:
(b) memory leaks

Question 33.
If ‘p’ is pointer to the variable ‘x’, which of the following statement are correct.
(a) ‘p’ and ‘x’ may be of different data types
(b) if ‘p’ points to ‘x’, then ‘x’ also points to ‘p’
(c) the statement x = &p is valid
(d) *p will give the value of the variable ‘x’
Answer:
(d) *p will give the value of the variable ‘x’

Question 34.
Suppose ‘m’ is a float variable. Another variable ‘ptr’ stores the address of it, which is 2001. After the operation ptr++, it will point to the location having address ………. Why?
(a) 2002
(b) 2003
(C) 2004
(d) 2005
Answer:
(d) 2005 (add the size of its data type, 2001 + 4)

Question 35.
State True of False.
‘Only integers can be added or subtracted from pointers’.
Answer:
True

Question 36.
Give the syntax for delete operator, delete operator
Answer:
delete pointer-variable;

Question 37.
What does the following statement mean?
delete [ ] marks;
Answer:
Deallocating marks array

Question 38.
Name the operator that returns the data value stored in the area being pointed to by the pointer.
Answer:
* Operator

Question 39.
If the name of an array is a pointer to its first ele¬ment, how to move to third element.
Answer:
Array name +2 gives the address of third element

Question 40.
What does the following statement mean?
int * ptr [10];
Answer:
An array of 10 integer pointers

Question 41.
State True of False.
‘It is faster to use an element pointer rather than an index when scanning arrays’.
Answer:
True.

Question 42.
S[0] is not equal to S[0]
(a) *(S+0)
(b) *S
(c) *(s)
(d) S
Answer:
(d) S

Question 43.
Which of the following declaration is valid?
(a) int*p = &a,
(b) int&p = *a,
(c) int &p = &a,
(d) int a = &p;
Answer:
(a) int*p = &a,

Question 44.
*p means ……….
(a) address of p
(b) value of p
(c) address of variable pointed to which p points
(d) value of virile to which p points
Answer:
(a) address of p

Question 45.
& p means ……..
(a) address of p
(b) value of p
(c) address of variable pointed to which p points
(d) value of variable to which p points
Answer:
(a) address of p

Question 46.
State True of False.
‘Only integers can be added or substracted from pointers’.
Answer:
True.

Question 47.
Read the following statements:
(i) While defining a structure in C++, tag may be omitted.

(ii) The data contained in a structure variable can be copied into another variable only if both of them are declared using the same structure tag.

(iii) Elements of a structure is referenced by structure_name.element

(iv) A structure can contain another structure.

Now, Choose the correct option from the following:
(a) Statements (i) and (ii) are true
(b) Statements (ii) and (iv) are true
(e) Statements (i), (ii) and (iv) are true
(d) Statements (i) and (iii) are true
Answer:
(d) Statements (i) and (iii) are true

Question 48.
Choose the correct syntax for accessing a structure element.
(a) structure_name.element_name;
(b) element_name.structure_name;
(c) structure_variable.element_name;
(d) element_name.structure_variable;
Answer:
(c) structure_variable.element_name;

Question 49.
Which one of the following is not correct about a pointer.
Answer:
(a) A pointer stores address
(b) A Pointer stores L-value of a variable
(c) A Pointer points to a memory location
(d) A pointer stores R-Value of a variable
Answer:
(d) A pointer stores R-Value of a variable

Question 50.
The following are some pointer variable declarations.
(a) float *ptr;
(b) int ptr*;
(c) short *ptr1,*ptr2;
(d) double ptr;
Choose the correct declaration
(i) a and b
(ii) a and c
(iii) a and d
(iv) a only
Answer:
(ii) a and c

Question 51.
The address of operator (&) is used for
(a) Obtaining R-Value of a variable
(b) Obtaining L-Value of a variable
(c) Obtaining content of a variable
(d) None of these
Answer:
(b) Obtaining L-Value of a variable

Question 52.
Differentiate static and dynamic memory allocation.
Answer:
When the amount of memory to be allocated is known in advance and memory is allocated during compilation itself, it is referred to as static memory allocation. When the amount of memory to be allocated is not known in advance and it is required to allocate memory as and when required during run time, it is known as dynamic memory allocation.

Question 53.
If a memory allocated using the dynamic memory allocation operator new, is not deallocated using the operator delete, a problem is occurred. Name that problem.
Answer:
Memory leak

Question 54.
There is no chance of memory leak in static memory allocation”. Say True/False. Justify.
Answer:
The operating system takes the responsibility of memory allocation and deallocation without users instruction.

Question 55.
Write a C++ code segment to declare an integer pointer variable named p and store the address of an integer variable n into the pointer variable p.
Answer:
int * p, n; p = &n;

Question 56.
Consider the following code segment in C++. If a is stored at memory location 4020 and size of int is 4, then predict the output.
int *x, a;
x = &a;
cout << x ++ << “\n”; .
cout << ++ x;
Answer:
4020
4028

Question 57.
Choose the operator which is not associated with a pointer variable.
(a) *
(b) .
(c) &
(d) →
Answer:
(b) .

Question 58.
The arrow (→) operator is used for accessing
(a) structure element using structure variable.
(b) structure element using a structure name.
(c) structure element using structure pointer variable.
(d) structure element using structure tag.
Answer:
(c) structure element using structure pointer variable.

Question 59.
Write a C++ statement to declare an array, using character pointer variable, which can store names of 60 students in a class.
Answer:
char*name[60];

Question 60.
Read the following statements.
(i) A structure which contains another structure is called self referential structure.
(ii) A structure which, contains pointer to another structure is called self referential structure
(iii) A structure which contains another structure is called nested structure.
(iv) A structure which contains pointer to same structure is called self referential structure
Now, choose the correct option from the following
(a) statements (i) and (ii) are true
(b) statements (i) and (iii) are correct
(c) statements (ii) and (iii) are correct
(d) statements (iii) and (iv) are correct
Answer:
(d) statements (iii) and (iv) are correct

Question 61.
Structure is a …………………. data type.
Fill up the blank by choosing correct option from the following.
(a) Fundamental
(b) Derived
(c) User defined
(d) Built in
Answer:
(c) User defined

Question 62.
Write the declaration statement of a student structure variable named ‘s’ which is initialized with admission number 19, name as ‘Aysha’ and mark as 75.
Answer:
student s = {19,“aysha”,75};

Question 63.
“The dot(.) operator cannot be used for Recessing structure member by using structure pointer variable” Say True or False. Justify your answer.
Answer:
False. Arrow (→) operator is used for accessing structure member by using structure pointer variable.

Plus Two Computer Science Structures and Pointers Two Mark Questions and Answers

Question 1.
Write C++ statement for the following.

To declare an integer variable named ‘x’ using new operator.
To initialize the integer pointer variable x with value 5.
To declare a dynamic array of ten integers named x.

Answer:

int*x = newint;
int*x = newint(5);
int *x = new int[10];

Question 2.
Consider the following C++ program code to store address of an integer variable a into a pointer variable p
int a = 5,*p;
P = a;
cout<< p;
Identify the error in the above program segment and correct the error.
Answer:
P = a is the error because p is an integer pointer and can store only address. The correct code is p = &a;

Question 3.
Correct the program code given below:
int a = 5;
float *p;
p = &a;
cout << p;
Answer:
float*p; is to be corrected as int *p;
or
int a = 5; is to be corrected as float a = 5;

Question 4.
Calculate the memory requirement for the following structure variable s1.
[Consider the size of int = 4, char = 1 and float = 4]
struct student
{
int mo;
char name[25];
float height;
}s1;
Answer:
Plus Two Computer Science Chapter Wise Questions and Answers Chapter 1 Structures and Pointers img1

Question 5.
Consider the following array initialisation, int ar[5]={-2,-4,-6,2,4}; If first element of array is stored at 2050 and size of int is 4, find the output of the following.

cout<< *ar+*(ar+3);
cout<< ar+*(ar+4);

Answer:
1.
*ar = -2
*(ar + 3) = 2 (fourth element)
Hence *ar + *(ar + 3) = -2 + 2 = 0
Hence output is 0

2.
Here ar = 2050 (starting address)
and * (ar + 4) = *(2050 + 4*4)
= *(2050+16)
= *(2066)
= 4 (fifth element)
Hence ar + * (ar + 4)
= 2050 + 4
= 2054

Question 6.
Match the following:
Plus Two Computer Science Chapter Wise Questions and Answers Chapter 1 Structures and Pointers img2
Answer:
a – iii
b – iv
c – ii
d – i

Question 7.
Complete the following table. The table is differentiating array and structure.
Plus Two Computer Science Chapter Wise Questions and Answers Chapter 1 Structures and Pointers img3
Answer:

Question 8.
Identify the errors in the following code fragment and correct it.
struct sample
{
int x;
float y;
} S1 ,S2;
void main()
{Struct test
int a;
float 6;
}T1,T2;
T1 = T2;
T1=S1;
}
Answer:
The assignment T1 = S1 is not correct. It is a variable of type test and S1 is a variable of type sample. So it cannot be assigned.

Question 9.
When a C++ program is executed, the primary memory allocated for it is organized in a particular manner to deal with runtime memory allocation, function calls, variables, etc. Show a diagrammatic representation of it with brief explanation.
Answer:
After compilation of C++ creates four distinct regions of memory used for distinct functions:

STACK 3
HEAP 4
Global	Program
Variables	Code
2	1

The first area (1) is used for storing the compiled code of the program. The second area (2) is used for storing global variables of the program. It remains in memory till the program ends. The third region (3) known as the stack is used for holding the return addresses of function calls, arguments passed to the function, etc. The last region heap is used for dynamic allocation.

Question 10.
Analyze the following C++ statements and predict
the output.
int k = 4 ;
int *p ;
p = &k ;
cout<< k ;
cout<< *p;
cout<< *p++ ;
cout<< *(&k);
Answer:
4, 4, 4, 4

Question 11.
A boy wrote a C++ program that uses several pointer variables out of which some are used for dynamic allocation. He did not use the ‘delete’ operator anywhere in the program. While running the program, the computer got stuck up after some time. Predict the reason for this kind of situation, in connection with the above program code
Answer:
The situation is called memory leak. The memory allocated using ‘new’ should be properly deallocated using ‘delete’. If any function forgets to deallocate the memory using ‘delete’ which was allocated using the ‘new’ operator, it consumes some memory in every execution.

Thus a part of memory disappears with its every execution and it has some adverse effect on the system. This situation is called memory leak.

Question 12.
Is there any difference between structure and array? Explain.
Answer:
Yes. An array can hold multiple elements of same datatype whereas a structure can hold multiple elements of different data types.

Question 13.
Write a C++ structure definition to store the student details given below:

Admission number
Name
Marks in six subjects

Answer:
Struct student
{
intAdmno;
charname[30];
int mark[6];
}

Question 14.
Observe the following layout of data,
Plus Two Computer Science Chapter Wise Questions and Answers Chapter 1 Structures and Pointers img5

Name the data type in C++ used to represent the above data format.
Write the definition for implementing the above layout.

Answer:
1. Structure

2. struct Account
{
int accno;
char name[30];
struct lastdate
{
int day;
int month;
int year;
} tdate;
float balanceamt;
}

Question 15.
Consider the following code segment.
struct student
{
int rollno;
char name[20];
struct date
{
int day;
int month;
int year;
} birthday;
char batch[30];
int score;
}
Write valid statements in C++ to accept the name and date of birth of a student.
Answer:
gets(S1 .name);
S1.birthday.day=25;
S1.birthday.month=5;
S1.birthday. year=1991;

Question 16.
Why do we need structure variables, once a structure is defined in program?
Answer:
Structure specification is merely a data type definition. Memory allocation does not take place during specification. To handle data, memory locations are needed. For memory allocation, variables are to be declared. Hence structure variable is essential for handling structured data in programs.

Question 17.
With an example explain how structures can be initialized.
Answer:
Student stud = {“Athul”, 17,101, “XII Science”};
The above statement initializes structure variable stud of type student which has data members name, age, roll no, batch.

Question 18.
Can we declare a structure within another structure? Explain.
Yes. We can declare a structure as a member of another structure. It is known as nested structure.

Question 19.
What does the following statement mean?
Answer:
It is accessing the element of a nested structure. S is a structure variable that contains another structure variable dob. ‘month’ is a member of inner structure variable dob. This statement assigns a value 10 to the member month.

Question 20.
Thomas says that name of an array is a pointer by default. Do you agree with him? Justify.
Answer:
Yes. C++ takes the array name as the address of its first element. That is, the name of an array is actually a pointer pointing to the first element of an array.

Question 21.
Distinguish between
int * ptr = new int (5);
int * ptr = new int [5];
Answer:
The first statement allocates memory of one integer to ptr and initialized it with value 5. The second statement allocates memory of 5 contiguous integers (ie. an array of 5 integers) and stores begining address in pointer ptr.

Question 22.
Give the syntax to declare an integer 2D array dy namically.
Answer:
int * ptr = new int [m*n];
where m and n are rows and columns

Question 23.
The array of character pointers is generally preferred over two dimensional array of characters. Justify.
Answer:
The array of character pointers is preferred. There are two reasons:
1. An array of pointers makes more efficient use of available memory by consuming lesser number of bytes to store the strings.

2. An array of pointers makes the manipulation of the strings much easier. One can easily exchange the positions of the strings in the array of pointers without actually touching their memory locations.

Question 24.
Predict the errors in the following code fragment in C++.
1. float f= 1.42;
float *a;
a = f;
2. int *p,*q;
p = q + 1;
Answer:

a=&f is correct
No values assigned for p and q.

Question 25.
Define a structure in C++ to store book details such as book id, book name, price, and author name.
Answer:
Structure is a group of different types of logically related data referenced by a single name,
struct book
{
int book id;
charbook_name[40];
float price;
chart author_name[40];
};

Question 26.
Complete the following structure definition, to store students details with appropriate values. Students details are roll number, name and height.
Plus Two Computer Science Chapter Wise Questions and Answers Chapter 1 Structures and Pointers img6
Answer:
struct students
{
short rno;
char name[25];
float height;
};

Question 27.
Correct the following structure definition to represent an employee with employee code, employee name, and salary.
Answer:
Employee struct
{
ecode int;
char name[25];
salary double;
}

Question 28.
Consider the array declaration. Assume that the first element of array is stored in 2020 location,
int a[] = {2,5,8,9,4},*p1 ,*p2;
p1 = a;
p2 = &a[0];
cout<< p1<< “\t”<< p2;
Predict the output and justify.
Answer:
p1 = a means, the array name is the starting address of an array.
That means p1 = 2020
p2 = &a[0] means p2 = 2020
Hence the output is 2020 2020

Plus Two Computer Science Structures and Pointers Three Mark Questions and Answers

Question 1.
Explain the type of structure defined below and calculate the number of bytes required to store the structure variable s1. [Considerthe size of int as 4 bytes and char as 1 byte]
struct student
{
int mo, mark;
char name[25];
struct date
{
int dd, mm, yy;
}doj;
}s1;
Answer:
A structure contains another structure, then it is called nested structure.
Plus Two Computer Science Chapter Wise Questions and Answers Chapter 1 Structures and Pointers img7

Question 2.
Define a structure to represent employee details. Employee details are employee code, employee name, date of joining and place of employee.
Answer:
struct employee
{
int empcode; ,
char name [40];
struct date
{
int day;
int month;
int year;
}doj;
char place [40];
};

Question 3.
“Self referential structure can be used for implementing linked list”. Justify the statement.
Answer:
A structure contains an element that is a pointer to the same structure is called self referential structure. Eg: struct date {
int dd, mm, yyyy;
date *ptr;
};
So self referential structure can be used for implementing linked list.

Question 4.
Write a program in C++ to copy one string to another without using strcpy() function.
Answer:
#include
using name space std; intmain()
{
charstr [ ] = “computer”;
char * sptr;
sptr = str;
cout<< str<< endl;
cout<< sptr<< endl;
}
This code copies the string str to sptr;

Question 5.
Define a structure in C++ to store students’ details such as admission number, name, address. Address contains house number, street name and city name.
Answer:
Structure is a group of different types of logically related data referenced by a single name,
struct student
{
int adm_No;
char name[40];
struct
{
int house_no;
charstreet_name[40];
charcity_name[40];
} address;
};

Question 6.
A structure variable can be declared along with structure definition.” Justify this statement with an ex-ample.
Answer:
Yes, it is possible that, a structure variable can be declared along with structure definition itself.
Syntax:
struct struet_name
{
Data members;
} structure variables;
Eg: struct date
{
intdd;
int mm;
intyyyy;
}d1,d2,d3;

Question 7.
Identify and correct the error in the following code fragment.
struct first {
int a; float b;
}S1;
struct second {
int a;
float b;
}S2;
S1 = S2;
Answer:
The assignment S1 = S2 is not right. Because S1 and S2 are different structure variables. C++ does not permit different structure variable assignment.

