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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.

1. To declare an integer variable named ‘x’ using new operator.
2. To initialize the integer pointer variable x with value 5.
3. To declare a dynamic array of ten integers named x.

Answer:

1. int*x = newint;
2. int*x = newint(5);
3. 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:

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.

1. cout<< *ar+*(ar+3);
2. 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:

Answer:
a – iii
b – iv
c – ii
d – i

Question 7.
Complete the following table. The table is differentiating array and structure.

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:

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:

1. Admission number
2. Name
3. Marks in six subjects

Answer:
Struct student
{
intAdmno;
charname[30];
int mark[6];
}

Question 14.
Observe the following layout of data,

1. Name the data type in C++ used to represent the above data format.
2. 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:

1. a=&f is correct
2. 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.

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.

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.

1. book no, book name, price
2. 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.

1. Correlate this situation with the memory allocation techniques in C++.
2. 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

1. How do you account for these two types of memory allocations?
2. 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

1. In C++, the ‘delete’ operator performs the reverse action of ‘new’ operator.
2. If ‘arr’ is the name of an array, then arr++ is not a valid statement
3. 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:

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.

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:


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:

Question 3.
How are the following related?

1. Structure and array
2. 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.

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:

The carrier wave is given by c (t) = 2 sing (8πt) volts.

1. Sketch the amplitude modulation wave form.
2. What is the modulation index?

Answer:
1. amplitude modulation wave form is as shown below:

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.

Answer:

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.

1. What is meant by modulation
2. 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

1. What are the different types of message signals used in a communication systems. (1)
2. Match the bandwidth given in Column B with the signal given in Column A.

Answer:
1. The different type of message signals are voice, music, picture and computer data.

2.

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.

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.

1. Will the radio receiver in the house get the signal from the radio transmitter? Justify.
2. In this situation, what are your suggestions for proper communication? Are Microwaves suitable there?

Answer:

1. Yes. Because wavelength of radio wave is larger than microwave.
2. 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.

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.

1. Identify the propagation modes in A, B, C
2. Why transmission of TV signals via sky wave is not possible?
3. 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

the wavelength band is 25m to 40m.

Question 5.
To transmit signals in the range 30 – 300 MHz space wave propagation is used

1. What is meant by space wave propagation?
2. Why it is called Line of Sight (LOS)?
3. Derive an expression for area covered by a sig¬nal if height of antenna is h

Answer:

1. In space wave propagation, wave travels in straight line from transmitting antenna to the receiving antenna.
2. Because the receiving antenna and transmitting antenna should be in line of sight of wave.
3. 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.

1. What do you mean by modulation? (1)
2. Why modulation is necessary? (2)
3. 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.

1. What are the names of (ω_c – ω_m) and (ω_c + ω_m). (1)
2. What is the value of amplitude of (ω_c – ω_m) if modulation is µ? (1)
3. 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.

3.

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.

Answer:

Question 2.
Some facts related to adsorption are given below. Classify them into those suitable for Physisorption and Chemisorption.

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
8. Heat of adsorption is 40-400 kJ/mol

Answer:

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.

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.

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,

1. Removing the gases already adsorbed and
2. 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.

1. Name the above phenomenon.
2. What do you mean by coagulating value?

Answer:

1. Coagulation.
2. 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.

1. Is there any mistake in the figure?
2. If a high potential is applied between the upper ends of the wire what will be the result?

Answer:

1. Yes, In this figure, the lower ends of the electrodes immersed in water are touching each other.
2. 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.

1. AS₂O₃ + 3H₂S → AS₂S₃ + 3H₂O
2. SO₂ + 2H₂S → 2S + 2H₂O
3. FeCl₃ + 3H₂O → Fe(OH)₃ + 3HCl

Answer:

1. Double decomposition
2. Oxidation
3. Hydrolysis

Question 4.
Emulsions are liquid-liquid colloidal systems.

1. What are the two types of emulsions?
2. Explain the difference between milk and butter.

Answer:

1. It is a colloidal form in which liquid is dispersed in liquid.
2. 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.”

1. Do you agree with this argument?
2. 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:

Answer:

Question 7.
Complete the following:

1. Lyophilic sol – starch; Lyophobic sol …………..
2. Deltas are formed where rivers and seawater meets. Do you agree with this statement? Explain.

Answer:

1. Lyophobic sol – gold sol (or any other correct example).
2. 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.

1. 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
   
2. Name the underlying principle.

Answer:

1. NaCl < CaSO₄ < Na₃(PO₄) < K₄[Fe(CN)₆.
2. 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

1. When a beam of light is passed through a colloidal sol?
2. An electrolyte, NaCl is added to hydrated ferric oxide sol?
3. Electric current is passed through a colloidal sol?

Answer:

1. Scattering of light by the colloidal particles takes place and the path of light become illuminated. This is called Tyndall effect.
2. The positively charged colloidal particles of Fe(OH)₃ sol get coagulated by the oppositely charged ion provided by the electrolyte.
3. 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:

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:

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.

1. Which method do you suggest?
2. Explain.
3. 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.

1. Is it correct?
2. What do you mean by emulsion?
3. How can we increase the stability of an emulsion?

Answer:

1. No.
2. It is a colloidal form in which a liquid is dispersed in another liquid.
3. 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.

1. Which statement is correct? Justify.
2. Name the law behind your opinion.
3. State the law.

Answer:

1. Statement 1. Precipitating power of an ion is directly proportional to its valency.
2. Hardy-Schulze rule.
3. 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?

1. When a beam of light is passed through a colloid.
2. When an electrolyte, NaCl is added to hydrated ferric hydroxide sol.

Answer:

1. Tyndall cone is observed due to Tyndall effect. Scattering of light by colloidal particles is called Tyndall effect.
2. The coagulation of ferric hydroxide sol is observed. This is in accordance with Hardy-Schulze rule.

Question 6.
Adsorption is a surface phenomenon.

1. Give an example for adsorption.
2. Name the following plot. What do x/m and P_s stands for?

Answer:

1. Adsorption of Cl₂ on charcoal.
2. 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:

1. Peptization
2. Electrophoresis

Answer:

1. Peptization – It is the process of converting a precipitate into a colloidal solution.
2. Electrophoresis – The migration of colloidal particles, under the influence of an electric field is called electrophoresis.

Question 8.

1. Define the terms ‘Kraft temperature’ and ‘Critical micelle concentration’.
2. 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.

1. What is this method known as? Why is it regarded as a dispersion method?
2. 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.

1. Give the reason for precipitation.
2. 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.

1. To which category of colloids does cheese belongs?
2. Explain how sols are classified depending upon the nature of the interaction between the dispersed phase and dispersion medium. Give suitable examples.

Answer:

1. Cheese belongs to gel (Liquid in solid type colloid).
2. 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.

1. Identify the phenomenon taking place in both cases.
2. 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,

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.

1. What is the role of gelatin here?
2. 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.

1. Do you agree with this statement?
2. What is the difference between dispersed phase and dispersion medium?
3. What do you mean by lyophilic colloid?

Answer:

1. Yes.
2. The medium in which dispersion takes place is called dispersion medium and the substance which undergo dispersion is called dispersed phase.
3. 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.

1. What are catalysts?
2. What do you mean by the terms ‘promoters’ and, ‘poisons’ in catalysis?
3. Suggest an example for a catalytic promoter.

Answer:

1. Catalysts are substances, which alter the rate of a chemical reaction and themselves remain unchanged.
2. Promoters are substances that enhance the activity of a catalyst. Catalytic poisons are substances which decrease the activity of a catalyst.
3. In Haber’s process for manufacture of ammonia, molybdenum (Mo) acts as a promoter for iron catalyst.

Question 16.

1. Classify the above reactions into Homogeneous and Heterogeneous catalysis.
2. What is shape-selective catalysis?

Answer:

1. (1) and (2) are heterogeneous catalysis and (3) is homogeneous catalysis.
2. 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.

1. Which of these would you prefer most to coagulate a sample of As₂S₃ sol?
2. Arrange the following electrolytes in the increasing order of coagulating power.
   NaCl, BaCl₂, AlCl₃
3. Name and state the rule based on which you have arrived at the coagulating power of these substances.

Answer:

1. AlCl₃
2. NaCl < BaCl₂ < AlCl₃
3. 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

1. When a beam of light is passed through a colloidal sol?
2. An electrolyte, NaCl is added to hydrated ferric oxide sol?
3. Electric current is passed through a colloidal sol?

Answer:

1. Scattering of light by the colloidal particles takes place and the path of light become illuminated. This is called Tyndall effect.
2. The positively charged colloidal particles of Fe(OH)₃ sol get coagulated by the oppositely charged ion provided by the electrolyte.
3. On passing direct current, colloidal particles move towards the oppositely charged electrode where they lose their charge and get coagulated.

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

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,

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

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

1. Each gate is a…….. gate
2. Complete the truth table of above circuit

Answer:
1. NAND gate

2.

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.

Answer:

Plus Two Physics Semiconductor Electronics: Materials, Devices, and Simple Circuits Two Mark Questions and Answers

Question 1.

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:

1. Collector
2. Emitter
3. EB junction
4. CB junction

Question 3.
Classify the following into conductors, insulators, and semiconductors.
Ga, As, Ni, Calcite, Graphite
Answer:

1. Conductor – Graphite, Ni
2. Insulator – Calcite
3. Semiconductor – Ga,As.

Question 4.
Construct truth table for following logic circuit.

Answer:

Question 5.
State whether true or false and justify.

1. Zener diode are used under forward bias.
2. In n-p-n transistor current conduction is primarily due to electrons.
3. Transistor amplifier do not strictly obey law of conservation of energy since output power is greater than input power.
4. In a transistor amplifier all the frequency will have exactly equal gain.

Answer:

1. False, Zener diodes are used under reverse bias
2. True
3. False
4. False

Question 6.
Match the following

Answer:

Plus Two Physics Semiconductor Electronics: Materials, Devices, and Simple Circuits Three Mark Questions and Answers

Question 1.

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.

Question 2.
A boy designs a circuit to study the input and output characteristics of an npn transistor

1. Identify the transistor configuration, input current and output current
2. 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.

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:

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

Question 4.
A circuit using two switches (A and B), cell and bulb is shown in the figure.

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 Semiconductor Electronics: Materials, Devices, and Simple Circuits Four Mark Questions and Answers

Question 1.
Forward biased pn junction diodes are shown in the figure.

1. Identify the figure, which shows the correct direction of flow of charges.
2. What do you mean by barrier potential and depletion region of a pn junction?
3. When forward bias is applied to a p-n junction, what happens to the potential barrier and the width of depletion region.

Answer:
1.

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.

1. Redraw the symbol and mark emitter, collector and base of the transistor.
2. Arrange the doping concentration and width of emitter, collector and base regions in ascending order.
3. What happens when both the emitter and the collector of a transistor are forward biased?

Answer:
1.

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.

1. What is meant by 1 and 0 in digital circuit?
2. Name logic gate X. Why is it used?
3. Name the logic gate Y?
4. 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.

Question 4.

The forward-biased diode is wrongly given above.

1. Redraw the above circuit correctly.
2. Draw the graph of current I with voltage v in forward bias.
3. Classify the following circuit into forward bias, reverse-bias, unbias.

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.

1. Which mode of bias will you suggest to connect zener diode voltage regulator?
2. Draw a circuit diagram of voltage regulation to help the boy.
3. Which device is essential for circuit diagram? Find the value of that device.

Answer:
1. Reversebias

2.

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.

1. Mention the charge carriers in the above cases.
2. Give the sketch of graph with V along X-axis and I along Y-axis for a metal at room temperature.
3. Give the physical significance of the slope of the graph.
4. If the above graph is drawn at 100°C, compared the nature of the graph with the graph at room temperature.

Answer:

1. Metals – electrons
   Semiconductor – Electrons and holes
2. Straight line graph
3. Slope gives conductance
4. 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

1. 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]
2. 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

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

1. Explain how its works? Also draw the output wave form
2. If the frequency of a.c. at the input is 50Hz what will be the output frequency?

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. 100Hz

Question 9.
Forward biased pn junction diodes are shown in the figure.

1. Identify the figure, which shows the correct direction of flow of charges. (1)
2. What do you mean by barrier potential and depletion region of a pn junction? (2)
3. 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.

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.

1. How these colours are obtained in a LED. (1)
2. Write any two uses of LED. (1)
3. 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.

1. Redraw the symbol and mark emitter, collector and base of the transistor. (1)
2. Arrange the doping concentration and width of emitter, collector and base regions in ascending order. (2)
3. What happens when both the emitter and the collector of a transistor are forward biased? (1)

Answer:
1.

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.

1. Identify the cut-off region, active region, saturation region from the figure.
2. In which of these regions, a transistor is said to be switched off.
3. 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?
4. 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.

