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Kerala Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions

Plus One Functions One Mark Questions and Answers

Question 1.
The process of dividing big programs into small programs are called __________.
Answer:
Modularization

Question 2.
The big programs are divided into smaller programs. This smaller programs are called __________.
Answer:
Functions

Question 3.
The execution of the program begins at __________ function.
Answer:
main function

Question 4.
One of the following is not involved in the creation and usage of a user defined function.
(a) Define a function
(b) Declare a function
(c) invoke a function
(d) None of these
Answer:
(d) None of these

Question 5.
The default data type returned by a function is ________.
(a) float
(b) double
(c) int
(d) char
Answer:
(c) int

Question 6.
After the execution of a function, it is returned back to the main function by executing _________ keyword.
Answer:
return

Question 7.
Supplying data to a function from the called func-tion by using _________.
Answer:
parameters (arguments)

Question 8.
_________ keyword is used to give a value back to the called function.
Answer:
return.

Question 9.
_________ key word is used to specify a function returns nothing.
Answer:
void

Question 10.
One of the following is not necessary in the function declaration. What is it?
(a) name of the function
(b) return type
(c) number and type of arguments
(d) name of the parameters
Answer:
(d) name of the parameters

Question 11.
A function declaration is also called _________.
Answer:
prototype

Question 12.
Consider the following declaration
int sum(int a , int b)
{
return a+b;
}
From the following which is the valid function call.
(a) n = sum(10)
(b) n = sum(10,20)
(c) n = sum(10,20,30)
(d) n = sum()
Answer:
(b) n = sum(10,20)

Question 13.
The ability to access a variable or a function from some where in a program is called ________.
Answer:
scope

Question 14.
A variable or a function declared within a function is have ______ scope.
Answer:
local

Question 15.
A variable or a function declared out side of all the functions is have __________ scope.
Answer:
global

Question 16.
State True/False

1. A local variable exist till the end of the program
2. A global variable destroyed when the sub function terminates

Answer:

1. False
2. False

Question 17.
consider the following declaration
int x;
void main()
{
……..
}
Here x is a _____ variable.
Answer:
global

Question 18.
consider the following declaration
void main()
{
int x;
}
Here x is a _______ variable
Answer:
local

Question 19.
__________ parameter is used when the function call does not supply a value for parameters
Answer:
default

Question 20.
The parameter used to call a function is called ____________.
Answer:
Actual parameter

Question 21.
The parameters appear in a function definition are ___________.
Answer:
formal parameters

Question 22.
After the distribution of answer scripts, the teacher gives the Photostat copy of the mark list to the students to check the marks. If the students make any change that do not affect the original mark list. There is a similar situation to pass the arguments to a function. What is this method?
(a) call by value
(b) call by reference
(c) call by address
(d) none of these
Answer:
(a) call by value

Question 23.
Your class teacher gives you the original mark list to check the mark. If you make any change it will affect the original mark list. There is a similar situation to pass the arguments to a function. What is this method?
(a) call by value
(b) call by reference
(c) call by function
(d) none of these
Answer:
(b) call by reference

Question 24.
Consider the following function declaration
int sum(int,a, int b)
{
Body
}
Here the arguments are passed by __________.
Answer:
call by value method

Question 25.
Consider the following function declaration
int sum(int &a, int &b)
{
Body
}
Here the arguments are passed by __________.
Answer:
call by reference method

Question 26.
A function calls it self is known as ____________.
Answer:
recursive function

Question 27.
Varun wants to copy a string by using strcpy() function. From the following which header file is used for this?
(a) cstdio
(b) cmath
(c) cstring
(d) cctype
Answer:
(c) cstring

Question 28.
__________ is a named group of statements to perform a job/task and returns a value.
Answer:
Function

Question 29.
In his C++ program Ajith wants to accept a lengthy text of more than one line. Which function in C++ can be used in this situation.
Answer:
gets() function can be used to accept a lengthy text.

Question 30.
A function can call itself for many times and return a result. What is the name given to such a function? (MARCH-2016)
Answer:
Recursive function

Question 31.
How many values can a C++ function return? (SCERT SAMPLE-I) (1)
Answer:
One Value

Question 32.
A function may require data to perform the task assigned to it. Which is the component of function prototype that serves this purpose? (SCERT SAMPLE – II) (1)
Answer:
Arguments (Parameters)

Question 33.
“Arguments used in call statement are formal arguments”. State true or false. (SAY-2016) (1)
Answer:
False

Question 34.
Name the built-in function to check whether a character is alphanumeric or not. (MARCH-2017)
Answer:
pisalnum()

Plus One Functions Two Mark Questions and Answers

Question 1.
How does C++ support modularity in programming C++
Answer:
The process of converting big programs into smaller programs is known as modularisation. This small programs are called modules or sub programs or functions. C++ supports modularity in programming called functions.

Question 2.
Consider the following code snippet
charch;
cout << “Enter an alphabet”;
cin>>ch;
cout<<toupper (ch);
What is the output of the above code? Give a sample output.
If the above code is used in a computer that has no cctype file, how will you modify the code to get the same output?
Answer:
It reads a character and convert it into upper case.
eg:
Enter an alphabet: a
The output is A.
If a computer has no cctype header file the code is as follows,
char ch;
cout<< “Enter an alphabet”;
cin»ch;
if (ch >= 97 && ch<<122)
cout<<ch – 32;

Question 3.
The following assignment statement will generate a compilation error.
char str[20];
str = “Computer”
Write a correct C++ statement to perform the same task
Answer:
char str[20] = “Computer”;

OR

char str[20];
strcpy(str,”Computer”); (The header file <string.h> should be included)

Question 4.
float area(const float pi = 3.1415, const float r)
{
r = 10;
return pi*r;
}
Is there any problem? If yes what is it?
Answer:
There is an error. The error is ‘r’ is a constant V must be initialised and cannot be changed during execution.

Question 5.
Match the following

Answer:
a. 3
b. 1
c. 5
d. 4
e. 2

Question 6.
What are the jobs of a return statement in a program?
Answer:
In the case of a sub function a return statement helps to terminate the sub function and return back to the main function or called function. But in the case of a main function it terminates the program.

Question 7.
How to invoke a function in C++ program?
Answer:
function can be called or invoked by providing the name of the function followed by the arguments in parenthesis.
eg: sum(m,n);

Question 8.
Briefly explain constant arguments
Answer:
By using the keyword const we can make argument (parameter) of a function as a constant argument. The value of the const argument cannot be modified within the function.

Question 9.
Short notes on header files
Answer:
A header file is a prestored file that helps to use some operators and functions. To write C++ pro-grams the header files are must. Following are the header files
alloc
iostream
iomanip
cstdio
cctype
cmath
cstring
The syntax for including a header file is as follows
#include<name of the header file>
eg: #include<iostream>

Question 10.
Identify the appropriate header files for the following.

1. setw()
2. cout
3. toupper()
4. exit()
5. strcpy()

Answer:

1. setw() – iomanip.h
2. cout – iostream
3. toupper() – cstring
4. exit() – process
5. strcpy() – cstring

Question 11.
Construct the function prototypes for the following functions.

1. The function Display() accepts one argument of type double and does not return any value.
2. Total () accepts two arguments of type int, float respectively and returns a float type value. (MARCH-2015) (2)

Answer:

1. void Display(double);
2. float Total(int, float);

Question 12.
Explain two types of variable according to its scope and life. (SAY-2015) (2)
Answer:
Scope and life of variables and functions
1. Local scope:
A variable declared inside a block can be used only in the block. It cannot be used any other block.
eg:
#include<iostream>
using namespace std;
int sum(int n1 ,int n2)
{
int s;
s = n1 + n2;
return (s);
}
int main()
{
int n1 ,n2;
cout<<“Enter 2 numbers cin>>n1>>n2;
cout<<“The sum is “<<sum(n1 ,n2);
}
Here the variable s is declared inside the function sum and has local scope;

2. Global scope:
A variable declared outside of all blocks can be used any where in the program.
#include<iostream>
using namespace std;
int s;
int sum(int n1,int n2)
{
s = n1 + n2;
return(s);
}
int main()
{
int n1 ,n2;
cout<<“Enter 2 numbers cin>>n1>>n2;
cout<<“The sum is “<<sum(n1,n2);
}
Here the variable s is declared out side of all functions and we can use variable s any where in the program.

Question 13.
There is a way to pass more than one value to the calling function from the called function. Explain how? (SCERT SAMPLE – I) (2)
Answer:
Methods of calling functions
Two types call by value and call by reference.
1. Call by value:
In call by value method the copy of the original value is passed to the function, if the function makes any change will not affect the original value.
Example
#include<iostream.h>
#include<conio.h> void
swap(int a, int b)
{
int temp;
temp = a;
a = b;
b = temp;
}
main()
{
clrscr();
int a,b;
cout<<“Enter values for a and b:-“;
cin>>a>>b;
cout<<“The values before swap a =”<<a<<“ and
b =”<<b;
swap(a,b);
cout<<“\nThe values before swap a =”<<a«“ and
b =”<<b;
getch();
}

2. Call by reference:
In call by reference method the address of the original value is passed to the function, if the function makes any change will affect the original value.
Example
#include<iostream.h>
#include<conio.h>
void swap(int &a, int &b)
{
int temp;
temp = a;
a = b;
b = temp;
}
main()
{
clrscrO;
int a,b;
cout<<“Enter values for a and b:- “; cin>>a>>b;
cout<<“The values before swap a =”<<a<<“ and
b=”<<b;
swap(a,b);
cout<<“\nThe values before swap a =”<<a<<“ and
b =”<<b;
getch();
}

Question 14.
A
Answer:
int sum (int N, int Start_No) .
{
int i, S = 0;
for (i = Start_No; i < = N + Start_No; i++)
S = S + i;
reurn S;
}

Question 15.
Differentiate between the string functions strcmp() and strcmpi(). (SAY-2016) (2)
Answer:
strcmp(): It is used to compare two strings and returns an integer.
Syntax:
strcmp(string1,string2)

• if it is 0 both strings are equal.
• if it is greater than 0(i.e. +ve) stringl is greater than string2
• if it is less than 0(i.e. -ve) string2 is greater than string1

strcmpi():
It is same as strcmp() but it is not case sensitive. That means uppercase and lowercase are treated as same.
eg: “ANDREA” and “Andrea” and “andrea” these are same.

Question 16.
Read the function definition given below. Predict the output, if the function is called as convert (7). (MARCH-2017) (2)
void convert (int n)
{
if (n>1)
convert (n/2);
cout<<n%2;
}
Answer:
The out put is 111. convert() is a recursive function.

Plus One Functions Three Mark Questions and Answers

Question 1.
Consider the following function declaration with optional (default) arguments and state legal or illegal and give the reasons

1. int sum(int x = 10, int y, int z)
2. int sum(int x = 10, int y = 20, int z)
3. int sum(int x = 10, int y = 20, int z = 30)
4. int sum(int x, int y = 20, int z)
5. int sum(int x, int y = 20, int z = 30)
6. inf sum(int x, int y, int z = 30)
7. int sum(int x = 10, int y, int z = 30)

Answer:
There is a rule to make an argument as default argument,i.e., to set an argument with a value that must be in the order from right to left. All the arguments in the right side of an argument must be set first to make an argument as a default argument.

1. illegal, because y and z are not have values
2. illegal, because z has no value
3. legal
4. illegal, because z has no value
5. legal
6. legal
7. illegal, because x has a value but y has no value

Question 2.
A list of C++ built-in functions are given. Classify them based on the usage and prepare a table with proper group names.

Answer:

Question 3.
Read the following program
#include<iostream>
using namespace std;
int main()
{
cout<<sum(2,3);
}
int sum(int x, int y)
{return (x + y);}
On compilation on the program, an error will be displayed. Identify and explain the reason. How can you rectify the problem
Answer:
The compilation of the program starts from the first line arid next line and so on( i.e. line by line). While compiling the line cout<<sum(2,3); The compiler does not understand the word sum(2,3) because it is not declared yet hence the error prototype required. To rectify this problem there are two methods

First method:
Give the function definition just before the main function as follows.
#include<iostream>
using namespace std;
int sum(intx, int y)
{return (x + y);}
int main()
{
cout<<sum(2,3);
}

Second Method:
Give the function declaration(prototype only) in the main function as follows.
#include<iostream>
using namespace std;
int main()
{
int sum(int.int);
cout<<sum(2,3);
}
int sum(int x, int y)
{return (x + y);}

Question 4.
Considering the following function definition.
{
for (int f = 1; n>0; n-)
f = f*n;
return f;
}
void main()
{
int a = 5, ans;
ans = fact (a);
cout<< a << “! = ” << ans;
}
The expected, desired output is 5! = 120
What will be the actual output of the program? It is not the same as above, why? What modification are required in the program to get the desired output.
Answer:
The output is 0! = 120
Because the address of variable ’a’ is given to the variable ‘n’ of the function fact(call by reference method). So the function changes its value (i.e. n-) to 0. Hence the result.
To get the desired result call the function as call by value method in this method the copy of the value of the variable ‘a’ is given to the function. So the actual value of ‘a’ will not changed. So instead of int fact(int &n) just write int fact(int n), i.e., no need of & symbol.

Question 5.
A function is defined as follows
int sum (int a, int b = 2)
{return (a + b);}
Check whether each of the following function calls is correct or wrong, Justify your answer

1. cout << sum (2,3);
2. cout << sum( 2) ;
3. cout << sum();

Answer:
Here the function is declared with one optional argument. So the function call with minimum one argument is compulsory.

1. It is valid. Here a becomes 2 and b becomes 3.
2. It is also valid . Here a becomes 2 and b takes the default value 2.
3. It is not a valid call. One argument is compulsory.

Question 6.
The factorial of a number, say N is the product of first N natural numbers. Thus, factorial of 5 can be obtained by taking the product of 5 and factorial of 4. Similarly factorial of 4 be found out by taking the product of 4 and factorial of 3. At last the factorial of 1 is 1 itself.

Which technique is applicable to find the factorial of a number in this fashion? Write a C++ function to implement this technique. Also explain the working of the function by giving the number 5 as input.
Answer:
A function calls itself is known as recursion.
#include<iostream>
using namespace std;
int fac(int);
int main()
{
int n;
cout<<“enter a number”;
cin>>n;
cout<<fac(n);
}
int fac(int n)
{
if (n == 1) return (1);
else
return (n × fac(n – 1));
}
The working of this program is as follows If the value of n is 5 then it calls the function as fa. The function returns value 5 × fac(4), That means this function calls the function again and returns 5 ×4 × fac(3). This process continues until the value n = 1. So the result is 5 × 4 × 3 × 2 × 1 = 120.

Question 7.
How do two functions exchange data between them? Compare the two methods of data transfer from calling function to called function
Answer:
There are two methods they are call by value and call by reference
1. call by value:
In call by value method, a copy of the actual parameters are passed to the formal parameters. If the function makes any change it will not affect the original value.

2. call by reference:
In call by reference method, the reference of the actual parameters are passed to the formal parameters. If the function makes any change it will affect the original value.

Question 8.
Read the following program
#include<iostream>
using namespace std;
int a = 0;
int main()
{
int showval(int);
cout<< a; a++;
cout << showval (a);
cout<< a;
}
int showval(int x)
{
int a = 5;
return (a + x);
}
Write down the value displayed by the output of the above program with suitable explanation. What are the inferences drawn regarding the scope of variables?
Answer:
The output is 061.
Global variable: A variable declared out side of all functions it is known as global variable.
Local variable: A variable declared inside of a function it is known as local variable.
If a variable declared inside a function(main or other) with the same name of a global variable. The function uses the value of local variable and does not use the value of the global variable.

Here int a = 0 is a global variable. In the main function the global variable ‘a’ is used. There is no local variable so the value of ‘a’, 0 is displayed. The statement ‘a++’ makes the value of ‘a’ is 1. It calls the function showval with argument ‘a = T.

The argument ‘x’ will get this value i.e. ‘x = 1 But in the function showval there is a local variable ‘a’ its value is 5 is used. So this function returns 6 and it will be displayed. After this the value 1 of the global variable ‘a’ will be displayed. Hence the result 061.

Question 9.
The following are function calling statements. Some of them will be executed, while some other generate compilation error. Write down your opinion on each of them with proper justification Function

1. har ch =
2. sqrt(25);
3. strcat (“Computer”, “Program”);
4. double num = pow(2,3,5)
5. putchar(getchar());

Answer:

1. getch get a character from the console(keyboard) but does not echo to the screen. So we can’t read a character from the console.
2. It returns the square root of 25.
3. It concatenates Program to computer, i.e. we will get a string “computer program”
4. The function pow should contains only two arguments. But here it contains 3 arguments so it is an error. We can write this function as follows Double num = pow(pow(2,3),5)
5. It reads a character from the console and display it on the screen.

Question 10.
Write down the operation performed by the following statements.

1. int l = strlen(“Computer Program”);
2. char ch[ ] = tolower(“My School”);
3. cout <<(strcmp(“High”, “Low”)>0 ? toupper(“High”):tolower(“Low”));

Answer:

1. The built in function strlen find the length of the string i.e. 16 and assigns it to the variable I.
2. This is an error because tolower is a character function.
3. This is also an error because tolower and toupper are character functions.

Question 11.
A line of given length with a particular character is to be displayed. For example, ********** js a line with ten asterisks (*). Define a function to achieve this output.
Answer:

Question 12.
Read the following function
int fib(int n)
{
if (n<3)
return 1;
else
return (fib(n – 1) + fib(n – 2));
}

1. What is the speciality of this function
2. How does it work?
3. What will be the output of the following code?

Answer:

1. This function is a recursive function. That means the function calls itself.
2. It works as follows
   • if i = 1, The function fib calls with value 1. i.e. fib(1) returns 1
   • if i = 2, The function fib calls with value 2. i.e. fib(2) returns 1
   • if i = 3, The function fib calls with value 3. i.e. fib(3) returns fib(2) + fib(1) i.e. it calls the function again.
     So the result is 1 + 1 = 2
   • if i = 4, The function fib calls with value 4. i.e. fib(4) returns fib(3) + fib(2) i.e. it calls the function again.
     So the result is 2 + 1 = 3
3. The output will be as follows 1 1 2 3

Question 13.
Explain scope rules of functions and variables in a C++ program
Answer:
Local variable or function:
A variable or function declared inside a function is called local variable or function. This cannot be accessed by the out-side of the function.
eg:
main()
{
int k; // local variable
cout<<sum(a,b); // local function
}

Global variable or function:
A variable or function declared out side of a function is called global variable or function. This can be accessed by any statements.
eg:
int k; // global variable
int sum(int a, int b); // global function
main()
{
}

Question 14.
Briefly explain default arguments Default arguments.
Answer:
A default value can be set for a parameter (argument) of a function. When the user does not give a value the function will take the default value. An important thing remember,is an argument cannot have a default value unless all arguments on its right side must have default value.
Functions with valid default arguments are given below

• float area(int x, int y, int z = 30);
• float area(int x, int y = 20, int z = 30);
• float area(int x = 10, int y = 20, int z = 30);

Functions with invalid default arguments are given below

• float area(int x = 10, int y, int z);
• float area(int x, int y = 20, int z);
• float area(int x = 10, int y = 20, int z);

Question 15.
How to pass an array to function.
Answer:
By using call by reference method we can send an array as argument. We have to send only the array name and its data type. The array name holds the starting address and will access the subsequent elements of an array.
The following example shows this

Question 16.
How to pass a structure to a function.
Answer:
A structure can be passed as argument to a function either call by value or call by reference method. An example is given below to read name and age of a structure and pass it to a function.

Question 17.
Write a program to read a character and check whether it is alphabet or not. If it is an alphabet check whether it is upper case or lower case?
Answer:

Question 18.
Write a program to read 2 strings and join them 2 string.
Answer:

Question 19.
Write a program to read 2 strings and compare it 2 string.
Answer:

Question 20.
Write a program to read a string and display the number of alphabets and digits and special characters.
Answer:

Question 21.
Write a program to perform the following opera-tions on a string

1. Length of a string
2. Search a character
3. Display the string

Answer:

Question 22.
Write short notes about conversion functions.
Answer:
The header file stdlib.h is used for conversion functions.Following are the important conversion functions
1. itoa():
It is used to convert an integer into a string.
eg:
int n = 100;
charstr[20];
itoa(n,str,10);

2. Itoa():
It is used to convert an long into a string.
eg:
long n = 1002345L
char str[20];
ltoa(n,str,10);

3. atoi():
It is used to convert a string into integer
eg:
int n;
str[] = “123”
n = atoi(str);

4. atoll():
It is used to convert a string into long
eg:
long n;
str[] = “123456”
n = atol(str);

Question 23.

1. Predict the output of both programs.
2. Justify your predictions.

Answer:
1. A. Output
value = 40

B. output
value = 0

2. In the first case (A) the argument x is passed by reference method. So the changes made in the function reflects in main()

In the second case (B) the argument x is passed by value method. So the changes made in the function will not reflect in main()

Question 24.
Consider the following C++ function.
int recfact(int x)
{
if(x == 0)
return 1;
else
return x * recfact(x – 1);
}

1. What type of function is this? Explain. (2 Scores)
2. What is the output of the above function when x = 5

Answer:

1. recursive function. The abilty of a function to call itself is called recursion. A function is said to be recursive if a statement in the body of the function calls itself.
2. Output 120

Question 25.
void initialise()
{
int k = 10;
}
int main()
{
int a, b;
float marks;
a = 20;
cout<<“First value =”<< a;
initialise();
b = k + a;
cout<<“New value =”<<b;
}

1. Identify the error in the above code and explain its reasons.
2. Correct the errors.

Answer:

1. K is a local variable in the function initialize 0. It is not accessible in main()
2. Making the variable K as global we can correct the error.

Question 26.
Write a program to read 2 strings and join them using string function.
Answer:
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int main()
{
char str1 [50],str2[50],str[100];
cout<<“Enter a string:”; gets(strl);
cout<<“Enter another string:”; gets(str2);
strcat(str,str1);
strcat(str,str2);
cout<<“The concatenated string is”<<str;
}

Question 27.
Consider the following code.
#include<iostream>
#include<cstdlib>
using namespace std;
int main()
{
char str1 [] = “123”,str2[] = “45”;
int a,b;
a = atoi(str1);
b = atoi(str2);
cout<<a + b;
}
What will be the output and give the reason?
Answer:
The output is 123 + 45 = 168. Here the conversion function atoi is used to convert the string “123” into integer 123 and “45” into integer 45 so the result.

Question 28.
Name the different methods used for passing arguments to a function. Write the difference between them with examples. (MARCH-2015)
Answer:
Methods of calling functions:
Two types call by value and call by reference.
1. Call by value:
In call by value method the copy of the original value is passed to the function, if the function makes any change will not affect the original value.
Example

2. Call by reference:
In call by reference method the address of the original value is passed to the function, if the function makes any change will affect the original value.
Example

Question 29.
Write a C++ program to display the simple interest using function. (SAY-2015) (3)
Answer:
include < iostream>
using namespace std;
float Simplelnt (float p, int n, float r)
{
return (p*n*r/100);
}
int main()
{
float p, r, SI;
int n;
cout<<“Enter values for p,n,r:”; cin>>p>>n>>r;
SI = Simplelnt (p,n,r); cout<<“simple interest is”<<SI;
}

Question 30.
Write a function’s definition of the above type to find the sum of matural numbers from 1 to N. (Hint: If the value of N is 5, the answer will be 1 + 2 + 3 + 4 + 5 = 15). (MARCH – 2016) (3)
Answer:
int sum (int n)
{
if (n == 0)
return 0;
else
return (n + sum (n – 1));
}

Question 31.
A program requires functions for adding 2 numbers, 3 numbers and 4 numbers, How can you provide a solution by writing a single functio0n? (SCERT SAMPLE -1) (3)
Answer:
It can be solved by writing function with default values.
eg: int add (int n1, int n2, int n3 = 0, int n4 = 0)
{
return (n1 + n2 + n3 + n4);
}
This function can be invoked by the following function calling
1. add (5,2);
2. add (5,2,7);
3. add (5,2,7,10);

The 1. Call returns 5 + 2 = 7
2. Call returns 5 + 2 + 7 = 14
3. Call returns 5 + 2 + 7 + 10 = 24

Question 32.
Explain the difference between call-by-value method and call-by-reference method with the help of examples. (SCERT SAMPLE – II) (3)
Answer:
Methods of calling functions:
Two types call by value and call by reference.
1. Call by value:
In call by value method the copy of the original value is passed to the function, if the ’ function makes any change will not affect the original value.
Example

2. Call by reference:
In call by reference method the address of the original value is passed to the function, if the function makes any change will affect the original value.
Example

Question 33.
Explain recursive functions with the help of a suitable example. (SAY-2016) (3)
Answer:
A function calls itself is called recursive function.
#include<iostream>
using namespace std;
void convert(int n)
{
if(n>1)
convert(n/2);
cout<<n % 2;
}
int main()
{
convert(7);
}
Here the function convert is a recursive function, that means it calls itself and output of the above program is 111.

Question 34.
Name the built-in function to check whether a character is alphanumeric or not. (MARCH-2017) (3)
Answer:
isalnum()

Question 35
Briefly explain the three components in the structure of a C++ program. (MARCH-2017) (3)
Answer:
1. Preprocessor directives:
A C++ program starts with the preprocessor directive i.e., # .include, #define, #undef, etc, are such a preprocessor directives. By using #include we can link the header files that are needed to use the functions. By using #define we can define some constants.
eg: #define x 100. Here the value of x becomes 100 and cannot be changed in the program. No semicolon is needed.

2. Header files:
A header file is a pre stored file that helps to use some operators and functions. To write C++ programs the header files are must. Following are the header files iostream

• iomanip
• cstdio
• cctype
• cmath
• cstring

The syntax for including a header file is as follows
#include<name of the header file>
eg: #include<iostream>

3. The main function:
The main function is the first function which is invoked at the time of execution and the program ends within main(). The other functions are invoke from main().

Question 36.
Explain the difference between call by value method and call by reference method with the help of examples. (MARCH-2017) (3)
Answer:
Two types are call by value and call by reference.
1. Call by value:
In call by value method the copy of the original value is passed to the function, if the function makes any change wilt not affect the original value.
Example

2. Call by reference:
In call by reference method the address of the original value is passed to the function, if the function makes any change will affect the original value.
Example

Plus One Functions Five Mark Questions and Answers

Question 1.
Short notes about character functions and string functions
Answer:
A. Character functions:
1. isalnum():
t is used to check whether a character is alphabet or digit. It returns a non zero value if it is an alphabet or digit otherwise it returns zero.

2. isalpha():
It is used to check whether a character is alphabet or not. It returns a non zero value if it is an alphabet otherwise it returns zero.

3. isdigit():
It is used to check whether a character is digit or not. It returns a non zero value if it is digit otherwise it returns zero.