Question 8.
Write structure definitions for structures containing the following.

book no, book name, price
item no, item name, price, monthly sales for last 4 months.

Answer:
1. struct book
{
long bookno;
char bookname[50];
float price;
};

2. struct item
{
int itemno;
char itemname[50];
float price;
int sale[4];
}

Question 9.
Assume that in a city, two football matches are going to be held. The coordinators of the match have decided to sell the tickets in two different styles. For the first match, the tickets will be sold one month in advance, but for the second one, tickets will be available from the stadium on the very same day of the match.

Correlate this situation with the memory allocation techniques in C++.
Compare the merits and demerits of the two in relation with the given context.

Answer:
1. Static and dynamic memory allocation.

2. Every data and instruction data is being executed must be allocated some area in the main memory. The main memory can be allocated in two methods.

Static memory allocation
Dynamic memory allocation

When the amount of memory to be allocated is known in advance and memory is allocated during compilation itself, it is referred to as static memory allocation. When the amount of memory to be allocated is not known in advance and it is required to allocate memory as and when required during run time, it is known as dynamic memory allocation.

Question 10.
When a C++ program is compiled, the compiler allocates 20 bytes of memory and when the same program is executed 10 more bytes of memory were allocated

How do you account for these two types of memory allocations?
Name the operator used for the second type of memory allocation and also write the syntax.

Answer:
1. Static and dynamic memory allocation.

2. ‘new’ operator is used for dynamic allocation of memory syntax,
datatype ‘pointer variable = new datatype;
eg: int *ptr = new int;

Question 11.
Justify the following statements

In C++, the ‘delete’ operator performs the reverse action of ‘new’ operator.
If ‘arr’ is the name of an array, then arr++ is not a valid statement
If ‘k’ is a long variable with address 1001 and it is stored by a pointer variable ‘p’, then ‘p + 1 ’ will point to the location with address 1005.

Answer:
1. In C++, the ‘delete’ operator performs the reverse action of ‘new’ operator. In C++, both ‘new’ and ‘delete’ operators are used for dynamic allocation and deallocation of memory respectively, ie. For allocation of memory at runtime new operator is used. For deallocation of memory that have been allocated by the new operator, ‘delete’ operator is used.

2. arr++ is not a valid statement. Because the base address of an array can not change. Array name gives the base address of the array.

3. Pointer variable hold the address of memory location where the data is stored. So incrementing the pointer variable is adding the size of the data type. Long data type takes 4 bytes of memory. So incrementing the pointer variable P makes it point to the address 1005.

Question 12.
Discuss the similarity and differences of a structure as compared to an array and a class.
Answer:

Structures	Array
1. It is a user defined data type	1. Predefined data type
2. It is a collection of different types of logically related data under one name.	2. Collection of data elements of same data type having a common name
3. Elements referenced using dot operator (.)	3. Elements references using its subscripts (position value)
4. When an element of a structure becomes another structure nested structure and complex structures are formed	4. When an element of another becomes another array, multidimensional arrays are formed.
5. Structures contain array as its elements	5. Array of structure can be formed

Question 13.
Considerthe following code fragment.
char nam[ ] = “COMPUTER”
for(int i = 0; nam[i]!= ‘\0’; i++)
cout << nam[i];
Rewrite the above code to get the same output using a character pointer variabe.
Answer:
char nam [ ] = “computer”
char*P;
for (P = nam; *P!= ‘\0’; P++)
cout<< *P;

Question 14.
Match the following.
Plus Two Computer Science Chapter Wise Questions and Answers Chapter 1 Structures and Pointers img8
Answer:
1 – b
2 – c
3 – d
4 – a

Question 15.
How is dynamic memory allocation different from static memory allocation?
Answer:
In the static memory allocation, the amount of memory to be allocated is predicted and pre known. This memory is allocated during the compilation itself. All the declared variables declared normally, are allocated memory statically.

In the dynamic memory allocation, the amount of memory to be allocated is not known beforehand. It is allocated during run time as and when required. The memory is dynamically allocated using new1 operator.

The objects that are allocated memory statically have the lifetime as their scope allows, as decided by the compiler. And the objects that are allocated memory dynamically have the lifetime as decided by the programmer. That is until the programmer explicitly deallocates the memory, such objects live in the memory.

Plus Two Computer Science Structures and Pointers Five Mark Questions and Answers

Question 1.
Read the following structure definition and answer the following questions:
struct Book
Computer Science
{
int book_no;
charbk_name [20];
struct
{
short dd;
short mm;
short yy;
} dt_of_purchase;
float price;
};
a) Write a C++ statement to declare a variable to refer to the details of a book. What is the memory requirement of this variable? Justify your answer.

b) Write a C++ statement to initialize this variable with the details of your Computer Science textbook.

c) Write C++ statement (s) to display the details of the book.

d) The missing of structure tag in the inner structure does not cause any error. state whether this is true or false. Give reason.
Answer:
a) Book b;
The memory requirements are as follows:
Plus Two Computer Science Chapter Wise Questions and Answers Chapter 1 Structures and Pointers img9

d) This is true. There is no need to mention the structure tag because it is declared inside the structure.

Question 2.
You are given an assignment by the teacher to write a C++ program that accepts the age of ’n’ people who attended a medical camp. The value of ‘n’ is not known beforehand. Its value is to be accepted at the time of running the program. The program should print the average age as the final output. Develop the code.
Answer:
Plus Two Computer Science Chapter Wise Questions and Answers Chapter 1 Structures and Pointers img11

Question 3.
How are the following related?

Structure and array
Structure and class

Answer:
1. Structure and array:
Structure and array are both derived data types in C++.: a Structure is a user defined data type while an array is a derived data type. Structure is a collection related data items that can be of different data types. Array is also a collection of elements of same data type Thus both array and structured allow several values to be treated together as a single data object.

2. Structure and class:
Structure and class are both user defined data types. Structure is a collection of related data items while class is a collection of data and functions. Hence structure plus related functions form a class.

Plus Two Physics Chapter Wise Questions and Answers Chapter 15 Communication Systems

April 13, 2021January 29, 2020 by Prasanna

Students can Download Chapter 15 Communication Systems Questions and Answers, Plus Two Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Chapter Wise Questions and Answers Chapter 15 Communication Systems

Plus Two Physics Communication Systems NCERT Text Book Questions and Answers

Question 1.
Frequencies in the UHF range normally propagate by means of:
(a) ground waves
(b) sky waves
(c) surface waves
(d) space waves
Answer:
(d) space waves.

Question 2.
Digital signals
(i) do not provide a continuous set of values
(ii) represent values as discrete steps,
(iii) can utilize only binary system, and
(iv) can utilize decimal as well as binary system. Which of the following options is True:

(a) Only (i) and (ii)
(b) Only (ii) and (iii)
(c) Only (i), (ii) and (iii) but not (iv)
(d) All the above (i) to (iv)
Answer:
(c) because decimal system implies continuous set of values.

Question 3.
Is it necessary for a transmitting antenna to be at the same height as that of the receiving antenna for line of sight communication? A T.V. transmitting antenna is 81 m tall. How much service area it can cover if the receiving antenna is at the ground level.
Answer:
No.

Question 4.
A modulating signal is a square wave as shown below:
Plus Two Physics Communication Systems NCERT Text Book Questions and Answers 1
The carrier wave is given by c (t) = 2 sing (8πt) volts.

Sketch the amplitude modulation wave form.
What is the modulation index?

Answer:
1. amplitude modulation wave form is as shown below:
Plus Two Physics Communication Systems NCERT Text Book Questions and Answers 2

2. µ = 0.5

Plus Two Physics Communication Systems One Mark Questions and Answers

Question 1.
For base station to mobile communication, the required frequency band is
(a) 540 – 1600 kHz
(b) 200 – 325 MHz
(c) 5.9 – 6.42 GHz
(d) 840 – 935 MHz
Answer:
(d) 840 – 935 MHz.

Question 2.
What do you mean by modulation index in AM?
Answer:
It is the ratio of the change in amplitude of the modulated carrier wave to the amplitude of the original carrier wave.

Question 3.
A communication between a fixed base station and several mobile units, located on ships or aircraft utilizing two way radio communication in the VHF and UHF is of frequency band
(a) 3 to 30 MHz
(b) 30 to 300 MHz
(c) 30 to 470 MHz
(d) 30 to 600 MHz
Answer:
(c) 30 to 470 MHz.

Question 4.
Which of the following frequency/frequencies will be suitable for beyond the horizon communication using sky waves?
(a) 10 kHz
(b) 10 MHz
(c) 1 GHz
(d) 1000 GHz
Answer:
(b) 10 MHz.

Question 5.
What are the different types of message signals used in a communication system?
Answer:
The different type of message signals are voice, music, picture and computer data.

Question 6.
What do you mean by modulation index in AM?
Answer:
It is the ratio of the change in amplitude of the modulated carrier wave to the amplitude of the original carrier wave.

Plus Two Physics Communication Systems Two Mark Questions and Answers

Question 1.
Arrange the contents in the table in the most appropriate way.

A	B
Low frequency ratio waves	Space wave propagation
Television transmission	Ground wave propagation
High frequency radio wave	Geo stationary satellite
Microwaves	Sky wave propagation

Answer:

A	B
Low frequency ratio waves	Ground wave propagation
Television transmission	Space wave propagation
High frequency radio wave	Sky wave propagation
Microwaves	Geo stationary satellite

Question 2.
What is the modulation index of AM, if the voltage amplitude of the carrier wave is 2kV and side band voltage amplitude is 200V.
Answer:

Question 3.
100 Hz is an audio frequency. Its wave length is 3000 km. What is the minimum length of the antenna to transmit this wave?
Answer:
C = νλ
3 × 10⁸ = 100 × λ
λ = $\frac{3 \times 10^{8}}{100}$m = 3 × 10⁶m
Length of antenna I = $\frac{\lambda}{4}=\frac{3 \times 10^{6}}{4}$ = 7.5 × 10⁵m.

Plus Two Physics Communication Systems Three Mark Questions and Answers

Question 1.
An expert said that it is necessary to modulate low frequency signals to sent long distances.

What is meant by modulation
What are the different types of modulation

Answer:
1. Modulation is the process of super posing a low frequency (audio signal) information on to a high frequency carrier wave.

2. Three types of modulation using sinusoidal carrier waves

Amplitude modulation
Frequency modulation
Phase modulation

Question 2.
The type of communication system needed fora given signal depends on the band of frequencies in the signal

What are the different types of message signals used in a communication systems. (1)
Match the bandwidth given in Column B with the signal given in Column A.

Column A	Column B
speech signals	6 MHz
Music signals	2.8 kHz
Video signals	20 kHz
TV signal	4.2 MHz

Answer:
1. The different type of message signals are voice, music, picture and computer data.

Column A	Column B
speech signals	2.8 kHz
Music signals	20 kHz
Video signals	4.2 MHz
TV signal	6 MHz

Plus Two Physics Communication Systems Four Mark Questions and Answers

Question 1.
The height of a transmitting antenna is 200m. Radius of earth is 6.4 × 10⁶m
1. Which one of the following is used in TV transmission?

Sky wave
Space wave
Groundwave
None of these

2. Find the range upto which the above antenna gives transmission signal.

3. Why is it necessary to use satellite for long distance transmission?
Answer:
1. Space wave

2.
Plus Two Physics Communication Systems Four mark Questions and Answers 4

3. Space waves can not be reflected back by iono sphere. Space wave can be reflected back to earth by making use of artificial satellite.

Question 2.
A radio transmitter, microwave transmitter, and a microwave receivers are placed in a hilly area.

Will the radio receiver in the house get the signal from the radio transmitter? Justify.
In this situation, what are your suggestions for proper communication? Are Microwaves suitable there?

Answer:

Yes. Because wavelength of radio wave is larger than microwave.
To have large coverage, increase the height of antenna or stations. Microwaves are not suitable for this kind of transmission.

Question 3.
The given block diagram shows general form of a communication system.
Plus Two Physics Communication Systems Four mark Questions and Answers 5
1. Identify the parts X and Yin the diagram.
2. What is the difference between

transmitter and receiver
Attenuation and Amplification

3. Repeaters are used;

To decrease the range of a communication system.
To increase the range of a communication system
Both (i) and (ii)

Answer:
1. X-transmitter, Y-receiver.

2. difference between:

A transmitter converts message signal into a form suitable for transmission. A receiver receives the transmitted signal and reconstructs the message signal.
The loss of strength of signal while passing through the medium is known as attenuation. The process of increasing the amplitude of a signal is known as amplification.

3. Repeaters are used to increase the range of a communication system.

Question 4.
The given figure shows the various propagation modes of e.m. waves for communication.
Plus Two Physics Communication Systems Four mark Questions and Answers 6

Identify the propagation modes in A, B, C
Why transmission of TV signals via sky wave is not possible?
A radio can tune to any station in 7.5 MHz. to 12 MHz. What is the corresponding wavelength band?

Answer:
1. A – ground wave, B – space wave, and C – sky wave.

2. TV signals are in the range of 100 to 220MHz. Ionosphere cannot reflect these waves.

3. Wave length corresponding to 100 MHz is

Wavelength corresponding to 220 MHz is
Plus Two Physics Communication Systems Four mark Questions and Answers 8
the wavelength band is 25m to 40m.

Question 5.
To transmit signals in the range 30 – 300 MHz space wave propagation is used

What is meant by space wave propagation?
Why it is called Line of Sight (LOS)?
Derive an expression for area covered by a sig¬nal if height of antenna is h

Answer:

In space wave propagation, wave travels in straight line from transmitting antenna to the receiving antenna.
Because the receiving antenna and transmitting antenna should be in line of sight of wave.
Range of signal from the antenna d = $\sqrt{2 \mathrm{Rh}}$

∴ Area A = πd² = π ($\sqrt{2 \mathrm{Rh}}$)²
A = π 2Rh.

Question 6.
In the broadcast mode of communication, modulation is necessary.

What do you mean by modulation? (1)
Why modulation is necessary? (2)
What do you mean by modulation index in AM? (1)

Answer:
1. Modulation is the process by which some characteristics, like amplitude, frequency or phase angle of a high frequency carrier wave is varied in accordance with the instantaneous value of the audio signal.

2. Modulation is necessary due to the following reasons.

The height of the transmitting antenna should be at least 1/4^th of the wave length. So to transmit audio frequency, we need very long antenna.
The power radiated by an antenna is inversely proportional to the square of the wavelength. In audio frequency, wavelength is higher.

3. It is the ratio of the change in amplitude of the modulated carrier wave to the amplitude of the original carrier wave.

Question 7.
The given figure shows the frequency spectrum of an amplitude-modulated wave.
Plus Two Physics Communication Systems Four mark Questions and Answers 9

What are the names of (ω_c – ω_m) and (ω_c + ω_m). (1)
What is the value of amplitude of (ω_c – ω_m) if modulation is µ? (1)
What is the modulation index of AM, if the voltage amplitude of the carrier wave is 2kV and side band voltage amplitude is 200V. (2)

Answer:
1. Lower side band frequency and upper side band frequency.

2. Value of the amplitude of the side band frequency is A_s = $\frac{\mu A_{c}}{2}$ where A_c is amplitude of carrier.

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 5 Surface Chemistry

April 13, 2021January 29, 2020 by Prasanna

Students can Download Chapter 5 Surface Chemistry Questions and Answers, Plus Two Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 5 Surface Chemistry

Plus Two Chemistry Surface Chemistry One Mark Questions and Answers

Question 1.
The process of removing an adsorbed substance from a surface on which it is adsorbed is called ___________.
Answer:
desorption

Question 2.
Which of the following statements is wrong?
(a) Physical adsorption of a gas is directly related to its critical temperature.
(b) Chemical adsorption decreases regularly as the temperature is increased.
(c) Adsorption is an exothermic process.
(d) A solid with a rough surface is a better adsorbent than the same solid with a smooth surface.
Answer:
(c) Adsorption is an exothermic process.