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.

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)

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.

3.

Question 3.
Following figure is an incomplete circuit of common emitter transistor in CE configuration with the input forward biased.

1. Identify the transistor an NPN or PNP.
2. Complete the above circuit diagram by giving proper bias in the output and connect load resistance of 4 KΩ.
3. When the base current changes by 20µ Afor V_BE = .02 V. What is the voltage gain of the amplifier, if I_c = 2mA
4. npn transistors are preferred in devices with very high-frequency source. Why?

Answer:
1. npn transistor

2.

3.

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.

1. Identify rectifier diode from the figure.
2. Draw the circuit diagram of a forward biased rectifier diode by using a battery.
3. Draw the forward and reverse bias characteristics of a rectifier diode and mark threshold voltage or cut in voltage.
4. What happens to the resistance of a semiconductor on heating?

Answer:
a.

b.

c. The forward and reverse characteristics of a silicon diode is as shown in the fig.

d. The resistance of a semiconductor decreases with rise in temperature.

Question 5.
A full wave rectifier circuit is shown in figure.

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?

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.

1. Name the type of biasing used here
2. Design a circuit diagram to draw the characteristics of the diode in above biasing.
3. Trace the characteristics curve if the polarity of battery is reversed

Answer:
1. forward biasing

2.

3.

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.

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?

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.

1. Identify rectifier diode from the figure. (1)
2. Draw the circuit diagram of a forward biased rectifier diode by using a battery. (1)
3. Draw the forward and reverse bias characteristics of a rectifier diode and mark threshold voltage or cut in voltage. (2)
4. What happens to the resistance of a semiconductor on heating? (1)

Answer:
1.

2.

3. The forward and reverse characteristics of a silicon diode is as shown in the fig.

4. The resistance of a semiconductor decreases with rise in temperature.

Question 9.
A full wave rectifier circuit is shown in figure.

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)

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.

(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)

(c) Classify solids into conductors, semiconductors, and insulators by drawing energy diagram. (3)
Answer:
(a)

(b) The gap between the top of the valence band and bottom of the conduction band is called the energy band gap.

(c) For conductors

For Insulators

For Semiconductors

Question 11.
The given figure shows an npn transistor.

1. Redraw the figure and show the biasing voltage, direction of current and direction of flow of electrons and holes. (2)
2. Draw the input and output characteristics of transistor connected in common emitter configuration. (2)
3. 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.

2. Input characteristic

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.

1. Identify the cut-off region, active region, saturation region from the figure. (1)
2. In which of these regions, a transistor is said to be switched off. (1)
3. 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)
4. 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.

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 13.
The basic building blocks of digital electronic circuits are called Logic Gates. Some logic gates and their names are given in the table.

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)

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.

3.

Question 14.

1. According to energy gap, Classify them as metal, Insulator, and semiconductor.
2. From which of the above material we can eject electrons with minimum effort Explain
3. In Photo electric effect, while we are measuring photo current by varying retarding potential the variations is as shown in graph.

Answer:

1. Insulator, semiconductor, and conductor
2. Metals
3. 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.

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).

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:

1. Gold movement
2. 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:

1. To purchase goods and services from abroad.
2. 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:

1. Trade services
2. Net transfers

Question 7.
Identify the situation mentioned below.

1. Rupee-dollar exchange rate change from 45 to 50.
2. Rupee-dollar exchange rate changed from 45 to 43.

Answer:

1. Depreciation
2. 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

1. b
2. a
3. m

Answer:

1. Marginal propensity to consume
2. Autonomous consumption
3. 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:

Question 12.
What happens to the aggregate demand due to

1. A leakage from the circular flow of income
2. 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.

1. Appreciation of domestic currency.
2. Depreciation of domestic currency
3. Increase in foreign direct investment.
4. Increase in import duty.

Answer:

1. Increase in Imports, Fall in Exchange Rate
2. Decrease in Imports, Rise in Exchange Rate
3. Increase in Imports, Fall in Exchange Rate
4. 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:

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:

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:

1. export and import of goods and services
2. tourism services
3. foreign investment incomes and out payments
4. private transfer payments
5. inter-government transfer payments.

Question 8.
What is trade deficit? Calculate the trade deficit from the following data.

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

1. How does these decisions affect the demand for foreign exchange?
2. 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 Open Economy Macroeconomics Five Mark Questions and Answers

Question 1.
Find the odd one out. Justify your answer.

1. tea, coir, tourism
2. rice, banking services, insurance service, transport services
3. foreign direct investment, foreign portfolio investment, remittances, borrowings
4. foreign investment, remittances, export earning from goods, export earning from services

Answer:

1. Tourism. Others are visibles
2. Rice. Others are invisibles
3. Remittances. Others are capital receipts
4. Remittances. Others are capital receipts.

Question 2.
Match the following.

Answer:

Question 3.
If c (marginal propensity to consume) = 0.8 and m (marginal propensity to import) = 0.3,

1. Find the open and closed economy multiplier
2. 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.

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.

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:

Question 12.
Complete the following flow chart.

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?

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.

Question 15.
Distinguish between appreciation and depreciation. Identify what happens to the exchange rate of rupees in 2015 compared to 2014.

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.

Show the effect of the following on the exchange rate.

1. The rate of interest of the country increases.
2. 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.

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.

1. Identify the Exchange rate system corresponding to each diagram
2. 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:

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.

1. Copper
2. Germanium
3. Zirconium

Answer:

1. Copper – Electrolytic refining
2. Germanium – Zone refining
3. 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:

Question 2.
Complete the table:

Answer:

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.

1. Name the two processes.
2. To which form the ore is converted through these processes?

Answer:

1. Calcination and Roasting
2. In both processes ore is converted into oxide form.

Question 5.
Match the following:

Answer:

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.

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:

Answer:

• Aluminium – Leaching – Bauxite
• Copper – Malachite – Brass
• Mond’s process – Nickel – CO

Question 11.
Match the following:

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)

1. Which ore is concentrated by froth floating process?
2. How is Haematite concentrated?
3. Which of the ores is concentrated by leaching?

Answer:

1. Cinnabar (HgS). Sulphide ores are concentrated by this process.
2. By magnetic separation.
3. Bauxite (Al₂O₃.2H₂O)

Question 2.
Some data are given below:
(Iron tank, Carbon lining, Cryolite, Carbon blocks, Electricity)

1. Identify the metal whose metallurgy is associated here.
2. Explain the extraction of this metal.

Answer:

1. Aluminium.
2. 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.

1. What is the role of cryolite in the metallurgy of aluminium?
2. Match the following :

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.

1. Name the chief ores of Aluminium and Iron.
2. 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.

1. How far do you agree with this statement?
2. Support your opinion with an example.

Answer:

1. 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.
2. 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.

1. Which method would you recommend for the purification of each of these metals?
2. What is ‘‘Copper matte”? How is it formed?

Answer:

1. Coper – Electrolytic refining
   Germanium – Zone refining
2. 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.

1. Give the name of this process.
2. Which type of metals are purified by this method?
3. Give example.

Answer:

1. Liquation
2. Metals with low melting point
3. 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.

1. What is slag?
2. Name the two gases formed when oxygen reacts with the impurities.
3. Name one element which is added to iron to make steel.

Answer:

1. Slag is a substance formed by the reaction of impurities with flux.
2. Carbon dioxide and Sulphur dioxide.
3. 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.

1. Name the method to obtain highly pure Si.
2. Ti is purified by using I₂. Name the process.
3. What is Mond’s process?

Answer:

1. Zone refining.
2. van Arkel Process.
3. 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:

Question 3.

1. 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?
2. 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.

2.

Question 4.

1. What is the importance of Ellingham diagram?
2. 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.

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.

1. What do you mean by an ore?
2. Name the method which is used to purify Bauxite.
3. 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.

1. What is the role of depressant in froth floatation process?
2. Explain with an example.

Answer:

1. Depressants prevent certain type of particles from forming froth during froth floatation process.
2. 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.

1. Which ore is used for the extraction of Al?
2. What do you mean by extraction of Aluminium?
3. Explain the process of purification of ore with chemical equations.

Answer:

1. Bauxite
2. Removal of earthy impurities (gangue) from bauxite ore and separation of metallic aluminium is called extraction of aluminium.
3. 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 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.

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.”

1. Do you agree with this statement?
2. 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.

1. Do you agree?
2. What do you understand by the term catenation?
3. List out some other elements which can show catenation?

Answer:

1. Yes.
2. Catenation is the self linking property of an atom. It is maximum for carbon, because C-C bond is very strong.
3. Sulphur and Phosphorus.

Question 4.
Some elements and their ores are given in table. Arrange them correctly.

Answer:

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.

1. Give reason.
2. Arrange the group 15 hydrides in the decreasing order of basic strength.

Answer:

1. Due to the presence of lone pair on the central atom which is available for donation.
2. NH₃ > PH₃ > ASH₃ > SbH₃ ≥ BiH₃.

Question 7.
According to VSEPR theory assign structure to XeOF₄.
Answer:
XeOF₄ possess square pyramidal shape.

Question 8.
Why does the reactivity of nitrogen differ from phosphorus?
Answer:

1. Nitrogen has a small size, high electronegativity, high ionisation enthalpy as compared to phosphorus.
2. Nitrogen does not contain vacant d-orbitals in its valence shell whereas phosphorus contains vacant d-orbitals in its valence shell.
3. 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

1. Concentrated H₂SO₄ is added to CaF₂?
2. 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.

1. Name the thermodynamically most stable allotrope.
2. Which allotrope of P is stored in water?

Answer:

1. Black phosphorus
2. 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 The p Block Elements Three Mark Questions and Answers

Question 1.
Consider the given reaction:
2A + 6SiO₂ → 6CaSiO₃ + B
B + 10C → D + 10CO

1. Identify A, B and D.
2. D is stored underwater. Why?
3. 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.

1. Suggest a method for the preparation of PCI₃.
2. 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.

1. Do you agree with this? Explain.
2. Write about the metallic character of p-block elements.
3. 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:

1. No. electronegativity of p-block elements increases along a period and decreases down a group.
2. 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.
3. 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.

Answer:

Question 5.
Account for the following:

1. H₂O is a liquid while H₂S is a gas.
2. H₂S is more acidic than H₂O
3. SF₆ is known while SH₆ is not known.

Answer:

1. There is intermolecular hydrogen bonding in H₂O molecule but there is no hydrogen bonding in H₂S .
2. 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.
3. 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.

1. What do you mean by allotropy?
2. How can you convert a – sulphur to b-sulphur?

Answer:

1. Certain elements can exist in different forms with different physical property and same chemical properties.
2. 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:

1. PH₃ has lower boiling point than NH₃.
2. Pentahalides are more covalent than trihalides.
3. ICI is more reactive than I₂.

Answer:

1. NH₃, PH₃ molecules are not associated through intermolecular hydrogen bonding in liquid state.
2. Higher the positive oxidation state of central atom, more will be its polarising power which, in turn, increases the covalent character.
3. 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.

1. Identify ‘X’.
2. Explain the laboratory preparation of X.

Answer:

1. Phosphine(PH₃).
2. By heating white phosphorus with concentrated NaOH solution in an inert atmosphere of CO₂.

Question 2.

1. How is bleaching powder prepared?
2. 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.

1. Why do noble gases form compounds with fluorine and oxygen only?
2. 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.

Question 4.

1. Interhalogen compounds are more reactive than halogens. Why?
2. Is there any exception to the above generalisation. Explain.

Answer:

1. This is because the bond in the interhalogen (X – X’) is weaker than X – X bond in the halogens.
2. 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.

1. Name the two oxoacids of nitrogen.
2. 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.

1. Arrange hydrides of group 16 in the order of increasing acidic strength.
2. Draw the structures of any three oxoacids of phosphorus and find out their basicity.

Answer:
1. H₂O < H₂S < H₂Se < H₂Te
2.

Question 7.

1. Noble gases have very low boiling points. Why?
2. 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.

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.

1. NH₃ is strongly basic while BiH₃ is weakly basic. Why?
2. 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.

1. Give a method for preparation of XeO₃.
2. 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

Question 11.
Account for the following:

1. BiH₃ is the strongest reducing agent among all the hydrides of group 15 elements.
2. Bleaching action of chlorine.

Answer:

1. It is because BiH₃ is least stable among the hydrides of group 15 elements.
2. 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.