4. islower():
It is used to check whether a character is lower case alphabet or not. It returns a non zero value if it is a lowercase alphabet otherwise it returns zero.

5. isupper():
It is used to check whether a character is upper case alphabet or not. It returns a non zero value if it is an upper case alphabet otherwise it returns zero.

6. tolower():
It is used to convert the alphabet into lower case

7. toupper():
It is used to convert the alphabet into upper case

B. String functions:

1. strcpy():
This function is used to copy one string into another

2. strcat():
This function is used to concatenate(join) second string into first string

3. strlen():
This function is used to find the length of a string.

4. strcmp():
This function is used to compare 2 strings. If the first string is less than second string then it returns a negative value. If the first string is equal to the second string then it returns a value zero and if the first string is greater than the second string then it returns a positive value.

Question 2.
Write functions to perform the following operations

1. sqrt()
2. power of 2 numbers
3. sin
4. cos

Answer:

Question 3.
Write a C++ program to display the roots of quadratic equation. (SAY-2015) (5)
Answer:
#include<iostream>
#include<cmath>

Question 4.
Explain ‘Call by Value’ and ‘Call by Reference’ methods of function calling with the help of a suitable example. (SAY-2016)
Answer:
Two types are call by value and call by reference.
1. Call by value:
In call by value method the copy of the original value is passed to the function, if the function makes any change will not affect the original value.
Example

2. Call by reference:
In call by reference method the address of the original value is passed to the function, if the function makes any change will affect the original value.
Example

Question 5.
Define network topology. Explain any four network topologies in detail. (SAY-2016)
Answer:
Network topologies:
Physical or logical arrangement of computers on a network is called structure or topology. It is the geometrical arrangement of computers in a network. The major topologies developed are star, bus, ring, tree and mesh

1. Star Topology:
A star topology has a server all other computers are connected to it. If computer A wants to transmit a message to computer B. Then computer A first transmit the message to the server then the server retransmits the message to the computer B.

That means all the messages are transmitted through the server. Advantages are add or remove workstations to a star network is easy and the failure of a workstation will not effect the other. The disadvantage is that if the server fails the entire network will fail.

2. Bus Topology:
Here all the computers are attached to a single cable called bus. Here one computer transmits all other computers listen. Therefore it is called broadcast bus. The transmission from any station will travel in both the direction. The connected computers can hear the message and check whether it is for them or not.

Advantages are add or remove computer is very easy. It requires less cable length and the installation cost is less. Disadvantage is fault detection is very difficult because of no central computer.

3. Ring Topology:
Here all the computers are connected in the shape of a ring and it is a closed loop. Here also there is no central computer. Here a computer transmits a message, which is tagged along with its destination computer’s address. The message travels in one direction and each node check

whether the message is for them. If not, it passes to the next node. FIt requires only short cable length. If a single node fails, at least a portion of the network will fail. To add a node is very difficult.

4. Hybrid Topology:
It is a combination of any two or more network topologies. Tree topology and mesh topology can be considered as hybrid topology.
(a) Tree Topology:
The structure of a tree topology is the shape of an inverted tree with a central node and branches as nodes. It is a variation of bus topology. The data transmission takes place in the way as in bus topology. The disadvantage is that if one node fails, the entire portion will fail.

(b) Mesh Topology:
In this topology each node is connected to more than one node. It is just like a mesh (net). There are multiple paths between computers. If one path fails, we can transmit data through another path.

Students can Download Chapter 8 Redox Reactions Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions

Plus One Chemistry Redox Reactions One Mark Questions and Answers

Question 1.
In which of the following, oxidation number of chlorine is +5?
a) Cl^–
b) ClO^–
c) ClO₂^–
d) ClO₃^–
Answer:
d) ClO₃^–

Question 2.
An oxidising agent is a substance which can
a) Gain electrons
b) Lose an electronegative radical
c) Undergo decrease in the oxidation number of one of its atoms
d) Undergo any one of the above changes
Answer:
d) Undergo any one of the above changes

Question 3.
The arrangement of metals in the order of decreasing tendency to lose electrons is called _________ .
Answer:
Activity series

Question 4.
When KMnO₄, reacts with acidified FeSO₄
a) Only FeSO₄ is oxidised
b) Only KMnO₄ is oxidised
c) FeSO₄ is oxidised and KMnO₄ is reduced
d) KMnO₄ is oxidised and FeSO₄ is reduced
Answer:
c) FeSO₄ is oxidised and KMnO₄ is reduced

Question 5.
In the disproportionation reaction, which of the following statements is not true?
a) The same species is simultaneously oxidised as well as reduced
b) The reacting species must contain an element having at least three oxidation states
c) The element in the reacting species is present in the lowest oxidation state
d) The element in the reacting species is present in the intermediate oxidation state
Answer:
c) The element in the reacting species is present in the lowest oxidation state

Question 6.
Find the oxidation state of oxygen in OF₂.
Answer:
The oxidation number of fluorine in its compounds is always taken as -1.
In OF₂
X+ (-1 × 2) = 0
X = +2

Question 7.
The oxidation numbers of chlorine atoms in bleaching powder is _________ .
Answer:
-1

Question 8
SO₂ can act as
a) Oxidising agent only
b) Reducing agent only
c) Both oxidising and reducing agents
d) Acid and a reducing agent only
Answer:
c) Both oxidising and reducing agents

Question 9.
In the reaction
2KMnO₄ +16HCl → 5Cl₂ + MnCl₂ + 2KCl + 8H₂O the reduction product is _________ .
Answer:
MnCl₂

Question 10.
The strongest reducing agent is .
a) K
b) Ba
c) Li
d) Na
Answer:
c) Li

Question 11.
Oxidation state of oxygen in H₂O₂ is _________ .
Answer:
-1

Plus One Chemistry Redox Reactions Two Mark Questions and Answers

Question 1.
Balance the following equation using oxidation number method:
MnO₂ + Cl^– → Mn²⁺ + Cl₂
Answer:
Assigning oxidation numbers:

Equating the increase and decrease in oxidation number:
MnO₂ + 2Cl^– → Mn²⁺ + Cl₂
Balancing hydrogen and oxygen atoms:
MnO₂ + 2Cl^– + 4H⁺ → Mn²⁺ + Cl₂ + 2H₂O

Question 2.
Balance the following equation using half reaction method:
Cu + NO₃^– → Cu²⁺ + NO₂
Answer:
Separating into half reactions:
Oxidation half: Cu → Cu²⁺
Reduction half: NO₃^– → NO₂
Balancing oxygen and hydrogen atoms:
NO₃^– + 2H⁺ → NO₂ + H₂O
Balancing charge by adding electrons and making the number of electrons equal in the two half reactions:
Cu → Cu²⁺ + 2e^–
2NO₃^– + 4H⁺ + 2e^– → 2NO₂ + 2H₂O
Adding the two half reactions to achieve the overall reaction:
Cu + 2NO₃ + 4H⁺ → Cu²⁺ + 2NO₂ + 2H₂O

Question 3.
Complete the following ionic equations:

1. Al³⁺ + 3e^– → …………….
2. MnO₄^2- → + e^–
3. K → K⁺ + ……………
4. Fe²⁺ → Fe³⁺ +

Answer:

1. Al³⁺ + 3e → Al
2. MnO₄^2- → MnO₄^–+ e^–
3. K → K⁺ + e^–
4. Fe²⁺ → Fe³⁺ + e^–

Question 4.
Find the oxidation number of P in the following compounds:

1. Na₂PO₄
2. H₃P₂O₇
3. PH₃
4. H₃PO₄

Answer:

1. Na₂PO₄, Oxidation state of P = +6
2. H₄P₂O₇, Oxidation state of P = +5
3. PH₃, Oxidation state of P = -3
4. H₃PO₄, Oxidation state of P = +5

Question 5.
Choose the correct oxidation number of sulphur in the compounds in column Afrom column B.

Answer:

Question 6.
Explain oxidation number and valency.
Answer:
Valency of an atom is its combining capacity and is denoted by a number without sign. The valency of an element is always a whole number.

Oxidation number is a net charge which an atom has or appears to have when the other atoms from the molecule are removed as ions assuming that the shared pair of electrons is with more electronegative atom.

Question 7.
Some rules related to oxidation number are given below. Correct the mistakes.

• Oxidation number of alkali metals and alkaline earth metals is +2.
• Oxidation number of hydrogen is always +1.
• Algebraicsum of oxidation number of all the atoms in an ion is not equal to the charge on the ion.

Answer:

• Oxidation number of alkali metals is +1.
• Oxidation number of alkaline earth metals is +2.
• Oxidation number of H is +1 except in metallic hydrides.

Question 8.
Match the following:

Answer:

Question 9.
1. Calculate the oxidation number of oxygen in OF₂ and KO₂.
2. When Zn rod is dipped in blue CuSO₄ solution ‘ the blue colour of CuSO₄ fades due to displacement reaction. Write the reaction and identify the following:
i) The substance oxidised and the substance reduced.
ii) The oxidant and the reductant.
Answer:
1. OF₂: x + (-1 × 2) = 0
x – 2 = 0
x = +2
KO₂: (+1 × 1) + 2x = 0
1 + 2x = 0
2x = -1
x = –\(\frac{1}{2}\)

2. Zn(s) + CuSO₄ (aq) → ZnSO₄(aq) + Cu(s)
i) Substance oxidised – Zn
Substance reduced – Cu
ii) Oxidant-Cu
Reductant – Zn

Question 10.
a) Calculate the oxidation number of C in CH₄ and in CH₃Cl.
b) The sum of oxidation numbers of all atoms in a molecule is …………
Answer:
a) CH₄:
x + (1 × 4) = 0
x + 4 = 0
x = -4
Oxidation number of C in CH₄ is -4.

CH₃Cl:
x + (3 × 1) + -1 = 0
x + 3 – 1= 0
x + 2 = 0
x = -2
Oxidation number of C in CH₃Cl is -2.

b) Zero

Question 11.
1. Write the oxidation state of each element and identify the oxidising agent and reducing agent in the following reaction:
H₂S(g) + Cl₂(g) → 2HCl(g) + S(s)
2. Fill in the blanks and classify the following reactions into oxidation and reduction:
i) Mn⁷⁺ + 5e^– → ……………
ii) Sn⁴⁺ + …………… → Sn²⁺
iii) Na → Na⁺ + ……………
iv) Fe³⁺ +…………… → Fe²⁺
Answer:
1.
Reducing agent – H₂S
Oxidising agent -Cl₂

2. i) Mn⁷⁺ + 5e^– → Mn²⁺
ii) Sn⁴⁺ + 2e^– → Sn²⁺
iii) Na → Na+ + e^–
iv) Fe³⁺ + e^– → Fe²⁺
Oxidation: Reaction (iii)
Reduction: Reactions (i), (ii) and (iv)

Question 12.
Dihydrogen undergoes redox reactions with many metals at high temperature.
a) Write the reaction between hydrogen with sodium.
b) Comment, whether the product formed, is covalent compound or ionic compound.
c) Which is the reducing agent in this reaction?
Answer:
1. 2Na + H₂ → 2NaH
2. Ionic compound is formed. When alkali metals react with hydrogen ionic hydrides are formed.
3. Na is the reducing agent.

Question 13.
1. Is it possible to keep copper sulphate solution in zinc pot? Why?
2. Assign oxidation numbers of the underlined elements.
i) NaH₂\(\underline { P } \)O₄
ii) NaH\(\underline { S } \)O₄
Answer:
1. No. Zn being more reactive will displace Cu from CuSO₄. Thus Cu will be deposited on the vessel.

2. i) NaH₂\(\underline { P } \)O₄
+1 +(+1 × 2) + x +(-2 × 4) = 0
1 + 2 + x – 8 = 0
x – 5 = 0
x = +5
ii) NaH\(\underline { S } \)O₄
+1 + 1 + x +(-2 × 4) = 0
+1 + 1 + x – 8 = 0
x – 6 = 0
x = +6

Question 14.
Identify the substance oxidised, reduced, oxidising agent and reducing agent in the reaction:
2Cu₂O + Cu₂S → 6Cu + SO₂
Answer:

In this reaction, Cu is reduced from +1 state to zero. oxidation state and S is oxidised from -2 state to +4 state. Cu₂O helps S in Cu₂S to increase its oxidation number. Therefore, Cu(l) is the oxidising agent. S of Cu₂S helps Cu both in Cu₂S itself and Cu₂O to decrease its oxidation number. Therefore, S of Cu₂S is the reducing agent.

Question 15.
Explain the following in terms of electron transfer concept:

1. Oxidation
2. Reduction
3. Oxidising agent
4. Reducing agent

Answer:

1. Oxidation: Loss of electron(s) by any species.
2. Reduction: Gain of electron(s) by any species.
3. Oxidising agent: Any species which accepts electrons).
4. Reducing agent: Any species which donates electron^).

Question 16.
Represent the following compounds using Stock notation:
Cu₂O, SnCl₄, MnO, Fe₂O₃, V₂O₅
Answer:

Question 17.
In a redox reaction, oxidation and reduction occur simultaneously.
a) Write the classical concept of oxidation and reduction.
b) Identify the species undergoing oxidation and reduction in the following reaction:
H₂S(S) + Cl₂(g) → 2HCl(g) + S(s)
Answer:
1. Oxidation:
addition of oxygen/electronegative element to a substance or removal of hydrogen/ electropositive element from a substance.

Reduction:
removal of oxygen/electronegative element from a substance or addition of hydrogen/ electropositive element to a substance.

2.Oxidised species:
H₂S. This is because a more electronegative element, Cl is added to H or a more electro positive element, H has been removed from S.

Reduced species:
Cl. This is due to addition of more electropositive element H to it.

Plus One Chemistry Redox Reactions Three Mark Questions and Answers

Question 1.
An equation is given below:
HNO₃+ l2 → HlO₃ + NO₂ + H₂O

• Find the oxidising agent and reducing agent.
• Balance the equation using half reaction method.

Answer:
Oxidising agent = HNO₃
Reducing agent = l₂
Skeletal equation:

Balancing the charge on the half reactions by adding electrons and equalising the number of electrons:

Question 2.
1. Define redox reactions.
2. Predict whether the following reaction is a redox reaction or not? Justify.
Cr₂O₇^2- + H₂O → 2CrO₄^2- + 2H⁺
Answer:
1. Redox reactions are those reactions are those reactions in which reduction and oxidation occur simultaneously. These reactions involve change in oxidation state of the interacting species.

2. No.
Because no element undergoes change in oxidation number.

Question 3.
a) Find out the oxidising agent and reducing agent in the following reaction:
Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s)
b) Balance the following redox reaction in acid medium using oxidation number method.
Cr₂O₇^2- + Fe²⁺ → Cr³⁺ + Fe³⁺
Answer:
1. Oxidising agent – Ag
ReducingAgent – Cu

2. Assigning oxidation number:

Question 4.
Calculate the oxidation number of sulphur, chromium and nitrogen in H₂SO₄, Cr₂O₇^2- and NO₃^–.
Answer:
H₂SO₄
(2 × +1) + x+ (4 × -2) = 0
+2 + x – 8 = 0
x – 6 = 0
x = +6

Cr₂O₇^2-
2x + 7 ×-2 = -2
2x = -2 + 14
2x = 12
∴ x = +6

NO₃^–
x + 3 × -2 = -1
x = -1 + 5
x = +4

Question 5.
1. Assign oxidation numbers
(i) P in NaH₂PO₄
(ii) Mn in KMnO₄
(iii) B in NaBH₄
(iv) S in H₂SO₄
2. Identify the oxidising and reducing agents in the following reaction:
CuO + H₂ → Cu + H₂O
Answer:
1. i) NaH₂PO₄
Na⁺¹H₂⁺¹PO₄^-2
1+2 + x- 8 = 0
3 + x – 8 = 0
x – 5 = 0
x =+ 5

ii) K⁺¹MnO₄^-2
1+ x – 8 = 0
x – 7 = 0
x = +7

iii) Na⁺¹BH₄⁺¹
1 + x + 1 × 4 = 0
x + 5 = 0
x = -5

iv) H₂⁺¹SO₄^-2
2 + x – 8 = 0
x – 6 = 0
x = +6

2.

Question 6.
A copper rod is dipped in silver nitrate solution.

1. What are the observations?
2. Write the displacement reaction.
3. Identify the species getting oxidised and reduced.

Answer:

1. The colour of the solution changes to blue. Silver is deposited on the copper rod.
2. Cu(s) +2AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s)
3. Oxidised species – Cu Reduced species – Ag⁺

Question 7.
1. Identify the oxidising and reducing agent in the reaction:
CuS + O₂ → Cu + SO₂
2. Determine the oxidation number of the underlined element in the following:

Answer:

Question 8.
1. Identify the substance oxidised, substance reduced, oxidising agent and reducing agent in the reaction:
Cl₂ + 2l^– → 2Cl^– +l₂

2. Calculate the oxidation number of underlined elements in the following compounds:
i) K₂\(\underline { Cr } \)₂O₇
ii) H\(\underline { H } \)O₃
Answer:
1. Cl₂ is reduced, therefore Cl₂ is the oxidising agent. I’ is oxidised, therefore I” is the reducing agent.

2. i) K₂\(\underline { Cr } \)₂O₇
(+1 × 2) + 2x +(-2 × 7) = 0
+2 + 2x – 14 =0
2x – 12 =0
2x = 12
x = +6
ii) H\(\underline { H } \)O₃
(+1 × 1) + x + (-2 × 3) = 0
1+ x – 6 = 0
x – 5 = 0
x = +5

Question 9.
Determine oxidation number of the elements underlined in each of the following.

Answer:

Plus One Chemistry Redox Reactions Four Mark Questions and Answers

Question 1.
Permanganate ion (MnO₄^–) reacts with bromide ion (Br^–) in basic medium to give manganese dioxide
(MnO₂) and bromate ion (BrO₃^–).
a) Write the balanced ionic equation for this reaction.
b) Identify the oxidising agent and reducing agent in this reaction.
Answer:

Question 2.
A redox reaction involves oxidation and reduction.
a) What do you understand by electrode potential?
b) Define a redox couple.
c) Explain the set-up for Daniell cell with a diagram.
d) Write the electrode reactions and overall cell reaction which occur in the Daniel cell.
Answer:
a) The potential difference between metal and its own ion is called electrode potential.

b) A redox couple is defined as the combination of oxidised and reduced forms of a substance taking part in an oxidation or reduction half reaction.

c) Take copper sulphate solution in a beaker and put a copper strip. Take zinc sulphate solution in another beaker and put a zinc rod. The two redox couples are represented as Zn²⁺/Zn and Cu²⁺/Cu. Put the beaker containing copper sulphate solution and beaker containing zinc sulphate side by side. Connect two solution by a salt bridge. The Zn and Cu rods are connected by a metalic wire with a provision for ammeter and switch. Transfer of electrons now does not take place directly from Zn to Cu²⁺, but through metallic wire. The electricity from solution in one beaker to solution in the other beaker flows by the migration of ions through the salt bridge.

Question 3.
Redox reactions are those in which oxidation and reduction takes place. Explain the different types of redox reactions with suitable examples.
Answer:
Combination Reactions: The reactions in which two substances combine together to form a new compound are called combination reactions. These can be denoted as A+ B → C where either A or B or both A and B should be in the elemental form.

Decomposition reactions:
The reactions in which a compound breaks up into two or more substances at least one of which is in elemental form are called decompositions reactions.

Displacement reactions:
The reactions of the type X + YZ → XZ + Y in which an atom or ion in a compound is displaced by an ion (atom) of another element, such that X and Y are in elemental form are called displacement reactions. They are of two categories:
1. Metal displacement reactions: Reactions in which a more electropositive metal displaces a less electropositive metal from its compound.

2. Non-metal displacement reactions: These are reactions in which a non-metal is displaced by another metal or non-metal.

Disproportionation reactions: These are special type of redox reactions in which an element in one oxidation state is simultaneously oxidised and reduced. Here one of the reactants should contain an element that should exist in at least three oxidation states. The element in the form of reacting substance is in the intermediate oxidation state; and both higher and lower oxidation states of that element are formed in the reaction.
e.g. The decomposition of hydrogen peroxide.

Here the oxygen of peroxide, which is present in -1 state, is converted to zero oxidation state in O₂ and to -2 state in H₂O.

Plus One Chemistry Redox Reactions NCERT Questions and Answers

Question 1.
Fluorine reacts with ice and results in the change :

Justify that this reaction is a redox reaction.
Answer:
In the given reaction O.N. of F₂ changes from zero to -1 in HF and HOF whereas O.N. of oxygen change from -2 in H₂O to zero in HOF. Thus, F₂ is reduced, whereas oxygen is oxidised and, therefore, it is a redox reaction.

Question 2.
Write formulas for the following compounds:

1. Mercury (II) chloride
2. Nickel (II) sulphate
3. Tin (IV) oxide
4.  Thallium (I) sulphate
5. Iron (III) sulphate
6. Chromium (III) oxide

Answer:

1. Hg(II)Cl₂
2. Ni(II)SO₄
3. Sn(IV)O₂
4. Tl₂(I)SO₄
5. Fe₂(III)(SO₄)₃
6. Cr₂(III)O₃

Question 3.
The compound AgF₂ is unstable. However, if formed, the compound acts as a very strong oxiding agent. Why?
Answer:
In AgF₂, oxidation state of Ag is + 2 which is very unstable. Since Ag can exist in a stable state of + 1 it quickly accepts an electron to form the more stable + 1 oxidation state.
Ag²⁺ + e^– → Ag⁺

Question 4.
Consider the reactions:

Why does the same reductant, thiosulphate react differently with iodine and bromine?
Answer:
The average O.N. of S in S₂O₃^2- is + 2 while in S₄O₆^2- it is + 2.5. The O.N. of S in SO₄^2- is+6. Since Br₂ is a stronger oxidising agent that l₂, it oxidises S of S₂O₃^2- to a higher oxidation state of + 6 and hence forms SO₄^2- ion. l₂, however, being a weaker oxidising agent oxidises S of S₂O₃^2- ion to a lower oxidation of + 2.5 in S₄O₆^2- ion.

Question 5.
Why does the following reaction occur?
XeO₆^4-(aq) + 2F^–(aq) + 6H⁺(aq) → XeO₃(s) + F₂(g) + 3H₂O(I)
What conclusion about the compound Na₄XeO₆ (of which XeO₆^4- is a part) can be drawn from the reaction?
Answer:
The balanced equation along with O.N. of the elements above their symbols will be as:

In the equation the, O.N. of Xe decreases from + 8 in XeO₆^4- to + 6 in XeO₃ while that of F increases from – 1 in F^– to 0 in F₂. Therefore, XeO₆^4- is reduced while F^– is oxidised. This reaction occurs because Na₄XeO₆ (0r XeO₆^4-) is stronger oxidising agent than F₂.

Students can Download Chapter 9 String Handling and I/O Functions Questions and Answers, Plus One 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 One Computer Science Chapter Wise Questions and Answers Chapter 9 String Handling and I/O Functions

Plus One String Handling and I/O Functions One Mark Questions and Answers

Question 1.
To read a single character for gender i.e. ‘m’ or ‘f’ ________ function is used.
Answer:
(a) getch()
(b) getchar()
(c) gets()
(d) getline()
Answer:
(b) getchar()

Question 2.
To use getchar(), putchar(), gets() and puts(), which header file is used?
(a) iostream
(b) cstdio
(c) input
(d) output
Answer:
(b) cstdio

Question 3.
To use cin and cout, which header file is needed?
(a) iostream
(b) cstdio
(c) input
(d) output
Answer:
(a) iostream

Question 4.
Predict the output of the following code snippet.
#include<cstdio>
int mainO
{
char name[ ] = “ADELINE”;
for(int i=0; name[i]!=’\0′;i++)
putchar(name[i]);
}
Answer:
The output is “ADELINE”.

Question 5.
From the following which is equivalent to the function getc(stdin).
(a) putchar()
(b) gets()
(c) getchar()
(d) puts()
Answer:
(c) getchar()

Question 6.
From the following which is equivalent to the function putc(ch, stdout).
(a) putchar(ch)
(b) ch = gets()
(c) ch = getchar()
(d) puts(ch )
Answer:
(a) putchar(ch)

Question 7.
To print a single character at a time which function is used?
(a) puts()
(b) putchar()
(c) gets()
(d) getchar()
Answer:
(b) putchar()

Question 8.
To read a string _______ function is used.
(a) puts()
(b) putchar()
(c) gets()
(d) getchar()
Answer:
(b) gets()

Question 9.
To print a string _______ function is used.
(a) puts()
(b) putchar()
(c) gets()
(d) getchar()
Answer:
(b) puts()

Question 10.
Consider the following code snippet.
main()
{
char str[80];
gets(str);
for(int i=0. len=0;str[il!=’\0′;i++.len++);
cout<<“The length of the string is ” <<len;
}
Select the equivalent forthe under lined statement from the following
(a) int len = strlen(str)
(b) int len = strcmp(str)
(c) int len = strcount(str)
(d) None of these
Answer:
(a) int len = strlen(str)

Question 11.
Arjun wants to read a string with spaces from the following which is suitable.
(а) cin>>
(b) cin.getline(str,80)
(c) str = getc(stdin)
(d) none of these
Answer:
(b) cin.getline(str,80)

Question 12.
State whether the following statement is true or false. The ‘<<‘ insertion operator stops reading a string when it encounters a space.
Answer:
True

Question 13.
_________ function is used to copy a string to another variable. (SAY-2016) (1)
Answer:
strcpy();

Question 14.

1. Write the declaration statement for a variable ‘name’ in C++ to store a string of maximum length 30.
2. Differentiate between the statement cin>>name and gets (name) for reading data to the variable ‘name’. (SAY-2016)

Answer:
1. char name[31];(One for null(\0) character).

OR

cin>> does not allows space. It will take characters up to the space and characters after space will be truncated . Here space is the delimiter. Consider the following code snippet that will take the input upto the space.
#include<iostream>
using namespace std;
int main()
{
char name[20];
cout<<“Enter your name:”;
cin>>name;
cout<<“Hello “<<name;
}
If you input a name “Alvis Emerin” then the output will be Hello Alvis. The string after space is truncated.

2. gets(): This function is used to get a string from the keyboard including spaces. Considerthe following code snippet that will take the input including the space.
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
char name[20];
cout<<“Enter your name:”;
gets(name);
cout<<“Hello “<<name;
}
If you input a name “Alvis Emerin” then the output will be Hello Alvis Emerin.

Question 15.
What is the advantage of using gets() function in the C++ program to input string data? Explain with an example.
Answer:
gets() function is used to get a string from the keyboard including spaces. Consider the following code snippet that will take the input including the space.
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
char name[20];
cout<<“Enter your name:”;
gets(name);
cout<<“Hello “<<name;
}
If you input a name “Alvis” then the output is Hello Alvis.