Question 3.
The substances that enhance the activity of a catalyst are called _____________.
Answer:
promoters

Question 4.
The enzyme present in malt which converts starch to maltose is ____________.
Answer:
diastase

Question 5.
The movement of colloidal particles under an applied electric potential is called ________________.
Answer:
electrophoresis

Question 6.
Hair cream is an example of ____________.
Answer:
Emulsion

Question 7.
Gold solution can be prepared by
Answer:
Reduction of gold (III) Chloride with formalin/ Bredig’s method

Question 8.
The correct ascending order of adsorption of the fol-lowing gases on the same mass of charcoal at the same temperature and pressure is
(a) CH₄ < H₂ < SO₂
(b) H₂ < CH₄ < SO₂
(c) SO₂ < CH₄ < H₂
(d) H₂ < SO₂ < CH₄
Answer:
(b) H₂ < CH₄ < SO₂

Question 9.
In the adsorption of a gas on solid, Freundlich iso-therm is obeyed. The slope of the plot is zero. Then the extent of adsorption is _________.
Answer:
Independent of pressure.

Question 10.
Adsorption is accompanied by _______ in enthalpy and _______ in entropy.
Answer:
decrease, decrease.

Plus Two Chemistry Surface Chemistry Two Mark Questions and Answers

Question 1.
Fill in the blanks.
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 5 Surface Chemistry two marks q1 img 1
Answer:

Question 2.
Some facts related to adsorption are given below. Classify them into those suitable for Physisorption and Chemisorption.

Weak attractive force between adsorbent and adsorbate.
Strong attractive force between adsorbent and adsorbate.
Reversible process.
Irreversible process.
Multilayers are formed.
Monolayer is formed.
Heat of adsorption is 20-40 kJ/mol
Heat of adsorption is 40-400 kJ/mol

Answer:

Physical adsorption	Chemical adsorption
1. Weak attractive force between adsorbent and adsorbate.	2. Strong attractive force between adsorbent and adsorbate.
3. Reversible process.	4. Irreversible process.
5. Multilayers are formed.	6. Monolayer is formed.
7. Heat of adsorption is 20-40 kJ/mol.	7. Heat of adsorption is 20-40 kJ/mol.

Question 3.
What is the influence of temperature in physical and chemical adsorption? Explain it with the help of a graph.
Answer:
In the case of physical adsorption, rate of adsorption x/m is inversely proportional to temperature.

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 5 Surface Chemistry two marks q3 img 3

In the case chemical adsorption, rate of adsorption intially increases with increase in temperature, reaches a maximum value and thereafter decreases with increase in temperature.

Question 4.
‘Activity’ and ‘Selectivity’ are two terms related to solid catalysts. Distinguish between these two terms with proper illustrations.
Answer:

Activity – It is ability of the catalyst to give the product easily.
Selectivity – It is the ability of a catalyst to direct to yield a particular product, e.g. starting with H₂ and CO, and using different catalysts, we get different products.

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 5 Surface Chemistry two marks q4 img 4

Question 5.
Give reason why a finely divided substance is more effective as an adsorbent.
Answer:
Adsorption is a surface phenomenon. Since finely divided substance has large surface area, hence, adsorption occurs to a greater extent.

Question 6.
What do you understand by activation of adsorbent? How is it achieved?
Answer:
Activation of adsorbent increases the adsorption power of the adsorbent. By increasing the surface area of the adsorbent,

Removing the gases already adsorbed and
Making the surface of adsorbent rough.

Question 7.
What role does adsorption play in heterogeneous catalysis?
Answer:
In heterogeneous catalysis, the reactants are generally gases whereas catalyst is a solid. The reactant molecules are adsorbed on the surface of the solid catalyst. As a result, the concentration of the reactant molecules on the surface increases and hence the rate of reaction increases.

Question 8.
Write notes on the Brownian movement and Tyndall effect.
Answer:
1. Brownian movement:

zig-zag movement of colloidal particles.

2. Tyndal effect:

Scattering of light by colloidal particles which illuminates the path of beam in the colloidal dispersion.

Question 9.
Gas mask is used in coal mines. Why? What is the principle behind it?
Answer:
In order to adsorb poisonous gases. The principle involved is adsorption.

Plus Two Chemistry Surface Chemistry Three Mark Questions and Answers

Question 1.
Colloids can be converted into precipitates on the addition of electrolytes.

Name the above phenomenon.
What do you mean by coagulating value?

Answer:

Coagulation.
The minimum concentration of an electrolyte in millimoles per liter required to cause precipitation of a sol in two hours is called coagulating value.

Question 2.
Observe the given figure.
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 5 Surface Chemistry three marks q2 img 5

Is there any mistake in the figure?
If a high potential is applied between the upper ends of the wire what will be the result?

Answer:

Yes, In this figure, the lower ends of the electrodes immersed in water are touching each other.
If a high potential is applied between the upper ends of the wire, the circuit will be complete. There will be no colloid formation.

Question 3.
Name the method of preparation in which the following examples are involved.

AS₂O₃ + 3H₂S → AS₂S₃ + 3H₂O
SO₂ + 2H₂S → 2S + 2H₂O
FeCl₃ + 3H₂O → Fe(OH)₃ + 3HCl

Answer:

Double decomposition
Oxidation
Hydrolysis

Question 4.
Emulsions are liquid-liquid colloidal systems.

What are the two types of emulsions?
Explain the difference between milk and butter.

Answer:

It is a colloidal form in which liquid is dispersed in liquid.
Milk is an oil in water type emulsion and butter is a water in oil type emulsion.

Question 5.
In a debate a student argued that “Colloid is neither a true solution nor a suspension.”

Do you agree with this argument?
List the properties of colloids.

Answer:
1. Yes
2. properties of colloids:

Colligative properties
Mechanical properties
Optical properties
Electrical properties

Question 6.
Match the following:
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 5 Surface Chemistry three marks q6 img 12
Answer:

Question 7.
Complete the following:

Lyophilic sol – starch; Lyophobic sol …………..
Deltas are formed where rivers and seawater meets. Do you agree with this statement? Explain.

Answer:

Lyophobic sol – gold sol (or any other correct example).
Yes. River water is a colloid of clay. Seawater contains a number of electrolytes. When river water meets the seawater, the electrolytes present in seawater coagulate the colloid of clay resulting in its deposition with the formation of delta.

Question 8.

Arrange the following electrolytes in the increasing order of coagulating power for the coagulation of Fe(OH)₃sol.
Na₃(PO₄)₂, NaCl, K₄[Fe(CN)₆], CaSO₄
Name the underlying principle.

Answer:

NaCl < CaSO₄ < Na₃(PO₄) < K₄[Fe(CN)₆.
Hardy-Schulze rule.

Question 9.
What is the difference between multimolecular and macromolecular colloids? Give one example of each. How are associated colloids differ from these two types of colloids?
Answer:
Multimolecular colloids are formed by the aggregation of a large number of small atoms/molecules. The aggregates thus formed have size in the colloidal range, e.g. Gold sol.

Macromolecular colloids contain large size molecules which have the dimensions of colloids, e.g. Starch. Associated colloids behave as electrolytes at low concentration but exhibit colloidal behaviour at higher concentration, e.g. Soap.

Question 10.
Explain what is observed

When a beam of light is passed through a colloidal sol?
An electrolyte, NaCl is added to hydrated ferric oxide sol?
Electric current is passed through a colloidal sol?

Answer:

Scattering of light by the colloidal particles takes place and the path of light become illuminated. This is called Tyndall effect.
The positively charged colloidal particles of Fe(OH)₃ sol get coagulated by the oppositely charged ion provided by the electrolyte.
On passing direct current, colloidal particles move towards the oppositely charged electrode where they lose their charge and get coagulated.

Plus Two Chemistry Surface Chemistry Four Mark Questions and Answers

Question 1.
A graph wrongly plotted between temperature and rate of adsorption for chemical adsorption is given below:
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 5 Surface Chemistry four marks q1 img 6

a. Draw it correctly.
b. Draw the graph of rate of adsorption and temperature for physical process.
c. Write three or four factors influencing the adsorption of a gas on a solid.
Answer:
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 5 Surface Chemistry four marks q1 img 7
c. Three or four factors influencing the adsorption of a gas on a solid

Nature of adsorbent
Nature of adsorbate
Temperature
Pressure

Question 2.
Gold sol is to be prepared from rods of gold.

Which method do you suggest?
Explain.
Distinguish between sol and gel with suitable examples for each.

Answer:
1. Electrical disintegration method – Bredig’s Arc Method.

2. An electric arc is struck between electrodes of the metal immersed in dispersion medium. The heat produced vapourises the metal which then condenses to form colloids.

3. Difference between sol and gel with suitable examples:

Sol – Colloids in which dispersed phase is solid and dispersion medium is liquid, e.g. Paints, Cell fluids.
Gel – Colloids in which dispersed phase is liquid and dispersion medium is solid, e.g. Cheese, Butter.

Question 3.
For a question, ‘What is emulsion? a student wrote ‘It is a colloidal form in which a liquid is dispersed in solid.

Is it correct?
What do you mean by emulsion?
How can we increase the stability of an emulsion?

Answer:

No.
It is a colloidal form in which a liquid is dispersed in another liquid.
Stability of emulsion can be increased by using emulsifying agents (certain substances like proteins, gums, soaps, detergents, lampblack, etc.)

Question 4.
Analyse the following statements:
Statement 1: Negatively charged Arsenic sulphide can be effectively precipitated using Al³⁺ ion than Na⁺ ion.
Statement 2: There is no difference in the precipitation of arsenic sulphide by Al³⁺ ion and Na⁺ ion.

Which statement is correct? Justify.
Name the law behind your opinion.
State the law.

Answer:

Statement 1. Precipitating power of an ion is directly proportional to its valency.
Hardy-Schulze rule.
A colloid can be precipitated by an ion having opposite charge to that of the colloid. Greaterthe valency of coagulating ion added greater its power to precipitate.
eg: AI³⁺ > Mg²⁺ > SO₄^2- > Na⁺ > PO₃⁺ > Cl^–

Question 5.
What is observed in the following situations?

When a beam of light is passed through a colloid.
When an electrolyte, NaCl is added to hydrated ferric hydroxide sol.

Answer:

Tyndall cone is observed due to Tyndall effect. Scattering of light by colloidal particles is called Tyndall effect.
The coagulation of ferric hydroxide sol is observed. This is in accordance with Hardy-Schulze rule.

Question 6.
Adsorption is a surface phenomenon.

Give an example for adsorption.
Name the following plot. What do x/m and P_s stands for?

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 5 Surface Chemistry four marks q6 img 8
Answer:

Adsorption of Cl₂ on charcoal.
Freundlich adsorption isotherm, x/m represents rate of adsorption (‘x’ is the mass of the gas adsorbed on mass ‘m’ of the adsorbent) and P_s is the pressure at which maximum adsorption takes place (saturation pressure).

Question 7.
Explain the following terms:

Peptization
Electrophoresis

Answer:

Peptization – It is the process of converting a precipitate into a colloidal solution.
Electrophoresis – The migration of colloidal particles, under the influence of an electric field is called electrophoresis.

Question 8.

Define the terms ‘Kraft temperature’ and ‘Critical micelle concentration’.
Explain the cleansing action of soap.

Answer:
1. Kraft temperature (T_k) – The temperature above which micelle formation occurs. Critical micelle concentration (CMC) – The concentration above which micelle formation occurs.

2. Soap is sodium or potassium salt of higher fatty acids. At a particular concentration the soap molecules will rearrange in such a way that its polar end is towards water and non-polar end towards oil.

These molecules aggregate to form ionic micelles around the oil/dirt particles. Since the polar groups can interact with water, the micelles are pulled in water and are removed from the dirty surface. Thus soap helps in emulsification and washing away of oils and fats.

Question 9.
Colloid of gold is prepared by striking an electric arc between gold electrodes kept in water.

What is this method known as? Why is it regarded as a dispersion method?
What is the charge of gold sol? How do particles acquire this charge?

Answer:
1. Bredig’s arc method. This process involves dispersion as well as condensation. When an electric arc is struck between the metal electrodes immersed in the dispersion medium, due to the intense heat produced the metal gets dispersed by vapourisation which then condenses to form particles of colloidal size.

2. Negative change. Due to electron capture by gold sol particles during electrodispersion of gold.

Question 10.
When excess of an electrolyte is added, the colloidal particles are precipitated.

Give the reason for precipitation.
Arrange the following ions in the increasing order of the flocculating power in coagulation of negative sol. K⁺, Ca²⁺, Fe³⁺

Answer:
1. A colloid can be precipitated by an ion having opposite charge to that of the colloid. Greater the valency of coagulating ion added greater its power to precipitate.
eg: Al³⁺ > Mg²⁺ > SO₄^2- > Na⁺ > PO₃⁺ > Cl^–

2. K⁺ < Ca²⁺ < Fe³⁺

3. These are arranged based on Hardy-Shulze rule.

Question 11.
Many food items are colloids in one form or the other.

To which category of colloids does cheese belongs?
Explain how sols are classified depending upon the nature of the interaction between the dispersed phase and dispersion medium. Give suitable examples.

Answer:

Cheese belongs to gel (Liquid in solid type colloid).
Based on the nature of interaction between dispersed phase and dispersion medium sols are divided into two categories:

a. Lyophilic colloids:
These are solvent-loving colloidal sols which are directly formed by mixing the dispersed phase with the dispersion medium. These are reversible, quite stable and cannot be easily coagulated, e.g. gum, gelatine, starch, rubber.

b. Lyophobic colloids:
These are solvent-hating sols which can be prepared only by special methods. These are irreversible, not stable and are easily coagulated on addition of small amounts of electrolytes. e.g. sols of metals and their sulphides.

Question 12.
Anhydrous CaCl₂ and silica gel differ in their behaviour towards moisture.

Identify the phenomenon taking place in both cases.
How is adsorption affected with pressure? Explain.

Answer:
1. Anhydrous CaCl₂ absorbs moisture. The phenomenon is absorption. Silica gel absorbs moisture. The phenomenon is adsorption.
2. The rate of adsorption increases with increase in pressure. Freudlich’s adsorption expression is given by
as $\frac{x}{m}$ = kp^1/n (n > 1)
The graphical representation is,
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 5 Surface Chemistry four marks q12 img 9

At high pressure rate of adsorption attains a maximum value and after that pressure is having no influence. At low pressure, rate of adsorption is directly proportional to the pressure.

Question 13.
Gelatin in icecream prevents the formation of large crystals of ice.

What is the role of gelatin here?
Justify the property of gelatin here as lyophilic colloid.

Answer:
1. Gelatin acts as a protective colloid.

2. Gelatin is a lyophilic (liquid loving) colloid and can act as a protective colloid. The colloid can be formed by simply mixing gelatin with ice cream. Also, it is very stable and difficult to get coagulated. Here gelatin forms a layer around the particles of ice cream and prevent the later from forming large crystals.

Question 14.
Milk is a colloid.

Do you agree with this statement?
What is the difference between dispersed phase and dispersion medium?
What do you mean by lyophilic colloid?

Answer:

Yes.
The medium in which dispersion takes place is called dispersion medium and the substance which undergo dispersion is called dispersed phase.
If there is strong attractive force between the dispersed phase and dispersed medium that colloid is called lyophilic colloid.

Question 15.
Many industrial processes use catalysts.

What are catalysts?
What do you mean by the terms ‘promoters’ and, ‘poisons’ in catalysis?
Suggest an example for a catalytic promoter.

Answer:

Catalysts are substances, which alter the rate of a chemical reaction and themselves remain unchanged.
Promoters are substances that enhance the activity of a catalyst. Catalytic poisons are substances which decrease the activity of a catalyst.
In Haber’s process for manufacture of ammonia, molybdenum (Mo) acts as a promoter for iron catalyst.

Question 16.
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 5 Surface Chemistry four marks q16 img 11

Classify the above reactions into Homogeneous and Heterogeneous catalysis.
What is shape-selective catalysis?

Answer:

(1) and (2) are heterogeneous catalysis and (3) is homogeneous catalysis.
The catalytic reaction that depends upon the pore structure of the catalyst and the size of the reactant and product molecules is called shape-selective catalysis.

Zeolites are good shape-selective catalysts because of their honeycomb-like structure. The reactions taking place in zeolites depend upon the size and shape of reactant and product molecules as well as upon the pores and cavities of the zeolites.

Question 17.
You are supplied with the following chemicals NaCl, BaCl₂, AlCl₃.

Which of these would you prefer most to coagulate a sample of As₂S₃ sol?
Arrange the following electrolytes in the increasing order of coagulating power.
NaCl, BaCl₂, AlCl₃
Name and state the rule based on which you have arrived at the coagulating power of these substances.