1. Noble gases have very low boiling points. Why?
2. 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.

They are readly hydrolysed even by traces of water,
e.g. 2XeF₂(s) + 2H₂O(l) → 2Xe(g) + 4HF(aq) + O₂(g)

Question 2.
Why does the reactivity of nitrogen differ from phosphorus?
Answer:

1. Nitrogen has a small size, high electronegativity, high ionisation enthalpy as compared to phosphorus.
2. Nitrogen does not contain vacant d-orbitals in its valence shell whereas phosphorus contains vacant d-orbitals in its valence shell.
3. 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

1. Small size and
2. 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).

Students can Download Chapter 13 Nuclei 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 13 Nuclei

Plus Two Physics Nuclei NCERT Text Book Questions and Answers

Question 1.
Obtain the binding energy of a nitrogen nucleus (\(_{ 7 }^{ 14 }{ N }\)) Given m \(_{ 7 }^{ 14 }{ N }\) = 14.00307 u.
Answer:
Here Z = 7 and A = 14, A – Z = 14 – 7 = 7
∴ Mass defect
= [Z m_H + (A – Z)m_n – M_n] u
= (7 × 1.00783 + 7 × 1.00867 – 14.00307) u
= (7.05481 + 7.06069-14.00307) u = 0.11243 u
Since I u = 931.5 MeV
∴ B.E. of 14N = 0.11243 × 931 Mev
= 104.7 MeV.

Question 2.
Obtain approximately the ratio of the nuclear radii of the gold isotope \(_{ 79 }^{ 197 }\)Au and the silver isotope \(_{ 47 }^{ 107 }\)Ag
Answer:
Here A₁ = 197 and A₂ = 107

Question 3.
How long an electric lamp of 100 W can be kept glowing by fusion of 2.0 kg of deuterium? The fusion reaction as.

Answer:
Number of deuterium atoms is 2 kg
= \(\frac{6.023 \times 10^{23}}{2}\) × 2000 = 6.023 × 10²⁶
Energy released when 6.023 × 10²³ nuclei of deuterium fuse together

= 15.42 × 10¹³J = 15.42Ws
Power of lamp = 100 W
If the lamp glows fortime t, then electric energy consumed = 100 t
∴ 100 t = 15.42 × 10³³
∴ t = 0.1542 × 10¹³s
\(=\frac{0.1542 \times 10^{13}}{365 \times 86400} y\)
or t = 4.0 × 10⁴y.

Question 4.
From the relation R = R₀ A^{\(\frac{1}{3}\)}, Where R₀ is a constant and A is the mass number of a nucleus show that the nuclear matter density is nearly constant (i.e. Independent of A).
Answer:
Density of nucleus(p) It is defined as the nuclear mass per unit volume.

Thus the nuclear density is of the order of 10¹⁷kgm^-3 and is independent of its mass number. Therefore, all nuclei have the same approximate density.

Plus Two Physics Nuclei One Mark Questions and Answers

Question 1.
Fusion reaction takes place at high temperature because
(a) nuclei break up at high temperature
(b) atoms get ionized at high temperature
(c) kinetic energy is high enough to overcome the coulomb repulsion between nuclei
(d) molecules break up at high temperature
Answer:
(c) kinetic energy is high enough to overcome the coulomb repulsion between nuclei.

Question 2.
Give the relation between half-life and mean life
Answer:
t_1/2 = 0.693τ.

Question 3.
Write down the expression for nuclear radius
Answer:
R = R₀A_1/3.

Question 4.
Complete the given nuclear reaction

Answer:

Plus Two Physics Nuclei Two Mark Questions and Answers

Question 1.
A beam of radio active radiation is unaffected in a combined electric and magnetic field in mutually perpendicular direction. A boy argues that, it is essentially a gamma ray. Do you agree with him. Justify your answer.
Answer:
It need not be gama ray. It can be α or β ray as the electric and magnetic field are in mutually perpendicular direction.

Question 2.
Pick the odd one out of the following.
a.

1. Curie
2. Roentgen
3. Becquerel
4. Rutherford

b.

1. γ – decay
2. β – decay
3. β⁺ – decay
4. α – decay

Answer:
a. odd one:
2. Roentgen.

b. odd one:
1. γ – decay

Question 3.
Figure below represents radiation coming out from a radioactive element.

1. Identify the radiation B. Give reason.
2. If the electric field is replaced by magnetic field perpendicular and into the plane of the paper, identify the particle deflecting towards right (C).

Answer:

1. Gamma-ray. It has no change
2. C which is β ray

Question 4.
Two radioactive substances P and Q have half-life 6 months and 3 months respectively. Find the ratio of the activity of these two materials after one year.
Answer:
Activity R = λN
R₁ = λ₁N₁
R₂ = λ₂N₂

R₁: R₂ = 1: 1

Question 5.
Match the following.

Answer:

Question 6.
Suppose you are a health physicist and you are being consulted about a spill occurred in a radio chemistry lab. The isotope spilled out in the lab was 500 micro cure of ¹³¹Ba which has half-life of 12 – days.

1. What is the decay constant of ¹³¹Ba
2. Your recommendation is to clear the lab until the radiation level is down to 1 micro curie. How length will the lab have to be closed.

Answer:
1.

2. 9 × 12 = 108 days.

Question 7.
Classify the following statement in to nuclear fission and nuclear fusion

1. Products are radio active
2. Can be controlled
3. Reaction is spontaneous
4. Can’t proceed as a chain reaction

Answer:

1. fission
2. fission
3. fission
4. fusion

Question 8.

Answer:

Plus Two Physics Nuclei Three Mark Questions and Answers

Question 1.
1. Nuclear fusion can liberate more energy than nuclear fission. But nuclear fusion is not commonly used in energy production. Why?
2. Match the following.

Answer:
1. Nuclear fusion is a thermonuclear reaction. It occurs at high temperature (10⁶k) and it is difficult to attain such a high temperature. Hence fusion is not commonly used in energy production.

2.

Question 2.
β – Particles does not exist inside a nucleus. But it is emitted from the nucleus!

1. What is β Particle
2. What happens to nucleus of a atom, when a particle is emitted? Explain
3. Why β particle is emitted from the nucleus?

Answer:

1. It is electron
2. When α is emitted, mass number decreases to 4 and atomic decreases by 2.
3. To get stability, β is emitted from the nucleus.

Question 3.
Nuclear radius depends on the mass number of the element.

1. Write down the expression for nuclear radius.
2. Prove that the density of the nucleus is independent of mass number A.

Answer:
1. R = R₀ A^1/3.

2. Density of nucleus = (mass of the nucleus)/ (volume of the nucleus)

This shows that nuclear density is independent of mass number.

Question 4.
Match the following.

Answer:

Question 5.
The fission of one nucleus of ₉₂U²³⁹ release 200 Mev of energy.

1. What is meant by fission
2. Express 200 Mev energy in joule
3. How many fission of ₉₂U²³⁹ should occur per send for producing a power of 1 Mev

Answer:
1. The splitting of heavy nucleus into two nucleus is called fission.

2. E = 200Mev
= 200 × 10⁶ev
E = 200 × 10⁶ × 1.6 × 10^-19J
= 3.2 × 10^-11J.

3. Number of fission

Question 6.
Match the following.

Answer:

Question 7.
Size of the nucleus increases with number of nucleons. As size increases, volume increase (ie mass number increases).

1. Based on the above facts find an expression for radius of nucleus.
2. Calculate radius of ₁₃Al²⁷ nucleus. The constant R₀ = 1.2 fermi.

Answer:

1. R = R₀A^1/3
2. R = 1.2 × (27)^1/3 fermi = 3.6 fermi.

Question 8.
Binding energy curve shows the variation of Binding energy per nucleon of nuclei with mass number.
1. Binding energy per nucleon is maximum for mass number……. (1)
2. The figure shows dis integration of Deuteron. (2)

What should be the frequency of the incident photon to break Deuteron into proton and neutron?
Mass of proton m_p = 1,007276u.
Mass of neutron m_n = 1,008665u
Mass of deuteron = 2.013553u
Answer:
1. 56

2. Mass defect = (1.007276 + 1.008665) – 2.013553 = 0.002388u
Binding Energy = 0.002388 × 931 MeV = 2.223
MeV Energy supplied to the photon
= 2.223 × 10⁶ × 1.6 × 10^-19
= 3.56 × 10^-13J
Frequency of photon

Question 9.
The figure shows the potential energy of a pair of nuclear particles and their distance of separation in Fermi (fm).

1. Fill in the blanks.

2. What conclusion do you obtain about the nature of nuclear force from the graph.
Answer:
1.

2. The nuclear force is a short range force.

Plus Two Physics Nuclei Four Mark Questions and Answers

Question 1.
Atomic mass of \(_{ 8 }^{ 16 }{ O }\) is found to be 16.0000u

1. what is the mass of \(_{ 8 }^{ 16 }{ O }\) nucleus
   (Hint: mass of an electron = 0.00055u)
2. determine the total mass of the constituents particles of the \(_{ 8 }^{ 16 }{ O }\) nuclei.
   (Hint : Mass of neutron = 1.00864 u, Mass of proton = 1.007274 u)
3. Give a general expression for mass defect and explain what is binding energy?
4. Binding energy per nucleon is lower for both very light nuclei (Z ≤ 10) and very heavy nuclei (Z ≥ 70) Justify a nuclear fission and fusion

Answer:
1. Mass of nucleus = (8 × 1.00864 + 8 × 1.007274) amu.

2. (8 × 1.00864 + 8 × 1.007214 + 8 × 0.0055)amu

3. ∆m = [ZM_p + (A-Z) M_n – M)
The energy equivalent to mass defect is called binding energy.

4. When two light nuclei are combined to form a heavy nucleus, the binding per nucleon increases. Hence the stability of atom increases. When heavy nucleus split into two light nuclei; B.E. for nucleon also increases.

Question 2.
Classify the following into alpha, beta, and gama

1. Similar to fast moving electron
2. It is an electromagnetic wave
3. Similar to helium nucleus
4. Travel with 1/10^th the velocity of light
5. Travel with 99/100^th the velocity of light
6. Travel with the velocity of light
7. Positively charged
8. Negatively charged

Answer:

1. beta particle – 1, 3, 8
2. Gama ray – 2, 6
3. alpha particle – 3, 4, 7

Question 3.
a. Two protons and two neutrons may bound to form a single particle it is called

1. α particle
2. β particle
3. deuteron
4. triton

b. If such a particle is emitted what changes will occur in the nucleus
c. The penetrating power of the particle is very small in air. Why?
d. If the particle is projected upward in a uniform magnetic field with direction perpendicular to plane in ward, towards which plate (A or B) it is deflected? To find this which law is applied?

Answer:
a.
1. α particle.

b. Mass number decreases to 4 and atomic number decreases to 2

c. penetrating power of the particle is very small in air:
(i) Mass of α particle is 4. Hence penetrating power of the particle is small.

d. Towards A (left) Flemings left hand rule is used.

Question 4.
Nuclear radius depends on the mass number of the element.

1. Write down the expression for nuclear radius. (1)
2. Prove that the density of the nucleus is independent of mass number A. (2)
3. Read the following statement and choose the correct option.

“Electric dipole moment is zero for nuclei in stationary state.” (1)
Assertion:
All nuclei have spherical symmetry about the centre of mass.
Reason:
The zero dipole moment for stationary nuclei is due to the symmetry about the centre of mass.

• Assertion and reason are true.
• Assertion is false reason is true
• Assertion is true reason is false
• Assertion and reason are false.

Answer:
1. R = R₀A^1/3.

2. Density of nucleus = (mass of the nucleus)/(volume of the nucleus)

This shows that nuclear density is independent of mass number.

3.
(ii) Assertion is false reason is true.

Question 5.

1. Which process is represented by the equation?
2. What happens to the parent nucleus after the process?
3. After this process the nucleus will be in a higher energy state. How will it come to ground state?
4. If a proton splits in a nucleus what are the changes. Represent it with an equation.

Answer:

1. β – emission.
2. Atomic increases to one unit. But mass number remains constant
3. The parent atom comes to ground state by emitting gamma ray.
4. P → n + e⁺ + ν
   Atomic number decreases by one unit. But mass number remains constant.

Plus Two Physics Nuclei Five Mark Questions and Answers

Question 1.
Rutherford and Soddy’s laws of radioactivity explain the rate of decay of radioactive material.

1. Arrive at the expression for the number of radio active atoms of a radioactive material remaining after an interval of time. (2)
2. Draw the curve showing the variation of log\(\left(\frac{N}{N_{0}}\right)\) with time. (1)
3. Two radioactive substances P and Q have half life 6 months and 3 months respectively. Find the ratio of the activity of these two materials after one year. (2)

Answer:
1. According to Law of Radioactive decay,

Integrating
InN = -λt + C ………(1)
C is the constant of integration. To get value of C, let us assume that initially (t=0) the number of nuclei be N₀.
∴ C = In N₀
Substituting for C in equation (1) we get,
In N – In N₀ = -λt

\(\frac{\mathrm{N}}{\mathrm{N}_{0}}\) e^-λt
N = N₀e^-λt

2.