Plus One String Handling and I/O Functions Two Mark Questions and Answers

Question 1.
In a C++ program, you forgot to include the header file iostream. What are the possible errors occur in that Program? Explain?
Answer:
Prototype error. To use cin and cout the header file iostream is a must.

Question 2.
Categorise the following into three according to their relationship
iostream, cstdio, gets(), puts(), getchar(), putchar(), getline(), write(), cin, cout.
Answer:

Question 3.
Pick the odd one out from the following and give reason.
gets(), getline(), getch() getchar().
Answer:
getline() – It is a stream function whereas the others are console functions.

Question 4.
My_name is a variable contains a string. Write two different C++ statements to display the string. (SAY-2016) (2)
Answer:

1. cout<<my_name;
2. puts(my_name);

Question 5.
Suggest most suitable built-in function in C++ to perform the following tasks: (MARCH-2016) (2)

1. To find the answer for 53
2. To find the number of characters in the string “KERALA” “HAPPY NEW YEAR”
3. To get back the number 10 if the argument is 100.

Answer:

1. pow(5,3);
2. strlen(“KERALA”)
3. tolower(‘M’)
4. sqrt(100);

Question 6.
Read the following C++ statements:
charstr[50];
cin>>str;
cout<<str;
During execution, if the string given as input is “GREEN COMPUTING”, the output will be only the word “GREEN”. Give reason for this. What modification is required to get the original string as output? (SCERT SAMPLE -1) (2)
Answer:
cin>>word;
cout<<word;
It displays “HAPPY” because cin takes characters upto the space. That is space is the delimiter for cin. The string after space is truncated. To resolve this use gets() function.

Because gets() function reads character upto the enter key.
Hence gets(word);
puts(word);
Displays “HAPPY NEW YEAR”

Question 7.
Suppose M[5][5] is a 2D array that contains the elements of a square matrix. Write C++ statements to find the sum of the diagonal elements. (2)
Answer:
gets() function is used to get a string from the keyboard including spaces. To use gets () function the header file cstdio must be included. It reads the characters upto the enter key pressed by the user.
eg:
char name[20];
cout << “Enter your name”;
gets(name);
cout<< “Hello”<< name;
When the user gives Alvis Emerin. It displays as “Hello Alvis Emerin”.

Plus One String Handling and I/O Functions Three Mark Questions and Answers

Question 1.
Suresh wants to print his name and native place using a C++ program. The program should accept name and native place first.
Name is: Suresh Kumar
Address is: Alappuzha
Answer:
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
char name[20],place[20];
cout<<“Enter your name”;
cin.getline(name,80);
cout<<“Enter your place”;
cin.getline(place,80);
cout<<“Your name is puts(name);
cout<<“Your place is puts(place);
}

Question 2.
“Programming is Fun”. Write a C++ program to read a string like this in lower case and print it in UPPER CASE. With out using toupper() library function.
Answer:
using namespace std;
#include<cstdio>
int main()
{
char line[80];
int i;
puts(Enter the string to convert”);
gets(line);
for(i=0;line[i]!=’\0′;i++)
if (Iine[i]>=97 && line[i]<=122)
line[i]=line[i] – 32;
puts(line);
}

Question 3.
An assignment Kumar has written a C++ program which reads a line of text and print the number of vowels in it. What will be his program code?
Answer:
#include<cstdio>
#include<cctype>
#include<iostream>
using namespace std;
int main()
{
char line[80];
int i,vowel=0;
puts(Enter a string”);
gets(line);
for(i=0;line[i]!=’\0′;i++)
switch(tolower(line[i]))
{
case ‘a’:
case ‘e’:
case ‘i’:
case ‘o’:
case ‘u’:
vowel++;
}
cout<<“The number of vowels is “<<vowel;

Question 4.
What will be the output of the following code if the user enter the value “GOOD MORNING”.
1. char string [80];
gets(string);
cout<<string;

2. char string [80];
cin>>string;
cout<<string;

3. charch;
ch = getchar();
cout<<ch;

4. char string [80];
cin.getline(string,9);
cout<<string;
Answer:

1. GOOD MORNING
2. GOOD
3. G
4. GOOD MORN

Question 5.
Consider the following code snippet.
int main()
{
int n;
cout<<“Enter a number”;
cin>>n;
cout<<“The number is “<<n;
}
Write down the names of the header files that must be included in this program
Answer:
Here cin and cout are used so the header file iostream must be included.

Question 6.
Write a program to display the following output.
A
BB
CCC
#include<iostream>
using namespace std;
int main()
{
char str[]=”ABC”;
int i,j;
for(i=0;i<3;i++)
{
for(j=0;j<=i;j++)
cout<<str[i];
cout<<endl;
}
}

Question 7.
Distinguish getchar and gets.
Answer:
getchar is a character function but gets is a string function. The header file cstdio.h must be included. It reads a character from the keyboard.
Eg.
char ch;
ch = getchar();
cout<<ch;
gets is used to read a string from the keyboard. It reads the characters upto enter key. The header file cstdio must be included.
char str[80J;
cout<<“Enter a string”;
gets(str);

Question 8.
Distinguish putch and puts.
Answer:
putch is a character function but puts is a string function. The header file cstdio must be included. It prints a character to the monitor.
Eg:
char ch;
ch = getc(stdin);
putch(ch);
puts is used to print a string. The header file stdio.h must be included.
charstr[80];
puts(“Entera string”);
gets(str);
puts(str);

Question 9.
Write a program to check whether a string is palindrome or not. (A string is said to be palindrome if it is the same as the string constituted by reversing the characters of the original string. eg: “MALAYALAM”, “MADAM”, “ARORA”, “DAD”, etc.)
Answer:
#include<iostream>
using namespace std;
int main()
{
char str[40];
int len,i,j;
cout<<“Enter a string:”;
cin>>str;
for(len=0;str[len]!-\0′;len++);
for(i=0,j=len-1;i<len/2;i++,j–)
if(str[i]!=str[j])
break;
if(i==len/2) .
cout<<str<<” is palindrome”;
else
cout<<str<<” is not palindrome”;
}

Question 10.
Explain multi-character function.
Answer:
getline() and write() functions are multi character functions:
1. getline() It reads a line of text that ends with a newline character. It reads white spaces also.
eg:
char line[80];
cin.getline(line,80);

2. write() It is used to display a string.
Eg.
char line[80];
cin.getline(line,80);
cout.write(line,80);

Question 11.
Read a string and print the number of vowels.
Answer:
#include<cstdio>
#include<cctype>
#include<iostream>
using namespace std;
int main()
{
char line[80];
int i,vowel=0;
puts(“Enter a string”);
gets(line);
for(i=0;line[i]!=’\0′;i++)
switch(tolower(line[i]))
{
case ‘a’:
case ‘e’:
case ‘i’;
case ‘o’:
case ‘u’:
vowel++;
}
cout<<“The number of vowels is in the string is “<< vowel;
}

Question 12.
Distinguish between get() and put() functions.
Answer:
get() function:
get() is an input function. It is used to read a single character and it does not ignore the white spaces and newline character.
Syntax is cin.get(variable);
eg: char ch;
cin.get(ch);

put() function:
put() is an output function. It is used to print a character.
Syntax is cout.put(variable);
eg:
charch;
cin.get(ch);
cout.put(ch);

Question 13.
Write a program to read a string and print the number of consonants.
Answer:
#include<iostream>
#include<cstdio>
#include<cctype>
using namespace std;
int main()
{
char str[40],ch;
int consonent = 0,i;
cout<<“Enter a string:”;
gets(str);
for(i=0;str[i]!=’\0′;i++)
{
ch = toupper(str[i]);
if(ch>=’B’ && ch<=’Z’)
if(ch!=’E’&& ch!=’I’&& ch!=’0’&& ch!=’U’)
consonent++;
}
cout<<“The number of consonents is “<<consonent;
}

Question 14.
Write a program to read a string and print the number of spaces.
Answer:
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
char str[40];
int space=0,i;
cout<<“Enter a string:”;
gets(str);
for(i=0;str[i]!=’\0′;i++)
if(str[i]==32)
space++;
cout<<“The number of spaces is “<<space;
}

Question 15.
Describe in detail about the unformatted console I/O functions.
Answer:
1. Single character functions: This function is used to read or print a character at a time,
(i) getchar():
It reads a character from the keyboard and store it in a character variable.
eg:
char ch;
ch=getchar();

(ii) putchar():
This function is used to print a character on the screen.
eg:
char ch;
ch = getchar();
putchar(ch);

2. String functions This function is used to read or print a string.
(i) gets():
This function is used to read a string from the keyboard and store it in a character variable.
eg:
charstr[80];
gets(str);

(ii) puts():
This function is used to display a string on the screen.
eg:
char str[80];
gets(str);
puts(str);

Question 16.
Write a program to input a string and find the number of uppercase letters, lowercase letters, digits, special characters and white spaces.
Answer:
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
char str[100];
int i,digit=0, Ualpha=0, Lalpha=0, special=0, wspace=0;
cout<<“Enter a string:”;
gets(str);
for(i=0;str[i]!=’\0′;i++)
if(str[i]>=48 && str[i]<=57)
digit++;
else if(str[i]>=65 && str[i]<=90)
Ualpha++;
else if(str[i]>=97 && str[i]<=122)
Lalpha++;
else if(str[i]==’ ‘ || str[i]==’\t’)
wspace++;
else
special++;
cout<<“The number of alphabets is “<<Ualpha+Lalpha<<
” the number of Uppercase letters is “<<Ualpha<< ” the number of Lowercase letters is “<<Lalpha<<” the number of digits is “<<digit<<” the special characters is “<<special<<” and the number of white spaces is “<<wspace;
}

Question 17.
Write a program to count the number of words in a sentence.
Answer:
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
int i,words=1;
char str[80];
cout<<“Enter a string\n”;
gets(str);
for(i=0;str[i]!=’\0′;i++)
if(str[i]==32)
words++;
cout<<“The number of words is “<<words;
}

Question 18.
Write a program to input a string and replace all lowercase vowels by the corresponding uppercase letters.
Answer:
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
char str[100];
int i;
cout<<“Enter a string:”;
gets(str);
for(i=0;str[i]!=\0′;i++)
if(str[i]>=65 && str[i]<=90 || str[i]> = 97 && str[i]<=122)
switch(str[i])
{
case ‘a’:
str[i] = str[i]-32;
break;
case ‘e’:
str[i] = str[i]-32;
break;
case ‘i’:
str[i] = str[i]-32;
break;
case ‘o’: .
str[i] = str[i]-32;
break;
case ‘u’:
str[i] = str[i]-32;
}
cout<<str;
}

Question 19.
Write a program to input a string and display its reversed string using console I/O functions only. For example if the input is “AND” the output should “DNA”.
Answer:
#include<iostream>
using namespace std;
int main()
{
char str[40],rev[40];
int len.ij;
cout<<“Enter a string:”;
cin>>str;
for(len=0;str[len]!=’\0′;len++);
for(i=0,j=len-1 ;i<len;i++,j–)
rev[il=str[j];
rev[i]=’\0′;
cout<<“The reversed string is “<<rev;
}

Question 20.
Write a program to input a word(say COMPUTER) and create a triangle as follows.
C
C O
C O M
C O M P
C O M P U
C O M P U T
C O M P U T E
C O M P U T E R
Answer:
#include<iostream>
#include<cstring>//for strlen()
using namespace std;
int main()
{
charstr[20];
cout<<“enter a word(eg.COMPUTER):”;
cin>>str;
int ij;
for(i=0;i<strlen(str);i++)
{
for(j=0;j<=i;j++)
cout<<str[j]<<“\t”;
cout<<endl;
}
}

Question 21.
Write a program to input a line of text and display the first characters of each word. Use only console I/O functions. For example, if the input is “Save Water, save Nature”, the output should be “SWSN”.
Answer:
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
int i;
charstr[80];
cout<<“Enter a string\n”;
gets(str);
if(str[0]!=32)
cout<<str[0];
for(i=0;str[i]!=’\0′;i++)
if(str[i]==32 && str[i+1]!=32)
cout<<str[i+1];
}

Students can Download Chapter 6 Data Types and Operators Questions and Answers, Plus One 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 One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators

Plus One Data Types and Operators One Mark Questions and Answers

Question 1.
___________ is the main activity carried out in computers.
Answer:
Data processing

Question 2.
The data used in computers are different. To differentiate the nature and size of data ___________ is used.
Answer:
Data types

Question 3.
Classify the following data types,
int, array, function, char, pointer, void, float, double, structure
Answer:

Question 4.
Sheela wants to store her age. From the following which is the exact data type.
(a) void
(b) char
(c) int
(d) double
Answer:
(c) int

Question 5
Integer data type uses ________ bytes of memory.
(a) 5
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 6.
char data type ________uses bytes of memory.
(a) 1
(b) 3
(c) 7
(d) 8
Answer:
(a) 1

Question 7.
From the following which data type uses 4 bytes of memory.
(a) float
(b) short
(c) char
(d) double
Answer:
(a) float

Question 8.
Full form of ASCII is ___________
Answer:
American Standard Code for Information Interchange

Question 9.
Ramu wants to store the value of From the following which is correct declaration.
(a) char pi = 3.14157
(b) int pi = 3.14157
(c) float pi = 3.14157
(d) long pi = 3.14157
Answer:
(c) float pi = 3.14157

Question 10.
From the following which is not true, to give a variable name.
(a) Starting letter must be an alphabet
(b) contains digits
(c) Cannot be a keyword
(d) special characters can be used
Answer:
(d) special characters can be used

Question 11.
Pick a valid variable name from the following.
(а) 9a
(b) float
(c) age
(d) date of birth
Answer:
(c) age.

Question 12.
To perform a unary operation how many number of operands needed?
(a) 2
(b) 3
(c) 1
(d) None of these.
Answer:
(c) 1 (Unary means one)

Question 13.
To perform a binary operation how many number of operands needed?
(а) 2
(b) 3
(c) 1
(d) None of these.
Answer:
(a) 2 (binary means two)

Question 14.
To perform a ternary operation how many number of operands needed?
(a) 2
(b) 3
(c) 1
(d) None of these.
Answer:
(b) 3 (eg: ternary means three)

Question 15.
In C++ 13 % 26 =
(a) 26
(b) 13
(c) 0
(d) None of these
Answer:
(b) % is a mod operator i.e. it gives the remainder. Here the remainder is 13.

Question 16.
In C++ 41/2 =
(a) 20.5
(b) 20
(c) 1
(d) None of these
Answer:
(b) 20. (The actual result is 20.5 but both 41 and 2 are integers so .5 must be truncated)

Question 17.
++ is a __________ operator.
(a) Unary
(b) Binary
(c) Ternary
(d) None of these
Answer:
(a) Unary.

Question 18.
Conditional operator is _________ operator.
(a) Unary
(b) Binary
(c) Ternary
(d) None of these
Answer:
(c) Ternary

Question 19.
% is a _______ operator
(a) Unary
(b) Binary
(c) Ternary
(d) None of these
Answer:
(b) Binary

Question 20.
State True/False

1. Multiplication, division, modulus have equal priority
2. Logical and (&&) has less priority than logical or ()

Answer:

1. True
2. False

Question 21.
_______is composed of operators and operands.
(a) expression
(b) Keywords
(c) Identifier
(d) Punctuators
Answer:
(a) expression

Question 22.
Supply value to a variable at the time of declaration is known as __________.
Answer:
Initialisation

Question 23.
From the following which is initialisation.
(a) int k;
(b) int k = 100;
(c) int k[10];
(d) None of these
Answer:
(b) int k = 100;

Question 24.
State True/False
In an expression all the operands having lower size are converted(promoted) to the data type of the highest sized operand.
Answer:
True

Question 25.
Classify the following as arithmetic/Logical expression.
(a) x + y * z
(b) x < y && y > z
(c) x / y
(d) x > 89 || y < 80
Answer:
(a) and (c) are Arithmetic, (b) and (d) are Logical

Question 26.
Suppose x = 5 and y = 2 then what will be cout<<(float)x/y
Answer:
2.5 The integer x is converted to float hence the result.

Question 27.
Consider the following.
a = 10;
a* =10;
Then a =
(a) a = 100
(b) a = 50
(c) a = 10
(d) a = 20
Answer:
(a) a = 100. This short hand means a = a*10

Question 28.
Consider the following.
a = 10;
a+ = 10;
Then a =
(a) a = 30
(b) a = 50
(c) a = 10
(d) a = 20
Answer:
(d) a = 20. This short hand means a = a + 10

Question 29.
Pick the odd one out
(a) structure
(b) Array
(c) Pointer
(d) int
Answer:
(d) int, it is fundamental data type the others are derived data types

Question 30.
From the following select not a character of C++ language
(a) A
(b) 9
(c) \
(d) @
Answer:
(d) @

Question 31.
Consider the following
float x = 25.56;
cout<<(int)x;
Here the data type of the variable is converted. What type of conversion is this?
(a) type promotion
(b) type casting
(c) implicit conversion
(d) None of these
Answer:
(b) type casting (explicit conversion);

Question 32.
Identify the error in the following C++ statement and correct it.
short population = 68000;
Answer:
The maximum number that can store in short type is less than 32767. So to store 68000 we have to use long data type.

Question 33.
Consider the following statements in C++ if(mark>=18)
cout<<“Passed”;
else
cout<<“Failed”;
Suggest an operator in C++ using which the same output can be produced.
Answer:
Conditional operator (?:)

Plus One Data Types and Operators Two Mark Questions and Answers

Question 1.
Analyses the following statements and write True or False. Justify

1. There is an Operator in C++ having no special character in it
2. An operator cannot have more than 2 operands
3. Comma operator has the lowest precedence
4. All logical operators are binary in nature
5. It is not possible to assign the constant 5 to 10 different variables using a single C++ expression
6. In type promotion the operands with lower data type will be converted to the highest data type in expression.

Answer:

1. True (sizeof operator)
2. False(conditional operator can have 3 operands)
3. True
4. False
5. False(Multiple assignment is possible)
   eg: a = b = c ==5
6. True

Question 2.
Consider the following declaration.
const int bp;
bp = 100;
Is it valid? Explain it?
Answer:
This is not valid. This is an error. A constant variable cannot be modified. That is the error and a constant variable must be initialised. So the correct declaration is as follows, const int bp = 100;

Question 3.
Consider the following statements in C++

1. cout<<41/2;
2. cout<<41/2.0;

Are this two statements give same result? Explain?
Answer:
This two statements do not give same results. The first statement 41/2 gives 20 instead of 20.5. The reason is 41 and 2 are integers. If two operands are integers the result must be integer, the real part must be truncated.

To get floating result either one of the operand must be float. So the second statement gives 20.5. The reason 41 is integer but 2.0 is a float.

Question 4.
If mark = 70 then what will be the value of variable result in the following
result = mark > 50? ’P’: ’F’;
Answer:
The syntax of the conditional operator is given below
Condition ? Value if true : Value if false;
Here the conditional operator first checks the condition i.e., 70 > 50 it is true. So ’P’ is assigned to the variable result.
So the result is ’P’;

Question 5.
Is it possible to initialise a variable at the time of execution. What kind of initialisation is this? Give an example
Answer:
Yes it is possible. This is known as Dynamic initialisation. The example is given below
eg: int a = 10, b = 5; int c = a*b; here the variable c is declared and initialised with the value 10*5.

Question 6.
Boolean data type is used to store True/False in C++. Is it true? Is there any data type called Boolean in C++?
Answer:
No there is no data type for storing boolean value true/false. But in C++ non -zero (either negative or positive) is treated as true and zero is treated as false

Question 7.
Consider the following
n=-15;
if (n)
cout<<“Hello”;
else
cout<<“hai”;
What will be the output of the above code?
Answer:
The output is Hello, because n = -15 a non zero number and it is treated as true hence the result.

Question 8.
Is it possible to declare a variable in between the program as and when the need arise? Then what is it?
Answer:
Yes it is possible to declare a variable in between the program as and when the need arise. It is known as dynamic initialisation.
eg. int x = 10, y = 20;
int z = x*y;

Question 9.
char ch;
cout<<“Enter a character”;
cin>>ch;
Consider the above code, a user gives 9 to the variable ‘ch’. Is there any problem? Is it valid?
Answer:
There is no problem and it is valid since 9 is a character. Any symbol from the key board is treated as a character.

Question 10.
“With the same size we can change the sign and range of data”. Comment on this statement.
Answer:
With the help of type modifiers we can change the sign and range of data with same size. The important modifiers are signed, unsigned, long and short.

Question 11.
Write short notes about C++ short hands?
Answer:
x = x + 10 can be represented as x + = 10, It is called shorthands in C++. It is faster. This is used with all the arithmetic operators as follows.

Question 12.
What is the role of ‘const’ modifier?
Answer:
This ‘const’ keyword is used to declare a constant.
eg: const int bp=100;
By this the variable bp is treated as constant and cannot be possible to change its value during execution.

Question 13.
Specify the most appropriate data type for handling the following data.

1. Roll no. of a student.
2. Name of an employee.
3. Price of an article.
4. Marks of 12 subjects

Answer:

1. short Roll no;
2. char name[20];
3. float price;
4. short marks[12];

Question 14.
Write C++ statement for the following,

1. The result obtained when 5 is divided by 2.
2. The remainder obtained when 5 is divided by 2.

Answer:

1. 5/2
2. 5 % 2

Question 15.
Predict the output of the following code. Justify.
int k = 5;
b = 0;
b = k++ + ++k;
cout<<b;
Answer:
Output is 12. In this statement first it take the value of k in 5 then increment it K++. So first operand for + is 5. Then it becomes 6. Then ++k makes it 7. This is the second operand. Hence the result is 12.

Question 16.
Predict the output.
1. int sum = 10, ctr = 5;
sum = sum + ctr–;
cout<<sum;

2. int sum = 10, ctr = 5;
sum = sum + ++ctr;
cout<<sum;
Answer:

1. 15
2. 16

Question 17.
Predict the output.
int a;
float b;
a = 5;
cout<<sizeof(a + b/2);
Answer:
Output is 4. Result will be the memory size of floating point number

Question 18.
Predict the output.
int a, b, c;
a = 5;
b = 2;
c = a/b;
cout<<c;
Answer:
Output is 2. Both operands are integers. So the result will be an integer

Question 19.
Explain cascading of i/o operations
Answer:
The multiple use of input or output operators in a single statement is called cascading of i/o operators.
eg: To take three numbers by using one statement is as follows
cin>>x>y>>z;
To print three numbers by using one statement is as follows
cout<<x<<y<<z;

Question 20.
Trace out and correct the errors in the following code fragments

1. cout<<“Mark =”45;
2. cin<<“Hellow World!”;
3. cout>>”X + Y;
4. Cout<<‘Good,<<‘Morning’

Answer:

1. cout<<“Mark = 45”;
2. cout<<“Hellow World!”;
3. cout<<X + Y;
4. Cout<<“Good Morning”;

Question 21.
Raju wants to add value 1 to the variable ‘p’ and store the new value in ‘p’ itself. Write four different statements in C++ to do the task.
Answer:

1. P=P+1;
2. p++;(post increment)
3. ++p;(pre increment)
4. p+=1;(short hand in C++)

Question 22.
Read the following code
char str [30];
cin>>str;
cout<<str;
If we give the input “Green Computing”, we get the output “Green”. Why is it so? How can you correct that? (2)
Answer:
The input statement cin>> cannot read the space. It reads the text up to the space, i.e. the delimiter is space. To read the text up to the enter key gets() or getline() is used.

Question 23.

Answer:

Question 24.
Write a C++ expression to. calculate the value of the following equation.

Answer:
x = (-b + sqrt(b*b – 4*a*c)/(2*a)

Question 25.
A student wants to insert his name and school address in the C++ program that he has written. But this should not affect the compilation or execution of the program. How is it possible? Give an example.
Answer:
He can use comments to write this information. In C++ comments are used to write information such as programmer’s name, address, objective of the codes etc. in between the actual codes. This is not the part of the programme. There are two types of comments

1. Single line (//) and
2. Multi-line (/* and *f)

1. Single line (if):
Comment is used to make a single line as a comment. It starts with //.
eg: /./programme starts here.

2. Multi-line (/* and */):
To make multiple lines as a comment. It starts with /* and ends with */.
Eg: /* this programme is used to find sum of two numbers */

Question 26.
Consider the following C++ statements:
char word [20];
cin>>word;
cout<<word;
gets(word);
puts(word);
If the string entered is “HAPPY NEW YEAR”, predict the output and jsutify your answer.
Answer:
cin>>word;
cout<<word;
It displays “HAPPY” because cin takes characters upto the space. That is space is the delimiter for cin. The string after space is truncated. To resolve this use gets () function. Because gets () function reads character upto the enter key.
Hence
gets(word);
puts(word);
Displays “HAPPY NEW YEAR”

Question 27.
Write the difference between x = 5 and x == 5 in C++.
Answer:
x = 5 means the value 5 of the RHS is assigned to the LHS variable x . Here = is the assignment operator. But x == 5, == this is the relational (comparison) operator. Here it checks whether the value of RHS is equal to the value of LHS and this expression returns a boolean value as a result. It is the equality operation.

Question 28.
1. What is the output of the following program?
# include <iostream.h>
void main ()
{
int a;
a = 5 + 3*5;
cout << a;
}

2. How do 9, ‘9’ and “9” differ in C++ program?
Answer:
Here multiplication operation has more priority than addition.
hence
1. a = 5 + 15 = 20

2. Here 9 is an interger
‘9’ is a character
“9” is a string

Question 29.
Read the following C++ program and predict the output by explaining the operations performed.
#include<iostream.h>
void main ()
{
int a = 5, b = 3;
cout<<a++ /–b;
cout<<a/ (float) b;
}
Answer:
Here a = 5 and b = 3
a++ /– b = 5/2 = 2
That is a++ uses the value 5 and next it changes its value to 6
So a/(float) b = 6/(float)2
= 6/2.0
= 3
So the output is 2 and 3

Question 30.
What is the preprocessor directive statement? Explain with an example.
Answer:
A C++ program starts with the preprocessor directive i.e., # include, #define, #undef, etc, are such a preprocessor directives. By using #include we can link the header files that are needed to use the functions. By using #define we can define some constants.
eg. #define x 100. Here the value of x becomes 100 and cannot be changed in the program. No semicolon is needed.