Answer:

AlCl₃
NaCl < BaCl₂ < AlCl₃
Hardy – Schulze rule. It states that greater the valence of the flocculating ion greater is its power to cause precipitation.

Plus Two Chemistry Surface Chemistry NCERT Questions and Answers

Question 1.
Give reason why a finely divided substance is more effective as an adsorbent.
Answer:
Adsorption is a surface phenomenon. Since finely divided substance has large surface area, hence, adsorption occurs to a greater extent.

Question 2.
What do you understand by activation of adsorbent? How is it achieved?
Answer:
Activation of adsorbent implies increase in the adsorption power of the adsorbent. It involves increase in the surface area of the adsorbent and is achieved by the following methods:

by finely dividing the adsorbent into small grains.
by removing the gases already adsorbed.
by making the surface of adsorbent rough by chemical or mechanical methods.

Question 3.
What role does adsorption play in heterogeneous catalysis?
Answer:
In heterogeneous catalysis, the reactants are generally gases whereas catalyst is a solid. The reactant molecules are adsorbed on the surface of the solid catalyst by physisorption or chemisorption. As a result, the concentration of the reactant molecules on the surface increases and hence the rate of reaction increases.

Alternatively, one of the reactant molecules undergo fragmentation on the surface of the solid catalyst producing active species which react faster. The product molecules in either case have no affinity for the solid catalyst and are desorbed making the surface free for fresh adsorption. This theory is called adsorption theory.

Question 4.
What is the difference between multimolecular and macromolecular colloids? Give one example of each. How are associated colloids different from these two types of colloids?
Answer:
Multimolecular colloids are formed by the aggregation of a large number of small atoms/molecules. The aggregates thus formed have size in the colloidal range, e.g. Gold sol.

Macromolecular colloids contain large size molecules which have the dimensions of colloids, e.g. Starch. Associated colloids are formed by surface active molecules having polar as well as non-polar ends.

They behave as electrolytes at low concentration but beyound critical micelle concentration and above the Kraft temperature, they associate together to form ionic micelles whose size lies in the colloidal range, e.g. Soap.

Question 5.
Explain what is observed

When a beam of light is passed through a colloidal sol?
An electrolyte, NaCl is added to hydrated ferric oxide sol?
Electric current is passed through a colloidal sol?

Answer:

Scattering of light by the colloidal particles takes place and the path of light become illuminated. This is called Tyndall effect.
The positively charged colloidal particles of Fe(OH)₃ sol get coagulated by the oppositely charged ion provided by the electrolyte.
On passing direct current, colloidal particles move towards the oppositely charged electrode where they lose their charge and get coagulated.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

April 13, 2021January 29, 2020 by Prasanna

Students can Download Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits Questions and Answers, Plus Two Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

Plus Two Physics Semiconductor Electronics: Materials, Devices, and Simple Circuits NCERT Text Book Questions and Answers

Question 1.
In an n-type silicon, which of the following statement is true.
(a) Electrons are majority carriers and trivalent atoms are the dopants.
(b) Electrons are minority carriers and pentavalent atoms are the dopants.
(c) Holes are minority carriers and pentavalent atoms are the dopants.
(d) Holes are majority carriers and trivalent atoms are the dopants.
Answer:
(c) Holes are minority carriers and pentavalent atoms are the dopants.

Question 2.
Carbon, silicon, and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (E_g)_c, (E_g)_Si and (E_g)_Ge. Which of the following statements is true?
(a) (E_g)_Si < (E_g)_Ge < (E_g)_c
(b) (E_g)_c < (E_g)_Ge < (E_g)_Si
(c) (E_g)_c > (E_g)_Si > (E_g)_Ge
(d) (E_g)_c = (E_g)_Si = (E_g)_Ge
Answer:
(c) (E_g)_c > (E_g)_Si > (E_g)_Ge

Question 3.
When a forward bias is applied to a p-n junction, it.
(a) raise the potential barrier.
(b) reduces the majority carrier current to zero.
(c) lowers the potential barrier.
(d) None of the above
Answer:
(c) lowers the potential barrier.

Question 4.
For transistor action, which of the following statements are correct:
(a) Base, emitter and collector regior should have similar size and dopin concentrations.
(b) The base region must be very thin are lightly doped.
(c) The emitter junction is forward biase and collector junction is reverse biased.
(d) Both the emitter junction as well as to collector junction are forward biased.
Answer:
(b) The base region must be very thin are lightly doped,
(c) The emitter junction is forward biase and collector junction is reverse biased.

Question 5.
For transistor amplifier, the volta gain.
(a) remains constant for all frequencies.
(b) is high at high and low frequencies.
(c) is low at high and low frequencies a constant in the middle frequency large.
(d) None of the above.
Answer:
(c) is low at high and low frequencies a constant in the middle frequency large.

Question 6.
In half-ware rectification, what is the output frequency if the input frequency is 50Hz. What is the output frequency of a fullwave rectifier for the same input frequency.
Answer:
Given Input frequency = 50Hz
Output frequency
For Halfwave rectifier = 50 Hz
For full wave rectifier = 50 × 2 = 100 Hz.

Question 7.
A p-n photodiode is fabricated from a semiconductor with band gap of 2.8eV. Can it detect a wavelength of 6000 nm?
Answer:
Band gap E_g = 2.8 eV
Energy band gap corresponding to wavelength 6000 nm is given by
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 1
Since E_g, < E_g, so it can not detect a wave length of 6000 nm.

Plus Two Physics Semiconductor Electronics: Materials, Devices, and Simple Circuits One Mark Questions and Answers

Question 1.
The zenerdiode works in______bias.
Answer:
reverse bias.

Question 2.
A transistor is operated in common emitter configuration at V_c = 2 V, such that a change in the base current from 100 µA to 200 µA produces a change in the collector current from 5 mA to 10 mA. The current gain is
(a) 100
(b) 150
(c) 75
(d) 50
Answer:
(d) 50
Explanation: Current gain,
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 2

Question 3.
The electrical conductivity of an intrinsic semiconductor at 0 K is
(a) less than that an insulator
(b) equal to zero
(c) equal to infinity
(d) more than that of an insulator
Answer:
(b) equal to zero.

Question 4.
The voltage between the terminals A and B is 17 V and Zener breakdown voltage is 9 V. Find the potential across R is
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 3
Answer:
The potential across R = 17V – 9V = 8V.

Question 5.
Hole is
(a) an antiparticle of electron.
(b) a vacancy created when an electron leaves a covalent bond.
(c) absence of free electrons.
(d) an artificially created particle.
Answer:
(b) a vacancy created when an electron leaves a covalent bond.

Question 6.
A circuit is constructed by using certain gates is given below
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 4
1. Each gate is a…….. gate
2. Complete the truth table of above circuit

Answer:
1. NAND gate

2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 6

Question 7.
In both p and n-type semiconductor, actually electrons are flowing. What difference do you observe in the motion of electrons in these semiconductors?
Answer:
Electrons in valence band are flowing. Electrons in conduction band are flowing.

Question 8.
Unidirectional property of diode; Rectification. Then the break down action of Zener diode:…….
Answer:
Voltage regulation.

Question 9.
For an input frequency 50Hz, the output frequency……..of hall wave rectifier is and the output of full
wave rectifier for the same input frequency is………
Answer:
50 Hz, 100 Hz

Question 10.
The following questions consists of two statements. Assertion: Zener diode works as a voltage regulator Reason: Zener voltage is independent of the Zener current variations and change of load resistance. Write the correct response from the following.
(a) Both assertion and reason are true and the rea¬son is not a correct explanation of the assertion.
(b) Assertion is true, but reason is false.
(c) Both assertion and reason are true and reason is correct explanation for the assertion.
(d) Both assertion and reason are false.
Answer:
(c) Both assertion and reason are true and reason is correct explanation for the assertion.

Question 11.
Assertion: Semiconductors have -ve temperature co-efficient of resistance.
Reason: As temperature of a semiconductor increases, number density of charge carriers also increases.
(a) Both assertion and reason are correct, but reason is not proper explanation.
(b) Both assertion and reason are correct and reason is proper explanation.
(c) Assertion is correct but reason is wrong.
(d) Assertion is correct, and reason also is correct.
Answer:
(a) Assertion js correct, but reason is incorrect

Question 12.
Correct the following CE amplifier circuit.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 7
Answer:

Plus Two Physics Semiconductor Electronics: Materials, Devices, and Simple Circuits Two Mark Questions and Answers

Question 1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 9
Answer:
Semiconductor – P-type, n-type.

Question 2.
Fill in the blanks with appropriate word given below. (Base, collector, emitter, bias-collector junction, collector-emitter junction, emitter bias junction) Structurally, a bipolar junction transfer consists of emitter, base and………Out of these regions……….is the most heavily doped. For proper functioning of a transistor………..is forward biased and………….is reverse biased.
Answer:

Collector
Emitter
EB junction
CB junction

Question 3.
Classify the following into conductors, insulators, and semiconductors.
Ga, As, Ni, Calcite, Graphite
Answer:

Conductor – Graphite, Ni
Insulator – Calcite
Semiconductor – Ga,As.

Question 4.
Construct truth table for following logic circuit.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 10
Answer:

Question 5.
State whether true or false and justify.

Zener diode are used under forward bias.
In n-p-n transistor current conduction is primarily due to electrons.
Transistor amplifier do not strictly obey law of conservation of energy since output power is greater than input power.
In a transistor amplifier all the frequency will have exactly equal gain.

Answer:

False, Zener diodes are used under reverse bias
True
False
False

Question 6.
Match the following
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 12
Answer:

Plus Two Physics Semiconductor Electronics: Materials, Devices, and Simple Circuits Three Mark Questions and Answers

Question 1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 14
An electric circuit containing a battery, a bulb, and two switches is give above
1. Identify the gate an alogues to the above electric circuit
2.

If the above two input signal are applied to the gate what will be the shape of out put wave Draw the out put wave.
Answer:
1. OR gate

2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 16

Question 2.
A boy designs a circuit to study the input and output characteristics of an npn transistor
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 17

Identify the transistor configuration, input current and output current
By keeping the output voltage constant, the boy measures the input current by varying the input voltage. If a graph is drawn, what is the nature of the input characteristic? Justify your answer.

Answer:
1. Common Emitter, i_b, and i_c

2. Input Characteristics (CE configuration):
The graph connecting base current with base emitter voltage (at constant VCE) is the input characteristics of the transistor.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 18
To study the input characteristics, the collector to emitter voltage (V_CE) is kept at constant. The base current I_B against V_BE is plotted in a graph. The ratio ∆ V_BE /∆_IB at constant V_CE is called the input resistance.
i.e.,Input resistance = $r_{i}=\frac{\Delta V_{B E}}{\Delta I_{B}}$.

Question 3.
Truth table of a logic gate is given below:
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 19
1. Identify the logic gate.
2. Explain the working of this gate using diode and battery.

3. If these input signals are applied across the gate, what will be the shape of the output wave form.
Answer:
1. OR gate

2. Out of syllabus
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 21

Question 4.
A circuit using two switches (A and B), cell and bulb is shown in the figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 22
1. The above circuit is equivalent to

OR gate
AND gate
NOT gate
NOR gate

2. Give symbol and truth table of the above gate
Answer:
1. OR gate

2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 23

Plus Two Physics Semiconductor Electronics: Materials, Devices, and Simple Circuits Four Mark Questions and Answers

Question 1.
Forward biased pn junction diodes are shown in the figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 24

Identify the figure, which shows the correct direction of flow of charges.
What do you mean by barrier potential and depletion region of a pn junction?
When forward bias is applied to a p-n junction, what happens to the potential barrier and the width of depletion region.

Answer:
1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 25

2. The potential developed across the junction, which tends to prevent the movement of electron from the n region into the p region, in semiconductor is called a barrier potential.

The space-charge region on either side of the junction at which there is no free charge carriers is known as depletion region.

3. The potential barrier decreases and depletion region gets reduced.

Question 2.
The symbol of a n-p-n transistor is shown in figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 26

Redraw the symbol and mark emitter, collector and base of the transistor.
Arrange the doping concentration and width of emitter, collector and base regions in ascending order.
What happens when both the emitter and the collector of a transistor are forward biased?

Answer:
1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 27
2. Doping concentration of base < doping concentration of collector < doping concentration of emitter. Width of base < Width of emitter < Width of collector.

3. The transistor will work as two p-n junctions with common base terminals.

Question 3.
A green house has an electric system, which automatically switches ON a heater if the air temperature in the green house drops too low. A manual switch is included so that the automatic system can be switched off.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 28

What is meant by 1 and 0 in digital circuit?
Name logic gate X. Why is it used?
Name the logic gate Y?
Construct a truth table of this electronic system by taking A and B as inputs and D as output.

Answer:
1. 1 – means maximum, 0 – means minimum voltage

2. NOT gate

3. AND gate

4.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 29

Question 4.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 30
The forward-biased diode is wrongly given above.

Redraw the above circuit correctly.
Draw the graph of current I with voltage v in forward bias.
Classify the following circuit into forward bias, reverse-bias, unbias.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 31
Answer:
1.

2.

3. circuit into forward bias, reversebias, unbias:

Reverse bias
Forward bias
Reverse bias
Unbias

Question 5.
A car stereo working at stabilized voltage supply of 9 v DC and has a zener diode of 9V, 0.25W. But the voltage supply inside the car is 12V DC.

Which mode of bias will you suggest to connect zener diode voltage regulator?
Draw a circuit diagram of voltage regulation to help the boy.
Which device is essential for circuit diagram? Find the value of that device.

Answer:
1. Reversebias

2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 34

3. Resistance: R can be calculated using the equation
V_s = IR + V_z , 12 = $\left(\frac{0.25}{9}\right)$R + 9, R = 108Ω.

Question 6.
Electric current is the flow of charges along a definite direction and take place through metals as well as semiconductors.

Mention the charge carriers in the above cases.
Give the sketch of graph with V along X-axis and I along Y-axis for a metal at room temperature.
Give the physical significance of the slope of the graph.
If the above graph is drawn at 100°C, compared the nature of the graph with the graph at room temperature.

Answer:

Metals – electrons
Semiconductor – Electrons and holes
Straight line graph
Slope gives conductance
When the temperature increases, resistance is also increased and hence slope decreases.

Question 7.
A transistor in the common-emitter mode can be used as an amplifier

Design a circuit to amplify an ac signal given in the input region
[Hint: Give forward biasing to input region, reverse biasing to output region and take output across a resistor]
Derive expressions for voltage gain, current gain and power gain in the above transistor configuration.

Answer:
1.

2. When we apply an AC signal as input, we get an AC base current denoted by i_B. Hence input AC voltage can be written as
V_i = i_Br ……..(1)
where ‘r’ is the effective input resistance.
This AC input current produces an AC output current (i_c) which can flow through a capacitor. Hence the output voltage can be written as
V₀ = i_c × output resistance
If we take output resistance as R_L then v₀ becomes
V₀ = i_c R_L
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 36
Substituting eq(1) and eq(2),in the above equation we get

Power gain:
The power gain A_p can be expressed as the product of the current gain and voltage gain.
ie. power gain A = β_ac × A_v.

Question 8.
The circuit diagram of a full wave rectifier is shown
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 38

Explain how its works? Also draw the output wave form
If the frequency of a.c. at the input is 50Hz what will be the output frequency?

Answer:
1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 39
Full wave rectifier consists of transformer two diodes and a load resistance R_L. Input a.c signal is applied across the primary of the transformer. Secondary of the transformer is connected to D₁, and D₂. The output is taken across R_L.
Working:
During the +ve half cycle of the a.c signal at secondary, the diode D₁ is forward biased and D₂ is reverse biased. So that current flows through D₁ and R_L.

During the negative half cycle of the a.c signal at secondary, the diode D₁ is reverse biased and D₂ is forward biased. So that current flows through D₂ and R_L.

Thus during both the half cycles, the current flows through R_L in the same direction. Thus we get a +ve voltage across R_L for +ve and -ve input. This process is called full wave rectification.

2. 100Hz

Question 9.
Forward biased pn junction diodes are shown in the figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 40

Identify the figure, which shows the correct direction of flow of charges. (1)
What do you mean by barrier potential and depletion region of a pn junction? (2)
When forward bias is applied to a p-n junction, what happens to the potential barrier and the width of depletion region. (1)

Answer:
1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 42

2. The potential developed across the junction .which tends to prevent the movement of electron from the n region into the p region, in semiconductor is called a barrier potential.

The space charge region on either side of the junction at which there is no free charge carriers is known as depletion region.