3. Activity R = λN
R₁ = λ₁N₁
R₂ = λ₂N₂

Question 2.
a. In the given figure a radioactive source is placed inside a lead block. Identify the rays incident on the photographic plates.

b. Which of the following statement is correct.

1. Gamma rays consist of high energy neutrons.
2. Alpha rays are equivalent to singly ionized He atoms.
3. Protons and neutrons have exactly the same mass.
4. Beta rays are same as cathode rays.

c. How many alpha and beta particles are emitted in the following reaction.

Answer:
a.

• 1 – Alpha
• 2-Gamma
• 3 – Beta

b.
4. Beta rays are same as cathode rays

c.

Difference in mass number = 32
Mass number of one alpha particle = 4
Hence number of alpha particles = 8
Change in atomic number = 10
Change in atomic number due to alpha particles = 16
Charge of beta particles = -1
Hence number of beta particles emitted = 6

Question 3.
The figure shows a nuclear reactor based on thermal neutron fission.

1. The energy of thermal neutrons is……..ev. (1)
2. Name the parts labelled X and Y in the figure. (1)
3. Write the function of X and Y (1)
4. The multiplication factor has great significance in nuclear reactor. Give reason. (2)

Answer:
1. 0.025eV.

2. Control rods.

3. Control rods are used in nuclear reactors to control the fission rate of Uranium and Plutonium.

4. The ratio, K, of number of fission produced by a given generation of neutrons to the number of fission of the preceding generation is called the multiplication factor, it is the measure of the growth rate of the neutrons in the reactor.

For K = 1, the operation of the reactor is said to be critical. Unless the factor K is brought down very close to unity, the reactor will become supercritical and can even explode.

Students can Download Chapter 5 The Government: Budget and The Economy 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 5 The Government: Budget and The Economy

Plus Two Economics The Government: Budget and The Economy One Mark Questions and Answers

Question 1.
Balanced budget multiplier will be always equal to:
(i) 0
(ii) 1
(iii) ∞
(iv) -1
Answer:
(ii) 1

Question 2.
Which among the following is a tax revenue?
(i) Fees
(ii) Fine
(iii) GST
(iv) Profit
Answer:
(iii) GST

Question 3.
IF MPC = 0.5,What is the value of tax multiplier
(i) 0.5
(ii) 0
(iii) 1
(iv) 2
Answer:
(iii) 1

Question 4.
Identify indirect tax from the following
(i) Income tax
(ii) Profession tax
(iii) Wealth tax
(iv) Customs duty
Answer:
(iv) Customs duty.

Question 5.
Which of the following is not a major item of revenue expenditure?
(i) defence service
(ii) subsidies
(iii) interest payment
(iv) acquisition of land
Answer:
(iv) acquisition of land

Plus Two Economics The Government: Budget and The Economy Two Mark Questions and Answers

Question 1.
Classify the following aspects into a table as Revenue Expenditure and Capital Expenditure
Subsidies, Grants to state, Loans to foreign government, Capital projects of plants, investment in shares, Loans to state government, Salary to government staff, interest payment.
Answer:

Question 2.
State the objectives of government budget.
Answer:

• To achieve economic growth
• To reduce inequalities of income and wealth
• To achieve economic stability

Question 3.
Find the odd man out and reason.

1. Excise duty, fees, fines, penalties.
2. Income tax, sales tax, customs duty, excise duty.

Answer:

1. Excise duty. Others are non-tax revenue
2. Income tax. Others are indirect taxes.

Question 4.
Which of the following is revenue receipt
(a) Taxes
(b) Revenue from fees, penalties
(c) All the above
(d) None of the above
Answer:
(c) All the above

Question 5.
Classify the following into revenue expenditure and capital expenditure. (2)
(a) Interest payment on debit.
(b) Investment in share
(c) Land and advances by central government to state.
(d) Grant given to state government.
Answer:

• a & d Revenue expenditure
• b & c Capital expenditure

Question 6.
MGNREGA aims to increase the purchasing power of the lower income group.

1. How does this affect the value of mpc and Aggregate Demand of our economy?
2. Can the government reduce economic inequality through this scheme? How?

Answer:

1. High value of mpc and AD increases
2. Yes. Income of poor sections increases

Question 7.
Which of the following diagrams represent proportional taxation? Substantiate your answer.

Answer:
C. Because of tax rate increases in proportion to the change in income.

Question 8.
Name the economic terms.

1. The receipts of the government which are non- redeemable
2. The receipts of the government which creates liability

Answer:

1. Revenue Receipts
2. Capital Receipts

Question 9.
Prove that the value of Balanced Budget Multiplier is equal to one.
Answer:
\(\frac{1}{1-c}+\frac{-c}{1-c}\)

Plus Two Economics The Government: Budget and The Economy Three Mark Questions and Answers

Question 1.
Do you agree to the saying the revenue expenditure should not exceed revenue receipts. Substantiate your answer.
Answer:
Yes, I do agree to the saying the revenue expenditure should not exceed revenue receipts. The fact is that the revenue expenditure is non-developmental expenditure. Therefore, borrowing for meeting revenue expenditure has to be avoided.

Question 2.
“The proportional income tax acts as an automatic stabilizer”. Do you agree? Substantiate.
Answer:
Yes, I agree with this statement. The proportional income tax acts as an automatic stabilizer – a shock absorber because it makes disposable income, and thus consumer spending, less sensitive to fluctuations in GDP. When GDP rises, disposable income also rises but by less than the rise in GDP because a part of it is siphoned off as taxes.

This helps limit the upward fluctuation in consumption spending. During a recession when GDP falls, disposable income falls less sharply, and consumption does not drop as much as it otherwise would have fallen had the tax liability been fixed. This reduces the fall in aggregate demand and stabilizes the economy.

Question 3.
‘Deficits are not desirable for a government’ discuss the issue of deficit reduction.
Answer:
Government deficit can be reduced by an increase in taxes or reduction in expenditure. In India, the government has been trying to increase tax revenue with greater reliance on direct taxes (indirect taxes are regressive in nature -they impact all income groups equally).

There has also been an attempt to raise receipts through the sale of shares in PSUs. However, the major thrust has been towards reduction in government expenditure. This could be achieved through making government activities more efficient through better planning of programmes and better administration.

Question 4.
The Government of India is spending huge amount of money on various activities. This paved the way for the growth of public expenditure and external borrowings. Do you support external borrowings for meeting public expenditure? Substantiate your answer.
Answer:
1. It is true that the Government of India is spending huge amount of money on various activities such as defence, urbanization, development programmes, welfare activities, etc.

2. It is desirable for the modem state to spend more for the welfare of the people. However, huge expenses have caused burden for the government.

3. Therefore, I do not support external borrowings for meeting public expenditure. This is because; borrowing from abroad will again bring another burden on the government in the form of interest payment.

Question 5.
Are fiscal deficits necessarily inflationary?
Answer:
Fiscal deficits are generally treated as inflationary. Increase in govt, expenditure and cuts in taxes both leads to government deficit. Increased govt, expenditure and reduced taxes tend to increase the aggregate demand. Generally firms are not able to produce higher quantities that are demanded at the going prices.

This leads to inflationary pressure. However, there is a solution to this inflationary pressure. Economy can utilizatfle unutilized resources and raise production. Therefore, the deficit cannot be inflationary when an economy has unutilized the resources.

Question 6.
Classify the following into public and private goods. List out two features of public goods. Road, cars, clothes, national defence.
Answer:

1. Road, national defence are public goods
2. Cars, clothes are private goods

Two features of public goods are nonrivalry in consumption and non excludability.

Question 7.
Reducing inequalities is an objective of Indian Five year plans. Observe the following tax proportions in the total tax revenue.

1. Which of the above tax proportions help to reduce the economic inequality in society? How?
2. Direct taxes are progressive and Indirect taxes are regressive in nature. Substantiate.

Answer:

1. A, The higher income group contribute more.
2. In progressive taxes more income contributed by the richer sections In regressive taxes more income contributed by the poor sections.

Question 8.
The government of India spent ₹27000 crores under the scheme MGNREGA in 2015-16. Suppose the mpc of India is 0.6. Calculate the impact of this spending on the equilibrium income of the economy.
Answer:
Equilibrium income = Autonomous Govt, expenditure × multiplier Here, multiplier (k) is
\(\frac{1}{1-\mathrm{mpc}}=\frac{1}{1-0.6}=\frac{1}{0.4}=2.5\)
∴ Equilibrium income = 27000 × 2.5 = 67500.

Plus Two Economics The Government: Budget and The Economy Five Mark Questions and Answers

Question 1.
Classify the following aspects into a table as tax revenue and nontax Revenue. Fees and fines, Dividends from PSU, Income tax, Special assessment, Escheats, Excise duties, Customs duties, wealth tax.
Answer:

Question 2.
If the marginal propensity to consume
a) Govt. Expenditure multiplier
b) Tax multiplier
Answer:

Question 3.
Match column B and C with column A.

Answer:

Question 4.
Find the odd one out

1. income tax, sales tax, wealth tax, excise duty
2. fine, corporate tax, fees
3. dividends, fees and fines, borrowings, grants
4. balanced budget, supplementary budget, surplus, budget.

Answer:

1. Income tax. Others are indirect taxes
2. Corporate tax. Others are non-tax revenues
3. Borrowings. Others are non-tax revenues
4. Supplementary budget. This is prepared during abnormal times.

Question 5.
Assume that marginal propensity to consume is 0.5, and change in government expenditure is ₹400. Calculate
a) government expenditure multiplier
b) equilibrium income
Answer:

Question 6.
Discuss the issue of deficit reduction.
Answer:
if the government increases taxes or decreases expenditure then the fiscal deficit gets reduced. Indian government is trying to reduce the fiscal deficit by increasing tax revenue by selling the share of PSUs and by reducing the government expenditure.

The deficit reduction influences the different sectors of an economy in different ways. The government is trying to fill the gap of reduced fiscal deficit by making government activities more efficient through better planning of programmes and better administration.

The cutting back government programmes in vital areas like agriculture, education, health, poverty alleviation has adverse effect on the economy. The same fiscal measures can lead to a large or small deficit government by the state of the economy. During recession period GDP falls which reduces tax revenue which increase the fiscal deficit.

Question 7.
Point out the main objectives of public expenditure.
Answer:
The main objectives of public expenditure are pointed out below

1. for satisfaction of collective needs of the people
2. forsmooth functioning of government machinery
3. for economic and social welfare of the people
4. for creation of infrastructure
5. for controlling depressionary tendencies
6. for accelerating the speed of economic development
7. for reducing regional disparities of growth.

Question 8.
Explain why the tax multiplier is smaller in absolute value than the government expenditure multiplier.
Answer:
The tax multiplier is smaller in absolute value compared to the govt, expenditure multiplier. This is because, the govt, expenditure directly affect the total expenditure and taxes enter the multiplier, process and put impact on the disposable income.

Disposable income influences the consumption expenditure of households. Therefore, the tax multiplier is always less in absolute value than the govt, expenditure multiplier.

For example, assume that MPC = 0.75, then
Govt, expenditure multiplier \(=\frac{1}{1-c}\), C stands for MPC
\(=\frac{1}{1-0.75}=\frac{1}{0.25}=4\)
Tax multiplier = \(\frac{-c}{1-c}=\frac{-0.75}{1-0.75}=\frac{-0.75}{0.25}\)
= -3 (absolute value is 3)
Thus, it is clear from the example, that tax multiplier is smallerthan govt. expenditure multiplier.

Question 9.

Answer:

Question 10.
Match column B and C with Column A.

Answer:

Question 11.
If MPC = 0.5, calculate
a) Tax multiplier
b) Govt, expenditure multiplier
Answer:

Question 12.
Classify the following budget receipts into revenue receipts and capital receipts.
Borrowing, tax revenue, fines and penalties, fees, Recovery of loans, disinvestment, special assessment, small savings.
Answer:
1. Revenue receipts

• Tax revenue
• Fees
• Fines and penalties
• Special assessment

2. Capital receipts

• Borrowings
• Disinvestment
• Recovery of loans
• Small Savings

Question 13.
Distinguish between deficit and debt. Asses to what extent a government can borrow.
Answer:
Budget deficit is the difference between the yearly revenue and expenditure of the government. Debt is the accumulated deficit. The government can finance the budgetary deficit by borrowing, taxation or printing new currencies. To what extent the government can borrow is based on two issues.

1. Whether the debt is a burden
2. How the debt is financed.

When the government borrows it transfers the burden of reduced consumption to future generation. The future tax would increase, hence a fall in disposable income and consumption in future. If the government borrows too much the saving available to private sector may fall.