Question 31.
The following C++ code segment is a part of a program written by Smitha to find the average of 3 numbers.
int a, b, c;
float avg;
cin>>a>>b>>c;
avg = (a + b + c)/3;
cout<<avg; .
What will be the output if she inputs 1, 4 and 5? How can you correct it?
Answer:
= (1 + 4 + 5)/3
= 10/3
= 3.3333
Instead of this 3.3333 the output will be 3. This is because if both operands are integers an integer division will be occurred, that is the fractional part will be truncated. To get the correct output do as follows
case 1: int a,b,c; is replaced by float a,b,c;

OR

case 2: Replace (a + b + c)/3 by (a + b + c)/3.0;

OR

case 3: Type casting.
Replace avg = (a + b + c)/3;
by avg = (float)(a + b + c)/3;

Plus One Data Types and Operators Three Mark Questions and Answers

Question 1.
In a panchayath or municipality all the houses have a house number, house name and members. Similar situation is in the case of memory. Explain
Answer:
The named memory locations are called variable. A variable has three important things

1. variable name: A variable should have a name
2. Memory address: Each and every byte of memory has an address. It is also called location (L) value
3. Content: The value stored in a variable is called content.lt is also called Read(R) value.

Question 2.
Briefly explain constants.
Answer:
A constant or a literal is a data item its value doe not change during execution. The keyword const is used to declare a constant. Its declaration is as follows
const data type variable name = value;
eg.const int bp = 100;
\const float pi = 3.14157;
const char ch = ‘a’;
const char[]=”Alvis”;

1. Integer literals:
Whole numbers without fractional parts are known as integer literals, its value does not change during execution. There are 3 types decimal, octal and hexadecimal.
eg: For decimal 100, 150, etc.
For octal 0100, 0240, etc.
For hexadecimal 0x100, 0x1A, etc.

2. Float literals:
A number with fractional parts and its value does not change during execution is called floating point literals.
eg: 3.14157, 79.78, etc

3. Character literal:
A valid C++ character enclosed in single quotes, its value does not change during execution.
eg: ‘m’, ‘f, etc.

4. String literal:
One or more characters enclosed in double quotes is called string constant. A string is automatically appended by a null character(‘\0’)
eg: “Mary’s”, “India”, etc.

Question 3.
Consider the following statements
int a = 10, x = 20;
float b = 45000.34, y = 56.78;
1. a = b;
2. y = x;
Is there any problem for the above statements? What do you mean by type compatibility?
Answer:
Assignment operator is used to assign the value of RHS to LHS. Following are the two chances
(a) The size of RHS is less than LHS. So there is no problem and RHS data type is promoted to LHS. Here it is compatible.

(b) The size of RHS is higher than LHS. Here comes the problem sometimes LHS cannot possible to assign RHS. There may be a chance of wrong answer. Here it is not compatible.
Here
1. a = b; There is an error since the size of LHS is 2 but the size of RHS is 4.
2. y = x; There is no problem because the size of LHS is 4 and RHS is 2.

Question 4.
A company has decided to give incentives to their salesman as perthe sales. The criteria is given below.
If the total sales exceeds 10,000 the incentive is 10%

1. If the total sales >= 5,000 and total sales <10,000, the incentive is 6 %
2. If the total sales >= 1,000 and total sales <5,000, the incentive is 3 %

Write a C++ program to solve the above problem and print the incentive after accepting the total sales of a salesman. The program code should not make use of ‘if’ statement.
Answer:
#include<iostream>
using namespace std;
int mainO
{
float sales,incentive;
cout<<“enter the sales”;
cin>>sales;
incentive = (sales>10000 ? sales*.10: (sales > =5000 ? sales * .06 : (sales >= 1000 ? sales * -03: 0)));
cout<<“\nThe incentive is ” << incentive;
}

Question 5.
A C++ program code is given below to find the value of X using the expression
\(x=\frac{a^{2}+b^{2}}{2 a}\)
where a and b are variables

#include<iostream>
using namespace std;
int main()
{
int a;b;
float x
cout<<“Enter the values of a and b;
cin>a>b;
x = a*a + b*b/2*a;
cout>>x;
}
Predict the type of errors during compilation, execution and verification of the output. Also write the output of two sets of input values

1. a = 4, b = 8
2. a = 0, b = 2

Answer:
This program contains some errors and the correct program is as follows.
#include<iostream>
using namespace std;
int main()
{
int a,b;
float x; .
cout<<“Enterthe values of a and b”;
cin>>ab;
x=(a*a + b*b)/(2*a);
cout<<x;
}
The output is as follows

1. a = 4 and b = 8 then the output is 10
2. a = 0 and b = 2 then the output is an error divide by zero error(run time error)

Question 6.
A list of data items are given below
45, 8.432, M, 0.124,8 , 0, 8.1 × 1031, 1010, a, 0.00025, 9.2 × 10120, 0471,-846, 342.123E03

1. Categorise the given data under proper headings of fundamental data types in C++.
2. Explain the specific features of each data type. Also mention any other fundamental data type for which sample

data is not given
Answer:
1.

2. The specific features of each data type.
(i) int data type:
It is used to store whole numbers without fractional (decimal point) part. It can be either negative or positive. It con¬sumes 4 bytes (32 bits) of memory.i.e. 232 . numbers. That is 231 negative numbers and 231 positive numbers (0 is considered as +ve ) So a total of 232 numbers. We can store a number in between -231 to + 231-1.

(ii) char data type:
Any symbol from the keyboard, eg. ‘A’ , ‘?’, ‘9’,…. It consumes one byte( 8 bits) of memory. It is internally treated as integers, i.e. 28 = 256 characters. Each character is having a ASCII code, ‘a’ is having ASCII code 97 and zero is having ASCII code 48.

3. float data type:
It is used to store real numbers i.e. the numbers with decimal point. It uses 4 bytes(32 bits) of memory.
eg: 67.89, 89.9 E-15.

4. double data type:
It is used to store very large real numbers. It uses 8 bytes(64 bits) of memory.

5. void data type:
void means nothing. It is used to represent a function returns nothing.

Question 7.
Write valid reasons after reading the following statements in C++ and comment on their correctness by give reasons.

1. char num = 66;
   char num = B’;
2. 35 and 35L are different
3. The number 14, 016 and OxE are one and the same
4. Char data type is often said to be an integer type
5. To store the value 4.15 float data type is preferred over double

Answer:

1. The ASCII number of B is 66. So it is equivalent.
2. 35 is of integer type but 35L is Long
3. The decimal number 14 is represented in octal is 016 and in hexadecimal is OXE.
4. Internally char data type stores ASCII numbers.
5. To store the value 4.15 float data type is better because float requires only 4 bytes while double needs 8 bytes hence we can save the memory.

Question 8.
Suggest most suitable derived data types in C++ for storing the following data items or statements

1. Age of 50 students in a class
2. Address of a memory variable
3. A set of instructions to find out the factorial of a number
4. An alternate name of a previously defined variable.
5. Price of 100 products in a consumer store
6. Name of a student

Answer:

1. Integer array of size 50
2. Pointer variable
3. Function
4. Reference
5. Float array of size 100
6. Character array

Question 9.
Considering the following C++ statements. Fill up the blanks

1. If p = 5 and q = 3 then q%p is _______
2. If E1 is true and E2 is False then E1 && E2 will be _______
3. If k = 8, ++k < = 8 will be ________
4. If x = 2 then (10* ++x) % 7 will be ________
5. If t = 8 and m = (n=3,t-n), the value of m will be ______
6. If i = 12 the value i after execution of the expres¬sion i+ = i- – + – -i will be ______

Answer:

1. 3
2. False
3. False(++k makes k = 9. So 9<=8 is false)
4. 2(++x becomes 3 ,so 10 * 3 = 30%7 = 2)
5. 5( here m = (n = 3,8-3) = (n = 3,5), so m = 5, The maximum value will take)
6. Here i = 12

i + = i- – + – -i
here post decrement has more priority than pre decrement. So “i- -” will be evaluated first. Here first uses the value then change so it uses the value 12 and i becomes 11
i + = 12 + – -i
now i = 11.
Here the value of i will be changed and used so “i- -” becomes 10
i + = 12 + 10 = 22
So, i = 22 +10
i = 32
So the result is 32.

Question 10.
The Maths teacher gives the following problem to Riya and Raju.
x = 5 + 3 * 6.
Riya got x = 48 and Raju got x = 23. Who is right and why it is happened? Write down the operator precedence in detail?
Answer:
Here the answer is x = 23. It is because of precedence of operators. The order of precedence of operators are given below.

Here multiplication has more priority than addition

Question 11.
Explain the data types’ in C++. (3)
Answer:
Fundamental data types:
It is also called built-in data type. They are int, char, float, double and void
1. int data type:
It is used to store whole numbers without fractional (decimal point) part. It can be either negative or positive. It consumes 4 bytes (32 bits) of memory. i.e. 232 numbers. That is 231 negative numbers and 231 positive numbers (0 is considered as +ve) So a total of 232 numbers. We can store a number in between -231 to + 2311.

2. char data type Any symbol from the keyboard, eg. ‘A’,‘9’, …. It consumes one byte( 8 bits) of memory. It is internally treated as integers, i.e. 28 = 256 characters. Each character is having an ASCII code, ‘a’ is having ASCII code 97 and zero is having ASCII code 48.

3. float data type:
It is used to store real numbers i.e. the numbers with decimal point. It uses 4 bytes(32 bits) of memory.
eg: 67.89, 89.9 E-15.

4. double data type:
It is used to store very large real numbers. It uses 8 bytes(64 bits) of memory.

5. void data type:
Void means nothing. It is used to represent a function returns nothing.

• User defined Data types: C++ allows programmers to define their own data type. They are Structure(struct), enumeration (enum), union, class, etc.
• Derived data types: The data types derived from fundamental data types are called Derived data types. They are Arrays, pointers, functions, etc

Question 12.
Predict the output of the following C++ statements.
int a = -5, b = 3, c = 4;
C+ = a++ + –b;
cout<<a<<b<<c;
Answer:
a = -4, b = 2 and c = 1.

Question 13.
Match the following

Answer:

Question 14.
Write any five unary operators of C++. Why are they called so?
Answer:
A unary operator is an operator that need only one operand to perform the operation. The five unary operators of C++ are given below.
Unary +, Unary -, ++, – – and ! (not)

Question 15.
Write C++ examples for the following:

1. Declaration statement
2. Assignment statement
3. Type casting

Answer:

1. int age;
2. age = 16;
3. avg = (float)a + b + c/3;

Plus One Data Types and Operators Five Mark Questions and Answers

Question 1.

• Name: Jose
• Roil no: 20
• Age: 17
• Weight: 45.650

Consider the above data, we know that there are different types of data are used in the computer. Explain different data types used in C++.
Answer:
1. int data type:
It is used to store whole numbers without fractional (decimal point) part. It can be either negative or positive. It consumes 4 bytes (32 bits) of memory, i.e. 232 numbers. That is 231 negative numbers and 231 positive numbers (0 is considered as +ve) So a total of 232 numbers. We can store a number in between -231 to + 2311.

2. char data type:
Any symbol from the keyboard, eg. A’,’9′,…. It consumes one byte( 8 bits) of memory. It is internally treated as integers, i.e. 28 = 256 characters. Each character is having a ASCII code, ‘a’ is having ASCII code 97 and zero is having ASCII code 48.

3. float data type:
It is used to store real numbers i.e, the numbers with decimal point. It uses 4 bytes(32 bits) of memory.
eg: 67.89, 89.9 E-15.

4. double data type:
It is used to store very large real numbers. It uses 8 bytes(64 bits) of memory.

5. void data type:
void means nothing. It is used to represent a function returns nothing.

Question 2.
Define an operator and explain operator in detail.
Answer:
An operator is a symbol that performs an operation. The data on which operations are carried out are called operands. Following are the operators
1. lnput(>>) and output(<<):
These operators are used to perform input and output operation.
eg: cin>>n;
cout<<n;

2. Arithmetic operators:
It is a binary operator. It is used to perform addition(+), subtraction(-), division (/), multiplication (*) and modulus (%- gives the remainder) operations.
eg: If x = 10 and y = 3 then

x/y = 3, because both operands are integer. To get the floating point result one of the operand must be float.

3. Relational operator:
It is also a binary operator. It is used to perform comparison or relational operation between two values and it gives either true(1) or false(O). The operators are <,<=,>,>=,== (equality)and !=(not equal to)
eg: If x = 10 and y = 3 then

4. Logical operators:
Here AND(&&), OR(||) are binary operators and NOT(!) is a unary operator. It is used to combine relational operations and it gives either true(1) or false(O). If x=True and y=False then

Both operands must be true to get a true value in the case of AND(&&) operation If x = True and y = False then

Either one of the operands must be true to get a true value in the case of OR(||) operation If x = True and y = False then

5. Conditional operator:
It is a ternary operator hence it needs three operands. The operator is ?:
Syntax: expression ? value if true : value if false. First evaluates the expression if it is true the second part will be executed otherwise the third part will be executed.
eg: If x = 10 and y = 3 then x>y ? cout<<x : cout<<y;. Here the output is 10

6. sizeof():
This operator is used to find the size used by each data type.
eg. sizeof(int) gives 2.

7. Increment and decrement operator:
These are unary operators.
(a) Increment operator (++): It is used to incre¬ment the value of a variable by one i.e., x++ is equivalent to x = x + 1;
(b) Decrement operator (–): It is used to decre¬ment the value of a variable by one i.e., x-is equivalent to x = x – 1.

8. Assignment operator (=):
lt is used to assign the value of a right side to the left side variable.
eg: x = 5; Here the value 5 is assigned to the variable x.

Students can Download Chapter 6 Thermodynamics Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Plus One Chemistry Thermodynamics One Mark Questions and Answers

Question 1.
Hot coffee in a thermos flask is an example of system.
Answer:
Isolated

Question 2.
Which of the following statements is incorrect about internal energy?
a) The absolute value of internal energy cannot be determined
b) The internal energy of one mole of a substance is same at any temperature or pressure
c) The measurement of heat change during a reaction by bomb calorimeter is equal to the internal energy change
d) Internal energy is an extensive property
Answer:
b) The internal energy of one mole of a substance is same at any temperature or pressure

Question 3.
For which of the following the standard enthalpy is not zero?
a) C (Diamond)
b) C (Graphite)
c) Liquid mercury
d) Rhombicsulphur
Answer:
a) C (Diamond)

Question 4.
Say TRUE or FALSE?
Any spontaneous process must lead to a net increase in entropy of the universe.
Answer:
TRUE

Question 5.
The ∆H fora reaction is-30 kJ. On the basis of this fact, we can conclude that the reaction
a) Gives off thermal energy
b) Is fast
c) Is slow
d) Is spontaneous
Answer:
a) Gives off thermal energy

Question 6.
Write the type of system in each of the following:

1. Hot water taken in an open vessel
2. Hot water taken in a closed metallic vessel
3. Hot water taken in a thermos flask

Answer:

1. Open system
2. Closed system
3. Isolated system

Question 7.
In a reversible process the total change in entropy is ∆s(universe) is
Answer:
Zero

Question 8.
For the reaction Ag₂O \(\rightleftharpoons \) 2Ag + \(\frac{1}{2}\)O₂(g) ∆S and ∆H are 66J K^-1mol^-1 and 30.56 Kg mol respectively. The reaction will not be spontaneous at.
Answer:
463K

Question 9.
One mole of methane undergoes combustion to form CO₂ and water at 25°C. The difference between ∆U & ∆H will be
Answer:
-2RT

Question 10.
A gas expands from 1 l to 6 l against a constant pressure of 1 atm and it absorbs 500J of heat ∆μ is
Answer:
-6.5J

Question 11.
Born Haber cycle is to find out __________
Answer:
lattice energy

Plus One Chemistry Thermodynamics Two Mark Questions and Answers

Question 1.
1. Explain enthalpy of fusion.
2. Give illustration of fusion of ice.
Answer:
1. It is the enthalpy change when one mole of a solid is converted into its liquid at its melting point.

2. Enthalpy of fusion of ice is 6 kJ/mol. From this it is clear that 6 kJ of energy is required to convert one mole of ice (18 g) into water at 0°C.

Question 2.
a) What do you meant by enthalpy of vapourisation?
b) Explain enthalpy of sublimation.
Answer:
a) It is the enthalpy change when one mole of a liquid is converted into its vapour at its boiling point.
b) It is the enthalpy change when one mole of a solid is converted into its vapour at its transition temperature.

Question 3.
One equivalent of an acid reacts completely with one equivalent of a base in dilute solution.
1. Which type reaction is this?
2. HCl + NaOH → NaCl + H₂O
On the basis of above equation, explain enthalpy of neutralisation.
Answer:
1. Nneutralisation.

2. When one equivalent of HCl (36.5 g) reacts completely with one equivalent of NaOH (40 g), 57.1 kJ energy is liberated.

Question 4.
1. What is the difference between system and surroundings?
2. There are different types of systems. What are they? Explain.
3. Give example for different types of systems.
Answer:
1. A system in thermodynamics refers to that part of universe in which observations are made. The remaining part of the universe other than the system constitutes the surroundings.

2. System is classified into the following three types. Open system: This is a system in which there is exchange of energy and matter between system and surroundings.
Closed system:
This is a system in which there is no exchange of matter, but exchange of energy is possible between system and the surroundings.

Isolated system:
This is a system in which there is no exchange of energy or matter between the • system and the surroundings.

3. Open system
Presence of reactants in an open beaker
Closed system
Presence of reactants in a closed vessel made of conducting material
Isolated system
Presence of reactants in a • thermos flask or any other closed insulated vessel

Question 5.
Match the following:

Answer:

Question 6.
1. What is meant by enthalpy?
2. Derive an equation for enthalpy change.
3. What is enthalpy change?
Answer:
1. Enthalpy is the sum of internal energy and pressure volume energy.
i.e.,H = U + pV

2. ∆H = ∆U + ∆pV)
∆H = ∆U + p∆V + V∆p
At constant pressure, ∆p=0
∆H = ∆U+p∆V
But ∆U=q+w
∆H=q+w+p∆V
w = -p∆V
i.e; ∆H= q – p∆V + p∆V
∆H = q_p

3. Enthalpy change is heat absorbed or released at constant pressure.

Question 7.
1. Find the enthalpy of the reaction,
C(graphite) + O₂(g) → CO₂(g)
Given,
i) C(graphite)+ ½ O₂(g) → CO(g); ∆H =-110.5 kJ mol^-1
ii) CO(g) + ½ O₂(g) → CO₂(g); ∆H =-283.0 kJ mol^-1
2. Melting of ice is a spontaneous process. What are the criteria for spontaneity of a process?
Answer:
1. Considerthe reaction,
C(grahite) + O₂(g) → CO₂ (g); ∆H = x
CO₂ can also be prepared through the following two steps:
i) C(graphite)+ ½ O₂(g) → CO(g); ∆H =110.5 kJ mol^-1
ii) CO(g) + ½ O₂(g) → CO₂(g); ∆H =-283.0 kJ mol^-1
Then by Hess’s law, x = (-110.5+-283.0) kJ=-393.5 kJ

2. Certain endothermic process are found to be spontaneous in nature. Hence, spontaneous behaviour of a process cannot be explained only on the basis of energy consideration.
For a spontaneous process ∆S_Total is +ve.
For a nonspontaneous process ∆S_Total is -ve.

Question 8.
Explain the following:

1. Enthalpy of atomization
2. Enthalpy of solution at infinite dilution

Answer:

1. It is the enthalpy change on breaking one mole of bonds completely to obtain atoms in the gas phase.
2. It is the enthalpy change observed on dissolving the substance in an infinite amount of solvent when the interactions between the ions or solute molecules are negligible.

Question 9.
The enthalpy change for the reaction,
N₂(g) + 3H₂(g) → 2NH₃(g) is -92.38 kJ at 298 K.
What is ∆U at 298 K?
Answer:
∆U = ∆H — ∆n_gRT
= -92.38 × 10³ J – [-2 × 8.314J K^-1 mol^-1 × 293 K)]
= -92.38 × 10³J +4.872 × 10³J
= -87.51 × 10³J
= – 87.51 kJ

Question 10.
What are the two types of heat capacities? How they are related?
Answer:
The two types of heat capacities are heat capacity at constant pressure (C_p) and heat capacity at constant volume (C_v). These two are related as C_p – C_v = R, where R is the universal gas constant.

Question 11.
Enthalpy and Entropy changes of two reactions are given below: Find out whether they are spontaneous or not at 27°C. Justify.
1. ∆H = 26 kJ/mole, ∆S = 8.3 J/K/mole
2. ∆H = -393.4 kJ/mole, ∆S = 6 J/K/mole
Answer:
1. ∆G = ∆H -T∆S
= 26000 – 300 × 8.3 = 23.510
Since ∆G is positive, the process is non-spontaneous.

2. ∆G = ∆H -T∆S
= -393400 – 300 × 6 = -391600
Since ∆G is negative, the process is spontaneous.

Question 12.
1. What is enthalpy of solution?
2. What is enthalpy of dilution?
Answer:
1. Enthalpy of solution is the enthalpy change associated with the addition of a specified amount of solute to the specified amount of solvent at a constant temperature and pressure.

2. Enthalpy of dilution is the heat withdrawn from the surroundings when additional solvent is added to the solution. It is dependent on the original concentration of the solution and the amount of solvent added.

Question 13.
What is the significance of the second law of thermodynamics in the spontaneity of exothermic and endothermic reactions?
Answer:
The second law of thermodynamics provides explanation for the spontaneity of chemical reactions. In exothermic reactions heat released by the reaction increases the disorder of the surroundings and overall 1 entropy change is positive which makes the reaction spontaneous.

In the case of endothermic reactions, since heat is ‘ absorbed by the system from the surroundings, the entropy change of the surroundings becomes negative (∆S_surr < 0). In this case the process will be spontaneous only if the entropy change of the reacting system is postive (∆S_sys > 0) and is also greater than ASsurr in magnitude so that the overall entropy change (∆S_total) is positive.

Question 14.
Explain the importance of third law of thermodynamics.
Answer:
The importance of third law of thermodynamics lies in the fact that it permits the calculation of absolute values of entropy of pure substances from thermal data alone. For a pure substance, this can be done by summing \(\frac { { q }_{ rev } }{ T } \) increments from 0 K to 298 K.

Question 15.
C₁₂H₂₂O₁₁ + 12O₂ → 12CO₂ + 11H₂O
Consider this equation and answer the following questions.
a) Thermodynamically, which type reaction is this?
b) What is enthalpy of combustion?
c) Give another example.
Answer:
a) Combustion.
b) It is the enthalpy change when one mole of a substance undergoes complete combustion in excess of air or oxygen.
c) C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

Question 16.
Bond dissociation energies of hydrogen and nitrogen are 430 kJ and 41.8 kJ respectively and the enthalpy of formation of NH₃ is – 46 kJ. What is the bond energy of N-Hbond?
Answer:
3H₂ + N₂ → 2NH₃; AH = -46 kJ
3 × ½H₂ + ½N₂ → 2 × ½NH₃
[(3× ½ × 430) + (½ × 941 .8)] – (3N-H) = – 46
[(3 × 215) + (470.9) + (46)] → [3N-H]
[645 + 470.9 + 46] = 3N-H
N-H = 387.3 kJ

Plus One Chemistry Thermodynamics Three Mark Questions and Answers

Question 1.
In 1840, G.H.Hess (a Russian chemist) proposed an important generalisation of thermochemistry which is known after his name as Hess’s law.
1. State Hess’s law.
2. Give illustration of Hess’s law.
Answer:
1. Enthalpy change in a chemical reaction is same whether it takes place in one step or in more than one step.

2. Considerthe formation of CO₂.
C + O₂ → CO₂; ∆H = x
CO₂ can be prepared through the following two steps:
C + ½O₂ → CO ; ∆H = y
CO + ½O₂ → CO₂; ∆H = Z
Then by Hess’s law,
x = y + z

Question 2.
∆U = q-p∆V. If the process is carried out at constant
volume, then ∆V=0. Answer the following questions.
1. Give the equation for ∆U.
2. 1000J was supplied to a system at constant volume. It resulted in the increase of temperature of the system from 45 °C to 50 °C. Calculate the change in internal energy.
Answer:
1. ∆U = q_v

2. Since the volume kept constant, ∆V=0
∴ ∆U = q_v = 1000J

Question 3.
Thermodynamics deals with macroscopic properties.
1. What is the difference between extensive and intensive properties?
2. Classify the following properties into extensive and intensive.
Pressure, Mass, Volume, Temperature, Density, Heat capacity, Viscosity, Surface tension, Internal • energy, Molar heat capacity, Refractive index, Enthalpy, Specific heat capacity
Answer:
1. Extensive properties are those properties whose value depend on the quantity or size of matter present in the system.
Intensive properties are those properties which do not dependent on the quantity or size of matter present in the system.

2. Extensive properties: Mass, Volume, Heat capacity, Internal energy, Enthalpy Intensive: Pressure, Temperature, Density, Viscosity, Surface tension, Molar heat capacity, Refractive index, Specific heat capacity.

Question 4.
1. What is meant by state of the system and state variables?
2. Give any four examples for state variables/state functions.
Answer:
1. The state of a system refers to the conditions of existence of a system when its macroscopic properties have definite values. If any of the macroscopic properties of the system changes, the state of the system will change. The measurable properties required to describe the state of a system are called state variables or state functions. A state function is a property of a system whose value depends only upon the initial and final states of the system and is independent of the path by which this state has been reached. Properties whose values depend on the path followed are called path functions.

2. State variables/State functions – Temperature, Pressure, Enthalpy, Entropy

Question 5.
1. Explain the Zeroth law of thermodynamics.
2. What are the important modes of transference of energy. Explain.
Answer:
1. The Zeroth law of thermodynamics states that if two bodies say, ‘A’ and ‘B’ are in thermal equilibrium with another body say, ‘C’, then the bodies A’ and ‘B’will also be in thermal equilibrium with each other. It provides the basis for the measurement of temperature.

2. The two important modes of transference of energy are heat and work.
Heat:
The exchange of energy, which is a result of temperature difference between system and surroundings is called heat (q).

Work:
The exchange of energy between system and surroundings can occur in the form of work which can be mechanical work, electrical work or pressure-volume work. The exchange of energy as pressure-volume work can occur if system consists of gaseous substance and there is a difference of pressure between system and surroundings.

Question 6.
1. Explain the symbols and sign conventions of heat and work.
2. Explain internal energy.
Answer:
1. Heat is represented by the symbol ‘q’. The ‘q’ is positive, when heat is transferred from the surroundings to the system and ‘q’ is negative when heat is transferred from system to the surroundings.
Work is represented by the symbol ‘w’. The ‘w’ is positive when work is done on the system and ‘w’ is negative when work is done by the system.

2. Every substance is associated with a definite amount of energy due to its physical and chemical constitution. This is called internal energy. It is the sum of the different types of energies such as chemical, electrical, mechanical etc.

Question 7.
Fill in the blanks.

1. If heat is released, ‘q’ is ……………
2. For exothermic process ‘∆H’is ………………
3. If work is done on the system, ‘w’ is ………………
4. For endothermic process ‘∆H’ is ………………
5. If work is done by the system, ‘w’ is ………………

Answer:

1. Negative
2. Negative
3. Positive
4. Positive
5. Negative

Question 8.
1. What is meant by enthalpy of formation?
2. What is the value of standard enthalpy of formation (∆_fH) of an element?
Answer:
1. Enthalpy of formation is the enthalpy change when one mole of a compound is formed from its elements in their most stable states of aggregation (i.e., reference state).