3. The potential barrier decreases and depletion region gets reduced.

Question 10.
LEDs that can emit red, yellow, orange, etc. commercially available.

How these colours are obtained in a LED. (1)
Write any two uses of LED. (1)
What are its advantages over ordinary bulbs? (2)

Answer:
1. Different colours are obtained by changing the concentration of arsenic and phosphors in Gallium Arsenide Phosphide.

2. LEDs find extensive use in remote controls, burglar alarm systems, optical communication, etc.

3. LEDs have the following advantages over conventional incandescent low power lamps:

Low operational voltage and less power.
Fast action and no warm-up time required.
The bandwidth of emitted light is 100A° to 500A° or in other words it is nearly (but not exactly) monochromatic.
Long life and ruggedness Fast on-off switching capability.

Question 11.
The symbol of a n-p-n transistor is shown in figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 43

Redraw the symbol and mark emitter, collector and base of the transistor. (1)
Arrange the doping concentration and width of emitter, collector and base regions in ascending order. (2)
What happens when both the emitter and the collector of a transistor are forward biased? (1)

Answer:
1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 44

2. Doping concentration of base < doping concentration of collector < doping concentration of emitter. Width of base < Width of emitter < Width of collector.

3. The transistor will work as two p-n junctions with common base terminals.

Plus Two Physics Semiconductor Electronics: Materials, Devices, and Simple Circuits Five Mark Questions and Answers

Question 1.
The transfer characteristic of n-p-n transistor in CE configuration is shown in the figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 45

Identify the cut-off region, active region, saturation region from the figure.
In which of these regions, a transistor is said to be switched off.
For a CE transistor amplifier, the audio signal voltage across collector resistance of 2.0 kΩ is 2.0V. Suppose the current amplification factor of the transistor is 100. What should be the value of R_B in series with V_BB supply of 2.0V, if DC base current has to be 10 times the signal current?
In the working of a transistor, the emitter-base (EB) junction is forward biased while collector-base (CB) junction is reverse biased. Why?

Answer:
1.

I – cut off region
II – active region
III-saturation region

2. Region I

3.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 46
I_dc =10^-5 × 10 = 10^-4A
V_bb = V_be + I_bR_b

4. Only forward biased emitter-base junction can send the majority charge carriers from emitter to base and only reverse biased collector can collect these majority charge carriers form the base region.

Question 2.
The basic building blocks of digital electronic circuits are called Logic Gates. Some logic gates and their names are given in the table.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 48
1. Match the symbols of logic gates with their names.
2. Draw the output wave form, from the given input wave form of a NAND gate as shown in figure (input terminals are A and B)
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 49
3. Write the truth table forthe given circuit.

Answer:
1. Match the symbols of logic gates with their names:

A – 5
B – 1
C – 2
D – 3
E – 4

2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 51

3.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 52

Question 3.
Following figure is an incomplete circuit of common emitter transistor in CE configuration with the input forward biased.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 53

Identify the transistor an NPN or PNP.
Complete the above circuit diagram by giving proper bias in the output and connect load resistance of 4 KΩ.
When the base current changes by 20µ Afor V_BE = .02 V. What is the voltage gain of the amplifier, if I_c = 2mA
npn transistors are preferred in devices with very high-frequency source. Why?

Answer:
1. npn transistor

2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 54

3.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 55

4. The resistance of a semiconductor decreases with rise in temperature.

Question 4.
Diodes are one of the building elements of electronic circuits. Some type of diods are shown in the figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 56

Identify rectifier diode from the figure.
Draw the circuit diagram of a forward biased rectifier diode by using a battery.
Draw the forward and reverse bias characteristics of a rectifier diode and mark threshold voltage or cut in voltage.
What happens to the resistance of a semiconductor on heating?

Answer:
a.

b.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 58

c. The forward and reverse characteristics of a silicon diode is as shown in the fig.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 59

d. The resistance of a semiconductor decreases with rise in temperature.

Question 5.
A full wave rectifier circuit is shown in figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 60
1. Draw the output wave form of the rectifier.
2. If a full wave rectifier circuit is operating from 50 Hz mains, the fundamental frequency in the ripple will be

50 Hz
70.7 Hz
100 Hz
25 Hz

3. In a zener regulated power supply, a Zener diode with V_z = 6.0V is used for regulation. The load current is to be 4.0 mA and the unregulated input 10.0V. What is the value of series resistor R?
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 61
Answer:
1.

2. 100 Hz

3. Input Voltage = 10V, Zener voltage V_z = 6V The voltage drop on the resistor should be 4V

Question 6.
A P N junction diode is connected to a cell as a shown in figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 64

Name the type of biasing used here
Design a circuit diagram to draw the characteristics of the diode in above biasing.
Trace the characteristics curve if the polarity of battery is reversed

Answer:
1. forward biasing

2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 65

3.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 66
In forward bias, current first increases very slowly up to a certain value of bias voltage. After this voltage, diode current increases rapidly. The diode offers low resistance in forward bias.

In reverse bias, current is very small. It remains almost constant upto break down voltage (called reverse saturation current). After this voltage reverse current increases sharply.

Question 7.
The circuit diagram of a full wave rectifier is shown.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 67
1. Explain how its works? Also draw the output wave form.
2. If another diode is connected in series with D₂, as shown below what will happen to the out put wave form?
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 68
3. If the frequency of a.c. at the input is 50Hz what will be the output frequency of full wave rectifier?
Answer:
1.

Full wave rectifier consists of transformer two diodes and a load resistance R_L. Input a.c signal is applied across the primary of the transformer. Secondary of the transformer is connected to D₁, and D₂. The output is taken across R_L.
Working:
During the +ve half cycle of the a.c signal at secondary, the diode D₁ is forward biased and D₂ is reverse biased. So that current flows through D₁ and R_L.

During the negative half cycle of the a.c signal at secondary, the diode D₁ is reverse biased and D₂ is forward biased. So that current flows through D₂ and R_L.

Thus during both the half cycles, the current flows through R_L in the same direction. Thus we get a +ve voltage across R_L for +ve and -ve input. This process is called full wave rectification.

2. Out put will be halfwave.

3. 100 Hz.

Question 8.
Diodes are one of the building elements of electronic circuits. Some type of diods are shown in the figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 70

Identify rectifier diode from the figure. (1)
Draw the circuit diagram of a forward biased rectifier diode by using a battery. (1)
Draw the forward and reverse bias characteristics of a rectifier diode and mark threshold voltage or cut in voltage. (2)
What happens to the resistance of a semiconductor on heating? (1)

Answer:
1.

2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 72

3. The forward and reverse characteristics of a silicon diode is as shown in the fig.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 73
4. The resistance of a semiconductor decreases with rise in temperature.

Question 9.
A full wave rectifier circuit is shown in figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 74
1. Draw the output wave form of the rectifier. (1)
2. If a full wave rectifier circuit is operating from 50 Hz mains, the fundamental frequency in the ripple will be (2)

50 Hz
70.7 Hz
100 Hz
25 Hz (2)

3. In a zener regulated power supply, a zenerdiode with V_z = 6.0V is used for regulation. The load current is to be 4.0 mA and the unregulated input 10.0V. What is the value of series resistor R?(2)
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 75
Answer:
1.

2. 100 Hz

3. Input Voltage = 10V, Zener voltage Vz = 6V The voltage drop on the resistor should be

Question 10.
The following diagram shows energy bands in a semiconductor.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 78
(a) Which diagram shows energy band positions at OK? (1)
(b) What do you mean by energy gap? Match the elements /compounds with their respective energy gap values. (1)

Diamond	6 eV
Aluminium	0.03 eV
Germanium	1.1 eV
Silicon	0.71 eV

(c) Classify solids into conductors, semiconductors, and insulators by drawing energy diagram. (3)
Answer:
(a)
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 79

(b) The gap between the top of the valence band and bottom of the conduction band is called the energy band gap.

Column I	Column II
A. Diamond	1. 1.1 eV
B. Aluminium	2. 0.71 eV
C. Germanium	3. 0.03 eV
D. Silicon	4. 6 eV

(c) For conductors
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 80
For Insulators

For Semiconductors

Question 11.
The given figure shows an npn transistor.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 83

Redraw the figure and show the biasing voltage, direction of current and direction of flow of electrons and holes. (2)
Draw the input and output characteristics of transistor connected in common emitter configuration. (2)
In a transistor, a change of 7.9mA is observed in the collector current for a change of 7.99mA in the emitter current. Determine the change in base current. (1)

Answer:
1.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 84

2. Input characteristic
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 85
Output characteristic

3. ∆I_E = ∆I_c + ∆I_B, ∆I_c = 7.9mA, ∆I_E = 7.99mA
∆I_B = ∆I_E – ∆I_c
= 7.99 – 7.9 = 0.09mA.

Question 12.
The transfer characteristic of n-p-n transistor in CE configuration is shown in the figure.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 87

Identify the cut-off region, active region, saturation region from the figure. (1)
In which of these regions, a transistor is said to be switched off. (1)
For a CE transistor amplifier, the audio signal voltage across collector resistance of 2.0 kΩ is 2.0V. Suppose the current amplification factor of the transistor is 100. What should be the value of R_B in series with V_BB supply of 2.0V, if DC base current has to be 10 times the signal current? (2)
In the working of a transistor, the emitter-base (EB) junction is forward biased while collector-base (CB) junction is reverse biased. Why? (1)

Answer:
1. cut-off region, active region, saturation region from the figure:

I – cut off region
II – active region
III – saturation

2. Region I

3.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 88

Question 13.
The basic building blocks of digital electronic circuits are called Logic Gates. Some logic gates and their names are given in the table.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 89
1. Match the symbols of logic gates with their names. (2)
2. Draw the output wave form, from the given input wave form of a NAND gate as shown in figure (input terminals are A and B) (1)
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 90
3. Write the truth table for the given circuit. (2)

Answer:
1. Match the symbols of logic gates with their names:

A – 5
B – 1
C – 2
D – 3
E – 4

2.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 93

3.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 94

Question 14.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 95

According to energy gap, Classify them as metal, Insulator, and semiconductor.
From which of the above material we can eject electrons with minimum effort Explain
In Photo electric effect, while we are measuring photo current by varying retarding potential the variations is as shown in graph.

Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 96
Answer:

Insulator, semiconductor, and conductor
Metals
This graph shows that photocurrent increases and reaches saturation with anode potential. This increase in current shows that electrons are emitted from the metal surface with different energies.

Question 15.
A diode can be properly doped at the time of its manufacture, so that it have a shape break down voltage
1. The above diode is called

Zener diode
Photo diode
Light emitting diode
Solar cell

2. Compare V-l Characteristics of above diode with that of an ordinary diode

3. Explain how the above diode can be used as an voltage regulator.

Answer:
1. Zener diode

2. Zener diode has sharp breakdown voltage than ordinary diode

3.
Plus Two Physics Chapter Wise Questions and Answers Chapter 14 Semiconductor Electronics Materials, Devices, and Simple Circuits - 97
The zener diode is connected to a fluctuating voltage supply through a resistor R_z. The out put is taken across R_L.
Working:
When ever the supply voltage increases beyond the breakdown voltage, the current through zener increases (and also through R_z).

Thus the voltage across R_z increases, by keeping the voltage drop across zener diode as a constant value. (This voltage drop across R_z is proportional to the input voltage)

Similarly, when supply voltage decreases beyond a certain value, the current through the zener diode decreases. Thus the voltage across R_z decreases, by keeping the voltage drop across zener diode as constant (Zener diode as a voltage regulator).

Plus Two Economics Chapter Wise Questions and Answers Chapter 6 Open Economy Macroeconomics

April 13, 2021January 29, 2020 by Prasanna

Students can Download Chapter 6 Open Economy Macroeconomics Questions and Answers, Plus Two Economics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus Two Economics Chapter Wise Questions and Answers Chapter 6 Open Economy Macroeconomics

Plus Two Economics Open Economy Macroeconomics One Mark Questions and Answers

Question 1.
Name the ‘balancing item’ affecting the inability to record all international transactions accurately?
Answer:
Errors and omissions

Question 2.
The amount of rupees required to buy one US$ is known as ………….
(i) Rupee dollar exchange rate
(ii) Dollar rupee exchange rate
(iii) Real exchange rate
(iv) Real effective exchange rate
Answer:
(i) Rupee dollar exchange rate

Question 3.
Which among the following is a component of BOP account?
Answer:
(i) Current account
(ii) Capital account
(iii) Official reserve
(iv) All the above
Answer:
(iv) All the above

Question 4.
WTO was formed in?
(i) 1948
(ii) 1964
(iii) 1991
(iv) 1995
Answer:
(iv) 1995

Question 5.
If export > imports, it represents
(i) Trade surplus
(ii) Trade balance
(iii) Trade deficit
(iv) None of these above
Answer:
(i) Trade surplus

Question 6.
Which of the following would be an appropriate policy to reduce a Balance of Payment / Deficit.
Answer:
(i) An increase in government spending.
(ii) A cut in the level of indirect taxes
(iii) An increase in interest rates
(iv) A decrease in interest rates
Answer:
(iii) An increase in interest rates

Plus Two Economics Open Economy Macroeconomics Two Mark Questions and Answers

Question 1.
How is the exchange rate determined under flexible exchange rate regime?
Answer:
In case of flexible exchange rates, the exchange rate changes to clear the market to equate the demand for and supply of foreign exchange.

Question 2.
List two items of the capital account of balance of payment account?
Answer:

Gold movement
Reserve, Monetary gold & SDR

Question 3.
When will balance of trade shows a deficit?
Answer:
Balance of trade shows a surplus when exports are greater than imports. That is, Surplus balance of trade = Exports > Imports

Question 4.
Name two sources of demand for foreign exchange.
Answer:
Two sources of demand for foreign exchange are:

To purchase goods and services from abroad.
To send gifts and grants to foreign countries.

Question 5.
The value of a country’s import of goods is ₹200 crore and value of export of goods is ₹250 crore. Find out its balance of trade.
Answer:
Balance of trade = Value of exports – value of imports = 250 – 200 = ₹50 crore

Question 6.
Identify the items to be included to trade balance to get current account balance.
Answer:

Trade services
Net transfers

Question 7.
Identify the situation mentioned below.

Rupee-dollar exchange rate change from 45 to 50.
Rupee-dollar exchange rate changed from 45 to 43.

Answer:

Depreciation
Appreciation

Question 8.
State the National Income identity for an open economy.
Answer:
Y = C + IG + (X – M)
Where,

Y = National Income
C = Comsumption
I = Investment
G = Government spending
X = Export
M = Import

Question 9.
The consumption function and import function of an economy can be given as,
C = a + b.y
M = M + m.y Identity the letters

Answer:

Marginal propensity to consume
Autonomous consumption
Marginal propensity to import

Question 10.
If MPC = 0.8, and increase in autonomous demand is 200, calculate multiplier and the national output.
Answer:
MPC = b – c = 0.8
Therefore multiplier is
$\begin{array}{l}
{\frac{1}{1-C} \text { or } \frac{1}{1-b}} \\
{\text { i.e. } \frac{1}{1-0.8}=\frac{1}{0.2}} \\
{=5}
\end{array}$
National output is 5 × 200 = 1,000

Question 11.
Calculate the open economy multiplier if c = 0.5, m = 0.3. Increase in autonomous demand = 200
Answer:
Plus Two Economics Chapter Wise Questions and Answers Chapter 6 Open Economy Macroeconomics img1

Question 12.
What happens to the aggregate demand due to

A leakage from the circular flow of income
A injection to the circular flow of income

Answer:
AD falls due to the leakage from circular flow of income and AD increases due to injection in too the circular flow of income.

Question 13.
List out the expert of services from the following……..
(a) India buys a new technology from France
(b) A Japaneese tourist visits India
(c) An Indian student registers for a UK exam
(d) an Indian doctor going to work in the US.
Answer:
b and d are examples of expert services from India.

Question 14.
Analyse the effect of the following on imports and exchange rate.

Appreciation of domestic currency.
Depreciation of domestic currency
Increase in foreign direct investment.
Increase in import duty.