It is also argued that the consumer spending is based not only on their current income but also on their expected future income. There is another argument, debt does not matter because we owe to ourselves. This is because although there is a transfer of resources between generation, purchasing power remain within the nation. Any debt from foreign involves a burden.

Question 14.
Calssify the following into allocation, distribution and stabilization function of the government. (5)

1. The government increases the defence expenditure. .
2. The government reduces the subsidies on chemical fertilizers.
3. The government increases the rate of interest.
4. The government increases investment.
5. The government spends on building new roads.
6. The government increases the income tax.

Answer:

• 1, 5 allocation function
• 2, 6 distribution function
• 3, 4 stabilization function

Allocation function refers to the provision of public goods by the government. Stabilization function refers to the reduction of fluctuations in the economy by influency of the aggregate demand.

Question 15.
“Provision of Public good is inevitable to maintain social justice.”
Evaluate the statement in light of the experience of our State in General Education and Public Health.
Answer:
Health and Education Indicators of Kerala. Government can achieve this which Market economy can’t.
\(=\frac{1-c}{1-c}=1\)

Plus Two Economics The Government: Budget and The Economy Eight Mark Questions and Answers

Question 1.
Calculate

1. fiscal deficit,
2. revenue deficit and
3. primary deficit

on the basis of the facts given below.
Borrowing – 40,000
Budget deficit – 60,000
Revenue receipts – 60,000
Interest payments – 20,000
Revenue expenditure – 80,000
Answer:
1. Fiscal deficit = Budget deficit + borrowing
= 60000 + 40000 = 1,00,000

2. Revenue deficit = Revenue expenditure – Revenue receipts
= 80000 – 60000 = 20,000

3. Primary deficit = Fiscal deficit – Interest payments
= 100000 – 20000 = 80,000

Question 2.
Construct a flow chart showing budget receipts and expenditure of the government.
Answer:
Budget Receipts and Expenditure of the government

Question 3.
Conduct a discussion on different concepts of deficits.
Answer:
When a government spends more than it collects by way of revenue, it incurs a budget deficit. There are various measures that capture government deficit and they have their own implications for the economy. The important concepts of deficits are discussed below.

1. Revenue Deficit:
The revenue deficit refers to the excess of government’s revenue expenditure over revenue receipts.

2. Fiscal Deficit:
Fiscal deficit is the difference between the government’s total expenditure and its total receipts excluding borrowing.
Gross fiscal deficit = Total expenditure – (Revenue receipts + Non-debt creating capital receipts

3. Primary Deficit:
We must note that the borrowing requirement of the government includes interest obligations on accumulated debt. To obtain an estimate of borrowing on account of current expenditures exceeding revenues, we need to calculate what has been called the primary deficit. It is simply the fiscal deficit minus the interest payments.
Gross primary deficit = Gross fiscal deficit – net interest liabilities.

Question 4.
Represent in the diagram the effect of the following on the equilibrium level of income

1. Effect of higher government expenditure
2. Effect of a reduction in taxes

Answer:
1. Effect of higher government expenditure
When there is higher government expenditure, the aggregate expenditure curve will shift upward. This will lead to an increase in equilibrium level of income. This effort is shown in the diagram.

2. Effect of a reduction in taxes
When there is a tax cut, there will be an increase in consumption and output. Therefore, the equilibrium level Of income increases as shown in the diagram.

Question 5.
The equilibrium of an economy is given below.

1. What is fiscal policy?
2. Identify the level of equilibrium in an economy and show the seperate diagram the effect of an increased govt, expenditure and an increased tax on the level of output.

Answer:
1. The policy of the government related to taxation and govt, expenditure is known as fiscal policy.
2. OY is the equilibrium level of output. The effects of an increased govt, expenditure is shown in the diagram below.

Govt, expenditure is one of the components of AD. AD = C + I + G + (x – m)
So the AD curve shifts upward when there is an increase in the govt, expenditure. As a result the equilibrium changes from e to e₁. The level of output increases from y to y₁.

It can be seen from the diagram that the increase in govt, expenditure is from A to A₁ But the output increased from y to y₁ This is because of multiplier effect. The effect of an increased tax on the level of output is given in the diagram below.

Students can Download Chapter 12 Atoms 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 12 Atoms

Plus Two Physics Atoms NCERT Text Book Questions and Answers

Question 1.
What is the shortest wavelength present in the Paschen series of spectral lines?
Answer:
Since

For Paschen series, n₁ = 3 and n₂ = ∞

Question 2.
The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10^-11 m. What are the radii of the n = 2 and m = 3 orbits?
Answer:
Given r₁ = 5.3 × 10^-11m
n₁ = 1, n₂ = 2, r₂ = ?
n₃ = 3, r₃ = ?
Since r ∝ n²
∴ r₁ ∝ n₁²
And r₂ ∝ n₂²

or r₂ = 4 × r₁ = 4 × 5.3 × 10^-11
or e₂ = 2.12 × 10^-10m
Also r₃ ∝ n₃²

or r₃ = 9r₁ = 9 × 5.3 × 10^-11
or r₃ = 4.77 × 10^-10m.

Question 3.
A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelength will be emitted?
Answer:
First excitation energy
E = E₂ – E₁ = -3.4 – (-13.6) = 10.2 eV
Second excitation energy
E = E₃ – E₁ = -1.51 -(-13.6) = 12.09 eV
Third excitation energy
E = E₄ – E₁ = -0.85 – (-13.6) = 12.75 eV
Second incident beam has energy = 12.5 eV
So only first two lines in the Lyman series of wave-length 103 nm and 122 nm will be emitted.
Also E₃ – E₂ = -1.51 – (-3.4) = 4.91 eV
i.e, first line in the Balmer series of wavelength 656nm will also be emitted.

Question 4.
In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1.5 × 10¹¹ m with orbital speed 3 × 10⁴ ms^-1.
Answer:
Here r= 1.5 × 10¹¹ m,
v = 3 × 10⁴ ms^-1, m = 6 × 10²⁴kg, n = ?

Plus Two Physics Atoms One Mark Questions and Answers

Question 1.
A radioactive nucleus emits beta particle. The parent and daughter nuclei are
(a) isotopes
(b) isotones
(c) isomers
(d) isobars
Answer:
(d) isobars

Question 2.
Consider an electron in the n^th orbit of a hydrogen atom in the Bohr model. The circumference of the orbit can be expressed in terms of de Broglie wavelength λ of that electron as
(a) (0.529) nλ
(b) \(\sqrt{n} \lambda\)
(c) (13.6)λ
(d) nλ
Answer:
(d) nλ

Question 3.
The ionization energy of hydrogen atom is 13.6 eV. Find the energy corresponding to a transition between 3^rd and 4^th orbit.
Answer:
E = E₄ – E₃

= -0.85 + 1.51 = 0.66 eV.

Question 4.
If 13.6 eV energy is required to ionize the hydrogen atom, then the energy required to remove an electron from n = 2 is
(a) 10.2 eV
(b) 0 eV
(c) 3.4 eV
(d) 6.8 eV
Answer:
(c) 3.4 eV
Explanation:

= 3.4 eV.

Question 5.
In the Bohr model of the hydrogen atom, the lowest orbit corresponds to
(a) infinite energy
(b) maximum energy
(c) minimum energy
(d) zero energy
Answer:
(c) minimum energy.

Question 6.
Write down the Balmerformula for wavelength of H_a line.
Answer:

Plus Two Physics Atoms Two Mark Questions and Answers

Question 1.
Given Rydberg constant as 1.097 × 10^-7m^-1. Find the longest and shortest wavelength limit of Baler Series.
Answer:
\(\bar{v}=\frac{1}{\lambda}=R_{H}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]\)
Longest wavelength n₁ = 2 and n₂ = 3

Shortest Wavelength n₁ = 2 and n₂ = α

Plus Two Physics Atoms Three Mark Questions and Answers

Question 1.
Bohr combined classical and early quantum concept and gave his theory in the form of three postulates.

1. The total energy of an electron in ground state of hydrogen atom is -13.6eV. What is the significance of negative sign?
2. The radius of innermost electron orbit of hydrogen atom is 5.3 × 10¹¹m. What are the radii of n = 2 and n = 3 orbits?

Answer:
1. Negative sign implies that the electrons are strongly bounded to the nucleus.

2. r_n = n²a₀ = 5.3 × 10^-11m
r₁ = a₀ = 5.3 × 10^-11m
r₂ = 4a₀ = 21.2 × 10^-11m
r₃ = 9a₀ = 47.7 × 10^-11m.

Students can Download Chapter 8 The d and f 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 8 The d and f Block Elements

Plus Two Chemistry The d and f Block Elements One Mark Questions and Answers

Question 1.
The element of the 3d series which is not regarded as a transition element is ____________ .
Answer:
Zinc

Question 2.
Say TRUE or FALSE:
Cr²⁺ compounds are strong oxidising agents
Answer:
False

Question 3.
Which of the following has the least magnetic moment?
(a) Cu²⁺
(b) Ni²⁺
(c) Co²⁺
(d) Fe²⁺
Answer:
(a) Cu²⁺

Question 4.
The oxidation state of Cr in K₂Cr₂O₇ is _________
Answer:
+ 6

Question 5.
Lanthanoid contraction implies
(a) Decrease in density.
(b) Decrease in mass.
(c) Decrease in ionic radii.
(d) Decrease in radioactivity.
Answer:
(c) Decrease in ionic radii.

Question 6.
Which of the following species are paramagnetic?
(a) Fe²⁺
(b) Zn
(c) Hg²⁺
(d) Ti⁴⁺
Answer:
(a) Fe²⁺

Question 7.
The approximate percentage of iron in misch metal is____
Answer:
5

Question 8.
The maximum oxidation state shown by Mn in its compounds is _____
Answer:
+ 7

Question 9.
The acidic, basic or amphoteric nature of Mn₂O₇, V₂O₅ and CrO are respectively
Answer:
acidic, amphoteric and basic

Question 10.
The correct order of oxidising power of the following is
(a) Cr₂O₇^2- > MnO₄^– > VO₂⁺
(b) MnO₄^– > Cr₂O₇^2- > VO₂⁺
(c) VO₂⁺ >MnO₄^– > Cr₂O₇^2-
(d) MnO₄^– > VO₂⁺ > Cr₂O₇^2-
(e) Cr₂O₇^2- > VO₂⁺ > MnO₄^–
Answer:
(b) MnO₄^– > Cr₂O₇^2- > VO₂⁺

Plus Two Chemistry The d and f Block Elements Two Mark Questions and Answers

Question 1.
Transition metals can form complexes.

1. Is the statement true?
2. Justify.
3. Give two examples of complexes.

Answer:

1. Yes
2. Transition elements form complex compounds due to the comparatively smaller size of the metal ion, their high ionic charges and the availability of d-orbital for the bond formation.
3. [Fe(CN)₆]^3-, [Cr(NH₃)₆]³⁺

Question 2.
Consider the statement: Scandium is a transition element but zinc is not.

1. Do you agree with his statement?
2. Justify your answer.

Answer:
1. Yes.

2. Scandium has incompletely filled 3d orbitals in its ground state (3d⁰), it is regarded as a transition element. On the other hand, zinc atom has completely filled d orbitals (3d¹⁰) in its ground state as well as in its oxidised state. Hence it is not regarded as a transition element.

Question 3.
Transition elements show variable oxidation state.

1. Give reason.
2. Illustrate with an example.

Answer:

1. The variable oxidation states of transition elements is due to the participation of ns and (n-1) d electrons in bonding,
2. E.g.: The outer configuration of Mn is 3d⁵,4s² it exhibits all the oxidation states from +2 to +7.

Question 4.
Cu+ is colourless while Cu²⁺ is coloured in aqueous solution.

1. Is it true?
2. What are the reasons for this?

Answer:

1. Yes.
2. In Cu+ it is 3d¹⁰,4s⁰ and there is no d-d transition. Hence no colour. But in Cu²⁺ it is 3d⁹, 4s⁰ it has d-d transition and hence it has colour.

Question 5.
The highest oxidation state of a metal is exhibited in its oxide or fluoride only.

1. Do you agree?
2. If yes, give reason.

Answer:

1. Yes.
2. Due to very high electronegativity and small size oxygen and fluorine can oxidise the metal to its highest oxidation state.

Question 6.
What is misch metal? Give its use.
Answer:
‘Misch metal’ is an alloy which consists of a lanthanoid metal (~ 95%) and iron (~ 5%) and traces of S, C, Ca and Al. It is used in Mg-based alloy to produce bullets, shell and lighter flint.