2. Zeno

Question 9.
First Law of thermodynamics is the law of conservation of energy.

1. Give the mathematical form of the first law.
2. Write the Gibb’s equation.
3. What is the sign for ∆G for a spontaneous process?

Answer:

1. ∆U=q+w w = work done
   q = heat absorbed
2. G = H- TS or ∆G= ∆H – T∆S
3.  In the case of spontaneous process ∆G = -ve

Question 10.
1. Predict the sign of ∆S for the reaction,
NH₃(g) + HCl(g) → NH₄Cl(s)
2. The reaction between gaseous hydrogen and chlorine is
H₂(g) + Cl₂(g) → 2HCl(g); ∆_rH = -1840 kJ
i) What is the enthalpy of formation of HCl?
ii) How much heat will be liberated at 298 K and 1 atm for the formation of 365 g of HCl?
Answer:
1. ∆S is negative

2. i) ∆_fH = \(\frac{-1840}{2}\) = -920 kJ mol^-1
ii) Heat liberated during the formation of 1 mole (36.5 g) of HCl = 920 kJ
∴ Heat liberated during the formation of 365 g of HCl = 9200 kJ

Question 11.
Derive the Meyer’s relationship.
Answer:
We have, q = C × ∆T
At constant volume, q_v = C_v × ∆T = ∆U
At constant pressure, q_p = C_p × ∆T = ∆H
For 1 mole of an ideal gas,
∆H = ∆U + ∆(pV) = ∆U + ∆(RT) = ∆U + R∆T
∴ ∆H = ∆U + R∆T
On putting the values of ∆H and ∆U,
C_p∆T = C_v∆T + R∆T
C_p = C_v + R
C_p – C_v = R, which is the Meyer’s relationship.

Question 12.
1. In a process 701J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?
2. What is free expansion? What is the work done during free expansion of an ideal gas?
Answer:
1. ∆U=q+w = 9 + w = 701J – 394 J = 307J

2. Expansion of a gas in vacuum is called free expansion. No work is done during free expansion of an ideal gas whether the process is reversible or irreversible.

Question 13.
1. Name the instrument used for measuring the ∆U of a process.
2. What is the value of ∆G for a reaction at equilibrium?
3. ∆H and ∆S of a reaction are 30.56 and 0.666 kJ/ mol respectively at 1 atm pressure. Calculate the temperature at which the reaction is in equilibrium.
Answer:
1. Bomb calorimeter
2. Zeno
3. ∆H-T∆S = 0 or ∆H = T∆S

Question 14.
Thermodynamic process differ based on the manner
in which it is carried out.
1. Distinguish between reversible and irreversible processes.
2. Calculate the amount of work done when 2 moles of a gas expands from a volume of 2 L to 6 L isothermally and irreversibly against a constant external pressure of 1 atm.
Answer:
1.

2. w = -p∆V = -1 × (6 – 2)
= – 4 L atm

Question 15.
1. What are thermochemical equations?
2. Give an example for a thermochemical equation.
Answer:
1. A balanced chem ical equation together with the value of its ∆_rH is called a thermochemical equation.
2.

Question 16.
1. Define lattice enthalpy of an ionic compound.
2. What is Born-Haber cycle?
Answer:
1. The lattice enthalpy of an ionic compound is the enthalpy change which occurs when one mole of an ionic compound dissociates into its ions in gaseous state.

2. It is a simplified method developed by Max Born and Fritz Haberto correlate lattice enthanpies of ionic compounds to otherthermodynamic data.

Question 17.
Predict what happens to entropy in the following changes:

1. Metal is converted into alloy.
2. Solute crystallizes from solution.
3. Hydrogen molecule dissociates.

Answer:

1. The entropy will increase.
2. The entropy will decrease.
3. The entropy will increase.

Question 18.
1. Give the relation between change in enthalpy and change in free energy.
2. Name the above relation.
3. What is the significance of the above relation?
Answer:
1. ∆G = ∆H – T∆S

2. Gibbs equation orGibbs-Helmholtz equation.

3. This relation is used to predict the spontaneity of a process based on the value of ∆G . If ∆G is negatve, the process is spontaneous. If ∆G is positive, the process is non-spontaneous.

Question 19.
1. Predict in each of the following whether entropy increases or decreases.
i) Sublimation of camphor
ii) 4Fe(s) + 3O₂(g) → 2Fe₂O₃(g)
2. The equilibrium constant for a reaction at 30 °C
2.5 x 10^-29. What will be the value of ∆G?
Answer:
1. i) entropy increases
ii) entropy increases
2.

Question 20.
1. Explain the effect of temperature on the spontaneity of a process based on Gibbs equation.
2. For a reaction 2A(g) + B(g) → 2D(g), enthalpy and entropy changes are – 20.5 kJ mol^-1 and – 50.4 J K^-1mol^-1 respectively. Predict whether the reaction occurs at 25 °C.
Answer:
1. If ∆H is -ve and ∆S is +ve, ∆G would certainly be -ve and the process will be spontaneous at all temperatures.
If both ∆H and ∆S are – ve ∆G would be -ve if ∆H > T∆S
If both ∆H and ∆S are + ve ∆G would be -ve if T∆S > ∆H
If ∆H is +ve and AS is -ve, ∆G would certainly be +ve and the process will be non-spontaneous at all temperatures.

2. According to Gibbs equation, ∆G = ∆H – T∆S
∆G = (-20.5 × 10³)-(298 ×-50.4)
= – 20500 + 15019.2 = – 5480.8 J mol^-1
Since ∆G is -ve, the process is spontaneous.

Plus One Chemistry Thermodynamics Four Mark Questions and Answers

Question 1.
1. Explain the first, second and third laws of thermodynamics.
2. What do you meant by entropy?
3. Explain the spontaneous process.
Answer:
1. First law:
Energy can neither be created nor be destroyed. Energy in one form can be converted into another form without any loss or gain.

Second law:
Entropy of the universe increases during a spontaneous process.

Third law:
Entropy of a perfect crystalline substance is zero at absolute zero of temperature.

2. Entropy:
Entropy is a measure of randomness or disorder of a system.

3. Spontaneous process:
A spontaneous process is defined as an irreversible process which has a natural tendency to occur either of its own or after proper initiation under the given set of conditions.

Question 2.
U₁, q, w, U₂ are given. U₁ is internal energy, q is absorbed heat, w is work done and U₂ is final energy.
a) Derive an equation for ∆U.
b) Give the equation for w.
c) Calculate the change in internal energy of a system which absorbs 200 J of heat and 315 J of work is done by the system.
Answer:
a) U₂ = U₁ + q +w
U₂ – U₁ = q + w
or ∆U = q + w
b) w= -p∆V
c) q = 200J
w = -315J
∆U = ?
∆U = q + w
= 200 + {315}
= 200-315 = -115J

Question 3.
a) Predict whether entropy increases or decreases in the following changes:
i) l₂(s) → l₂(g)
ii) Temperature of a crystalline solid is raised from 0 Kand 115 K.
iii) Freezing of water
b) Calculate the enthalpy of combustion of methane. Given that standard enthalpies of formation of CH₄, CO₂ and H₂O are -75.2, -394 and -285.6 kJ/mol respectively.
Answer:
a) i) Entropy increases
ii) Entropy increases
iii) Entropy decreases
b) The required equation is,

Plus One Chemistry Thermodynamics NCERT Questions and Answers

Question 1.
In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process? (2)
Answer:
Heat absorbed by the system, (q) = + 701 J
Work done by the system (w) = – 304 J
Change in internal energy (∆U) = q + w
= 701 – 394
= 307 J

Question 2.
The reaction of cyanamide, NH₂CN (s) with oxygen was carried out in a bomb calorimeter and AU was found to be – 742.7 kJ mol1 at 298 K. Calculate the enthalpy change for the reaction at 298 K. (2)
Answer:
NH₂CN(S) + 3/2O₂(g) → N₂(g) + CO₂(g) + H₂O(l)
∆U = 742.7 kJ mol^-1;
∆n(g) =2 – 3/2= + 0.5
R = 8.314 × 10^-3kJ K^-1 mol^-1;
T = 298K
According to the relation, ∆H = ∆U + ∆n_gRT
∆H = -742.7 kJ + 0.5 mol × 8.314 × 10^-3kJ K^-1 mol^-1 × 298 K
=-742.7 kJ + 1.239kJ
=-741.5 kJ

Question 3.
Calculate the number of kJ of heat necessary to raise the temperature of 60 g of aluminium from 35 °C to 55 °C. Molar heat capacity of Al is 24 J mol^-1 K^-1. (2)
Answer:
Moles of Al (n) = \(\frac { 60{ g } }{ 27{ g }{ mol }^{ -1 } } \) = 2.22 mol
Molar heat capacity (Cm) = 24 J mol^-1 K^-1
∆T = 55 °C – 35 C° = 20C° or 20 K
Now, q = C_m × n × ∆T
= 24.0 J mol^-1 K^-1 × 2.22 mol × 20 K
= 1065.6 J
= 1.067 kJ

Question 4.
The enthalpy of formation of CO(g), CO₂ (g), N₂O (g), N₂O₄ (g) are -110, – 393, 81 and 9.7 kJ mol^-1 respectively. Find the value of ∆_rH for the reaction:
N₂O₄ (g) + 3CO(g) → N₂O(g) + 3CO₂(g)
Answer:

Question 5.
The equilibrium constant for the reaction is 10. Calculate the value of ∆G ; Given R = 8 J K^-1 mol^-1; T = 300 K.
Answer:

Question 6
Calculate the entropy change in surroundings when 1.0 mol of HzO (I) is formed under standard conditions. Given ∆_fH = – 286 kJ mol^-1.
Answer:

Question 7.
Comment on the thermodynamic stability of NO(g) and NO₂ (g) given :

Answer:

Students can Download Chapter 5 States of Matter Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Plus One Chemistry States of Matter One Mark Questions and Answers

Question 1.
When a gas is compressed at constant temperature
a) The speeds of the molecules increase
b) The collisions between the molecules increase
c) The speed of the molecules decrease
d) The collisions between the molecules decrease
Answer:
b) The collisions between the molecules increase

Question 2.
The temperature at which a real gas obeys ideal gas law over an appreciable range of pressure is called ___________
Answer:
Boyle temperature or Boyle point

Question 3.
The compressibility factor is given by the expression

Answer:
a) \(\frac{p V}{n R T}\)

Question 4.
Two flasks of equal volume contain CO₂ and SO₂ respectively at 298 K and 1.5 atm pressure. Which of the following is equal in them?
a) Masses of the two gases
b) Rates of effusion
c) Number of molecules
d) Molecular structures
Answer:
c) Number of molecules

Question 5.
With rise in temperature, viscosity of a liquid
a) Increases
b) Decreases
c) Remains constant
d) May increase or decrease
Answer:
b) Decreases

Question 6.
The unit of‘b’ in VanderWaals equation of state.
Answer:
l mol^-1

Question 7.
Most probable velocity, average velocity, and root mean square velocity are related by
Answer:
1 : 1.128 : 1.224

Question 8.
The volume of 2.8g of CO at 27°C and 0.821 atm pressure is (R = 0.0821 l atm Km^-1 ol^-1)
Answer:
3L

Question 9.
The density of gas at 27°C and 1 atm is d. Pressure remaining constant at which of the following temp will its density become 0.75d?
Answer:
400K

Question 10.
The rms velocity of an ideal gas at 27°C is 0.3ms^-1. ‘ Its rms velocity at 927°C in (ms^-1) is
Answer:
0.6m/s

Plus One Chemistry States of Matter Two Mark Questions and Answers

Question 1.
Find out the relation between the first pair and complete the second pair.
a) Boyle’s law: Temperature
Charles’ law : ……………….
b) Avagadro’slaw: V α n
Ideal gas equation: ……………..
Answer:
a) Pressure
b) pV=nRT

Question 2.
The graphs of Boyle’s law as plotted by Student 1 (Graph 1) and Student 2 (Graph 2) are given below:

1. Which is the correct graph?
2. Justify your answer.

Answer:

1. Both the graphs are correct.
2. According to Boyle’s law, v α 1/p or p α 1/v. This is clear from graph 1. Also, according to Boyle’s law, pv is a constant at constant n and T. This is clear from graph 2.

Question 3.
The rate of diffusion of hydrogen is less than that of oxygen.

1. Do you agree?
2. Which law is applied here?
3. State the law.

Answer:

1. Yes.
2. Graham’s law of diffusion.
3. Rate of diffusion of a gas is inversely proportional to the square root of its molecular mass.

Question 4.
The ideal gas equation has been modified for real gases by applying pressure and volume corrections.

1. What is the corrected equation known as?
2. Write the equation and explain the terms.

Answer:

1. The van derWaals’equation.
2. \(\left(p+\frac{a n^{2}}{V^{2}}\right)\) (V — nb) = nRT
   where p – pressure, V – volume, n – no. of moles of the gas, a & b – van der Waals’ constants, R – universal gas constant and T – absolute temperature.

Question 5.
‘Moist soil grains are pulled together.’

1. Name the related phenomenon.
2. Justify.

Answer:

1. Surface tension.
2. This is because the surface area of thin film of water in moist soil is reduced due to surface tension.

Question 6.
1. What is aqueous tension?
2. What is its significance in the determination of pressure of a dry gas?
Answer:
1. The pressure exerted by saturated water vapour at a given temperature is called aqueous tension at that temperature.

2. Pressure of dry gas can be calculated by subtracting aqueous tension from the total presssure of the moist gas.
P_{dry gas}=P_Total-Aqueous tensi0n

Question 7.
A balloon filled with air, when kept in sunlight bursts after some time.

1. Name the related law.
2. Justify.

Answer:

1. Charles’ law
2. According to Charles’ law, volume a Temperature. Therefore, the volume increases when temperature increases, When the volume of the gas inside the baloon expanded more than that the balloon could afford, it bursted.

Question 8.
Define surface energy. What is its SI unit?
Answer:
Surface energy is defined as the energy required to rise the surface area of the liquid by one unit. The SI unit of surface energy is J m^-2.

Question 9.
a) Based on Boyle’s law how will you show that at a constant temperature, pressure is directly proportional to the density of a fixed mass of the gas?
b) Give the relation between density and molar mass of a gaseous substance.
Answer:
a) According to Boyle’s law,

Question 10.
The isotherm of carbon dioxide at various temperatures is given below:

1. What is the significance of the shaded area?
2. Identify the pressure at which liquid CO₂ appears for the first time when cooled form 30.98 °C. What is this pressure called?
Answer:

1. At any point in the dome shaped shaded area liquid and gaseous CO₂ exists in equilibrium.
2. 73 atm. Critical pressure (p_c).

Question 11.
Certain properties of liquids are given below: Classify them on the basis of effect of temperature on them,

1. Evaporation
2. Vapour pressure
3. Surface tension
4. Viscosity

Answer:
Properties which increase with increase in temperature: Evaporation & Vapour pressure Properties which decrease with increase in temperature: Surface tension & Viscosity

Question 12.
The size of the water bubbles increases on moving to the surface.
1. Name the law responsible for this.
2. What is your justification?
Answer:
1. Boyle’s law.

2. According to Boyle’s law volume is inversely proportional to pressure. At the bottom of the pond, pressure is greater. So the volume (size) of the bubble was the least. But on coming up, pressure decreases and hence size of the bubble increases.

Question 13.
What are the properties of liquid state?
Answer:
Vapour pressure, Boiling point, Viscosity and Surface tension.

Plus One Chemistry States of Matter Three Mark Questions and Answers

Question 1
Analyse the following graph :

1. Name the gas law associated with this graph.
2. State the law.
3. Give the mathematical expression of this law.
Answer:
1. Boyle’s law.

2. It states that at constant temperature, pressure of a fixed amount of gas varies inversly with its volume.

3. Mathematically, Boyle’s law can be written as p ∝ \(\frac{1}{V}\) (at constant T and n)
Or p = k × \(\frac{1}{V}\) where k is the proportionality
constant which depends upon the amount of the gas, temperature of the gas and the units in which p and V are expressed.

Question 2.
1. What are van der Waals’ forces?
2. Which are the different types of van der Waals’ forces?
3. Arrange the van der Waal’s forces in the increasing order of their strength.
Answer:
1. The attractive intermolecular forces are known as van der Waals’ forces.

2. Dispersion forces/London forces, Dipole-Dipole forces, Dipole-Induced dipole forces and Hydrogen bonding.

3. Dispersion forces/London forces < Dipole-Induced dipole forces < Dipole-Dipole forces < Hydrogen bonding

Question 3.
A graphical representation of Charles’ law is given below:

1. What is the temperature corresponding to the point‘A’called? .
2. What will be the temperature at that point A’ in degree Celsius?
3. What is the significance of this temperature?

Answer:

1. Absolute zero temperature
2. -273.15 °C
3. Absolute zero is the lowest hypothetical or imaginary temperature at which gases are supposed to occupy zero volume.

Question 4.
Assume that two gases X and Y at the same temperature and pressure have the same volume.
1. Which of the following is correct?
No. of moles of X= No.of moles of Y
No. of moles of X ≠ No.of moles of Y
2. Which law helped you to find the answer?
3. State the law.
Answer:
1. No. of moles of X= No.of moles of Y

2. Avogadro’s law

3. Equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules.

Question 5.
During a seminar session in the class, the presenter argued that equal amounts of both H2and N2 on heating at constant pressure will expand in the same rate. Another student objected this argument by saying that they will expand differently since their molecular masses are different.

1. Who is correct in your opinion?
2. Which law helped you to reach the answer?
3. State the law and give its mathematical expression.

Answer:
1. The argument of the presenter is correct.

2. Charles’ law

3. Charles’ law states that pressure remaining constant, the volume of a fixed mass of a gas is directly proportional to its absolute temperature.

Mathematically, V ∝ T (at constant n and P) Or \(\frac{V}{T}\) = K, where K is the proportionality constant which depends on the pressure of the gas, its amount and the unit in which V is expressed.

Question 6.
1. What is an ideal gas?
2. Give the ideal gas equation and explain the terms.
3. Derive the ideal gas equation.
Answer:
1. A gas that follows Boyle’s law, Charles’ law and Avogadro law strictly at all conditions is called an ideal gas.

2. The ideal gas equation is pV = nRT where p = pressure, V = volume, n = number of moles, R = universal gas constant and T = absolute temperature.

3. According to Boyle’s law:
V ∝ \(\frac{1}{p}\) at constant T and n.
According to Charles’ law:
V ∝ T. at constant n and p.
According to Avogadro law,
V ∝ n , at constant p and T Combining the above three equations,
V ∝ \(\frac{nT}{p}\)
⇒ V = R\(\frac{nT}{p}\), where R is the proportionality constant known as universal gas constant.
Or pV = nRT, the ideal gas equation.

Question 7.
Partial pressure of a vessel containing Cl₂, CO₂ and CO is the sum of the partial pressures of Cl₂, O₂ and CO.
1. If so, is it correct to say partial pressure of a vessel containing NH₃ and HCl gases is the sum of their partial
pressures? Justify.
2. Which law helped you to answer this?
3. State the law.
Answer:
1. No. NH₃ reacts with HCl to form NH₄CI. Since they are not non-interacting gases, their sum of partial pressures may not be equal to the total pressure.

2. Dalton’s law of partial pressures.

3. It states that the total pressure exerted by the mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases.

Question 8.
The average kinetic kenergy of the gas molecules is directly proportional to the absolute temperature.

1. Which theory is related to this assumption?
2. Write the other postulates of this theory.

Answer:
1. Kinetic moleculartheory of gases.

2.

• The volume of a gas molecule is negligible when compared to the whole volume of the gas.
• There is no force of attraction between the particles of a gas at ordinary temperature and pressure.
• The gas molecules are in random motion.
• During motion, they collide with each other and also with the walls of the container.
• Gravity has no influence in the movement of gas molecules.
• Pressure of a gas is due to the collision of gas molecules with the walls of the container.
• At any particular time, different particles in the gas have different speeds and hence different kinetic energies.

Question 9.
Three mateorological baloons filled with equal amount of helium, rising in the atmosphere are shown below: (Assume that temperature remains constant).

1. Which of the baloons will be at the lowest altitude?
2. Which law helped you to find the answer?
3. State the law.

Answer:

1. C –
2. Boyle’s law.
3. At low altitudes pressure is high. According to Boyle’s law for a given mass of a gas, greater the pressure lower is the volume at constant temperature.

Question 10.
‘All the postulates of the kinetic molecular theory of gases are correct.’

1. Do you agree with the statement?
2. If no, write the wrong postulates of this theory.
3. Give justification.

Answer:
1. No.

2.

• Volume of the molecules of a gas is negligibly small in comparison to the space occupied by the gas.
•  There is no force of attraction between the molecules of a gas.

3. If assumption i) is correct, the p vs V graph of experimental data (real gas) and that theoritically calculated from Boyle’s law (ideal gas) should coincide. But this never happens.
If assumption ii) is correct, the gas will never liquify. But gases do liquify when cooled and compressed.

Question 11.
Two gases with equal molecular mass will have the same rate of diffusion.’

1. Do you agree?
2. Explain.
3. Substantiate your answer with an example.

Answer:
1. Yes.

2. According to Graham’s law of diffusion the rate of diffusion depends only on the molecular mass. So if the molecular masses are the same, their rate of diffusion is same.

3. Both CO and N₂ have the same molecular mass (28 g mol^-1)
Rate of diffusion of CO = Rate of diffusion of N₂

Question 12.
Water can be boiled more quickly on the top of a mountain.

1. Do you agree?
2. What is the reason?
3. What is called boiling point of a liquid?
4. How normal boiling point and standard boiling point differ?

Answer:
1. Yes.

2. As we move to the top of a mountain atmospheric pressure decreases and hence boiling point decreases. So water boils quickly.

3. Boiling point of a liquid is the temperature at which the vapour pressure of a liquid is equal to the external pressure or atmospheric pressure.

4. The boiling point at 1 atm pressure is called normal boiling point. The boiling point at 1 bar pressure is called standard boiling point.

Question 13.
Ethanol flows faster than honey.

1. Name the related phenomenon.
2. Explain this phenomenon.
3. What is the effect of temperature on this?

Answer:
1. Viscosity.

2. Viscosity is a measure of resistance to flow which arises due to the internal friction between layers of fluid as they slip past one another while liquid flows.

3. Viscosity of liquids decreases as the temperature rises because at high temperature molecules have high kinetic energy and can overcome the intermolecular forces to slip past one another between the layers.

Question 14.
Liquid drops attain spherical shape.

1. Which property of liquids is responsible for this?
2. Explain the phenomenon and justify.
3. Suggest another consequence of this phenomenon.

Answer:
1. Surface tension.

2. Surface tension is defined as the force acting per unit length perpendicular to the line drawn on the surface of liquid. The lowest energy state of the liquid will be when surface area is minimum. Spherical shape satisfies this condition.

3. Fire polishing of glass – On heating, the glass melts and the surface of the liquid tends to take the rounded shape at the edges due to surface tension, which makes the edges smooth.

Question 15.
Vapour pressure is an important property of liquids.

1. What is vapour pressure?
2. How boiling point and vapour pressure are related?
3. Pressure cooker is used for cooking food at higher altitudes. Give reason.

Answer:
1. Vapour pressure of a liquid is the pressure exerted by the vapour which is in equilibrium with liquid at a given temperature.

2. Boiling point of a liquid is the temperature at which the vapour pressure of liquid becomes equal to the atmospheric pressure. Thus, lower the vapour pressure of a liquid higher will be its boiling point and vice-versa.

3. At high altitudes atmospheric pressure is low. Therefore, liquids at high altitudes boil at lower temperatures in comparison to that at sea level. In a pressure cooker, the internal pressure is greater than atmospheric pressure. Hence, in a pressure cooker water boils at a temperature higher than its normal boiling point of 100 °C. Thus, cooking becomes more effective.

Question 16.
Assume that ‘A’, ‘B’ and ‘C’ are three non-reacting gases kept in a vessel at a constant temperature.
Then, P_Total= P_A + P_B + P_C
1. Name the related law.
2. How can you explain the above law on the basis of kinetic molecular theory of gases?
Answer:
1. Dalton’s law of partial pressures.

2. In the absence of attractive forces, the particles of the gas behave independent of one another. The same is true even if there are more than one type of molecules. Thus, the number of molecules colloding the unit area of the wall per second at a given temperature, fora fixed amount of the gas issame.lt implies that the partial pressure of the gas will be unaffected by the presence of the molecules of other gases. But, the total pressure exerted is duet the impact of molecules of all the gases. Hence, the total pressure would be the sum of the partial pressures of the gases.

Question 17.
1. Write the general equation which relates the different variables of a gas used to describe the state of any ideal gas.
2. A flask at 295 K contains a gaseous mixture of N₂ and O₂ at a total pressure of 1.8 atm. If 0.2 moels of N₂ and 0.6 moles of O₂ are present, find the partial pressures of N₂ and O₂.
3. What is meant by Boyle temperature or Boyle point?
Answer:
1. PV = nRT
2.

3. It is the temperature at which a real gas obeys ideal gas law over an appreciable range of pressure.

Question 18.
1. Liquid tries to rise or fall in the capillary. Name the related phenomenon.
2. What is the effect of temperature on the above phenomenon?
3. WhatistheSI unit of the above phenomenon.
Answer:
1. Surface tension.

2. The magnitude of surface tension of a liquid depends on the attractive forces between the molecules. When the attractive forces are large, the surface tension is large. Increase in temperature increases the kinetic energy of the molecules and effectiveness of intermolecular attraction decreases, so surface tension decreases as the temperature is raised.

3. N m^-1.

Question 19.
1. Define critical temperature (T_c ).
2. CO₂ cannot be liquified above 31.1°C. Why?
3. The critical temperatures of ammonia and carbon dioxide are 405.5 Kand 304.10 K respectively. On cooling, which of these gases will liquify first? Justify.
Answer:
1. It is the highest temperature at which a gas can
be liquified by applying external pressure.

2. The critical temperature (T_c ) of CO₂ is 30.98°C. This is the highest temperature at which liquid CO₂ is observed. Above this temperature it is gas.

3. Ammonia. This is because, on cooling, critical temperature of ammonia will be reached first. Liquefaction of carbon dioxide will require more cooling.

Question 20.
a) Will water boils at higher temperature at sea level or at top of a mountain. Explain.
b) A vessel of 120 mL capacity contains a certain amount of gas at 35 °C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35 °C. What would be its pressure?
Answer:
a) When atmospheric pressure decreases boiling point of the liquid also decreases. So the boiling point of water at sea level is not same as that at the top of a mountain. Atmospheric pressure decreases from sea level as we go high. Hence, the boiling point at the top of the mountain is less than that at sea level.