Answer:

Increase in Imports, Fall in Exchange Rate
Decrease in Imports, Rise in Exchange Rate
Increase in Imports, Fall in Exchange Rate
Decrease in Imports, Fall in Exchange Rate

Plus Two Economics Open Economy Macroeconomics Three Mark Questions and Answers

Question 1.
Classify the following into visible and invisible.
Steel, computer software, shipping services, wheat, machinery, food articles, banking, IT- enabled services, crude oil, shipping, textiles, Online business.
Answer:

Visibles	Invisibles
Steel	Shipping Services
Textiles	Banking
Wheat	IT Enabled Services
Machinery	Shipping
Food Articles	Insurance
Crude Oil	Online Business

Question 2.
Classify the following into current account and capital account.
Foreign direct investment, borrowing from abroad, export earning from merchandise, export earnings from banking services, earning from tourism, foreign portfolio investment.
Answer:

Current Account	Capital Account
Export earnings from banking services	Foreign direct investment
Earning from tourism	Borrowing from abroad
Export earnings from merchandise	Foreign portfolio investment

Question 3.
The open economy multiplier is smaller than that in a closed economy. Do you agree? Give reason.
Answer:
Yes, I do agree with this statement.
The open economy multiplier is smaller than that in a closed economy because a part of domestic demand falls on foreign goods. An increase in autonomous demand thus leads to a smaller increase in output compared to a closed economy. It also results in a deterioration of the trade balance

Question 4.
Complete the following table.

Export > Import
Export = Import
Export < Import

Answer:

Export > Import – Trade surplus
Export = Import – Trade balance
Export < Import – Trade deficit

Question 5.
Export promotion is one of the key factors for correcting disequilibrium in BOR Is there any other measure for correcting BOP? If yes, suggest 3 measures.
Answer:
Measures to correct BOP disequilibrium,

Increase in production
Reduction in imports
Encouraging foreign investment
Promotion of exports

Question 6.
What is the MPM (Marginal propensity to import) When M = 60 + 0.67?
Answer:
Marginal propensity to import (MPM) is the fraction of an additional currency of income spent on imports. The concept of MPM is same as the marginal propensity to consume (MPC). Thus, demand for imports is to depend on income and have an autonomous component.

Question 7.
Point out the items included in current account transactions of BOP.
Answer:
The current account includes receipts and payments on account of:

export and import of goods and services
tourism services
foreign investment incomes and out payments
private transfer payments
inter-government transfer payments.

Question 8.
What is trade deficit? Calculate the trade deficit from the following data.
Plus Two Economics Chapter Wise Questions and Answers Chapter 6 Open Economy Macroeconomics img2
Answer:
Trade deficit is the difference between export of goods and import of goods in trade in goods in current account. It is the situation where import is greater than export.
Export of goods = 90,660
Import of goods = 1,20,364 – 90,660 = 29,704 crores

Question 9.
a. Recently the government of UK decided to relax the visa norms to Indian visitors.
b. The government of India approved a purchase of weapon for Indian defence from rest of the world for an amount of 82000 crores

How does these decisions affect the demand for foreign exchange?
Analyse the consequences in the foreign exchange market with the help of a diagram.
(supply curve of foreign exchange remain the same)

Answer:
1. Demand for foreign exchange increases

2. diagram
Plus Two Economics Chapter Wise Questions and Answers Chapter 6 Open Economy Macroeconomics img3

Plus Two Economics Open Economy Macroeconomics Five Mark Questions and Answers

Question 1.
Find the odd one out. Justify your answer.

tea, coir, tourism
rice, banking services, insurance service, transport services
foreign direct investment, foreign portfolio investment, remittances, borrowings
foreign investment, remittances, export earning from goods, export earning from services

Answer:

Tourism. Others are visibles
Rice. Others are invisibles
Remittances. Others are capital receipts
Remittances. Others are capital receipts.

Question 2.
Match the following.

A	B
Bretton Woods system	1944
SDR	1967
Fixed Exchange Rate	Pegged Rate
Triffin Dilemma	Dollar accumulation
Flexible Exchange Rate	Floating Rate

Answer:

A	B
Bretton Woods system	Pegged Rate
SDR	Dollar accumulation
Fixed Exchange Rate	1944
Triffin Dilemma	Floating Rate
Flexible Exchange Rate	1967

Question 3.
If c (marginal propensity to consume) = 0.8 and m (marginal propensity to import) = 0.3,

Find the open and closed economy multiplier
If domestic autonomous demand increases by 100, find the output level in a closed and an open economy.

Answer:
1. Closed economy multiplier
$=\frac{1}{1-c}=\frac{1}{1-0.8}=\frac{1}{0.2}=5$
Open economy multiplier
$=\frac{1}{0.5}=\frac{1}{1-0.8+0.3}=\frac{1}{1-0.5}=\frac{1}{0.5}=2$

2. If domestic autonomous demand increases by 100, in a closed economy output increases by 500 whereas it increases by only 200 in an open economy.

Question 4.
Differentiate between fixed exchange rate and flexible exchange rate.
Answer:
In a system of flexible exchange rates (also known as floating exchange rates), the exchange rate is determined by the forces of market demand and supply. Countries have had flexible exchange rate system ever since the breakdown of the Bretton Woods system in the early 1970s. Prior to that, most countries had fixed or what is called pegged exchange rate system, in which the exchange rate is pegged at a particular level.

Sometimes, a distinction is made between fixed and pegged exchange rates. Under a fixed exchange rate system, such as the gold standard, adjustment to BoP surpluses or deficits cannot be brought about through changes in the exchange rate.

Question 5.
What do you mean by managed floating? How far it is a mixture of fixed exchange rate and flexible exchange rates?
Answer:
Without any formal international agreement, the world has moved on to what can be best described as a managed floating exchange rate system. It is a mixture of a flexible exchange rate system (the floating part) and a fixed rate system (the managed part).

Under this system, also called dirty floating, central banks intervene to buy and sell foreign currencies in an attempt to moderate exchange rate movements whenever they feel that such actions are appropriate. Official reserve transactions are, therefore, not equal to zero.

Question 6.
Distinguish between the nominal exchange rate and the real exchange rate. If you were to decide whether to buy domestic goods or foreign goods, which rate would be more relevant?
Answer:
The price of one currency in terms of the other is known as the exchange rate. Nominal exchange rates are bilateral in the sense that they are exchange rates for one currency against another and they are nominal because they quote the exchange rate in money terms, i.e. so many rupees per dollar or per pound.

However, the real exchange rate is the ratio of foreign to domestic prices, measured in the same currency. It is defined as Real exchange rate = ePf/P where P and Pf are the price levels here and abroad, respectively, and e is the rupee price of foreign exchange (the nominal exchange rate).

The real exchange rate is often taken as a measure of a country’s international competitiveness. Therefore, real exchange rate is considered to be more relevant.

Question 7.
Balance of payment is a broader concept than balance of trade. Give explanation to this view.
Answer:
Balance of trade is the record of a country’s visible export and visible imports. It includes only visible trade and excludes invisible trade of services. However, balance of payment is a more comprehensive term which denoted a country’s total monetary transactions with the rest of the world. It includes both visible and invisible trade of goods and, services.

The balance of payments (BoP) records the transactions in goods, services and assets between residents of a country with the rest of the world. There are two main accounts in the BoP the current account and the capital account.

Question 8.
The current account is differentiated from capital account. Do you agree? Give explanation.
Answer:
Yes. The current account balance is the sum of the balance of merchandise trade, services and net transfers received from the rest of the world. The capital account balance is equal to capital flows from the rest of the world, minus capital flows to the rest of the world.

Question 9.
Illustrate the method of determining equilibrium under flexible exchange rate system. Also, show the effect of increase in demand for imports in the foreign exchange markets.
Answer:
In a system of flexible exchange rates, the exchange rate is determined by the forces of market demand and supply. In this case of flexible exchange rates without central bank intervention, the exchange rate moves to clear the market, to equate the demand for and supply of foreign exchange. In the following figure equilibrium exchange rate is e* which is determined by the forces of demand and supply.
Plus Two Economics Chapter Wise Questions and Answers Chapter 6 Open Economy Macroeconomics img4
At the initial equilibrium exchange rate e*, suppose there is now an excess demand for foreign exchange. To clear the market, the exchange rate must rise to the equilibrium value e₁ as shown in the following figure.
Plus Two Economics Chapter Wise Questions and Answers Chapter 6 Open Economy Macroeconomics img5
The rise in exchange rate (depreciation) will cause the quantity of import demand to fall since the rupee price of imported goods rises with the exchange rate. Also, the quantity of exports demanded will increase since the rise in the exchange rate makes exports. less expensive to foreigners. At the new equilibrium, the supply and demand for foreign exchange is again equal.

Question 10.
Differentiate between devaluation and depreciation.
Answer:
Devaluation means increase in exchange rate. Devaluation is said to occur when the exchange rate is increased by social action under a pegged exchange rate system. Devaluation is used as a tool to bridge the gap of trade deficit.

On the other hand, change in the price of foreign exchange under flexible exchange rate, when it becomes cheaper as compared to domestic currency is known as depreciation.

Question 11.
Compare balance of trade (BOT) and balance of payments (BOP).
Answer:
Balance of trade is the difference between money value of imports and exports of material goods only whereas BOP is the difference between a country’s receipts and payments in foreign exchange. The difference between the two can be summarized as follows:

BOT	BOP
1. It records only merchandise transactions	1. It records transactions relating to both goods and services
2. It does not record transactions of special nature.	2. It records transactions of capital nature.
3. It is a narrow concept because it is only one part of BOP account	3. It is wider concept because it includes balance of trade, balance of Services, balance of unrequired transfers and balance of capital transactions.
4. It may be favorable, un favorable or equilibrium	4. It always remains in balance in accounting sense because receipt side is always made to be equal to payment side

Question 12.
Complete the following flow chart.
Plus Two Economics Chapter Wise Questions and Answers Chapter 6 Open Economy Macroeconomics img6
Answer:

Question 13.
Distinguish between autonomous and accommodating transactions?
Answer:
International economic transactions are called autonomous when transactions are made independently of the state of the BOP. These items are called above the line.

On the other hand, accommodating transactions are determined by the net consequences of the autonomous items, that is whether the BOP is in surplus or deficit. These items are called ‘below the line.

Question 14.
Suppose the equilibrium exchange rate is shown in the figure. What happens to this equilibrium situation when there is increase in demand for foreign exchange?
Plus Two Economics Chapter Wise Questions and Answers Chapter 6 Open Economy Macroeconomics img8
Answer:
When the demand for foreign exchange increases, there is rise in exchange rate (depreciation). At the higher exchange rate, more quantity of foreign exchange will be transacted. This is shown below.
Plus Two Economics Chapter Wise Questions and Answers Chapter 6 Open Economy Macroeconomics img9

Question 15.
Distinguish between appreciation and depreciation. Identify what happens to the exchange rate of rupees in 2015 compared to 2014.

Year	Rupee dollar exchange rate
2014	50.
2015	60.

Answer:
Appreciation refers to the increase the exchange rate of a currency. Depreciation refers to the decrease in the rate of exchange of currency. Both appreciation depreciation of exchange rate occurs due to the changes in the supply and demand of currencies. Compared to 2014 there is a depreciation of currency exchange rate in 2015.

Question 16.
The diagram below shows how the rate of exchange is determined in a free market.
Plus Two Economics Chapter Wise Questions and Answers Chapter 6 Open Economy Macroeconomics img10
Show the effect of the following on the exchange rate.

The rate of interest of the country increases.
The rate of inflation of the nearby countries.

Answer:
1. When the rate of interest increases the rate of exchange will increase. This is because an increased rate of interest would attract more depositors into the country, the demand for the currency would increase and the rate of interest also will increase as shown in the diagram below.
Plus Two Economics Chapter Wise Questions and Answers Chapter 6 Open Economy Macroeconomics img11
2. When the inflation of the nearby countries increases the people around would prefer to buy goods from this country. So the demand for the currency would increase leading to an increase in the rate of exchange.

Question 17.
Exchange rate is determined through different methods. Diagrams related with exchange rate are given below.
Plus Two Economics Chapter Wise Questions and Answers Chapter 6 Open Economy Macroeconomics img12

Identify the Exchange rate system corresponding to each diagram
Distinguish between the two.

Answer:
1. Diagram A is Flexible Exchange Rate System and Diagram B is Fixed Exchange Rate System

2. In a system of flexible exchange rates (also known as floating exchange rates), the exchange rate is determined by the forces of market demand and supply. Countries have had flexible exchange rate system ever since the breakdown of the Bretton Woods system in the early 1970s.

Prior to that, most countries had fixed or what is called pegged exchange rate system, in which the exchange rate is pegged at a particular level. Sometimes, a distinction is made between fixed and pegged exchange rates. Under a fixed exchange rate system, such as the gold standard, adjustment to BoP surpluses or deficits cannot be brought about through changes in the exchange rate.

Plus Two Economics Open Economy Macroeconomics Eight Mark Questions and Answers

Question 1.
Determine the equilibrium level of income based on the following information.
C = 100 + 0.75 (Y – T)
I = 200 – 2000;
G = 100
T = 80 + 0.20Y
X = 50
M = 20 + 0.10Y
Answer:
Plus Two Economics Chapter Wise Questions and Answers Chapter 6 Open Economy Macroeconomics img13

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 6 General Principle and Processes of Isolation of Elements

April 13, 2021January 29, 2020 by Prasanna

Students can Download Chapter 6 General Principle and Processes of Isolation of Elements Questions and Answers, Plus Two Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 6 General Principle and Processes of Isolation of Elements

Plus Two Chemistry General Principle and Processes of Isolation of Elements One Mark Questions and Answers

Question 1.
Siderite is chemically _______________.
Answer:
Iron carbonate (Fe₂CO₃)

Question 2.
A mineral is called an ore if
(a) The metal present in the mineral is costly
(b) A metal can be extracted from it
(c) A metal can be profitably extracted from it
(d) A metal cannot be extracted from it.
Answer:
(c) A metal can be profitably extracted from it

Question 3.
Predict whether the following statement is true or false? Calcination is done in presence of plenty of air.
Answer:
false

Question 4.
The Ellingham diagram is a plot of
(a) ∆_fG° vs T
(b) ∆_fH° vs T
(c) ∆_fS° vs T
(d) ∆_fG° vs ∆_fH°
Answer:
(a) ∆_fG° vs T

Question 5.
Which of the following metals can be refined using van Arkel method?
(a) Ni
(b) Si
(c) Cu
(d) Zr
Answer:
(d) Zr

Question 6.
Arrange the five elements which together constitute more than 90% of earth’s crust in the decreasing order of their abundance.
Answer:
Oxygen > Silicon > Aluminium > Iron > Calcium.

Question 7.
Suggest the method for the refining of following metals.

Copper
Germanium
Zirconium

Answer:

Copper – Electrolytic refining
Germanium – Zone refining
Zirconium – van Arkel method

Question 8.
Which one of the following does not occurs as sulphide ore
(a) Zn
(b) Cr
(c) Ag
(d) Fe
(e) Hg
Answer:
(b) Cr

Question 9.
Refining of zirconium is by __________________ method
Answer:
Van Arkel method.

Question 10.
Sphalerite is concentrated by ___________.
Answer:
Froth floation

Question 11.
Litharge is an ore of ___________.
Answer:
Lead

Question 12.
The process used for the extraction of sodium is
Answer:
Down’s process

Plus Two Chemistry General Principle and Processes of Isolation of Elements Two Mark Questions and Answers

Question 1.
Explain the terms
Calcination and Roasting with example.
Answer:
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 6 General Principle and Processes of Isolation of Elements two mark q1 img 8

Question 2.
Complete the table:

A	B
1. Iron	Haematite
2. Sodium	…………………
3. Chromium	………………….
4 ……………..	SnO₂
5 ……………..	CuFeS₂

Answer:

A	B
1. Iron	Haematite
2. Sodium	Rock salt
3. Chromium	Chromite ore
4. Tin	SnO₂
5. Copper	CuFeS₂

Question 3.
Why is the reduction of a metal oxide easier if the metal formed is in liquid state at the temperature of reduction?
Answer:
The entropy is higher if the metal is in liquid state than in solid state. The value of (∆S) of the reduction process is +ve when the metal formed is in liquid state the metal oxide being reduced is in solid state. The value of ∆G^Θ becomes more on -ve side and the reduction becomes easier.

Question 4.
Before final metallurgical operations the concentrated ore is subjected to some preliminary chemical treatments. Two processes employed for this purpose are carried out in reverberatory furnace.

Name the two processes.
To which form the ore is converted through these processes?

Answer:

Calcination and Roasting
In both processes ore is converted into oxide form.

Question 5.
Match the following:

Process	Metal Purified
1) Mond’s Process	Zirconium
2) van Arkel process	Silicon
3) Zone refining	Zinc
4) Distillation	Nickel

Answer:

Process	Metal Purified
1) Mond’s Process	Nickel
2) van Arkel process	Zirconium
3) Zone refining	Silicon
4) Distillation	Zinc

Question 6.
What is Ellingham diagram? Mention its application.
Answer:
It is a graph showing the variation of ∆_rG°forthe formation of oxides with temperature. It helps in the choice of reducing agent in the reduction of oxides.