Question 7.
Some ions of transition metals are given below. Categorize them into those which are coloured or not. Justify your answer.
Ti³⁺, V²⁺, Sc³⁺, Mn²⁺, Ti⁴⁺, Cu²⁺, Zn²⁺, Cu⁺
Answer:

• In Ti³⁺, V²⁺, Mn²⁺, Ca²⁺ d-d transition is possible. Hence they have colour.
• In Sc³⁺ and Ti⁴⁺(3d⁰), Zn²⁺ and Cu⁺(3d¹⁰) d-d transition is not possible.

Question 8.
Why are Mn²⁺ compounds more stable than Fe²⁺ toward oxidation to their +3 state?
Answer:
The electronic configuration of Mn²⁺ is 3d⁵ , which is stable due to half-filled sub-shell. Hence, Mn²⁺ is not easily oxidised to Mn³⁺. On the other hand, Fe²⁺ has electronic configuration 3d⁶ . After losing one electron it changes to Fe³⁺ which has stable 3d⁵ electronic configuration. Hence, Fe²⁺ is relatively easily oxidised to Fe³⁺

Question 9.
Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?
Answer:
Except Sc, the most common oxidation state of the first row transition elements is +2 which arises from the loss of two 4s electrons. The +2 state becomes more and more stable in the first half of first transition elements with increasing atomic number because 3d orbitals acquires only one electron in each of the five 3d orbitals.

Question 10.
In what way is the electronic configuration of the transition elements different from that of the nontransition elements?
Answer:
Transition elements have partially filled d-dubshell belonging to penultimate energy level whereas non-transition elements do not have any partially filled d-subshell. In non-transition elements, the last electron enters the s or p-subshell whereas in transition elements the last electron enters the d-subshell of penultimate energy level.

Question 11.
Predict which of the following will be coloured aqueous solution?
Ti³⁺ , V³⁺ cu⁺, Sc³⁺ , Mn²⁺, Fe³⁺ and CO²⁺ and MnO₄^–. Give reasons for each.
Answer:
Any ion that has partially filled d-orbitals is coloured due to d-d transition in visible light. MnO₄^– has purpule colour due to charge-transfer. Cu⁺ (d¹⁰ ) and Sc³⁺ (d⁰) are white since there is no d-d transition.

Question 12.
Use Hund’s rule to derive the electronic configuration of Ce³⁺ ion and calculate its magnetic moment on the basis of ‘spin-only’ formula.
Answer:
Ce (Z = 58) = ₅₄[Xe]4f¹5d¹6s²
Ce³⁺ = ₅₄[Xe]4f¹
It has one unpaired electron.
µ = \(\sqrt{n(n+2)}\) B.M.
µ = \(\sqrt{1(1+2)}=\sqrt{3}\) B.M = 1.732 B.M.

Question 13.
Among the following transition metal compounds which are coloured. Why?
TiO, TiO₂, ZnSO₄, MnCl₂, CrCl₃
Answer:

• TiO → Coloured
• MnCl₂ → Coloured
• CrCl₃ → Coloured

Question 14.
In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest. Why?
Answer:
Zinc has weaker interparticle forces due to fully filled d-subshell (3d¹⁰).

Question 15.
Fill in the blanks:

Answer:
i. 5

ii. 4.90 BM

Plus Two Chemistry The d and f Block Elements Three Mark Questions and Answers

Question 1.
Explain why

1. Transition metals have higher density than alkaline earth metals?
2. Transition metals are less electropositive than alkaline earth metals?
3. Transition elements exhibit higher enthalpies of atomisation?

Answer:
1. Transition metals have smaller atomic size and stronger interparticle attractive forces (metallic bond) than alkaline earth metals and hence have higher density than alkaline earth metals.

2. Transition metals due to their smaller atomic size and greater effective nuclear charge, have higher ionisation energy than alkaline earth metals and hence are less electropositive.

3. Transition elements have very strong interatomic bonds due to their small size and also due to the presence of large number of unpaired electrons in their atoms. As a result, they have very high enthalpies of atomisation.

Question 2.
Two students wrote the electronic configuration of Chromium (Z = 24) as shown below:
Student A: [Ar] 4s², 3d⁴
Student B : [Ar] 4s¹ 3d⁵

1. Pick out the correct electronic configuration and justify.
2. Calculate the spin only magnetic moments of Sc³⁺, Mn²⁺, Fe³⁺ and find out which is more paramagnetic.

Answer:
1. [Ar] 3d⁵4s¹ . Due to the extra stability of the half-filled 3d level.

2. Sc³⁺ – n = 0 µ = 0 BM
Mn²⁺ – n = 5 µ = \(\sqrt{5(5+2)}=\sqrt{35}\) = 5.92 BM
Fe³⁺ – n = 5 µ = \(\sqrt{5(5+2)}=\sqrt{35}\) = 5.92 BM

Plus Two Chemistry The d and f Block Elements Four Mark Questions and Answers

Question 1.
Some information regarding a particular compound is given below.
A deep purple black crystal which is moderately soluble in water at room temperature and it is prepared from pyrolusite.

1. Identify the compound.
2. Give the chemical reactions involved in the preparation of this compound.
3. Upon heating to 746 K, what happens to this compound? Give the equation.

Answer:

1. KMnO₄
2. 2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O 3MnO₄^2- + 4H⁺→ 2MnO₄^– + MnO₂ + 2H₂O
3. Potassium permanganate on strong heating gives potassium manganate, manganese dioxide and oxygen.
   2KMnO₄→ K₂MnO₄ + MnO₂ + O₂

Question 2.
Consider the statement: All transition elements form coloured compounds.

1. Do you agree?
2. Name three transition elements which cannot form coloured compounds.
3. What is the reason for colour of transition metal compounds?

Answer:

1. No.
2. Zn, Cd and Hg.
3. The colour of transition compound is due to d-d transition. The colour observed corresponds to the complementary colour of the light absorbed.

Question 3.
Complete the table:

Answer:

Question 4.
The size of the atoms from Lanthanum to Lutetium, shows a steady and slow decrease in atomic size.

1. Do you agree with it?
2. What are the causes for this?
3. What are the consequences of this phenomenon?

Answer:
1. Yes.

2. It is due to Lanthanoid contraction. In Lanthanoid series the 4f electrons are being added in the antipenultimate shell. The 4f electrons have very poor shielding effect because of diffused shape of 4f orbitals and the effective nuclear charge experienced by each 4f electrons increases. Hence there is a regular decrease in radii with increase in atomic number.

3. Consequences of lanthanoid contraction are:

• Difficulty in separation of lanthanoids due to similarity in chemical properties.
• Similarity in size of elements belonging to same group of second & third transition series.

Question 5.
Zn, Cd, Hg are pseudo transition elements.

1. What do you mean by pseudo transition elements?
2. Write the outer electronic configuration of transition elements.
3. Write any four properties of transition elements,

Answer:
1. Pseudo transition elements have completely filled d orbitals and will not exhibit the characteristic properties of true transition elements.

2. The outer electronic configuration of transition elements is (n-1) d^{1 – 10} ns^{1 – 2}.

3. Four properties of transition elements.

• They form coloured compounds.
• They exhibit variable oxidation state.
• They form complex.
• They are good catalysts.

Question 6.
In a demonstration experiment, the teacher prepared the following solutions.
Solution 1: ZnSO₄ in water
Solution 2: CuSO₄ in water

1. Which of the above solutions is coloured?
2. Explain the Chemistry behind it.

Answer:
1. Solution 2: CuSO₄ in water is coloured.

2. In CuSO₄, Cu is in +2 oxidation state. Cu²⁺ has 3d⁹ electronic configuration and it can undergo d-d transition and hence has colour(blue). In ZnSO₄, zinc is in +2 oxidation state. Zn²⁺ has the 3d¹⁰ configuration. It has no d-d transition and hence it is colourless.

Question 7.

1. What is the effect of heat on KMnO₄
2. Give any two examples to show the oxidising property of KMnO₄.

Answer:
1. On heating, KMnO₄ decomposes at 513 K into potassium manganate and oxygen gas is evolved.

2. Two examples to show the oxidising property of KMnO₄

• In acid medium it oxidises iodide to iodine.
  10I^– + 2MnO₄^– + 16H⁺ → 2Mn²⁺ + 8H₂O + 5I₂
• In neutral or faintly alkaline solutions it oxidises iodide to iodate.
  2MnO₄^– + H₂O + I^– → 2MnO₂ + 2OH^– + IO₃^–

Question 8.
(a) Hf and Zr have similar ionic radii though they belongs to the 3^rd and 4^th transition series. Explain the reason for the anomalous behaviour.
(b) MnO₄^– & MnO₄^2- are the ions of manganese. They differ in colour and magnetic nature.

1. How do this difference occured?
2. Support your answer with their structure.

Answer:
(a) This is a consequence of lanthanide contraction.

(b) Differ in colour and magnetic nature.
1. This is due to difference in number of electrons in d- subshell.
2.

Question 9.
Pyrolusite on heating with KOH in the presence of air gives a dark green compound (A). The solution of (A) on treatment with H₂SO₄ gives a purple coloured compound (B).
a) Identify ‘A’ and ‘B’ and write the balanced chemical equation.
b) Write two reactions to show the oxidising nature of ‘B’.
Answer:

Question 10.
Give the method of preparation and two examples for the oxidizing property of K₂Cr₂O₇.
OR
Some d-block elements are given below:
Cr, Mn, Fe, Co, Ni
Identify the element which shows maximum paramagnetic behaviour.
Answer:
K₂ Cr₂ O₇ is prepared from chromate ore Fe Cr₂ O₄.
Step I: The powdered ore is heated with molten alkali to form soluble sodium chromate.

4 Fe Cr₂ O₄ + 16 NaOH + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8H₂O

Step II: Sodium chromate (Na₂CrO₄) is filtered and acidified with dil.H₂SO₄ to form sodium dichromate.

2Na₂CrO₄ + H₂SO₄ → Na₂Cr₂O₇ + Na₂SO₄ + H₂O

Step III: Na₂Cr₂O₇ solution is treated with KCI to form

K₂ Cr₂ O₇.
Na₂Cr₂O₇ + 2 KCI —» K₂ Cr₂ O₇ + 2NaCI

OR

Magnetic moment, µ = \(\sqrt{n(n+2)}\) BM
Cr; n=6 µ = \(\sqrt{48}\) = 6.93 BM
Mn; n=5 µ = \(\sqrt{35}\) = 5.92 BM
Fe; n=4 µ = \(\sqrt{24}\) = 4.90 BM
Co; n=3 µ = \(\sqrt{15}\) = 3.87BM
Ni; n=2 µ = \(\sqrt{8}\) = 2.83 BM
∴ Cr has maximum paramagnetic behaviour.

Question 11.

1. Even though usually transition elements exhibit +2 oxidation states, it changes from 0 to +8 in different compounds. What is the reason for it?
2. In the first transition series the atomic radius is decreasing from Sc to Cr and remains constant from Cr to Cu. From Cu to Zn it increases. Why?
3. The atomic radii of second and third transition series are same. Why?

Answer:
1. The variable valency is due to the following reason. The energy difference between inner ‘d’ level and outers level is so small. Therefore electrons from ‘d’ level as well as ‘s’ level can take part in reaction.

2. Sc to Cr – Atomic radius decreases due to increase in nuclear charge. Cr to Cu – Due to a balance between the increased nuclear charge and increased screening effect atomic radii become almost constant.

Cu to Zn – Repulsive interactions between the paired electrons in d-orbitals become very dominant towards the end of the period and cause the expansion of electron cloud and thus result in increased atomic size.

3. Lanthanoid, Contraction.

Question 12.
a) Calculate the magnetic moments of the following ions and predict its magnetic property.
Ti³⁺ (Z = 22)
Ni²⁺ (Z = 25)
Fe³⁺ (Z = 26)
Zn²⁺ (Z = 30)
Answer:

Question 13.
TiO, TiO₂, ZnSO₄, MnCl₂, CrCl₃
Arrange the above metal ions in the increasing order of their magnetic moment. (3)
Answer:

The increasing order of magnetic moment is TiO₂ = ZnSO₄ < TiO < CrCl₃, MnCl₂

Question 14.
How magnetic moment is calculated? How will you identify magnetism of an element from its electronic configuration?
Answer:
By using the ‘spin-only’ formula, n = \(\sqrt{n(n+2)}\) BM where ‘n’ is the number of unpaired electrons and ‘µ’ is the magnetic moment in Bohr Magneton (BM). If the element contains an unpaired electron it is paramagnetic. If all the electrons are paired and

Plus Two Chemistry The d and f Block Elements NCERT Questions and Answers

Question 1.
Why are Mn²⁺ compounds more stable than Fe²⁺ toward oxidation to their +3 state?
Answer:
The electronic configuration of Mn²⁺ is 3d⁵, which is stable due to half-filled sub-shell. Hence, Mn²⁺ is not easily oxidised to Mn³⁺. On the other hand, Fe²⁺ has electronic configuration 3d⁶. After losing one electron it changes to Fe³⁺ which has stable 3d⁵ electronic configuration. Hence, Fe²⁺ is relatively easily oxidised to Fe³⁺

Question 2.
Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?
Answer:
Except Sc, the most common oxidation state of the first row transition elements is +2 which arises from the loss of two 4s electrons. The +2 state becomes more and more stable in the first half of first row transition elements with increasing atomic number because 3d orbitals acquire only one electron in each of the five 3d orbitals (i.e., each d-orbital remains half-filled) and inter electronic repulsion is the least and nuclear charge increases.