Question 21.
Real gases deviate from ideal behaviour.
1. What are the two wrong postulates of kinetic theory of gases, responsible for deviation of real gases from ideal behaviour?.
2. When do real gases deviate from ideal behaviour?
Answer:
1.

• Volume of the molecules of a gas is negligibly small in comparison to the space occupied by the gas.
• There is no force of attraction between the molecules of a gas.

2. Real gases deviate from ideal behaviour at high pressure and low temperature, when the gas molecules are very close to each other.

Question 22.
1. What is meant by compressibility factor, Z?
2. What is the significance of compressibility factor?
3. A plot of pV/nRT of oxygen gas against p is as follows:

a) Is the gas ideal or real?
b) Write the equation of state of the above gas.
Answer:
1. Compressibility factor (Z) is the ratio of product pV and nRT. Mathematically, Z = \(z=\frac{p V}{n R T}\).

2. The deviation of real gases from ideal behaviour can be measured in terms of compressibility factor.

3. a) Real gas.
\(\left[p+\frac{a n^{2}}{V^{2}}\right]\)[V-nb] = nRT (van der Waals’ equation)

Question 23.
a) What is the difference between gas and vapour?
b) Analyse the vapour pressure vs temperature curve shown below:

Arrange the compounds shown in the graph in the decreasing order of their normal boiling points,

c) The density of a gas was found to be 2.92 g L1 at 27 °C and 2.0 atm. Calculate the molar mass of the gas.
Answer:
a) A gas below its critical temperature can be liquified by applying pressure. Under these conditions, it is called vapour of the substance.

b) water > ethyl alcohol > carbon tetrachloride > diethyl ether

Question 24.
1. How will you account for the observation that automobile tyre is inflated with lesser air in summer than in winter?
2. A sample of gas occupies 250 mLat27 °C. What volume will it occupy at 35 °C if there is no change in pressure?
Answer:
1. This can be explained on the basis of Gay Lussac’s law, according to which at constant volume, pressure of a fixed amount of a gas varies directly with the temperature. In summer season the temperature will be higher. Hence, pressure will increase and the tyre may burst if filled with more air. But during winter temperature is low and hence pressure will below.

2. According to Charles’ law,

Question 25.
Real gases behave ideally at low temperature and high pressure.

1. Is the above statement correct or not?
2. Justify.
3. Write the van der Waals’ equation for 1 mole of a real gas.

Answer:
1. The statement is wrong.

2. This is because real gases behave ideally at high temperature and low pressure, when the gas molecules are far apart.

3. \(\left(p+\frac{a}{V^{2}}\right)\)(V – b) = RT

Question 26.
1. Distinguish between real gas and ideal gas.
2. Explain the deviation of the following gases from ideal behaviouron the basis of the pV vs. p plot. CO, CH₄, H₂ and He.
Answer:
1. Real gas do not follow, Boyle’s law, Charles law, and Avagadro law perfectly under all conditions. Ideal gas follow, Boyle’s law, Charles’ Law and Avagadro law strictly under all conditions.

2.

It can be seen that at constant temperature pV vs p plots for these gases are not straight lines. Two types of curves are seen. In the curves for H₂ and He, as the pressure increases the value of pV also increases. These gases show positive deviation from ideal behaviour at all pressures. The second type of plot is seen in the case of CO and CH₄. For these gases the pV value decreases with increase in pressure and reaches to a minimum value characteristics of the gas. After that PV value starts increasing. The curve then crosses the line for ideal gas and after that shows positive deviation continuously.

Question 27.
1. What is meant by laminar flow?
2. Derive the expression for the force responsible for flow of layers of a liquid.
Answer:
1. When a liquid flows overa fixed surface, the layer of molecules in the immediate contact of surface is stationary. The velocity of upper layers increases as the distance of layers from the fixed layer increases. This type of flow in which there is a regular gradation of velocity in passing from one layer to the next is called laminar flow.

2. Consider three layers of a flowing liquid as shown below:

For any layer, the layer above it accelerates its flow and the layer below this retards its flow. If the velocity of the layer at a distance dz is changed by a value du then velocity gradient is given by \(\frac{du}{dz}\). A force is required to maintain the
flow of layers. This force is proportional to the area of contact (A) of layers and velocity gradient.

where η is a proportionality constant called coefficient of viscosity.

Question 28.
1. What are London forces?
2. What is the relation between London forces and the distance between the particles?
Answer:
1. The attractive force between two temporary dipoles is known as London forces or Dispersion forces.

2. London forces are always attractive and the interaction energy is inversely proportional to the sixth power of the distance between two interacting particles.

Question 29.
1. What is meant by thermal energy and thermal motion?
2. Can oxygen exist as a gas at -273.15°C? Write the significance of this temperature.
Answer:
1. Thermal energy is the energy of a body arising from motion of its atoms or molecules. The movement of particles due to thermal energy is called thermal motion.

2. At – 273.15 °C oxygen will not exist as a gas. In fact all the gases get liquified before this temperature is reached. It is the absolute zero of temperature, which is the lowest hypothetical or imaginary temperature at which gases are supposed to occupy zero volume.

Question 30.
Molecues of a gas are in a state of continuous motion
1. What is most probable speed?
2. Give the equation for average speed of molecules.
Answer:
1. Most probable speed is the speed possessed by the maximum fraction of molecules of the gas at a given temperature.

2. If there are ‘n’ molecules in a sample and their individual speeds are u₁, u₂ ……… u_n, then, average speed of molecules, u_av is given by the equation:
\(u_{a v}=\frac{u_{1}+u_{2}+\ldots+u_{n}}{n}\)

Plus One Chemistry States of Matter Four Mark Questions and Answers

Question 1.
1. State the Avogadro law.
2. Give the mathematical expression of this law.
3. What is the value of molar volume of an ideal gas at 273.15 K and 1 bar?
4. Show that, at constant temperature and pressure, the density of an ideal gas is proportional to its molar volume.
Answer:
1. It states that equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules.

2. v ∝ n (at constant P and T)
⇒ V = k × n where k is a proportionality constant.

3. 22.71098 L

4. According to Avogadro law, for n moles of an ideal gas,
V = k × n
But n = \(\frac{m}{M}\) where m is the mass of the gas and M is its molar mass.
Thus, V = k × \(\frac{m}{M}\)
On rearranging the above equation,
M = k × \(\frac{m}{V}\) = k × d
⇒ M ∝ d

Question 2.
The speed of molecules is a measure of their average kinetic energy.
a) What is root mean square speed?
b) Give the equation for root mean square speed.
c) Calculate the following:
i) Root mean square speed of methane molecule at 27°C.
ii) Most probable speed of nitrogen molecule at 25°C
Answer:
a) It is the square root of the mean of the squares of speeds of various molecules of the gas at a given temperature.

Question 3.
The graph A is drawn at high temperature and low pressure and graph B is drawn at low temperature and high pressure.

a) Which graph represents ideal behaviour?
b) Give the equation for combined gas law.
c) A baloon occupies volume of 700 mL at 25°C and 760 mm of pressure. What will be its volume at higher attitude when temperature is 15°C and pressure is 600 mm Hg.
Answer:

Question 4.
1. Give the relationship among the three types of molecular speeds.
2. Drawthe Maxwell-Boltzmann distribution showing all the molecularspeeds.
3. Which of the following molecules will have the higher value of most probable speed at the same temperature, N₂ or Cl₂? Justify.
Answer:
1. Root mean square speed, average speed and the most probable speed have the following relationship:
U_rms > U_av > U_mp
2.

3. N₂. This is because at the same temperature, gas molecules with heavier mass have slower speed than lighter gas molecules. At the same temperature, lighter nitrogen molecules move faster than heavier chlorine molecules. Hence, at any given temperature, nitrogen molecules have higher value of most probable speed than the chlorine molecules.

Plus One Chemistry States of Matter NCERT Questions and Answers

Question 1.
What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 at 30 °C? (2)
Answer:
p₁ = 1 bar p₂ = ?
V₁ = 500 dm³ V₂ = 200 dm³
Temperature remains constant.
According to Boyle’s law,

Question 2.
Using the equation of state pV=nRT show that at a given temperature, the density of the gas is proportional to the gas pressure p. (2)
Answer:
According to ideal gas equation :

Question 3.
The density of a gas is found to be 5.46 g/dm³ at 27°C and under 2 bar pressure. What will be its density at STP. (3)
Answer:

Question 4.
Calculate the volume occupied by 8.8 g of CO₂ at 31.1°C and 1 bar pressure (R = 0.083 bar L K^-1 mol^-1). (2)
Answer:

Question 5.
Explain the significance of van der Waal parameters. (2)
Answer:
The van der Waal parameter ‘a’ is a measure of the magnitude of intermolecular forces. The van der Waal parameter ‘b’ which is also called co-volume is a measure of effectve size of the gas molecules.

Students can Download Chapter 5 Introduction to C++ Programming Questions and Answers, Plus One 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 One Computer Science Chapter Wise Questions and Answers Chapter 5 Introduction to C++ Programming

Plus One Introduction to C++ Programming One Mark Questions and Answers

Question 1.
IDE means _____________
Answer:
Integrated Development Environment

Question 2.
We know that C++ is a high level language. From the following which statement is true.
(a) C++ contains English like statements.
(b) C++ contains mnemonics
(c) C++ contains only 0 and 1
(d) None of these
Answer:
(a) C++ contains English like statements.

Question 3.
C++ is a ______ language.
(a) High level
(b) Low level
(c) Middle level
(d) None of these
Answer:
(a) High level

Question 4.
C++ was developed at ___________
(a) AT & T Bell Laboratory
(b) Sanjose Laboratory
(c) Kansas University Lab
(d) None of these
Answer:
(a) AT & T Bell Laboratory

Question 5.
C++ is a successor of ___________ language
(a) C#
(b) C
(c) java
(d) None of these
Answer:
(b) C

Question 6.
The most adopted and popular approach to write programs is __________
Answer:
Structured programming

Question 7.
From the following which uses OOP concept
(a) C
(b) C++
(c) Pascal
(d) Fortran
Answer:
(b) C++

Question 8.
______________ is the smallest individual unit
Answer:
Token

Question 9
Pick the odd one out
(a) float
(b) void
(c) break
(d) Alvis
Answer:
(d) Alvis, the others are keywords.

Question 10.
Reserved words for the compiler is ____________
(a) Literals
(b) Identifier
(c) Keywords
(d) None of these
Answer:
(c) Keywords

Question 11.
Pick an identifier from the following
(а) auto
(b) age
(c) float
(d) double
Answer:
(b) age

Question 12.
Pick the invalid identifier
(a) name
(b) Date of birth
(c) age
(d) joining time
Answer:
(b) Date of birth, because it contains space.

Question 13.
Pick the octal integer from the following
(a) 217
(b) 0 X 217
(c) 0217
(d) None of these
Answer:
(c) 0217, an octal integer precedes 0

Question 14.
Pick the hexadecimal integer from the following
(a) 217
(b) 0 × 217
(c) 0217
(d) None of these
Answer:
(b) 0 × 217, a hexadecimal integer precedes 0×

Question 15.
From the following pick a character constant
(a) ‘A’
(b) ‘ALL’
(c) ‘AIM’
(d) None of these
Answer:
(a) ‘A’, a character enclosed between single quote

Question 16.
Non graphic symbol can be represented by using ___________
Answer:
Escape Sequence

Question 17.
Manish wants to write a program to produce a beep sound. Which escape sequence is used to get an alert (sound)
(a) \a
(b) \d
(c) Vs
(d) None of these
Answer:
(a) \a.

Question 18.
Ajo wants to print a matter in a new line. Which escape sequence is used for this?
(a) \a
(b) \n
(c) \s
(d) None of these
Answer:
(b) \n

Question 19.
To represent null character is used ______
(a) \n
(b) \0
(c) \f
(d) As
Answer:
(b) \0

Question 20.
State True/ False a string is automatically appended by a null character.
Answer:
True

Question 21.
From the following pick a string constant
(a) ‘a’
(b) “abc”
(c) ‘abc’
(d) None of these
Answer:
(b) “abc”, a character constant must be enclosed between double quotes.

Question 22.
C++ was developed by __________
(a) Bjarne Stroustrup
(b) James Gosling
(c) Pascal
(d) None of these
Answer:
(a) Bjarne stroustrup

Question 23.
From the following which is not a character constant.
(а) ‘c’
(b) ‘e’
(c) ‘d’
(d) “c”
Answer:
(d) “c”, It is a string constant the others are character constant.

Question 24.
From the following which is a valid declaration.
(a) int 91;
(b) int x;
(c) int 9x;
(d) int “x”;
Answer:
(b) int x;

Question 25.
Symbols used to perform an operation is called ____________
(a) Operand
(b) Operator
(c) Variable
(d) None of these
Answer:
(b) Operator

Question 26.
Consider the following
C = A + B. Here A and B are called
(a) Operand
(b) Operator
(c) Variable
(d) None of these,
Answer:
(b) Operand

Question 27.
The execution of a program starts at ________ function
Answer:
main()

Question 28.
The execution of a program ends with ________ function
Answer:
main()

Question 29.
______ is used to write single line comment
(a) //
(b) /*
(c) */
(d) None of these
Answer:
(a) //

Question 30.
const k = 100 means
(a) const float k = 100
(b) const double k = 100
(c) const int k = 100
(d) const char k = 100
Answer:
(c) const int k = 100

Question 31.
Each and every statement in C++ must be end with ________
(а) Semicolon
(b) Colon
(c) full stop
(d) None of these
Answer:
(a) Semicolon

Question 32.
From the following select the input operator
(а) >>
(b) <<
(c) >
(d) <
Answer:
(a) >>

Question 33.
From the following select the output operator
(a) >>
(b) <<
(c) >
(d) <
Answer:
(b) <<

Question 34.
From the following which is known as a string terminator.
(а) ‘\0’
(b) ‘\a’
(c) ‘As’
(d) ‘\t’
Answer:
(a) ‘\0’

Question 35.
Adeline wrote a C++ program namely sum.cpp and she compiled the program successfully with no error. Some files are generated. From the following which file is a must to run the program
(a) sum.exe
(b) sum.obj
(c) sum.vbp
(d) sum.htm
Answer:
(a) sum.exe

Question 36.
Adeline wrote a C++ program namely sum.cpp and she compiled the program successfully with no error. Some files are generated namely sum.obj and sum.exe. From this which file is not needed to run the program
Answer:
sum.obj is not needed and can be deleted.

Question 37.
From the following which is ignored by the compiler
(a) statement
(b) comments
(c) loops
(d) None of these
Answer:
(b) comments

Question 38.
To write a C++ program, from the following which statement is a must
(a) sum()
(b) main()
(c) #include<iostream>
(d) #include<iomanip>
Answer:
(b) main(). A C++ program must contains at least one main() function

Question 39.
State True / False
Comment statements are ignored by the compiler
Answer:
True

Question 40.
More than one input/output operator in a single statement is called _______
Answer:
Cascading of I/O operator

Question 41.
Is 0X85B a valid integer constant in C++? If yes why?
Answer:
Yes. It is a hexadecimal number.

Plus One Introduction to C++ Programming Two Mark Questions and Answers

Question 1.
Mr. Dixon declared a variable as follows
int 9age. Is it a valid identifier. If not briefly explain. the rules for naming an identifier.
Answer:
It is not a valid identifier because it violates the rule
The rules for naming an identifier is as follows:

1. It must be start with a letter(alphabet)
2. Under score can be considered as a letter
3. White spaces and special characters cannot be used.
4. Key words cannot be considered as an identifier

Question 2.
How many bytes used to store ‘\a’.
Answer:
To store ‘\a’ one byte is used because it is an escape sequence. An escape sequence is treated as one character. To store one character one byte is used.

Question 3.
How many bytes used to store “\abc”.
Answer:
A string is automatically appended by a null character.

• Here one byte for \a (escape sequence).
• One byte for character b.
• One byte for character c.
• And one byte for null character.
• So a total of 4 bytes needed to store this string.

Question 4.
How many bytes used to store “abc”.
Answer:
A string is automatically appended by a null character.

• Here one byte for a.
• One byte for character b.
• One byte for character c.
• And one byte for null character.
• So a total of 4 bytes needed to store this string.

Question 5.
Consider the following code
int main()
{
cout<<“welcome to C++”;
}
After you compile this program there is an error called prototype error. Why it is happened? Explain.
Answer:
Here we used the output operator cout<<. It is used to display a message “welcome to C++” to use this operator the corresponding header file <iostream> must be included. We didn’t included the header file hence the error.

Question 6.
In C++ the size of the string “book” is 5 and that of “book\n” is 6. Check the validity of the above statement. Justify your answer.
Answer:
A string is automatically added by a null character(\0). The null character is treated as one character. So the size of string “book” is 5. Similarly, a null character (\0) is also added to “book\n”. \n and \0 is treated as single characters. Hence the size of the string “book\n” is 6.

Question 7.
Pick the odd man out. Justify
TOTSAL, TOT_SAL, totsal5, Tot5_sal, SALTOT, tot.sal
Answer:
tot.sal. Because it contains a special character dot(.). An identifier cannot contain a special character. So it is not an identifier. The remaining satisfies the rules of naming identifier. So they are valid identifiers.

Question 8.
Write a C++ statement to print the following sentence. Justify.
“\ is a special character”
answer:
cout<<“\\ is a special character”
\\ is treated as an escape sequence.

Question 9.
A student type a C++ program and saves it in his personal folder as Sample.cpp. After getting the output of the program, he checks the folder and finds three files namely Sample.cpp, Sample.obj and Sample.exe. Write the reasons for the generation of the two files in the folder.
Answer:
After the compilation of the program sample.cpp, the operating system creates two files if there is no error. The files are one object file (sample.obj) and one executable file(sample.exe). Now the source file(sample.cpp) and object file(sample.obj) are not needed and can be deleted. To run the program sample.exe is only needed.

Question 10.
Mention the purpose of tokens in C++. Write names of any four tokens in C++. (2)
Answer:
Token: It is the smallest individual units similar to a word in English or Malayalam language. C++ has 5 tokens.

1. Keywords
2. Identifier
3. Literals (Constants)
4. Punctuators
5. Operators

Question 11.
The following are some invalid identifiers. Specify its reason.

1. Sum of digits
2. 1 year
3. First jan
4. For

Answer:

1. Sum of digits → space not allowed hence it is invalid
2. 1 year → First character must be an alphabet hence it is invalid
3. First.jan → special characters such as dot (.) not allowed hence it is invalid.
4. For → It is valid. That is it is not the keyword for

Question 12.
Some of the literals in C++ are given below. How do they differ?(5, ‘5’, 5.0, “5”)
Answer:

• 5 – integer literal
• ‘5’ – Character literal
• 5.0 – floating point literal
• “5”- string literal

Question 13.
Identify the invalid literals from the following and write reason for each:

1. 2E3.5
2. “9”
3. ‘hello’
4. 55450 (2)

Answer:
1. 2E3.5 → The mantissa part (3.5) will not be a floating point number. Hence it is invalid

3. ‘hello’ → It is a string hence it must be enclosed in double quotes instead of single quotes. It is invalid.

Question 14.
Which one of the following is a user defined name?
(a) Keyword
(b) Identifier
(c) Escape sequences
(d) All of these
Answer:
(b) Identifier

Question 15.
Identify whether the following are valid identifiers or not? If not give the reason.

1. Break
2. Simple.interest (2)

Answer:

1. Break – It is valid( break is the keyword, not Break);
2. Simple.interest – It is not valid, because dot(.) is used.

Question 16.
Identify the invalid literals from the following and write a reason for each:

1. 2E3.5
2. “9”
3. ‘hello’
4. 55450 (2)

Answer:
1. Invalid, because exponent part should not be a floating point number

2. valid

Plus One Introduction to C++ Programming Three Mark Questions and Answers

Question 1.
Rose wants to print as follows
\n is used for New Line. Write down the C++ statement for the same.
Answer:
#include<iostream>
using namespace std;
int main()
{
cout<<“\\n is used for New Line”;
}

Question 2.
Alvis wants to give some space using escape sequence as follows
Welcome to C++. Write down the C++ statement for the same
Answer:
#include<iostream>
using namespace std;
int main()
{
cout<<“Welcome to \t C++”;
}

Question 3.
We know that the value of pi = 3.14157, a constant (literal). What is a. constant? Explain it.
Answer:
A constant or a literal is a data item its value doe not change during execution.
1. Integer literals:
Whole numbers without fractional parts are known as integer literals, its value does not change during execution. There are 3 types decimal, octal and hexadecimal.
Eg. For decimal “100, 150, etc

• For octal 0100, 0240, etc
• For hexadecimal 0 × 100, 0x1 A, etc

2. Float literals:
A number with fractional parts and its value does not change during execution is called floating point literals.
eg: 3.14157,79.78, etc

3. Character literal:
A valid C++ character enclosed in single

Question 4.
Write a program to print the message “TOBACCO CAUSES CANCER” on screen.
Answer:
#include<iostream>
using namespace std;
int main()
{
cout<<” TOBACCO CAUSES CANCER”;
}

Question 5.
You are supplied with a list of tokens in C++ program, Classify and Categorise them under proper headings.
Explain each category with its features. tot_mark, age, M5, break, (), int, _pay, ;, cin.
Answer:

Question 6.
Write a program to print the message “SMOKING IS INJURIOUS TO HEALTH” on screen.
Answer:
#include<iostream>
using namespace std;
int mainO
{
cout<<” SMOKING IS INJURIOUS TO HEALTH”;
}

Plus One Introduction to C++ Programming Five Mark Questions and Answers

Question 1.
Consider the following code
The new line character is \n. The output of the following code does not contain the \n. Why it is happened? Explain.
Answer:
\n is a character constant and it is also known as escape sequence. This is used to represent the non graphic symbols such as carriage return key(enter key), tab key, backspace, space bar, etc. It consists of a backslash symbol and one more characters.

Question 2.
You are about to study the fundamentals of C++ programming Language. Do a comparative study of the basics of the new language with that of a formal language like English or Malayalam to familiarize C++? Provide sufficient explanations for the compared items in C++ Language.
Answer:
1. Character set:
To study a language first we have to familiarize the character set. For example, to study English language first we have to study the alphabets. Similarly here the character set includes letters(A to Z & a to z), digits(0 to 9), special characters(+, -, ?, *, /, …..) white spaces(non printable), etc.

2. Token:
It is the smallest individual units similar to a word in English or Malayalam language. C++ has 5 tokens

• Keywords: These are reserved words for the compiler. We can’t use for any other purposes.
  Eg: float is used to declare variable to store numbers with decimal point. We can’t use this for any other purpose
• Identifier: These are user defined words. Eg: variable name, function name, class name, object name, etc…
• Literals (Constants): Its value does not change during execution
  eg: In maths % = 3.14157 and boiling point of water is 100.
• Punctuators: In English or Malayalam language punctuation mark are used to increase the readability but here it is used to separate the tokens.
  eg:{,}, (,), ……..
• Operators: These are symbols used to perform an operation(Arithmetic, relational, logical, etc…).

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

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure

Plus One Chemistry Chemical Bonding and Molecular Structure One Mark Questions and Answers

Question 1.
The octet rule is not valid for
a) CO₂
b)H₂O
c) O₂
d) CO
Answer:
d) CO

Question 2.
The stability of an ionic crystal depends principally on
a) High electron gain enthalpy of the anion forming species
b) The lattice enthalpy of the crystal
c) Low ionization enthalpy of the cation forming species
d) Low heat of sublimation of cation forming solid
Answer:
b) The lattice enthalpy of the crystal

Question 3.
Which of the following molecules has highest dipole moment?
a) H₂S
b)CO₂
c) CCl₄
d) BF₃
Answer:
a) H₂S

Question 4.
The d-orbital involved in sp3d hybridization is .
Answer:
d_z²

Question 5.
Which of the following is paramagnetic and has a bond order of ½?

Answer:
H₂⁺

Question 6.
Dipole moment µ electric charge ‘e’ and bond length ‘d’are related by the equation.
Answer:
M = e x d

Question 7.
In which of the following carbon atom is sp2 hybridised?
a) CO₂
b) C₂H₆
c) C₆H₆
d) HCN
e) \(C{ H }_{ 3 }-C\equiv CH\)
Answer:
C₆H₆

Question 8.
AgF is ionic whereas Agcl is covalent. This can be explained by
Answer:
Faja’ens Rule

Question 9.
The shape of covalent molecule CIF₃ is _________
Answer:
T – shaped

Question 10.
The C – O bond order in CO₃^2- is
Answer:
1.33

Plus One Chemistry Chemical Bonding and Molecular Structure Two Mark Questions and Answers

Question 1.
The order of repulsion of electron pairs as written by student is given below:
lone pair-lone pair repulsion < lone pair-bond pair repulsion>bond pair-bond pair repulsion.
1. Can you see anything wrong in this?
If yes, correct it.
2. Name the theory behind this.
Answer:
1. Yes.
Repulsion decreases in the order: lone pair-lone pair repulsion>lone pair-bond pair repulsion> bond pair-bond pair repulsion,

2. VSEPR theory

Question 2.
During a small group discussion in the class room a student argued that in acetylene both the carbon atoms are in sp3 hybridised state.

1. What is your opinion?
2. What is the bond angle between carbon atoms in acetylene?

Answer:

1. The student’s argument is wrong. In acetylene both the carbon atoms are triple bonded and are in sp hybridised state,
2. 180°

Question 3.
Classify the following compounds according to their hybridisation.
CH₄, BF₃, C₂H₄, BeF₂, C₂H₂
Answer:
sp³ hybridisation – CH₄
sp² hybridisation – BF₃, C₂H₄
sp hybridisation – BeF₂, C₂H₂

Question 4.
A student arranged the halide ions in the increasing order of polarisability as: F < l < CI < Br
1. Is this the correct order? If not write it in correct order.
2. Justify.
Answer:
1. No.
Polarisability increases in the order: F^– < Cl^– <Br^– < l^–

2. Polarisability increases when the size of anion increases.

Question 5.
Give any two differences between sigma and pi bonds.
Answer:
Sigma bond (σ) is formed when two atomic orbitals under head-on overlapping. It is a strong bond. Pi(π) bond is formed when atomic orbitals undergo lateral (sidewise) overlapping. It is a weak bond.

Question 6.
Write the type of hybridisation of each carbon in the compound.
CH₃-CH=CH-CN
Answer:
Carbon 1 → sp³
Carbon 2 → sp²
Carbon 3 → sp²
Carbon 4 → sp

Plus One Chemistry Chemical Bonding and Molecular Structure Three Mark Questions and Answers

Question 1.

1. What is meant by the above picture?
2. Which type of bond is present here?
3. Which type of overlapping leads to the formation of π bond?

Answer:

1. s-s overlapping
2. A strong sigma bond
3. This type of covalent bond is formed by the lateral or sidewise overlap of half-filled atomic orbitals.