Question 7.
Although thermodynamically feasible, in practice, magnesium metal is not used for the reduction of alumina in the metallurgy of Aluminium. Why?
Answer:
The process would be uneconomic because Mg itself is a costly metal. Moreover, there is one technological difficulty also. The reaction between Mg and Al₂O₃ is exothermic. If the temperature increases to 2000 K then the reverse reaction becomes feasible, i.e., Al starts reducing MgO.
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 6 General Principle and Processes of Isolation of Elements two mark q6 img 1

Question 8.
Distinguish between mineral and ore.
Answer:

Mineral: Various compounds of metals which are found in earth’s crust.
Ores: The minerals from which metal can be easily and economically extracted.

Question 9.
Which flux can be used to remove a metal oxide impurity from a sulphide ore of noble metal? Substantiate.
Answer:
Silica, SiO₂. Generally, metal oxides are basic in nature. To remove basic impurities an acidic flux like SiO₂ is used.

Question 10.
Match the following:
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 6 General Principle and Processes of Isolation of Elements two mark q10 img 2
Answer:

Aluminium – Leaching – Bauxite
Copper – Malachite – Brass
Mond’s process – Nickel – CO

Question 11.
Match the following:
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 6 General Principle and Processes of Isolation of Elements two mark q11 img 3
Answer:

Mond’s process – Vapour phase refining – Nickel
Sulphide ore – Zinc blende – Froth floatation
Germanium – Zone refining – Semiconductor
Calamine – ZnCO₃ – Calcination

Question 12.
Differentiate Cast Iron and pig iron.
Answer:
1. Cast iron:

It is a form of iron obtained from pig iron.
lt has 3% carbon content.

2. Pig iron:

It is the least pure form of iron obtained directly from the blast furnace.
It contains about 4% carbon and many impurities in smaller amount.

Question 13.
How is leaching carried out in case of low grade copper ores?
Answer:
Copper is leached out using acid or bacteria. The solution containing Cu²⁺ ions is treated with iron scrap or H₂ to recover copper.
Cu²⁺(aq) + Fe(s) → Cu(s) + Fe²⁺(aq)
Cu²⁺(aq) + H₂(g) → Cu(s) + 2H⁺(aq)

Question 14.
Why is the extraction of copper from pyrites more difficult than that from its oxide ore through reduction?
Answer:
Carbon is a poor reducing agent for sulphide ores whereas it is good reducing agent for oxide ores.

Question 15.
What is the role of graphite rod in the electrometallurgy of aluminium?
Answer:
Graphite rod acts as anode in the electrometallurgy of aluminium. Graphite anode facilitates reduction of Al₂O₃ to aluminium by electrolysis. Carbon reacts with oxygen liberated at anode producing CO and CO₂

Question 16.
Name the common elements present in the anode mud in electrolytic refining of copper. Why are they so present?
Answer:
The elements antimony, selenium, gold, silver, platinum, etc. are present in the anode mud during refining of copper. These impurities being less electropositive do not undergo oxidation at anode and hence settle down as such.

Plus Two Chemistry General Principle and Processes of Isolation of Elements Three Mark Questions and Answers

Question 1.
The following are some ores. Calamine (ZnCO₃), Haematite (Fe₂O₃), Cinnabar (HgS), Bauxite (Al₂O₃.2H₂O)

Which ore is concentrated by froth floating process?
How is Haematite concentrated?
Which of the ores is concentrated by leaching?

Answer:

Cinnabar (HgS). Sulphide ores are concentrated by this process.
By magnetic separation.
Bauxite (Al₂O₃.2H₂O)

Question 2.
Some data are given below:
(Iron tank, Carbon lining, Cryolite, Carbon blocks, Electricity)

Identify the metal whose metallurgy is associated here.
Explain the extraction of this metal.

Answer:

Aluminium.
The alumina is dissolved in a mixture of molten cryolite. It is then electrolysed in a rectangular steel tank, with carbon lining, which serves as cathode. Anode is a set of thick carbon rods suspended from top into the fused Al₂O₃. The temperature is maintained as 1200 Kand 1310 K. Oxygen is evolved at anode which reacts with carbon of anode producing CO and CO₂. Aluminium formed at the cathode gets collected.

Question 3.

What is the role of cryolite in the metallurgy of aluminium?
Match the following :

Metal	Process
1. Al	Mond’s process
2. Si	van Arkel process
3. Zr	Zone refining
4. Ni	Leaching

Answer:
1. Cryolite is used as a solvent to dissolve alumina.
2.

Al → Leaching
Si → Zone refining
Zr → van Arkel process
Ni → Mond’s process

Question 4.

Name the chief ores of Aluminium and Iron.
What methods are employed for the concentration of these ores?

Answer:
1. The chief ores of Aluminium and Iron

Al → Bauxite
Iron → Haematite

2. Bauxite is concentrated by leaching and haematite is concentrated by magnetic separation.

Question 5.
The choice of reducing agents in a particular case depends on the thermodynamic factor.

How far do you agree with this statement?
Support your opinion with an example.

Answer:

The statement is true. Choice of reducing agents depends strongly on factors like ΔH, ΔS, ΔG and T for the formation of the oxide to be reduced.
Electropositive metals like Al, K etc. can be extracted using electricity. Whereas CO is used for reducing haematite in the extraction of iron.

Question 6.
You are provided with samples of impure copper and germanium.

Which method would you recommend for the purification of each of these metals?
What is ‘‘Copper matte”? How is it formed?

Answer:

Coper – Electrolytic refining
Germanium – Zone refining
The copper in the furnace that contains Cu₂S and FeS is called copper matte. It is formed when copper ore is heated in reverberatory furnace after mixing with silica.

Question 7.
As a part of a field trip, students visited a metallurgical plant. They saw that metal is heated in a slopping floor of the furnace.

Give the name of this process.
Which type of metals are purified by this method?
Give example.

Answer:

Liquation
Metals with low melting point
Lead, Tin etc.

Question 8.
Blast furnace produces molten iron which contains impurities such as carbon and sulphur. To make steel, oxygen is blown into the surface of the molten iron. Other elements are then added to give the type of steel required.

What is slag?
Name the two gases formed when oxygen reacts with the impurities.
Name one element which is added to iron to make steel.

Answer:

Slag is a substance formed by the reaction of impurities with flux.
Carbon dioxide and Sulphur dioxide.
Carbon.

Question 9.
What do you mean by refining? Mention the methods also.
Answer:
The process of removal of impurities from the crude metal is called refining. The methods are:

Distillation
Liquation
Electrolytic refining
Zone refining
Van Arkel process
Mond’s process
Chromatographic methods

Question 10.
Copper can be extracted by hydrometallurgy but not zinc. Explain.
Answer:
Metals occupying low positions in the electrochemical series can be extracted by hydrometallurgy. The metals occupying higher positions in the electrochemical series cannot be extracted by hydrometallurgy because such metal ions are difficult to be reduced.

Copper can be extracted by hydrometallurgy because it occupies lower position in the electrochemical series but Zn occupies higher position.

Plus Two Chemistry General Principle and Processes of Isolation of Elements Four Mark Questions and Answers

Question 1.
The metals such as Ge, Ga, Si etc. are used as semiconductors. So they are to be obtained at high degree of purity.

Name the method to obtain highly pure Si.
Ti is purified by using I₂. Name the process.
What is Mond’s process?

Answer:

Zone refining.
van Arkel Process.
For the refining of nickel. In this process, nickel is heated in a stream of CO forming a volatile complex, Ni(CO)₄. It is decomposed at high temperature giving pure nickel.

Ni + 4CO → Ni(CO)₄ → Ni + 4CO

Question 2.
Some ores are given below:
(ZnS, Al₂O₃, Fe₂O₃,Cu₂S)
Make a table containing ores, methods of concentration, name of the metal and alloy of the metal.
Answer:
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 6 General Principle and Processes of Isolation of Elements four mark q2 img 5

Question 3.

The value of enthalpy of formation for Cr₂O₃ is -540 kJ/mol and that of Al₂O₃is -827 kJ/mol. Is the reduction of Cr₂O₃ possible with Al?
Name the metallurgical refining techniques used for Ge and Ni.

Answer:
1. Yes. From the enthalpy of formation values of the concerned oxides it is celar that Al is a strong reducing agent than Cr.

Element	Metallurgical technique
Ge	Zone refining
Ni	Mond’s process

Question 4.

What is the importance of Ellingham diagram?
Using the following Ellingham diagram select the suitable reducing agents that can be used for the reduction of Fe₂O₃ in blast furnace above and below 1000 K.

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 6 General Principle and Processes of Isolation of Elements four mark q4 img 6

Answer:
1. Ellingham diagram help us in predicting the feasibility of thermal reduction of ore. The criteria is that at a given temperature Gibbs energy of reaction should be negative.

2. Below 1000 K CO is the good reducing agent while above 1000 K carbon is the good reducing agent. This is because below 1000 K the (CO, CO₂) line is below the (Fe, FeO) line. But, above 1000 Kthe (C, CO₂) line is below the (Fe, FeO) line.

Question 5.
Bauxite is ore of Aluminium.

What do you mean by an ore?
Name the method which is used to purify Bauxite.
Write two examples for ores and their purification methods.

Answer:
1. The mineral from which metal can be easily and economically extracted is called ore.
2. Leaching
3. Two examples for ores and their purification methods

Hematite → Magnetic separation
Cinnabar → Froath floatation

Question 6.

What is the role of depressant in froth floatation process?
Explain with an example.

Answer:

Depressants prevent certain type of particles from forming froth during froth floatation process.
NaCN acts as a depressant for ZnS but not for PbS. Thus, when an ore containing PbS and ZnS is subjected to froth floatation process NaCN selectively prevents ZnS from coming to the froth but allows PbS to come with the froth. In this way, PbS can be separated from ZnS.

Question 7.

Which ore is used for the extraction of Al?
What do you mean by extraction of Aluminium?
Explain the process of purification of ore with chemical equations.

Answer:

Bauxite
Removal of earthy impurities (gangue) from bauxite ore and separation of metallic aluminium is called extraction of aluminium.
Bauxite is treated with NaOH solution and sodium meta aluminate is formed. The aluminate solution is neutralised by passing CO₂ gas and hydrated Al₂O₃ is precipitated by seeding with freshly prepared samples of hydrated Al₂O₃. Hydrated alumina is filtered, dried and heated to obtain pure Al₂O₃

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 6 General Principle and Processes of Isolation of Elements four mark q7 img 7

Plus Two Chemistry General Principle and Processes of Isolation of Elements NCERT Questions and Answers

Question 1.
Copper can be extracted by hydrometallurgy but not zinc. Explain.
Answer:
Metals occupying low positions in the electrochemical series can be extracted by hydrometallurgy because the metal ions (Mn⁺) of such metals can be easily reduced by treatment with some more electropositive metal. The metals occupying higher positions in the electrochemical series cannot be extracted by hydrometallurgy because the metal ions of such metals are difficult to be reduced.

Copper can be extracted by hydrometallurgy because it occupies quite lower position in the electrochemical series. On the other hand, zinc cannot be extracted by hydrometallurgy because it occupies higher position in the series and has large negative reduction potential.

Question 2.
How is leaching carried out in case of low grade copper ores?
Answer:
Copper is leached out using acid or bacteria. The solution containing Cu²⁺ ions is treated with iron scrap or H₂ to recover copper.
Cu²⁺(aq) + Fe(s) → Cu(s) + Fe²⁺(aq)
Cu²⁺(aq) + H₂(g) → Cu(s) + 2H⁺(aq)

Question 3.
Why is the extraction of copper from pyrites more difficult than that from its oxide ore through reduction?
Answer:
Carbon is a poor reducing agent for sulphide ores whereas it is good reducing agent for oxide ores.

Question 4.
What is the role of graphite rod in the electrometallurgy of aluminium?
Answer:
Graphite rod acts as anode in the electrometallurgy of aluminium. Graphite anode facilitates reduction of Al₂O₃ to aluminium by electrolysis. Carbon reacts with oxygen liberated at anode producing CO and CO₂
At anode:
C (solid) + O^2- (melt) → CO(g) + 2e^–
C(solid) + 2O^2-(melt) → CO₂(g) + 4e^–
At cathode:
Al³⁺(melt + 3e^– → Al(I)

Question 5.
Name the common elements present in the anode mud in electrolytic refining of copper. Why are they so present?
Answer:
The elements antimony, selenium, gold, silver, platinum, etc. are present in the anode mud during refining of copper. These impurities being less electropositive do not undergo oxidation at the anode and hence settle down as such.

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 7 The p Block Elements

April 13, 2021January 28, 2020 by Prasanna

Students can Download Chapter 7 The p Block Elements Questions and Answers, Plus Two Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 7 The p Block Elements

Plus Two Chemistry The p Block Elements One Mark Questions and Answers

Question 1.
The gas obtained by the thermal decomposition of barium azide is _________.
Answer:
Nitrogen

Question 2.
The number of replaceable hydrogen atoms in ortho phosphorous acid (H₃PO₃) is
(a) 3
(b) 2
(c) 1
(d) 0
Answer:
(b) 2

Question 3.
The stable allotrope of sulphur at room temperature is ___________ .
Answer:
Rhombic sulphur

Question 4.
Which element would readily replace oxygen from an oxide?
(a) N
(b) S
(c) F
(d) Cl
(Answer:
(c) F

Question 5.
The geometry of XeF₄ is.
Answer:
Square planar

Question 6.
The shape and hybridisation of XeF₄ molecule
Answer:
Square planar and sp³d²

Question 7.
The least stable hydride of 15^th group elements is.
Answer:
(BiH₃)

Question 8.
Choose the weak monobasic acid among the following.
(a) H₃BO₃
(b) H₃PO₂
(c) H₃PO₄
(d) HNO₃
Answer:
(a) H₃BO₃

Question 9.
The bond enthalpy is highest for ________
Answer:
H₂

Question 10.
Magnetic moment of an atom with atomic no. 24 in aqeous solution is ____________.
Answer:
4.90 B.M.

Plus Two Chemistry The p Block Elements Two Mark Questions and Answers

Question 1.
In the class, a student argued that “heterogenous” catalysis and Le-Chatlier’s principles are applied in contact process.”

Do you agree with this statement?
Justify.

Answer:
1. Yes

The reaction is exothermic, reversible and the forward reaction leads to a decrease in volume. Therefore, low temperature and high pressure are the favourable conditions for maximum yield according to Le Chatelier’s principle.

2. Contact process is an example of heterogeneous catalysis since the catalyst V₂O₅ is a solid while the reactants are gases.

Question 2.
ClF₃ exists, but FCl₃ doesn’t, why?
Answer:
In the valence shell of Cl vacant 3d orbitals are available. Hence it can form ClF₃. But, F has no vacant d orbital to show higher oxidation state and FCl₃ is not possible.

Question 3.
In a classroom discussion, a student argued that carbon has the maximum catenation property.

Do you agree?
What do you understand by the term catenation?
List out some other elements which can show catenation?

Answer:

Yes.
Catenation is the self linking property of an atom. It is maximum for carbon, because C-C bond is very strong.
Sulphur and Phosphorus.

Question 4.
Some elements and their ores are given in table. Arrange them correctly.

A	B
a) Calcium	Galena
b) Potassium	Magnetite
c) Lead	Fluorapatite
d) Tin	Gypsum
e) Phosphorus	Cassiterite
	Carnallite

Answer:

A	B
a) Calcium	Gypsum
b) Potassium	Carnallite
c) Lead	Galena
d) Tin	Cassiterite
e) Phosphorus	Fluorapatite

Question 5.
Analyse the statement: ‘Ionisation enthalpy of halogens decreases with a decrease in atomic size’. Is it true? Justify your answer.
Answer:
No. On moving from top to bottom in a given group, the size of the atom increases and hence ionisation enthalpy decreases.

Question 6.
Hydrides of group 15 are Lewis bases.

Give reason.
Arrange the group 15 hydrides in the decreasing order of basic strength.

Answer:

Due to the presence of lone pair on the central atom which is available for donation.
NH₃ > PH₃ > ASH₃ > SbH₃ ≥ BiH₃.