In second half of first row transition elements, electron starts pairing up in 3d orbitals and hence there is inter electronic repulsion. (Ti²⁺ to Mn²⁺ electronic configuration changes from 3d² to 3d⁵ but in second half i.e., Fe²⁺ to Zn²⁺ it changes from 3d⁶ to 3d¹⁰.)

Question 3.
In what way is the electronic configuration of the transition elements different from that of the non-transition elements?
Answer:
Transition elements have partially filled d-subshell belonging to penultimate energy level whereas non-transition elements do not have any partially filled d- subshell. In non-transition elements, the last electron enters the s or p-subshell whereas in transition elements the last electron enters the d-subshell of penultimate energy level.

Question 4.
Predict which of the following will be coloured in aqueous solution?
Ti³⁺, V³⁺, Cu⁺, Sc³⁺, Mn²⁺, Fe³⁺ and CO²⁺ and MnO^4-. Give reasons for each.
Answer:
Any ion that has partially filled d-orbitals is coloured due to d-d transition in visible light. Ti³⁺ (d¹), V³⁺ (d²) Mn²⁺ (d⁵), Co²⁺ (d⁸) are coloured. MnO^4- has purpule colour due to charge-transfer. Cu⁺ (d¹⁰) and Sc³⁺ (d⁰) are white since there is no d-d transition.

Question 5.
Use Hund’s rule to derive the electronic configuration of Ce³⁺ ion and calculate its magnetic moment on the basis of ‘spin-only’ formula.
Answer:
Ce (Z = 58) = ₅₄[Xe]4f¹5d¹6s²
Ce³⁺ = ₅₄[Xe]4f¹
It has one unpaired electron.
µ = \(\sqrt{n(n+2)}\) B.M.
where n = number of unpaired electrons.
∴ Magnetic moment of Ce³⁺,
μ = \(\sqrt{1(1+2)}\) = \(\sqrt{3}\) BM
= 1.732 BM

Students can Download Chapter 11 Dual Nature of Radiation and Matter 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 11 Dual Nature of Radiation and Matter

Plus Two Physics Dual Nature of Radiation and Matter NCERT Text Book Questions and Answers

Question 1.
Find the

1. Maximum frequency, and
2. Minimum wavelength of X-rays produced by 30 kV electrons.

Answer:
Given V_o = 30 kV = 30 × 10³ V
v_max = ?
λ_max = ?
1. Since k_max = eV_o
So hv_max = eV_o
or

= 7.24 × 10¹⁸ Hz

2.

= 0.041 × 10^-9
or λ_min = 0.041 nm.

Question 2.
The work function of caesium metal is 2.14 eV. When light of frequency 6 × 10¹⁴Hz is incident on the metal surface, photoemission of electrons occurs. What is the

1. maximum kinetic energy of the emitted electrons,
2. maximum speed of the emitted photoelectrons?

Answer:
Given W₀ = 2.14 eV
= 2.14 × 1.6 × 10^-19 J
= 3.424 × 10^-19J
v = 6 × 10¹⁴ Hz

1. K_max = hv – W₀
= 6.62 × 10^-34 × 6 × 10¹⁴ – 3.424 × 10¹⁹
= 3.972 × 10^-19 – 3.424 × 10^-19
= 0.54 × 10^-19 J.

2. Since eV₀ =k_max

or v₀ = 0.34 V
Since \(\frac{1}{2}\) mV²_max = K_max

or v_max = 0.344 × 10⁶ ms-1 = 344 kms^-1

Question 3.
The photoelectronic cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
Answer:
Given V₀ = 1.5 V
Since k_max = eV₀
= 1.5 eV
= 1.5 × 1.6 × 10^-19 J = 2.4 × 10^-19 J.

Question 4.
In an experiment on photoelectric effect, the slope of the cutoff voltage versus frequency of incident light is found to be 4.12 × 10^-15Vs. Calculate the value of Planck’s constant.
Answer:
Given, slope of graph = 4.12 × 10^-15 Vs
since, slope of graph = \(\frac{h}{e}\)
∴ h = e × slope of graph
= 1.6 × 10^-19 × 4.12 × 10^-15 = 6.59 × 10^-34 Js.

Question 5.
The threshold frequency fora certain metal is 3.3 × 10¹⁴ Hz. If light of frequency 8.2 × 10¹⁴ Hz is incident on the metal, predict the cutoff voltage for the photoelectric emission.
Answer:
Given v₀ = 3.3 × 10¹⁴Hz
v = 8.2 × 10¹⁴Hz
Since eV₀ = hv – hv₀
So V₀ = \(\frac{h}{e}\) (v – v₀)
\(=\frac{6.62 \times 10^{-34}}{1.6 \times 10^{-19}}\) × (8.2 × 10¹⁴ – 3.3 × 10¹⁴)
= 4.14 × 10^-15 × 4.9 × 10¹⁴
= 2.02 V = 2.0 V

Question 6.
Light of frequency 7.21 × 10¹⁴Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 10⁵m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?
Answer:
Given v = 7.21 × 10¹⁴Hz
u_max = 6.0 × 10¹⁴ms^-1
v₀ = ?
Since K_max = hv – hv₀
∴ v₀ = v – K_max

Plus Two Physics Dual Nature of Radiation and Matter One Mark Questions and Answers

Question 1.
Find out the wrong statement
(i) As frequency increases photo current increases
(ii) As frequency increase KE increases
(iii) As frequency increase velocity of electrons increases
(iv) As frequency increase stopping potential increases
(v) As frequency is below a certain value photo electrons are not emitted.
Answer:
(i) As frequency increases photo current increases.

Question 2.
Read the following statements and write whether true or false.

1. During photo electric effect photon share its energy with a group of electrons.
2. Intensity is directly proportional to square of amplitude

Answer:

1. False
2. True

Question 3.
In photoelectric emission the number of photoelectrons emitted per second depends on
(a) wavelength of incident light
(b) frequency of incident light
(c) intensity of incident light
(d) work function of the material
Answer:
(c) intensity of incident light

Question 4.
What is de-Broglie wave?
Answer:
The wave associated with material particle is called de-Broglie wave.

Question 5.
In photoelectric emission process from a metal of work function 1.8 eV, the kinetic energy of most energetic electrons is 0.5 eV. The corresponding stopping potential is
(a) 1.8 V
(b) 1.3 V
(c) 0.5 V
(d) 2.3 V
Answer:
(c) 0.5 V
Explanation:
The stopping potential V_s is related to the maximum kinetic energy of the emitted electrons K_max through the relation
K_max = eV_s
0.5 eV = eV_s or V_s = 0.5 V

Question 6.
Name the experiment, which establish the wave nature of moving electrons.
Answer:
Davisson Germer experiment.

Question 7.
Pick the odd one out of the following,
(a) Interference
(b) Diffraction
(c) Polarization
(d) Photoelectric effect
Answer:
(d) Photoelectric effect

Question 8.
Find out the wrong statement

1. If two particles have same momentum then they have same de-Broglie wave length.
2. If two particles have same KE the lighter particle has smallerwave length
3. As velocity of a given mass decreases wave-length increases

Answer:

1. True
2. False
3. True

Question 9.
Pick the odd one out from the following x-rays, visible light, matter waves, radio waves
Answer:
Matter waves.

Question 10.
If the electrons are accelerated by a potential of 50V, calculate the de-Broglie wavelength of electrons.
Answer:

Plus Two Physics Dual Nature of Radiation and Matter Two Mark Questions and Answers

Question 1.
Classify the following properties of the waves into de Broglie wave, em wave, and sound wave.

1. Associated with the moving particle.
2. Longitudinal wave
3. Electric field and magnetic field are perpendicular to each other.
4. Can produce photo electric effect.
5. Wave length is inversely proportional to mass of the moving particle.
6. Velocity in vacuum is 3×10⁸ m/s.

Answer:

1. de-Broglie wave – 1,5
2. Em wave – 3,4,6
3. Sound wave – 2

Question 2.
Table given below gives the work function of certain elements.

Identify the element in which photoelectric effect occurs easily. Justify your answer.
Answer:
Na – 2.70eV, work function is least.

Question 3.
“Louis De Broglie suggested existence of matter waves based on a hypothesis”

1. What do you mean by matter wave?
2. The objects in our daily life do no exhibit wave like properties. Why?

Answer:

1. The wave associated with material particle is called matterwave.
2. objects in ourdaily life have large mass. Hence λ is very small (In the order of 10^-34 cm) to be neglected.

Question 4.
A body of mass 1 Kg is moving with a velocity 1 m/s, a wave is associated with this body

1. Name the wave
2. Can you measure wave length of this wave. Explain?

Answer:

1. Matterwave
2. No. wave length of matterwave is very small.

Plus Two Physics Dual Nature of Radiation and Matter Three Mark Questions and Answers

Question 1.
In figure below represents the variation of current with potential fora metal

1. Identify the law governing it.
2. Even when the potential is zero, there is current. Explain.
3. Current is zero for a particular potential. How does this potential help in determining the velocity of electrons.

Answer:
1. Laws of photoelectric emission

• For a given frequency of radiation, number of photoelectrons emitted is proportional to the intensity of incident radiation.
• The K.E. of photoelectrons depends on the frequency of incident light but is independent of the light intensity.

2. When radiation falls on metal, photo electrons are emitted with certain velocity even if accelerating potential is zero.

3. At slopping potential (v₀), photocurrent is zero.
ie. 1/2 mv²_max = ev₀

Question 2.
A metal whose work function is 2 eV is illuminated by light of wavelength 3 × 10^-7 m. Calculate

1. threshold frequency
2. maximum energy of photoelectrons
3. the stopping potential.

Answer:
F₀ = h ν₀ – 2eV

1. F₀ = h ν₀

2. 1/2mv² = h(ν – ν₀)
= 6.63 × 10^-34 (10¹⁵ – 4.8 × 10¹⁴)
= 3.44 × 10^-19J

3. eV₀ = h(ν – ν₀)

Question 3.
The wave nature of electron was experimentally verified by diffraction of electron by Nickel crystal.

1. Name the experiment which establish wave nature of moving electron.
2. An electron and a proton have same kinetic energy which of these particles has shortest de-Broglie wave length?

Answer:
1. Davisson and Gemner experiment

2.

Mass of alpha particle is more than that of proton, hence it has shortest wavelength.

Question 4.
“Moving particles of matter shows wave like properties under suitable conditions’’

1. Who put forward this hypothesis?
2. A proton and an electron have been accelerated through same potential. Which one have higher matter wave length. Write the reason

Answer:
1. De broglie

2. \(\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}}\)
The mass of proton is higher than electron. Hence wave length of proton is less than electron.

Question 5.
Three light beams of same frequency and different intensity I₁, I₂ and I₃ are incident on the same metal. I₁ >I₂ >I₃.

1. Which beam produce maximum photocurrent?
2. Which beam produce electrons of maximum speed and KE?
3. Draw a graph showing variation of photocurrent with intensity in same speed.

Answer:
1. I₃.

2. Frequency is same. Hence electron emitted from the metals will be same.

3.

Plus Two Physics Dual Nature of Radiation and Matter Four Mark Questions and Answers

Question 1.
The graph shows photoelectric current with anode potential.

1. The potential at ‘O’ is called

• accelerating potential
• retarding potential
• stopping potential
• saturation potential

2. Why current becomes constant in the region BC?

3.

Is the above graph possible? Justify your answer
Answer:

1. Stopping potential
2. The whole electrons emitted from the cathode will reach at anode. Hence current becomes saturation at BC.
3. This graph is not possible. Stopping potential is directly proportional to frequency of incident light, ie. when frequency of incident light increases, stopping potential also increases. Hence we expect a high stopping potential for v₂.

Question 2.
The magnification of an electron microscope is much larger than that of an optical microscope, because electron beams are used instead of light beams in an electron microscope.

1. Which property of electrons is used in the construction of the electron microscope?
2. Obtain expression for the wavelength of de-Droglie waves associated with an electron accelerated through a potential of V volts.

Answer:
1. Wave nature.

2. ev = 1/2mv²
mv² = 2 eV
m²v² = 2eVm

Question 3.
The wavelengths of violet and red ends of visible spectrum are 390nm and 760 nm respectively.
1. Evaluate the energy range of the photons of the visible light in electron volts.
2. The work function of four different materials is given in the table below.