Question 2.
‘In ethane there are 6 covalent bonds. Five are strong σ bonds and the remaining one is a weak π bond.’

1. Do you agree with this?
2. How is a bond different from π bond in the mode of formation?

Answer:

1. Yes.
2. Sigma bond is formed by the end to end overlapping of bonding orbitals along the internuclear axis, π bond is formed by the lateral or sidewise overlap of half filled atomic orbitals.

Question 3.
Choose the correct molecules from the given clues: H₂O, SF₆, BF₃
1. Clue -1 The central atom is in sp² hybridised state and the molecule has trigonal planar in shape. Clue-2 The bond angle is 120°
2. Clue-1 The number of electron pairs in this molecule is 6.
Clue -2 It has octahedral geometry.
3. Clue-1 The bond angle is reduced from 109° 28′ to 104.5°
Clue-2 It has a bent shape.
Answer:

1. BF₃
2. SF₆
3. H₂O

Question 4.
Give theoretical explanation for the following statements.
1. H₂S is acidic while H₂O is neutral.
2. Hydrogen chloride gas dissolves in water.
Answer:
1. S-H bond energy is less than that of O-H bond energy. So H⁺ can be easily generated from H₂S.

2. When HCl is treated with H₂O it undergoes hydrolysis as per the following reaction and dissolves.
HCl + H₂O → H₃O+ +Cl^–

Question 5.
The potential energy level diagram forthe formation of hydrogen molecule as drawn by a student is given below:

1. Resketch the graph with correct labelling.
2. How can you determine the radius of one hydrogen atom?
Answer:
1.

2. Bond length in hydrogen molecule is the intermolecular distance between two hydrogen atoms. So, half of the bond length is taken as the radius of hydrogen atom.

Question 6.
Ionisation enthalpy is one of the factors favoring the
formation of ionic bonds.
1. Will you agree with this statement?
2. Explain how?
3. Write anotherfactorfavouring the formation of ionic bonds.
Answer:
1. Yes.

2. In the formation of the ionic bond, a metal atom losses electrons to form cation. This process requires energy equal to the ionisation enthalpy. Lesser the ionisation enthalpy of the metal atom, easier will be the removal of electron from the atom to form cation and hence greater will be the tendency to form ionic bond.

3. Electron gain enthalpy of the element forming anion.

Question 7.
Complete the following table:

Answer:

Question 8.
a) The dipole moment of BF₃ is zero even though the B – F bonds are polar. Justify.
b) Give the hybridisation involved in the following compounds

1. NH₃
2. C₂H₄
3. SF₆
4. PCl₅

c) o-nitro phenol has a lower boiling point than its para isomer. Why?
Answer:
a) In BF₃, dut to the symmetric trigonal planar geometry of the molecule, the B – F bond are oriented at an angle of 120°to one another. The three bond moments give a net sum of zero as the resultant of any two is equal and opposite to the third.

b)

1. sp³
2. sp²
3. sp³d²
4. sp³d

c) In o-nitrophenol, intramolecular hydrogen bonding is present and there is no association of molecules whereas in p-nitrophenol there is inter-molecular hydrogen bonding which causes association of molecules.

Question 9.
1. How many a and n bonds are there in the following molecules i) ethane ii) acetylene?
2. BF₃ and NH₃ are tetra atomic molecules. But the shape of BF₃ is different from that of NH₃. Explain this using hybridisation.
Answer:
1. i) Ethane – 7σ bonds
ii) Acetylene-3σ bonds and 2 π bonds

2. In BF₃ molecule, the ground state electronic configuration of central boron atom is 1s²2s²2p¹. In the excited state, one of the 2s electrons is promoted to vacant 2p orbital. As a result, boron has three unpaired electrons. These three orbitals (one 2s and two 2p) hybridise to form three sp2 hybrid orbitals oriented in a trigonal planar arrangement and overlap with 2 p orbitals of F to form three B – F bonds. Therefore, BF₃ molecule has a planar geometry with FBF bond angle of 120°.

In NH₃, the valence shell electronic configuration of N in ground state is 2s² \(2p_{ x }^{ 1 }2{ p }_{ y }^{ 1 }2{ p }_{ z }^{ 1 }\). These four orbitals undergo sp³ hybridisation to form four sp³ hybrid orbitals, three of them containing unpaired electrons and the fourth one containing lone pair. The three hybrid orbitals overlap with 1 s orbitals of hydrogen atoms to form three N – H sigma bonds. Since, the bp-lp repulsion is greater than the bp-bp repulsion, the molecule gets distorted and the bond angle is reduced to 107° from 109.5°. Thus, the geometry of NH₃ molecule is trigonal pyramidal.

Question 10.
Covalent bond is formed by the overlaping of atomic orbitals.
1. What is meant by orbital overlapping?
2. What are the 3 types of overlapping?
Answer:
1. Orbital overlapping is the partial interpenetration or merging of atomic orbitals. It results in the pairing of electrons. Greater the overlap the stronger is the bond formed between two atoms.

2. s-s overlapping
s-p overlapping
p-p overlapping

Question 11.
1. Which among the following will exist He₂ or He₂⁺? Explain.
2. H₂S is a gas at ordinary condition, while H₂O is liquid. Account for the above statement.
3. State the hybridisation in the following molecules,
i) PF₆
ii) C₂H₆
Answer:
1. He₂⁺
Helium molecule contains 4 electrons. Out of this 4 electrons, 2 are present in the bonding molecular orbital and the remaining 2 are present in the anti-bonding molecular orbital.
Bond order = ½ (N_b-N_a)
= ½ (2-2) = 0
Hence, He₂ cannot exist. The molecular orbital diagram is given below:

He₂⁺ contains 3 electrons. Out of these 3 electrons, 2 are present in the σ_1s level and the remaining one is present in the σ* 1s level.
Bond order = ½ (N_b-N_a)
= ½ (2-1) = ½
Since the bond order is half the molecular ion exists but possesses low stability.

2. In H₂S, there is no hydrogen bonding whereas in water, molecular association is possible due to intermolecular hydrogen bonding.

3. i) PF₅ = sp³d hybridisation
ii) C₂H₆ = sp³ hybridisation

Question 12.
Bond order is a term commonly used in MO theory.
1. How is it calculated?
2. How is it related to bond length and bond energy?
Answer:
1. Bond order = ½ (N_b-N_a)
where N_b = No. of electrons occupying bonding orbitals and N_a = No. of electrons occupying antibonding orbitals.

2. As the bond order increases, bond length decreases and bond energy increases, i.e., bond order is directly proportional to bond energy and inversely proportional to bond length.

Question 13.
1. Explain the hybridisation and geometry of ethyne.
2. What is the difference between bonding molecular orbital and antibonding molecular orbital?
3. How the magnetic nature of a molecule is related to its electronic structure?
Answer:
1. In the formation of ethyne (C₂H₂), both the carbon atoms undergo sp hybridisation having two unhybridised orbitals (2p_x and 2p_y). One sp hybrid orbital of one carbon atom overlaps axially with sp hybrid orbital of the other carbon atom to form C-C sigma bond, while the other hybridised orbital of each carbon atom overlaps axially with the half filled s orbital of hydrogen atoms forming o bonds. Each of the two unbybridised p orbitals of both the carbon atoms overlaps sidewise to form two K bonds between the carbon atoms. Thus, ethyne has a linear geometry with π bond angle of 180°.

2. The molecular orbital which has lower energy than the atomic orbital is called bonding molecular orbital and the molecular orbital which has greater energy than the atomic orbital is called anti bonding molecular orbital.

3. If all the molecular orbitals in a molecule are doubly occupied (i.e., paired), the substance is diamagnetic (repelled by magnetic field). It one or more molecular orbitals are singly occupied (i.e., unpaired) it is paramagnetic (attracted by magnetic field).

Question 14.
Molecular Orbital Theory (MOT) is an advanced theory
of chemical bonding.
1. Write the salient features of MOT.
2. What is meant by LCAO? Illustrate using hydrogen molecule.
3. What are the conditions for the combination of atomic orbitals?
Answer:
1.

• The electrons in a molecule are present in the
  various molecular orbitals.
• The atomic orbitals of comparable energies and proper symmetry combine to form molecular orbitals.
• iii) The electron in a molecular orbital is influenced by two or more nuclei depending upon the number of atoms in the molecule.
• The number of molecular orbitals formed is equal to the number of combining atomic orbitals. When two atomic orbitals combine, two
  molecular orbitals are formed. One is known as bonding molecular orbital while the other is called antibonding molecular orbital.
• The bonding molecular orbital has lower energy and hence greater stability than the corresponding antibonding molecular orbital.
• The electron probability distribution around a group of nuclei in a molecule is given by a molecular orbital.
• The molecular orbitals are filled in accordance with the Aufbau principle obeying the Pauli’s exclusion principle and the Hund’s rule.

2. LCAO refers to the linear combination of atomic orbitals. It is an approximate method used to explain the formation of molecular obritals. Consider hydrogen molecule consisting of two atoms A and B. Each hydrogen atom has one electron in the 1s orbital. The atomic orbitals of these atoms can be represented by the wave functions ψ_A and ψ_B. Mathematically, the formation of molecular orbitals can take place by addition and by subtraction of wave functions of individual atomic orbitals.
ψ_MO = ψ_A ± ψ_B
Therefore, the two molecular orbitals σ and σ* are formed as:
σ* = ψ_A – ψ_B
The molecular orbital σ formed by the addition of atomic orbitals is called the bonding molecular orbital while the molecular orbital σ* formed by the subtraction of atomic orbital is called antibonding molecular orbital. The energy level diagram is shown below:

3.

• The combining atomic orbitals must have the same or nearly same energy.
• The combining atomic orbitals must have the same symmetry about the molecular axis.
• The combining atomic orbitals must overlap to the maximum extent.

Question 15.
Consider a reaction PCl₅(g) → PC₃(g) + Cl₂(g)
1. What is the change in hybridisation state of phosphorus?
2. Explain why does PCl₅ decomposes easily?
Answer:
1. When PCl₅ decomposes to PCl₃, the hybridisation of P changes from sp³d to sp³.

2. In PCl₅, the five sp³d orbitals of P overlap with the singly occupied p orbitals of Cl atoms to form five P-CI sigma bonds. Three P-Cl bonds which lie in one plane and make an angle of 120° with each other are called equatorial bonds. The remaining two P – Cl bonds, called axial bonds, one lie above and the other lie below the equatorial plane, make an angle of 90° with the plane. As the axial bond pairs suffer more repulsive interaction from the equatorial bond pairs, axial bonds are slightly longer and hence slightly weaker than the equatorial bonds. This makes PCl₅ molecule more reactive and hence it decomposes easily.

Question 16.
The electron dot structure (Lewis structure) of ammonia molecule is shown below:

1. Write the number of bond pairs of electrons and lone pairs of electrons in ammonia molecule.
2. The structures of o-nitrophenol and p-nitrophenol are shown in the figure. The former is a steam volatile liquid whereas the latter is a solid. Justify your answer giving reason.

Answer:
1. Ammonia molecule contains one lone pair of electrons and 3 bond pair of electrons,

2. In o-nitrophenol, there is intramolecular hydrogen bonding and there is no molecular association. But in p-nitro phenol intermolecular hydrogen bonding is present and hence molecular association is possible.

Plus One Chemistry Chemical Bonding and Molecular Structure Four Mark Questions and Answers

Question 1.
Hydrogen bonding is present in NH₃ and H₂O.
1. What is hydrogen bond?
2. What are different types of hydrogen bonds?
3. Explain the effect of hydrogen bonding.
Answer:
1. Hydrogen bond is defined as the attractive force which binds hydrogen atom of one molecule with the electronegative atom (F, O, N) of the same or another molecule.

2. Intermolecular hydrogen bond and Intramolecular hydrogen bond.

3. Compounds containing hydrogen bonds show higher melting and boiling points. Compounds whose molecules can form hydrogen bonds with water molecules are soluble in water.

Question 2.
Classify the following compounds according to their
shape.
BeF₂, BeCl₂, CH₄, BF₃, PCl₅, SF₆, SbCl₅ NH₄⁺, SiF₄, AlCl₃.
Answer:
Linear – BeF₂, BeCl₂
Trigonal planar-AlCl₃, BF₃
Tetrahedral – CH₄, NH₄⁺, SiF₄
T rigonal bipyramidal – PCl₅, SbCl₅

Question 3.
Benzene is an example of a compound exhibiting resonance.
1. What is meant by resonance?
2. Explain the resonance of ozone.
Answer:
1. When a molecule cannot be represented by a single structure but its characteristic properties can be described by two or more different structures, then the actual molecule is said to be a resonance hybrid of these canonical structures.

2. The resonance in ozone can be represented by the following structures:

According to structures I and II, there is one single bond and one double bond in ozone molecule. But experiments show that both the oxygen- oxygen bonds are equal and the bond length 128 pm) is intermediate between single (148 pm) and double bond (121 pm) lengths. Hence it is assumed that ozone is a resonance hybrid (structure III) of structures I and II.

Question 4.
Match the following:

Answer:

Question 5.
In the formation of methane, carbon undergoes sp³ hybridisation.

1. What do you mean by sp³ hybridisation?
2. Give the % s-character and p-character of an sp³ hybrid orbital.
3. What is the bond angle in methane?
4. What is the geometry of methane molecule?

Answer:

1. sp³ hybridisation involves mixing up of one – s and three-p orbitals of the valence shell of an atom to form four sp³ hybrid orbitals of equivalent energies and shape.
2. Each sp³ hybrid orbital has 25% s-character and 75% p-character.
3. The angle between the sp3 hybrid orbitals in methane is 109°28′.
4. Tetrahedron.

Question 6.
Complete the following table:

Answer:

Question 7.
Fill in the blanks:

Answer:

Question 8.
Dipole moment is used to predict the shape of
molecules.
1. Justify the statement based on the shapes of CO₂ and H₂O.
2. Which is having high dipole moment? NH₃ or NF₃? Why?
Answer:
1. Carbon dioxide is a linear molecule in which the two C=0 bonds are oriented in the opposite directions at an angle of 180°. Hence the two C=0 bond dipoles cancel each other and the resultant dipole moment of CO₂ is zero. Thus CO₂ is non¬polar molecule

On the other hand, water molecule has a bent structure in which two O-H bonds are oriented at an angle of 104.5°. Therefore, the bond dipoles of two O-H bonds do not cancel each other and the molecule will have a net dipole moment (1.85D).

2. The dipole moment of NH₃ is higher than that of NF₃. In both cases, the central N atom has a lone pair whose orbital dipole points away from the N atom. In NH₃ the orbital dipole due to the lone pair is in the same direction as the resultant bond dipole of the three N-H bonds. On the other hand, in the case of NF₃, the resultant dipole of the three N-F bonds is in the opposite direction to the orbital dipole due to the lone pair. Thus, the orbital dipole due to the lone pair decreases the effect of the resultant N-F bond moments, which results in the low dipole moment of NF₃.

Question 9.
The geometry of a covalent molecule is related to the hybridisation involved in the central atom. Complete the following table:

Answer:

Question 10
Depending upon the type of overlapping, covalent bonds are of two types.
a) Name them and give any two difference between them.
b) Find the total number of these two types of bonds ’ in propane and 2-butene.
Answer:
a) Sigma (σ) bond and pi (π) bond.
Sigma (σ) bond:
This type of covalent bond is formed by the end to end overlapping of half-filled atomic orbitals along the internuclear axis. The overlap is also known as head on overlap or axial overlap. The electrons constituting sigma bond are called sigma electrons.

Pi (π) bond:
This type of covalent bond is formed by the lateral or sidewise overlap of half-filled atomic orbitals. The atomic orbitals overlap in such a way that their axes remain parallel to each other and perpendicularto the internuclear axis.
b) 10 σ bond in propane
11 σ bond and 1π bond in 2-butene

Plus One Chemistry Chemical Bonding and Molecular Structure NCERT Questions and Answers

Question 1.
Explain the formation of a chemical bond. (2)
Answer:
According to Kossel-Lewis approach, the formation of chemical bond between the two atoms takes place either by the transference of electrons or by mutual sharing of electrons. However, according to the modem view the formation of chemical bond between the two approaching atoms occurs only if there is a net decrease of energy because of attractive and repulsive forces.

Question 2.
Write the favourable conditions for the formation of ionic bond. (2)
Answer:
Ionic bond is formed by transference of electrons from one atom to another. The favourable conditions for its formation are:

• Low ionisation enthalpy of element forming cation.
• More negative value of electron gain enthalpy of element forming the anion and
• High value of lattice enthalpy of the compound formed.

Question 3.
Although geometries of NH₃ and H₂O molecules are distorted tetrahedral, bond angle in water is less than that in ammonia. Discuss. (2)
Answer:
The difference in bond angles is due to the different numbers of lone pairs and bond pairs in the two species. In NH₃, the N atom has two lone pairs and three bond pairs while in H₂O, the O atom has two lone pairs and two bond pairs. The repulsive interactions of lone pairs and bond pairs in water are relatively more than those in NH₃. Hence, bond angle around central atom in water is relatively smaller (104.5°) than that in NH₃ molecule (107°).

Question 4.
Is there any change in the hybridisation of B and N atoms as a result of the following reaction? (2)
BF₃ + NH₃ → F₃B.NH₃
Answer:
During combination of species BH₃ and NH₃, N atom of NH₃ is donor and B atom of BF₃ is acceptor. The hybrid state of B in BF₃ is sp² and that of N in NH₃ is sp³. In the compound F₃B⁺-NH₃^– both N and B atoms are surrounded by four bond pairs. Thus, the hyrid state of both is sp³. Hence during the reaction the hybrid state of B changes from sp² to sp³ but that of N remains the same.

Question 5.
Define hydrogen bond. Is it weaker or stronger than the van der Waals’ forces? (2)
Answer:
Hydrogen bond can be defined as the attractive force which binds hydrogen atom of one molecule with the electronegative atom (F, O or N) of another molecule. Hydrogen bond is stronger than van der Waals’forces because it is a strong dipole-dipole interaction. The van der Waals’ forces, on the other hand, are weak dispersion forces.

Students can Download Chapter 2 Data Representation and Boolean Algebra Questions and Answers, Plus One 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 One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra

Plus One Data Representation and Boolean Algebra One Mark Questions and Answers

Question 1.
___________ is a collection of unorganized fact.
Answer:
Data

Question 2.
Data can be organized into useful ____________
Answer:
Information

Question 3.
___________ is used to help people to make decision.
Answer:
Information

Question 4.
Processing is a series of actions or operations that convert inputs into __________
Answer:
Output

Question 5.
The act of applying information in a particular context or situation is called ____________
Answer:
Knowledge

Question 6.
What do you mean by data processing?
Answer:
Data processing is defined as a series of actions or operations that converts data into useful information.

Question 7.
Odd man out and justify your answer.
(a) Adeline
(b) 12
(3) 17
(d) Adeline aged 17 years is in class 12.
Answer:
(d) Adeline aged 17 years is in class 12. This is information. The others are data.

Question 8.
Raw facts and figures are known as _______
Answer:
data

Question 9.
Processed data is known as _______
Answer:
Information

Question 10.
Which of the following helps us to take decisions?
(a) data
(b) information
(c) Knowledge
(d) intelligence
Answer:
(b) information

Question 11.
Manipulation of data to get information is known as ___________
Answer:
Data processing

Question 12.
Arrange the following in proper order
Process, Output, Storage, Distribution, Data Capture, Input.
Answer:

1. Data Capture
2. Input
3. Storage
4. Process
5. Output
6. Distribution

Question 13.
Pick the odd one out and give reason:
(a) Calculation
(b) Storage
(c) Comparison
(d) Categorization
Answer:
(b) Storage. It is one of the data processing stage the others are various operations in the stage Process

Question 14.
Information may act as data. State true or False.
Answer:
False

Question 15.
Complete the Series.

1. (101)₂, (111)₂, (1001)₂, ……….
2. (1011)₂, (1110)₂, (10001)₂, ………

Answer:

1. 1011, 1101
2. 10101, 10111

Question 16.
What are the two basic types of data which are stored and processed by computers?
Answer:
Characters and number

Question 17.
The number of numerals or symbols used in a number system is its _______________
Answer:
Base

Question 18.
The base of decimal number system is ________
Answer:
10

Question 19.
MSD is ________
Answer:
Most Significant Digit

Question 20.
LSD is _________
Answer:
Least Significant Digit

Question 21.
Consider the number 627. Its MSD is _________
Answer:
6

Question 22.
Consider the number 23.87. Its LSD is __________
Answer:
7

Question 23.
The base of Binary number system is ___________
Answer:
2

Question 24.
What are the symbols used in Binary number system?
Answer:
0 and 1

Question 25.
Complete the following series,
(101)₂, (111)₂, (1001)₂
Answer:
1011, 1101

Question 26.
State True or False. In Binary, the unit bit changes either from 0 to 1 or 1 to 0 with each count.
Answer:
True

Question 27.
The base of octal number system is ________
Answer:
8

Question 28.
Consider the octal number given below and fill in the blanks.
0, 1, 2, 3, 4, 5, 6, 7, __
Answer:
10

Question 29.
The base of Hexadecimal number system is ________
Answer:
16

Question 30.
State True or False.
In Positional number system, each position has a weightage.
Answer:
True

Question 31.
In addition to digits what are the letters used in Hexadecimal number system.
Answer:
A(10), B(11), C(12), D(13), E(14), F(15)

Question 32.
Convert (1110.01011)₂ to decimal.
Answer:
1110.01011 = 1 × 23 + 1 × 22 + 1 × 21 + 0 × 20 + 0 × 2 – 1 + 1 × 2 – 2 + 0 × 2 – 3 + 1 × 2 – 4 + 1 × 2 – 5
= 8 + 4 + 2 + 0 + 0 + 0.25 + 0 + 0.0625 + 0.03125
= (14.34375)10

Question 33.
1 KB is bytes.
(a) 2⁵
(b) 2¹⁰
(c) 2¹⁵
(d) 2²⁰
Answer:
(b) 2¹⁰

Question 34.
The base of hexadecimal number system is ________.
Answer:
16

Question 35.
A computer has no _________
(a) Memory
(b) l/o device .
(c) CPU
(d) IQ
Answer:
(d) IQ

Question 36.
Pick the odd man out.
(AND, OR, NAND, NOT)
Answer:
NOT

Question 37.
Select the complement of X + YZ.
(a) \(\bar{x}+\bar{y}+\bar{z}\)
(b) \(\bar{x} .\bar{y}+\bar{z}\)
(c) \(\bar{x} \cdot(\bar{y}+\bar{z})\)
(d) \(\bar{x}+\bar{y} \cdot \bar{z}\)
Answer:
(c) \(\bar{x} \cdot(\bar{y}+\bar{z})\)

Question 38.
Select the expression for absorption law.
(a) a + a = a
(b) 1 + a = 1
(c) o . a = 0
(d) a + a . b = a
Answer:
(a) a + a . b = a

Question 39.
What is the characteristic of logical expression?
Answer:
Logical expressions yield either true or false values

Question 40.
Name the table used to define the results of Boolean operations.
Answer:
Truth Table

Question 41.
According to ________ law, \(\bar{x}+\bar{y}=\overline{x y}\) and \(\overline{x y}=\bar{x}+\bar{y}\)
Answer:
De Morgan’s law

Question 42.
A NOR gate is ON only when all its inputs are
(a) ON
(b) Positive
(c) High
(d) OFF
Answer:
(d) OFF

Question 43.
The only function of a NOT gate is __________
Answer:
Invert an output signal

Question 44.
NOT gate is also known as _________
Answer:
Inverter

Question 45.
What is the relation between the following statements.
x + 0 = x and x . 1 = x
Answer:
One is the dual of the other expression.

Question 46.
The algebra used to solve problems in digital systems is called __________
Answer:
Boolean Algebra

Question 47.
Pick the one which is not a Basic Gate.
(AND, OR, XOR, NOT)
Answer:
XOR

Question 48.
Select the universal gates from the list. (NAND, NOR, NOT, XOR)
Answer:
NAND, NOR

Question 49.
Which is the final stage in data processing?
Answer:
Distribution of information is the final stage in data processing

Question 50.
Fill up the missing digit.
(41)₈ = ( )₁₆
Answer:

• Step 1: Divide the number into one each and write down the 3 bits equivalent.
• Step 2: Then divide the number into group of 4 bits starting from the right then write its equivalent hexa decimal.

Answer:
So the answer is 21.

Question 51.
Real numbers can be represented in memory by using __________
Answer:
Exponent and Mantissa

Question 52.
Consider the number 0.53421 x 10^-8 Write down the mantissa and exponent.
Answer:
Mantissa: 0.53421
Exponent: -8

Question 53.
Characters can be represented in memory by using _________
Answer:
ASCII Code

Question 54.
ASCII Code of A’ is __________
Answer:
(100 0001)₂ = 65

Question 55.
ASCII Code of’a’ is __________
Answer:
(110 0001)₂ = 97

Question 56.
Define the term ‘bit’?
Answer:
A bit stands for Binary digit. That means either 0 or 1.

Question 57.
Find MSD in the decimal number 7854.25
Answer:
Because it has the most weight

Question 58.
ASCII stands for __________.
Answer:
American Standard Code for Information Interchange

Question 59.
List any two image file formats.
Answer:
BMP, GIF

Question 60.
Name the operator which performs logical multiplication.
Answer:
AND

Question 61.
Name a gate which is ON when all its inputs are OFF .
Answer:
NAND or NOR

Question 62.
Specify the laws applied in the following cases.

1. a (b + c) = ab + ac
2. (a + b) + c = a + (b + c)

Answer:

1. Distribution law
2. Associative law

Question 63.
Pick the correct Boolean expression from the following.
(a) \(A +\bar{A}=163.\)
(b) \(\text { A. } \bar{A}=1\)
(c) \(A \cdot \overline{A B}=A + B\)
(d) A + AB = A
Answer:
(a) & (d)

Question 64.
1’s complement of the binary number 110111 is _________
Answer:
Insert 2 zeroes in the left hand side to make the binary number in the 8 bit form 00110111
To find the 1’s complement, change all zeroes to one and all ones to zero. Hence the answer is 11001000

Plus One Data Representation and Boolean Algebra Two Mark Questions and Answers

Question 1.
Why do we store information?
Answer:
Normally large volume of data has to be given to the computer for processing so the data entry may be taken more days, hence we have to store the data. After processing these stored data, we will get information as a result that must be stored in the computer for future references.

Question 2.
What is source document.
Answer:
Acquiring the required data from all the sources for the data processing and by using this data design a document, that contains all relevant data in proper order and format. This document is called source document.

Question 3.
Briefly explain data, information and processing with real life example.
Answer:
Consider the process of making coffee. Here data is the ingredients – water, sugar, milk and coffee powder
Information is the final product i.e. Coffee Processing is the series of steps to convert the ingredients into final product, Coffee. That is mix the water,sugar and milk and boil it. Finally pour the coffee powder.