Question 7.
According to VSEPR theory assign structure to XeOF₄.
Answer:
XeOF₄ possess square pyramidal shape.
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 7 The p Block Elements Two mark q7 Img 2

Question 8.
Why does the reactivity of nitrogen differ from phosphorus?
Answer:

Nitrogen has a small size, high electronegativity, high ionisation enthalpy as compared to phosphorus.
Nitrogen does not contain vacant d-orbitals in its valence shell whereas phosphorus contains vacant d-orbitals in its valence shell.
Nitrogen has the ability to form triple bond ( N ≡ N) as a result of which its bond enthalpy (941.4 kJ mol^-1) is very high making it less reactive.

Question 9.
Why does NH₃ form hydrogen bond but PH₃ does not?
Answer:
In NH₃, the nitrogen atom forms hydrogen bond because of the following reasons:

Small size of nitrogen
High electronegativity (3.0) of nitrogen

N-H bond is polar forming hydrogen bond.
P-H bond is almost purely covalent due to larger size and lesser electronegativity.

Question 10.
The HNH angle is higher than HPH, HAsH and HSbH angles. Why?
Answer:
Because in NH₃ is sp³ hybridised. Due to lone pair of electrons the bond angle contracts from 109° 28’ to 106.5°. The sp³ hybridisation becomes less and less distinct with increasing size of the central atom. Thus, the bond angle of the hydrides of group 15 decreases as.

Question 11.
Can PCl₃ act as an oxidising as well as a reducing agent? Justify.
Answer:
This is because in PCl₃ phosphorus is in the intermediate oxidising state of -3.
1. As reducing agent:
The following reactions support the reducing behaviour of PCl₃.

PCl₃ + SO₂Cl₂ → PCl₅ + SO₂
PCl₃ + SO₃ → POCl₃ + SO₂

2. As an oxidising agent:
It oxidises metals to their respective chlorides.

12Ag + 4PCl₃ → 12AgCl + P₄
6Na + PCl₃ → 3NaCl + Na₃P

Question 12.
Explain why inspite of nearly the same electronegativity, nitrogen forms hydrogen bonding while chlorine does not.
Answer:
Due to its larger size (99 pm) as compared to oxygen (66 pm).

Question 13.
Ozone is a strong oxidising agent. Why?
Answer:
O₃ undergoes dissociation to form nascent oxygen. This nascent oxygen is responsible for oxidising property of ozone.

O₃ → O₂ + [O]

Question 14.
White phosphorus is more reactive than other solid phases of phosphorus. Give reason.
Answer:
This is beacuse of angular strain in the P₄ molecule where the angles are only 60°.

Question 15.
What happens when

Concentrated H₂SO₄ is added to CaF₂?
SO₃ is passed through water?

Answer:
1. It form hydrogen fluoride.
CaF₂ + H₂SO₄ → CaSO₄ + 2HF

2. It dissolves SO₃ to give H₂SO₄.
SO₃ + H₂O → H₂SO₄

Question 16.
Give two important fluorides of Xenone. Predict their structure.
Answer:

XeF₂ – linear
XeF₄ – Square planar
XeF₆ – Distorted octahedral

Question 17.
Why is N₂ less reactive at room temperature?
Answer:
Due to the presence of triple bond between N atoms N₂ has high bond dissociation energy and is less reactive.

Question 18.
Arrange the following in the increasing order of thermal stability.
ASH₃, NH₃, PH₃
Answer:
Thermal stability of hydrides, decreases from nitrogen to bismuth: NH₃ > PH₃ > AsH₃.

Question 19.
The 3 allotropic forms of Pare white, red and black.

Name the thermodynamically most stable allotrope.
Which allotrope of P is stored in water?

Answer:

Black phosphorus
White phosphorus

Question 20.
Nitrogen and Phosphorus are in the same group. PCI₅ is known but NCI₅ is not known. Why?
Answer:
In phosphorus vacant d-orbitals are present. It can form pentahalides. But due to absence of d-orbitals nitrogen cannot form NCI₅.

Question 21.
Why is helium used in diving apparatus?
Answer:
Helium is used in diving apparatus because of its very low solubility in blood.

Question 22.
Why has it been difficult to study the chemistry of radon?
Answer:
Radon is radioactive with very short half-life which makes the study difficult.

Question 23.
Account for the following:
SO₃ is more covalent than SO₂.
Answer:
Due to high charge and small size of sulphur in +6 oxidation state in SO₃ it is more covalent than SO₂, in which sulphur is in +4 oxidation state.

Question 24.
I₂ is more soluble in KI than in water.
Answer:
I₂ combines with KI to form the soluble complex, KI₃.
KI + I₂ → KI₃

Question 25.
H₃PO₃ is dibasic while H₃PO₄ is tribasic.
Answer:
There are three ‘P-OH’ bonds that are ionisable in H₃PO₄. So, it is tribasic. In H₃PO₃ there are only 2 ionisable ‘P-OH’ bonds. Hence it is dibasic.
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 7 The p Block Elements Two mark q25 Img 4

Plus Two Chemistry The p Block Elements Three Mark Questions and Answers

Question 1.
Consider the given reaction:
2A + 6SiO₂ → 6CaSiO₃ + B
B + 10C → D + 10CO

Identify A, B and D.
D is stored underwater. Why?
Give two examples of oxoacids of D and compounds of D.

Answer:
1. A – Ca₃(PO₄)₂
B – P₄O₁₀
D – P₄

2. Because P₄ readily catches fire in air.

3. Two examples of oxoacids of D and compounds of D

Oxoacids of D:
Oxoacids – Phosphorous acid(H₃PO₃), Phosphoric acid (H₃PO₄).
Compounds of D:
phosphine (PH₃), PCI₅, PCI₄, P₄O₆

Question 2.

Suggest a method for the preparation of PCI₃.
Why does PCI₃ fume in moist air?

Answer:
1. By passing dry chlorine overheated white phosphorus.
P₄ + 6CI₂ → 4PCI₃

2. PCI₃ hydrolyses in the presence of moisture giving fumes of HCI.
PCI₃ + 3H₂O → H₃PO₃ + 3HCI

Question 3.
A student argued that electronegativity of p-block elements decrease along period and increases down the group.

Do you agree with this? Explain.
Write about the metallic character of p-block elements.
Arrange the following of p-block elements in the order of decreasing oxidising power.

F₂/F-( E° =+2.85 V), Br/Br^–( E° =+1.07 V), CI₂/CI^– (E° =+1.36V), I₂/I^–( E° =+0.57V).
Answer:

No. electronegativity of p-block elements increases along a period and decreases down a group.
Most of the p-block elements are non-metallic in nature. Among p-block elements the metallic character decreases along a period and increases down a group.
F₂ > CI₂ > Br₂ > I₂, because the standard electrode potential values decreases in the same order.

Question 4.
The shape and hybridisation of some interhalogen compounds are given below in wrong order. Match them correctly.
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 7 The p Block Elements Three mark q4 Img 5
Answer:

Question 5.
Account for the following:

H₂O is a liquid while H₂S is a gas.
H₂S is more acidic than H₂O
SF₆ is known while SH₆ is not known.

Answer:

There is intermolecular hydrogen bonding in H₂O molecule but there is no hydrogen bonding in H₂S .
The S-H bond is weaker than O-H bond because size of S atom is greater than that of O atom. Hence H₂S can dissociate to give H⁺ ions in aqueous solution.
In the higher oxidation state, S can combine only with highly electronegative elements like F.

Question 6.
Sulphur forms many allotropes such as a – sulphur, β -sulphur etc.

What do you mean by allotropy?
How can you convert a – sulphur to b-sulphur?

Answer:

Certain elements can exist in different forms with different physical property and same chemical properties.
b -sulphur is prepared by melting a -sulphur in a dish and cooling, till crust is obtained. Holes are then pierced into the crust, and the liquid is taken out. On removing the crust, needle shaped β -sulphur is obtained.

Question 7.
Account for the following:

PH₃ has lower boiling point than NH₃.
Pentahalides are more covalent than trihalides.
ICI is more reactive than I₂.

Answer:

NH₃, PH₃ molecules are not associated through intermolecular hydrogen bonding in liquid state.
Higher the positive oxidation state of central atom, more will be its polarising power which, in turn, increases the covalent character.
Bond energy of I-CI bond is less than that fo I-I bond.

Plus Two Chemistry The p Block Elements Four Mark Questions and Answers

Question 1.
My name is ‘X’. I am a poisonous colourless gas with the smell of rotten fish.

Identify ‘X’.
Explain the laboratory preparation of X.

Answer:

Phosphine(PH₃).
By heating white phosphorus with concentrated NaOH solution in an inert atmosphere of CO₂.

Question 2.

How is bleaching powder prepared?
Give the composition of bleaching powder.

Answer:
1. By treating Cl₂ with dry slaked lime.
2Ca(OH)₂ + 2CI₂ → Ca(OCI)₂ + CaCI₂ + 2H₂O

2. Ca(OCI)₂ CaCI₂.Ca(OH)₂.2H₂O.

Question 3.

Why do noble gases form compounds with fluorine and oxygen only?
Does the hydrolysis of XeF₆ lead to a redox reaction?

Answer:
1. Fluorine and oxygen are small atoms with high value of electronegativity. Fluorine is also highly reactive in nature. It is for this reason they form compounds with noble gases.

2. No, the products of hydrolysis are XeOF₄ and XeO₂F₂ where the oxidation states of all the elements remain the same as it was in the reacting state.
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 7 The p Block Elements four mark q3 Img 8

Question 4.

Interhalogen compounds are more reactive than halogens. Why?
Is there any exception to the above generalisation. Explain.

Answer:

This is because the bond in the interhalogen (X – X’) is weaker than X – X bond in the halogens.
Yes. F₂ is more reactive than interhalogen compounds. This is because the F – F bond is weaker than X – X’ bond in interhalogen compounds.

Question 5.
Nitrogen cannot extend its covalency beyond four but it forms wide variety of oxides and also forms oxoacids.

Name the two oxoacids of nitrogen.
Action of nitric acid with metals depends on many factors. Justify.

Answer:
1. HNO₂, HNO₃

2. The product of oxidation depends upon the concentration of the acid, temperature and the nature of the material undergoing oxidation, e.g.

a. 3Cu + 8HNO₃(dilute) → 3Cu(NO₃)₂ + 2NO + 4H₂O
Cu + 4HNO₃(conc.) → Cu(NO₃)₂ + 2NO₂ + 2H₂O

b. Zinc reacts with dilute nitric acid to give N₂O and with concentrated acid to give NO₂.
4Zn + 4HNO₃(dilute) → 4Zn(NO₃)₂ + 5H₂O + N₂O
Zn + 10HNO₃(conc.) → Zn(NO₃)₂ + 2H₂0 + 2NO₂

Some metals (e.g., Cr, Al) do not dissolve in concentrated nitric acid due to the formation of a passive film of oxide on the surface.

Question 6.

Arrange hydrides of group 16 in the order of increasing acidic strength.
Draw the structures of any three oxoacids of phosphorus and find out their basicity.

Answer:
1. H₂O < H₂S < H₂Se < H₂Te
2.
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 7 The p Block Elements four mark q6 Img 9

Question 7.

Noble gases have very low boiling points. Why?
XeF₂ and XeF₄ are important Xe compounds. How can we prepare them and what is the action of water on them?

Answer:
1. Noble gases being monoatomic have no interatomic forces except weak dispersion forces and therefore, they are liquefied at very low temperatures. Hence, they have low boiling points.

2. XeF₂ and XeF₄ are formed by the direct reaction of elements under the specific conditions.
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 7 The p Block Elements four mark q7 Img 11
They are readly hydrolysed even by traces of water,
e.g. 2XeF₂(s) + 2H₂O(l) → 2Xe(g) + 4HF(aq) + O₂(g)

Question 8.
It is greenish yellow gas with an offensive smell used in water purification. It partially dissolves in water to give a solution which turns blue litmus to red. When it is passed through NaBr solution bromine is formed.

Identify the gas.
Identify the group to which it belongs.
Write the electronic configuration.
Write the equation showing its reaction with water.

Answer:

Chlorine
Halogen family (17^th group)
1s² 2s² 2p⁶ 3s² 3p⁵
CI₂ + H₂O → HCI + HOCI

Question 9.
Basic character of hydrides of group 15 is due to the presence of lone pair on the central atom.

NH₃ is strongly basic while BiH₃ is weakly basic. Why?
Differentiate between allotropic forms of phosphorus.

Answer:
1. Due to the presence of lone pair of electrons on nitrogen atom of NH₃. But in BiH₃, due to the large size of Bi the electron density decreases and hence is less basic.

2. The allotropic forms of P are White, Red and Black phosphorus. White phosphorus consists of tetrahedral P₄ molecules. Red phosphorus is polymeric in structure consisting of chains of P₄ tetrahedra linked together. Black phosphorus has a layer type structure and has α and β forms.

Question 10.

Give a method for preparation of XeO₃.
Deduce the molecular shape of BrF3 on the basis of VSEPR theory.

Answer:
1. XeO₃ is prepared by hydrolysis of xenon hexa fluride in presence of water.
XeF₆ + 3H₂0 → XeO₃ + 6HF

2. In BrF₃ the central atom Br is sp³d hybridised with 3 bond pairs and 2 lone pairs. According to VSEPR theory the expected geometry is trigonal bipyramidal. But due to strong Ip – Ip and Ip – bp repulsions compared to weak bp – bp repulsion it Assumes a bend T – shape as shown below
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 7 The p Block Elements four mark q10 Img 12

Question 11.
Account for the following:

BiH₃ is the strongest reducing agent among all the hydrides of group 15 elements.
Bleaching action of chlorine.

Answer:

It is because BiH₃ is least stable among the hydrides of group 15 elements.
Bleaching action of chlorine is due to oxidation. The nacent oxygen ([O]) produced is responsible for the bleaching action.

CI₂ + H₂O → 2HCI + [O]
Coloured substance + [O] → Colourless substance

Plus Two Chemistry The p Block Elements NCERT Questions and Answers

Question 1.

Noble gases have very low boiling points. Why?
XeF₂ and XeF₄ are important Xe compounds. How can we prepare them and what is the action of water on them?

Answer:
1. Noble gases being monoatomic have no interatomic forces except weak dispersion forces and therefore, they are liquefied at very low temperatures. Hence, they have low boiling points.

Question 2.
Why does the reactivity of nitrogen differ from phosphorus?
Answer:

Nitrogen has a small size, high electronegativity, high ionisation enthalpy as compared to phosphorus.
Nitrogen does not contain vacant d-orbitals in its valence shell whereas phosphorus contains vacant d-orbitals in its valence shell.
Nitrogen has the ability to form triple bond ( N ≡ N) as a result of which its bond enthalpy (941.4 kJ mol^-1) is very high making it less reactive.

Question 3.
Why does NH₃ form hydrogen bond but PH₃ does not?
Answer:
In NH₃, the nitrogen atom forms hydrogen bond because of the following reasons:

Small size of nitrogen
High electronegativity (3.0) of nitrogen

Due to more difference of electronegativity between N and H atom the N-H bond is polar forming hydrogen bond. On the contrary, in PH₃ the P-H bond is almost purely covalent due to larger size and lesser electronegativity (2.11) of phosphorus and hence does not form hydrogen bond.

Question 4.
The HNH angle is higher than HPH, HAsH and HSbH angles. Why?
Answer:
Because in NH₃ is sp³ hybridised. Due to lone pair of electrons the bond angle contracts from 109° 28’ to 106.5°. The decreased bond angle in other hydrides is because of the fact that the sp³ hybridisation becomes less and less distinct with increasing size of the central atom i.e., pure p-orbitals are utilised in M-H bonding or in simple words the s- orbital of H atom overlaps with orbital having almost pure p-character. Thus, the bond angle of the hydrides of group 15 decreases as

Question 5.
Can PCI₃ act as an oxidising as well as a reducing agent? Justify.
Answer:
Yes, This is because in PCI₃ phosphorus is in the intermediate oxidising state of -3.
1. As an reducing agent:
The following reactions support the reducing behaviour of PCI₃.

PCl₃ + SO₂Cl₂ → PCl₅ + SO₂
PCl₃ + SO₃ → POCl₃ + SO₂

2. As an oxidising agent:
It oxidises metals to their respective chlorides.

12Ag + 4PCl₃ → 12AgCl + P₄
6Na + PCl₃ → 3NaCl + Na₃P

Question 6.
Explain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not.
Answer:
Oxygen atom can form hydrogen bond whereas chlorine does not. The tendency for hydrogen bonding depends upon

Small size and
High electronegativity values

Although the electronegativities of O and Cl are nearly the same yet chlorine does not form hydrogen bond due to its larger size (99 pm) as compared to oxygen (66 pm).