Pick out the suitable metal/ metals for the construction of the photo cell which is to operate with visible light.
3. Calculate the threshold frequency of the selected metal/metals.
Answer:
1. 1.65ev to 3.1ev

2. Cs

3.

Question 4.
“To emit a free electron from a metal surface a minimum amount of energy must be supplied”.

1. It is called……..
2. Give three method to supply energy to a free electron
3. For metal A (tungsten) – work function is 4.52 eV for metal B (thoriated tungsten) it is 2.6 ev, for metallic (oxide coated tungsten) it is 1 eV. Which will you prefer as a good electron emitter and why?

Answer:

1. Threshold energy
2. Give light energy or heat energy
3. Work function for metallic oxide coated tungsten is small (1 ev.) Hence this material is good electron emitter.

Question 5.
Louis de Broglie argued that electron in circular orbit as proposed by Bohr, must be seen as a Particl wave.

1. From Bohr’s postulate of angular quantization momentum, arrive at an expression for wave length of an orbital electron. (2)
2. Comment on the above result (2)

Answer:
1.

2. Since λ = \(\frac{2 \pi r}{n}\), length of the first orbit is the de-Broglie wavelength of the orbit.

Plus Two Physics Dual Nature of Radiation and Matter Five Mark Questions and Answers

Question 1.
Schematic diagram of an experimental set up to study the wave nature of electron is shown below.

1. Identify the experiment.
2. In the experiment the intensity of electron beam is measured for different values of ‘q’. At 54V accelerating potential and q = 50°, a sharp diffraction maximum is obtained. What is the wave length associated with the electron.
3. A particle is moving three times as fast as electron. The ratio of debroglie wavelength of the particle to that electron is 1.813 × 10^-14. Calculate the mass of the particle.

Answer:
1. Davisson and German experiment

2. \(\lambda=\sqrt{\frac{150}{v}}=\sqrt{\frac{150}{54}}=1.66 \mathrm{A}^{0}\)

3.

Question 2.
An electron moves under a potential difference of 300V

1. The wave associated with electron is called……….
2. Derive an expression for its wave in terms of charge of particle.
3. Calculate the wavelength of above electron.

Answer:
1. matter wave

2.

3.

Question 3.
Figure below shows a version of Young’s Experiment performed by directing a beam of electrons on double slit. The screen reveals a pattern of bright and dark fringes similar to an interference as pattern produced when a beam of light is used.

1. Which property of electron is revealed in this observation?
2. If the electrons are accelerated by a p.d. of 54V, what is the value of wavelength associated with electrons.
3. In similar experiment if the electron beam is replaced by bullets fired form a gun, no interference pattern is observed. Give reason.

Answer:
1. Dual nature or wave nature.

2.

3. λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\) Since the mass of the bullet is very large compared to the mass of electron, the de Broglie wavelength is not considerable.

Question 4.
A particles moving with KE 5Mev. Its mass is 1.6 × 10^-27 Kg

1. What is the energy of particle in joule?
2. Derive an equation of find De Broglie wave length in terms of KE
3. Calculate De-Broglie wave length of above particle.

Answer:
1. KE = 5 Mev = 5 × 10⁶ × 1.6 × 10^-19J

2.

3.

Question 5.
Einstein got Nobel Prize in 1921 for his explanation of photoelectric effect.

1. In order to start photoelectric emission, the minimum energy acquired by free electron in the metal is called as…… (1)
2. The minimum energy forthe emission of an electron from metallic surface is given below Na: 2.75eV K: 2.3eV Mo:4.17eV Ni:5.15eV Select the metal which is more photo sensitive. Why? (1)
3. Draw variation of photoelectric current with applied voltage for radiation of intensities I₁ and I₂ (I₁ > I₂). Comment on the relatiion between intensity of light and photoelectric current. (2)
4. Does Light from a bulb falling on an iron table emit photoelectron? Justify your answer. (1)

Answer:
1. Work function/ Threshold energy.

2. K is more photosensitive because it has less work function.

3.

As intensity increases photoelectric current also increases.

4. No. The work function of iron is very large.

Question 6.
Lenard and Hallwachs investigated the phenomenon of photoelectric effect in details during 1886-1902 through experiments
1. Write any two characteristic features observed in the above experiment? (2)
2. Explain with reason

• Green light emit electron from certain metal surface while yellow light does not
• When the wavelength of incident light is decreased, the velocity of emitted photo electrons increases (2)

3. Complete the following statement about photoelectric effect.
The radiations having minimum frequency called…….falls on a metallic surcace, electrons are emitted from it. The metal which emits photoelectrons are called………The kinetic energy of photoelectrons emitted by a metal depends on………of the radiations, while intensity of the incident radiations depends on……… (1)
Answer:
1. Any two statement of laws of photoelectric effect.

2. Explain with reason:

• Energy of incident photon is inversely proportional to its wavelength. Since λ of green light is less than that of yellow, it has larger energy. So it can emit photoelectrons
• As the wavelength decreases, frequency and hence energy of incident radiation increases and hence kinetic energy of photo electrons increases.

3. Threshold frequency, photosensitive frequency, number of photons.

Question 7.
Figure below shows variation of stopping potential (V₀) with frequency(?) of incident radiations for two different metals A and B.

1. Write down the values of work function A and B.

2. What is the significance of slope of the above graph? (1)

3. The value of stopping potential for A and B for a frequency γ₀₁ (which is greaterthan γ₀₂) of incident radiations are V₁ and V₀ respectively. Show that the slopes of the lines is equal to \(\frac{v_{1}-v_{2}}{\gamma_{01}-\gamma_{02}}\). (3)
Answer:
1. Work function of A, Φ₀₁ – hν₀₁
Work function of B, Φ₀₁ – hν₀₂.

2. The slope of the graph gives value of h/e.

3. For the metalA, hν₁ = hν₀₁ + eV₁………..(1)
For the metal B, hν₁ = hν₀₂ + eV₂…………..(2)
From equation (1) and (2)
hν₀₁ + eV₁ = hν₀₂ + eV₂
e(V₁ – V₂) = h(ν₀₂ – ν₀₁)
\(\frac{h}{e}=\frac{V_{1}-V_{2}}{v_{02}-v_{01}}\)

Question 8.
Albert Einstein proposed a radically new picture of electromagnetic radiation to explain photoelectric effect.
1. Identify Einstein’s photoelectric equation? (1)
2. With the help of Einstein’s photoelectric equation explain the following facts.

• Kinetic energy of photoelectrons is directly proportional to frequency not on intensity.
• Existence of threshold frequency for a given photosensitive material. (2)

3. A metal whose work function is 2 eV is illuminated by light of wavelength 3 × 10^-7m. Calculate

• threshold frequency
• maximum energy of photoelectrons
• the stopping potential. (3)

Answer:
1. hν = hν₀ + 1/2 mv²

2. Einstein’s photoelectric equation:
a. hν ∝ 1/2 mv²
1/2 mv² ∝ ν
Hence kinetic energy is proportional to frequency.

b. hν – hν₀ = 1/2 mv²
h(ν – ν₀) = 1/2 mv²
ν should be greater than ν₀ otherwise h(ν – ν₀) is negative and is not possible.

3.

a. Φ₀ = hν₀
ν₀ = \(\frac{2 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}}\)
= 4.8 × 10¹⁴ Hz

b. 1/2 mv² = h(ν – ν₀)
= 6.63 × 10^-34 (10¹⁵ – 4.8 × 10¹⁴)
= 3.44 × 10^-19J

c. ev₀ = h(ν – ν₀)

Question 9.
The wave nature of electron was experimentally verified by diffraction of electron by Nickel crystal.

1. Name the experiment which establish wave nature of moving electron. (1)
2. With a neat diagram explain the existence of matter wave associated with an electron. (3)
3. An electron and a proton have same kinetic . energy which of these particles has shortest de-Broglie wave length? (2)

Answer:
1. Davisson and Germer experiment.

2.

Aim:
To confirm the wave nature of electron.
Experimental setup:
The Davisson and Germer Experiment consists of filament ‘F’, which is connected to a low tension battery. The Anode Plate (A) is used to accelerate the beam of electrons. A high voltage is applied in between A and C. ’N’ is a nickel crystal. D is an electron detector. It can be rotated on a circular scale. Detector produces current according to the intensity of incident beam.

Working:
The electron beam is produced by passing current through filament F. The electron beam is accelerated by applying a voltage in between A (anode) and C. The accelerated electron beam is made to fall on the nickel crystal.

The nickel crystal scatters the electron beam to different angles The crystal is fixed at an angle of Φ = 50° to the incident beam. The detector current for different values of the accelerating potential ‘V’ is measured. A graph between detector current and voltage (accelerating) is plotted. The shape of the graph is shown in figure.

Analysis of graph:

The graph shows that the detector current increases with accelerating voltage and attains maximum value at 54V and then decreases. The maximum value of current at 54 V is due to the constructive interference of scattered waves from nickel crystal (from different planes of crystal). Thus wave nature of electron is established.

Experimental wavelength of electron:
The wave length of the electron can be found from the formula
2d sinθ = n λ ………..(1)
From the figure, we get
θ + Φ + θ = 180°
2θ = 180 – Φ, 2θ = 180 – 50°
θ = 65°
for n = 1
equation (1) becomes
λ = 2d sinθ ………(2)
for Ni crystal, d = 0.91 A°
Substituting this in eq. (2), we get
wavelength λ = 1.65 A°

Theoretical wave length of electron:
The accelerating voltage is 54 V
Energy of electron E = 54 × 1.6 × 10^-19J
∴ Momentum of electron P = \(\sqrt{2 \mathrm{mE}}\)
P = \(\sqrt{2 \times 9.1 \times 10^{-31} \times 54 \times 1.6 \times 10^{-19}}\)
= 39.65 × 10^-25 Kg ms^-1
∴ De- Broglie wavelength λ = \(\frac{h}{P}\)

The experimentally measured wavelength is found in agreement with de-Broglie wave length. Thus wave nature of electron is confirmed.

3.

Mass of alpha particle is more than that of proton, hence it has shortest wavelength.

Question 10.
In Geiger-Marsden Scattering experiment alpha particles of 5.5 MeV is allowed to fall on a thin gold foil of thickness 2.1 × 10^-7m.
1. Draw Schematic diagram of above experimental arrangement.
2.

In the above graph nearly 10⁷ particles were detected when scattering angle is Zero. What do you understand by it?
3. Why gold foil is used in this experiment?
4. Does there exist any relation between impact parameter and scattering angle? If yes, explain your answer.
Answer:
1.

2. Most of the alpha particles get unscattered means that most of the space in an atom is empty.

3. Atomic number of gold 79, so number of protons is very high. Hence scattering between alpha and nucleons is larger. Gold foil can be made very thin so that the alpha particles suffer not more than one scattering.

4. Yes.
As impact Parameter increases, scattering angle decreases.

Question 11.
The study of emission line spectra of a material serve as a fingerprint for identification of the gas.

1. Name different series of lines observed in hydrogen spectrum. (1)
2. Draw energy level diagram of hydrogen atom? (2)
3. Write down the Balmer formula for wavelength of H_α line. (1)
4. Given Rydberg constant as 1.097 × 10⁷m^-1. Find the longest and shortest wavelength limit of Baler Series. (2)

Answer:
1. Lyman series, Balmer series, Paschen series, Bracket series, Pfund series

2.

3.

4.

Longest wavelength n₁ = 2 and n₂ = 3

Shortest Wavelength n₁ = 2 and n₂ = α

Question 12.
Bohr combined classical and early quantum concept and gave his theory in the form of three postulates.

1. State three postulates of Bohr Model of atom? (2)
2. The total energy of an electron in ground state of hydrogen atom is -13.6eV. What is the significance of negative sign? (1)
3. The radius of innermost electron orbit of hydrogen atom is 5.3 × 10¹¹m. What are the radii of n = 2 and n = 3 orbits? (2)

Answer:
1. Bohr postulates:
Bohr combined classical and early quantum concepts and gave his theory in the form of three postulates.

• Electrons revolve round the positively charged nucleus in circular orbits.
• The electron which remains in a privileged path cannot radiate its energy.
• The orbital angular momentum of the electron is an integral multiple of h/2π.
• Emission or Absorption of energy takes place when an electron jumps from one orbit to an other.

2. Negative sign implies that the electrons are strongly bounded to the nucleus.

3.

• r_n = n²a₀ = 5.3 × 10^-11m
• r₁ = a₀ = 5.3 × 10^-11m
• r₂ = 4a₀ = 21.2 × 10^-11m
• r₃ = 9a₀ = 47.7 × 10^-11m.