Question 4.
ASCII is used to represent characters in memory. Is it sufficient to represent all characters used in the written languages of the world? Propose a solution. Justify.
Answer:
No It is not sufficient to represent all characters used in the written languages of the world because it is a 7 bit code so it can represent 2⁷ = 128 possible codes. To represent all the characters Unicode is used because it uses 4 bytes, so it can represent 232 possible codes.

Question 5.
The numbers in column A have an equivalent number in another number system of column B.
Find the exact matvh

Answer:

Question 6.

1. Name various number systems commonly used in computers.
2. Include each of the following numbers into all possible number systems
   123 569 1101

Answer:

1. The number system are binary, octal, decimal and hexa decimal
2. All possible number systems are
   • 123 Octal, decimal and hexa decimal
   • 569 Decimal, hexa decimal
   • 1101 Binary, Octal, Decimal, Hexa decimal

Question 7.
Fill up the missing digit. (Score 2)
If (220)_a = (90)_b then (451)_a = (  )₁₀
Answer:
It contains 2 & 9, so a and b 2, b 8. The values of a can be 8 or 10. The values of b can be 10 or 16. L.H.S > R.H.S. a < b and a b also.
The possible values of a and b are given below.

Question 8.
Convert (106)₁₀ = (  )₂?
Answer:

Question 9.
Convert (106)₁₀ = (  )₈?
Answer:

Question 10.
(106)₁₀ = (  )₁₆?
Answer:

Question 11.
Convert (55.625)₁₀ = (  )₂?
Answer:
First convert 55, for this do the following

Write down the remainders from bottom to top.
(55)₁₀ = (110111 )₂
Next convert 0.625, for this do the following.

Write down the remainder from top to bottom. So the answer is
(55.625)₁₀ = (110111.101)₂

Question 12.
Convert (55.140625)₁₀ = (  )₈?
Answer:
First convert 55, for this do the following.

Write down the remainders from bottom to top.
(55)₁₀ = (67)₈
Next convert 0.140625, for this do the following.

Write down the remainders from top to bottom. So the answer is
(55.140625)₁₀ = (67.11 )₈

Question 13.
Convert (55.515625)₁₀ = (  )₁₆
Answer:
First convert 55, for this do the following.

Write down the remainders from bottom to top.
ie. (55)₁₀ = (37)₁₆
Next convert .515625

So the answer is
(55.515625)₁₀ = (37.84)₁₆

Question 14.
Convert (101.101)₂ = ( )₁₀?
Answer:
101.101 = 1 × 2² + 0 × 2¹ + 1 × 2⁰ + 1 × 2^-1 + 0 × 2^-2 + 1 × 2^-3 = 4 + 0 + 1 + 1/2 + 0 + 1/8 = 5 + 0.5 + 0.125
(101.101)₂ = (5.625)₁₀

Question 15.
Convert (71.24)₈ = (  )₁₀?
Answer:
71.24 = 7 × 8¹ + 1 × 8⁰ + 2 × 8^-1 + 4 × 8⁼²
= 56 + 1 + 2/8 + 4/82
= 57 + 0.25 + 0.0625 (71.24)8
(71.24)₈ = (57.3125)₁₀

Question 16.
Convert (AB.88)₁₆ = (  )₁₀?
Answer:

= 160 + 11 + 0.5 + 0.03125
(AB.88)₁₆ = (171.53125)₁₀

Question 17.
Convert (1011)₂ = (  )₈?
Answer:
Step I: First divide the number into groups of 3 bits starting from the right side and insert necessary zeroes in the left side.
0 0 1 | 0 1 1

Step II: Next write down the octal equivalent.

So the answer is (1011)₂ = (13)₈.

Question 18.
Convert (110100)₂ = (  )₁₆
Answer:

• Step I: First divide the number into groups of 4 bits starting from the right side and insert necessary zeroes in the left side.
• Step II: Next write down the hexadecimal equivalent.

So the answer is (110100)₂ = (34)₁₆.

Question 19.
(72)₈ = ( )₂?
Answer:
Write down the 3 bits equivalent of each digit.

So the answer is (72)₈ = (111010)₂.

Question 20.
Convert (AO)₁₆ = (  )₂ ?
Answer:
Write down the 4 bits equivalent of each digit

So the answer is (AO)₁₆ = (1010 0000)₂.

Question 21.
Convert (67)₈ = (  )₁₆?
Answer:
Step I: First convert this number into binary equivalent for this do the following

Step II: Next convert this number into hexadecimal equivalent for this do the following.

So the answer is (67)₈ = (37)₁₆

Question 22.
Convert (A1)₁₆ = (  )₈?
Answer:
Step I: First convert this number into binary equivalent. For this do the following

Step II: Next convert this number into octal equivalent. For this do the following.

So the answer is (A1)₁₆ = ( 241)₈.

Question 23.
What is the use of the ASCII Code?
Answer:
ASCII means American Standard Code for Information Interchange. It is a 7 bit code. Each and every character on the keyboard is represented in memory by using ASCII Code.
eg: A’s ASCII Code is 65 (1000001), a’s ASCII Code is 97 (1100001)

Question 24.
Pick invalid numbers from the following.

1. (10101)₈
2. (123)₄
3. (768)₈
4. (ABC)₁₆

Answer:

1. (10101)₈ – Valid
2. (123)₄ – Valid
3. (768)₈ – Invalid. Octal number system does not contain the symbol 8
4. (ABC)₁₆ – Valid

Question 25.
Convert the decimal number 31 to binary.
Answer:

(31)₁₀ = (11111)₂

Question 26.
Find decimal equivalent of (10001 )₂
Answer:

= 1 × 24 + 0 × 23 + 0 × 22 + 0 × 21 + 1 × 20
= 16 + 0 + 0 + 0 + 1
= (17)₁₀

Question 27.
If (X)₈ =(101011 )₂ then find X.
Answer:
Divide the binary number into groups of 3 bits and write down the corresponding octal equivalent.

X = 53

Question 28.
Fill the blanks:
(a) (………..)₂ = (AB)₁₆
(b) (——D—–)₁₆ = (1010 1000)₂
(c) 0.25₁₀ = (—–)₂

Answer:
Write down the 4bit equivalent of each digit
(a) (………..)₂ = (AB)₁₆

= (10101011)₂
(b) (——D—–)₁₆ = (1010 1000)₂
(A D 8)₁₆ =(1010 1101 1000)₂

(c) 0.25₁₀ = (—–)₂

0.25₁₀ = (0.01)₂

Question 29.
Which is the MSB of representation of -80 in SMR?
Answer:
It is 1 because In SMR if the number is negative then the MSB is 1.

Question 30.
Write 28.756 in Mantissa exponent form.
Answer:
28.756 = .28756 × 100
= .28756 × 102
= .28756 E + 2

Question 31.
Represent -60 in 1’s complement form.
Answer:
+60 = 00111100
Change all 1 to 0 and all 0 to 1 to get the 1’s complement.
-60 is in 1’s complement is 11000011

Question 32.
Define Unicode.
Answer:
The limitations to store more characters is solved by the introduction of Unicode. It uses 16 bits so 2¹⁶ = 65536 characters(i.e, world’s all written language characters) can store by using this.

Question 33.
Substract 1111 from 10101 by using 2’s complement method.
Answer:
To subtract a number from another number find the 2’s complement of the subtrahend and add it with the minuend. Here the subtrahend is 1111 and minuend is 5 bits. So insert a zero. So subtrahend is 01111. First take the 1’s complement of subtrahend and add 1 to it

Here is a carry. So ignore the carry and the result is +ve.
So the answer is 110

Question 34.
You purchased a soap worth Rs. (10010)₂ and you gave Rs. (10100)₂ and how much rupees will you get back in binary.
Answer:
Substract (10010)₂ from (10100)₂

You will get rupees (10)₂

Question 35.
Draw the logic circuit diagram for the following Boolean expression.
Answer:

Question 36.
Simplify the expression using basic postulates and laws of Boolean algebra.

1. \(\bar{x}+x \cdot \bar{y}\)
2. \(x(y+y . z)+y(\bar{x}+x z)\)

Answer:

1. \(\bar{x}+\bar{y}\)
2. y

Question 37.
Show \(A(\bar{B}+C)\) using NOR gates only.
Answer:

Question 38.
The following statement Demorgan’s theorem of Boolean algebra. Identify and state ‘Break the line, change the sign’.
Answer:
Demorgan’s theorems,
Demorgan’s first theorem,
\(\overline{x + y}\) = \(\bar{x} . \bar{y}\)
Demorgan’s second theorem,
\(\overline{x – y}\) = \(\bar{x} + \bar{y}\)

Question 39.
Prove algebraically that
\(x \cdot y + x \cdot \bar{y} \cdot z\) = x . y + x . z
Answer:
\(x \cdot y + x \cdot \bar{y} \cdot z\) = \(x(y+\bar{y} . z)\)
= x . (y + z) = x . y + x . z
Hence proved.

Question 40.
State which of the following statements are logical statements.
(a) AND is a logical operator
(b) ADD 3 to y
(c) Go to class
(d) Sun rises in the west.
(e) Why are you so late?
Answer:
(a) and (d) are logical statements because these statements have a truth value which may be true or false.

Question 41.
Express the integer number -39 in sign and magnitude represnetation.
Answer:
First find the binary equivalent of 39 for this do the following

In sign and magnitude representation -39 in 8 bit form is (10100111)₂.

Question 42.

1. Which logic gate does the Boolean expression \(\overline{\mathrm{AB}}\) represent?
2. Some NAND gates are given. How can we construct AND gate, OR gate and NOT gate using them?

Answer:
1. NAND

2. AND gate

Question 43.
Perform the following number conversions.

1. (110111011.11011)₂ = (….)₈
2. (128.25)₁₀= (…..)₈

Answer:
1. 110111011.11011
Step 1: Insert a zero in the right side of the above number and divide the number into groups of 3 bits as follows
110 111 011 . 110 110

Step 2: Write down the corresponding 3 bit binary equivalent of each group
6 7 3 .6 6
Hence the result is (673.66)₈

2. It consists of 2 steps.
Step 1: First convert 128 into octal number for this do the following

Write down the remainders from bottom to top.
(128)₁₀ = (200)₈

Step 2: Then convert .25 into octal number for this do the following

(0.25)₁₀ = (0.2)₈.
Combine the above two will be the result.
(128.25)₁₀= (200.2)₈

Question 44.
Represent -38 in 2’s complement form.
Answer:
+38 = 00100110
First take the 1 ’s complement for this change all 1 to 0 and all 0 to 1

2’s complement of -38 is (11011010)₈.

Plus One Data Representation and Boolean Algebra Three Mark Questions and Answers

Question 1.
Differentiate manual data processing and electronic data processing?
Answer:
In manual data processing human beings are the processors. Our eyes and ears are input devices. We get data either from a printed paper, that can be read using our eyes or heard with ears. Our brain is the processor and it can process the data, and reach in a conclusion known as result. Our mouth and hands are output devices.

In electronic data processing the data is processing with the help of a computer. In a super market, key board and hand held scanners are used to input data, the CPU process the data, monitor and printers (Bill) are output devices.

Question 2.
Complete the series.

1. 3248, 3278 ,3328, …., ….
2. 5678, 5768, 605s, ……, …..

Answer:
1. (324)₈ = 3 × 82 + 2 × 81 + 4 × 80
= 3 × 64 + 2 × 8 + 4 × 1
= 192 + 16 + 4 = (212)10

(327)₈ = 3 × 82 + 2 × 81 + 7 × 80
= 192 + 16 + 7 = (215)₁₀

(332)₈= 3 × 82 + 3 × 81 +2 × 80
= 192 + 24 + 2 = (218)₁₀
So the missing terms are (221)₁₀, (224)₁₀ we have to convert this into octal.

So the missing terms are (335)₈, (340)₈

2. (567)₈ = 5 x 82 + 6 x 81 + 7 x 80
= 5 x 64 + 6 x 8 + 7 x 1
= 320 + 48 + 7 = (375)₁₀

(576)₈ = 5 x 82 + 7 x 81 + 6 x 80
= 320 + 56 + 6 = (382)₁₀

(605)₈ = 6 x 82 + 0 x 81 + 5 x 80
= 6 + 64 + 0 + 5
= 384 + 0 + 5 = (389)₁₀
So the missing terms are (396)₁₀, (403)₁₀ we have to convert this into octal.

So the missing terms are (614)₈, (623)₈

Question 3.
Fill up the missing digits.

1. (4……)₈ = (……110)₂
2. (…….7……)₈ = (100…….110)₂

Consider the following

Answer:
1. 4…….100 and 110……….6
So (46)₈ = (100 110)₂

2. 100…….4
7……111
110………6
So (476)₈ = (100 111 110)₂

Question 4.
Fill up the missing numbers.

1. (A…….)₁₆ = (……..1001)₂
2. (…….B…….)₁₆ = (1000………1111)₂

Consider the following:


Answer:
1. A……..1010
1001………9
So (A9)₁₆ = (1010 1001)₂

2. B…….1011
1000………8
1111………F
So (8BF)₁₆ = (1000 1011 1111)₂

Question 5.
Complete the Series.

1. 6ADD, 6ADF, 6AE1, ……., …….
2. 14A9, 14AF, 14B5, …….., ……

Answer:
1. Consider the sequence
6ADD, 6ADF, 6AE1, ………….
Here the ‘numbers’ are
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, 10, 11,——–
The difference between 6ADD & 6ADF is 2

Similarly 6ADF & 6AE1 is 2

So Add 2 to 6AE1 we will ge 6AE3 Then add 2 to 6AE3 we will get 6AE5 Therefore the missing terms 6AE3, 6AE5

2. Consider the sequence.
14A9, 14AF, 14B5, ———
The difference between 14A9 and 14AF is 6.
The normal sequence is

The difference between 14AF and 14B5 is also 6.

The normal sequence is

Similarly the next 6 terms in the sequence are given below.

Similarly the next 6 terms are

So the missing terms are 14BB and 14C1

Question 6.
Find the octal numbers corresponding to the following numbers using shorthand method.

1. (ADD)₁₆
2. (DEAD)₁₆

Answer:
1. (ADD)₁₆
Step 1: Write down the 4 bit binary equivalent of each digit

Step 2: Divide this number into groups of 3 bits starting from the right and write down the octal equivalent.

(ADD)(ADD)₁₆ = (5335)(ADD)₈

2. (DEAD)₁₆
Step 1: Write down the 4 bit binary equivalent of each digit.

Step 2: Divide this number into groups of 3 bits starting from the right and write down the octal equivalent.

(DEAD)₁₆ = (157255)₈

Question 7.
If (126)x = (56)y, then find x and y.
Answer:
L.H.S contains 2 & 6, so x ≠ 2
R.H.S contains 5 & 6, so y ≠ 2
L.H.S > R.H.S
So x < y and x ≠ y also The possible values of x and y are given below.

Case I:
Let x = 8 then y = 10

It is grater than (56)₁₀
so when x = 8 then y ≠ 10

case II:
let x = 8 and y = 16

Question 8.
If (102)x = (42)y then (154)x = (  )y.
Answer:
L.H.S contains 2, so x ≠ 2
R.H.S contains 5 & 4, so y ≠ 2
L.H.S > R.H.S
So x < y and x ≠ y also
The possible values of x and y are given below.

case I:
let x = 8 and y = 10

So when x = 8 then y ≠ 10

case II:
let x = 8 and y = 16

So x = 8 and y = 16
then we have fo find the hexadecimal equivalent of (154)₈ For this first convert this into binary thus again convert it into hexadecimal. First write down the 3 bit equivalent of 154.

Then divide this number into groups of 4 bits starting from the right and write down the hexa decimal equivalent.

so the result is (154)₈ = (6C)₁₆

Question 9.
Fill up the missing digit.
If (121)_a = (441)_b then (121)_b = (  )₁₀
Answer:
L.H.S. contains 2, so a ≠ 2
R.H.S. contains 4, so b ≠ 2
L.H.S. < R.H.S. So a > b and a b also.
Hence the values of a can be 10 or 16.
The values of b can be 8 or 10.
The possible values of a and b are given below.

Case I:
Let a = 16 and b = 10
(121 )₁₆ = (289)_10, so b ≠ 10

Case II:
Let a = 16 and b = 8
(121)₁₆ = (289)₁₀
(441)₈ = 4 × 8² + 4 × 8¹ + 1 × 8⁰
= 256 +32 + 1
= (289)₁₀.
So a = 16 and b = 8.
Then (121)₈ = 1 × 82 + 2 × 81 + 1 × 80
= 64 + 16 + 1 = (81)₁₀

Question 10.
Fill up the missing digit. (Score 3)
If (128)_a = (450)_b then (16)_a = (  )₁₀
Answer:
L.H.S. contains 2 & 8, so a 2 and a ≠ 8.
R.H.S. contains 4 and 5, so b ≠ 2.
L.H.S. < R.H.S. so a > b and a ≠ b also.
The possible values of a and b are given below.

Case I:
a = 16 and b = 8
(128)₁₆ = (296)₁₀
(450)₈ = (296)₁₀ So a = 16 and b = 8.
Then (16)₁₆ = 1 × 16 + 6 × 16⁰ = (22)₁₀

Question 11.
Fill up the missing digit.
(3A.6D)₁₆ = (  )₈
Answer:
Step I: Write down the 4 bits equivalent of each digit.

Step II: Divide this number into groups of 3 bit starting from the right side of the left side of the decimal point and starting from the left side of the right side of the decimal point.
So 00/111/010.011/011/010

Step III: Write the octal equivalent of each group. SO we will get. (72.332)₈.
(3A.6D)₁₆ = (72.332)₈

Question 12.
What are the various ways to represent integers in computer?
Answer:
There are three ways to represent integers in computer. They are as follows:

1. Sign Magnitude Representation (SMR)
2. 1’s Complement Representation
3. 2’s Complement Representation

1. SMR:
Normally a number has two parts sign and magnitude, eg: Consider a number +5. Here + is the sign and 5 is the magnitude. In SMR the most significant Bit (MSB) is used to represent the sign. If MSB is 0 sign is +ve and MSB is 1 sign is – ve.
eg: If a computer has word size is 1 byte then

Here MSB is used for sign then the remaining 7 bits are used to represent magnitude. So we can represent 27 = 128 numbers. But there are negative and positive numbers. So 128 + 128 = 256 number. The numbers are 0 to +127 and 0 to -127. Here zero is repeated. So we can represent 256 – 1 = 255 numbers.

2. 1 ‘s Complement Representation: To get the 1’s complement of a binary number, just replace every 0 with 1 and every 1 with 0. Negative numbers are represented using 1’s complement but +ve number has no 1’s complement.
eg: To find the 1’s complement of 21 +21 = 00010101
To get the 1 ‘s complement change all 0 to 1 and all 1 to 0.
-21 = 11101010
1’s complement of 21 is 11101010

3. 2’s Complement Representation: To get the 2’s complement of a binary number, just add 1 to its 1’s complement +ve number has no 2’s complement.
eg: To find the 2’s complement of 21 +21 =00010101
First take the 1’s complement for this change all 1 to 0 and all 0 to 1

2’s complement of 21 is 1110 1011

Question 13.
Write short notes about Unicode (3)
Answer:
It is like ASCII Code. By using ASCII, we can represent limited number of characters. But using Unicode we can represent all of the characters used in the written languages of the world.
eg: Malayalam, Hindi, Sanskrit .

Question 14.
Match the following.

Answer:
1 – b, 2 – f, 3 – a, 4 – c, 5 – d, 6 – e

Question 15.
Find the largest number in the list.

1. (1001)₂
2. (A)₁₆
3. (10)₈
4. (11)₁₀

Answer:
Convert all numbers into decimal
1. (1001)₂ = 1 × 2³ + 0 × 2² + 0 × 2¹ + 1 × 2⁰
= 8 + 0 + 0 + 1
= (9)₁₀

2.) (A)₁₆ = (10)₁₀

3. (10)₈ = 1 × 8¹+0 × 8⁰
= (8)₁₀
So the largest number is 4 – (11)₁₀

Question 16.
Subtract 10101 from 1111 by using 2’s complement method.
Answer:
To subtract a number from another number find the 2’s complement of the subtrahend and add it with the minuend. Here subtrahend is 10101 and minuend is 1111 First take the 1’s complement of subtrahend and add 1 to it.
1’s complement of 10101 is 01010 add 1 to it

Here is no carry. So the result, is -ve and take the 2’s complement of 11010 and put a -ve symbol. So 1’s complement of 11010 is 00101 add 1 to this

So the result is -00110

Question 17.
Mr. Geo purchased (10)₂ kg sugar @Rs. (110 10)₂ and (1010)₂ kg Rice @Rs. (10100)₂. So how much rupees he has to pay in decimal.
Answer:
Convert each into decimal number system multiply and sum it up.
(10)₂ = (2)₁₀

(11010)₂ = 1 × 2⁴ + 1 × 2³ + 0 × 2² + 1 × 2¹ + 0 × 2¹
= 16 +8 + 0 +2 + 0
= (26)₁₀

(1010)₂ = 1 × 2³ + 0 × 2² + 1 × 2¹ +0 × 2⁰
= 8 + 0 + 2 + 0
= (10)₂

(10100)₂ = 1 × 2⁴ + 0 × 2³ + 1 × 2² +0 × 2¹ + 0 × 2⁰
= 16 + 0 + 4 + 0 + 0
= (20)₂
therfore 2 × 26 + 10 × 20
= 52 + 200
= 252
So Mr. Geo has to pay Rs. 252/-

Question 18.
Mr. Vimal purchased a pencil @ Rs. (101)2, a pen @ Rs. (1010)2 and a rubber @ Rs. (10)2. So how much rupees he has to pay in decimal.
Answer:
Add 101 + 1010 + 10

then convert (10001)₂ into decimal
(10001)₂ = 1 × 2⁴ + 0 × 2³ + 0 × 2² + 0 × 2¹ + 1 × 2⁰
= 16 + 0 + 0 + 0 + 1
= (17)₁₀
So Mr. Vimal has to pay Rs. 17/-

Question 19.
Mr. Antony purchased 3 books worth Rs. a total of (1100100)₂. Atlast he returned a book worth Rs. (11001)₂. So how much amount he has to pay for the remaining two books in decimal number sys¬tem.
Answer:

then convert (1001011 )₂ into decimal
(1001011)₂ = 1 × 2⁶ + 0 × 2⁵ + 0 × 2⁴ + 1 × 2³ + 0 × 2² + 1 × 2¹ + 1 × 2⁰
=64 + 0 + 0 + 8 + 0 + 2 + 1
= (75)₁₀
So he has to pay Rs. 75/-

Question 20.
Mr. Leones brought two products from a super market a total of Rs. (11010010)₂ and he got a dicount of Rs. (1111)₂ So how much he has to pay for this products in decimal number system.
Answer:
Substract (1111)₂ from (11010010)₂

then convert (11000011)₂ into decimal
(11000011)₂ = 1 × 2⁷ + 1 × 2⁶ + 0 × 2⁵ + 0 × 2⁴ + 0 × 2³ + 0 × 2² + 1 × 2¹ + 1 × 2⁰
=128 + 64 + 0 + 0 + 0 + 0 + 2 + 1
= (195)₁₀

Question 21.
A textile showroom sells shirts with a discount of Rs. (110010)₂ on all barads. Mr. Raju wants to buy a shirt worth Rs. (11111000)₂. So after discount how much amount he has to pay in decimal.
Answer:
Substract (110010)₂ from (11 111 000)₂

then convert (11000110)₂ into decimal
(11000110)2 = 1 × 2⁷ + 1 × 2⁶ + 0 × 2⁵ + 0 × 2⁴ + 0 × 2³ + 1 × 2² + 1 × 2¹ + 0 × 2⁰
= 128 + 64 + 0 + 0 + 0 + 4 + 2 + 0
= (198)₁₀

Question 22.
Mr. Lijo purchased a product worth Rs. (1110011)₂ and he has to pay VAT @ Rs. (1100)₂. Then calculate the total amount he has to pay in decimal.
Answer:
Add (1110011)₂ and (1100)₂

then convert (1111111)₂ into decimal
(1111111)₂ = 1 × 2⁶ + 1 × 2⁵ + 1 × 2⁴ + 1 × 2³ + 1 × 2² + 1 × 2¹ + 1 × 2⁰
= 64 + 32 + 16 + 8 + 4 + 2 + 1
= (127)₁₀

Question 23.
By using truth table, prove the following laws of Boolean Algebra.

1. Idempotent law
2. Involution law

Answer:
1. A + A = A
A = A = A

2. (A¹)¹ = A

Question 24.
Consider the logical gate diagram.

1. Find the logical expression for the circuit given.
2. Find the compliment of the logical expression.
3. Draw the circuit diagram representing the compliment.

Answer:
1. \((x+\bar{y}) \cdot z\)

2. \((\bar{x} . y)+\bar{z}\)
3.

Question 25.
Draw the logic circuit diagram for the following Boolean expression.
\(A \cdot(\bar{B} + \bar{C})+\bar{A} \bar{B} \bar{C}\)
Answer:

Question 26.
Consider a bulb with three switches x, y and z. Write the Boolean expression representing the following states.

1. All the switches x, y and z are ON
2. x is ON and y is OFF or Z is OFF
3. Exactly one switch is ON.

Answer:

1. xy . z
2. \(x \bar{y}+\bar{z}\)
3. \(x . \bar{y} . \bar{z}+\bar{x} . y . \bar{z}+\bar{x} . \bar{y} . \bar{z}\)

Question 27.
Match the following.

Answer:
i – f, ii – e, iii – d, iv – c, v – b, vi – a

Question 28.
Explain the principle of duality.
Answer:
It states that, starting with a Boolean relation, another Boolean relation can be derived by

1. Changing each OR sign (+) to a AND sign (.)
2. Changing each AND sign (.) to an OR sign (+)
3. Replacing each 0 by 1 and each 1 by 0.

The relation derived using the duality principle is called the dual of the original expression,
eg: x + 0 = x is the dual of x . 1 = x

Question 29.
Draw the circuit diagram for \(F=A \bar{B} C+\bar{C} B\) using NAND gate only.
Answer:
\(F=A \bar{B} C+\bar{C} B\)
= (A NAND (NOT B) NAND C) NAND ((NOT C) NAND B)

Question 30.
Draw a logic diagram for the function f = YZ + XZ using NAND gates only.
Answer:
f = YZ + XZ
= (Y NAND Z) NAND (X NAND Z)

Question 31.
How do you make various basic logic gates using NAND gates.
Answer:
1. AND operation using NAND gate,
A.B = (A NAND B) NAND (A NAND B)

2. OR operation using NAND gate,
A + B = (A NAND A) NAND (B NAND B)

3. NOT operation using NAND gate,
NOT A = (A NAND A)