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Students can Download Computer Science Chapter 11 Functions Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Computer Science Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Computer Science Solutions Chapter 11 Functions

Samacheer Kalvi 11th Computer Science Functions Text Book Back Questions and Answers

PART – 1
I. Choose The Correct Answer

Question 1.
Which of the following header file defines the standard I/O predefined functions?
(a) stdio.h
(b) math.h
(c) string.h
(d) ctype.h
Answer:
(a) stdio.h

Question 2.
Which function is used to check whether a character is alphanumeric or not?
(a) isalpha()
(b) isdigit()
(c) isalnum()
(d) islower()
Answer:
(c) isalnum()

Question 3.
Which function begins the program execution?
(a) isalpha()
(b) isdigit()
(c) main()
(d) islower()
Answer:
(c) main()

Question 4.
Which of the following function is with a return value and without any argument?
(a) x = display(int, int)
(b) x = display()
(c) y = display(float)
(d) display(int)
Answer:
(b) x = display()

Question 5.
Which is return data type of the function prototype of add(int, int);?
(a) int
(b) float
(c) char
(d) double
Answer:
(a) int

Question 6.
Which of the following is the scope operator?
(a) >
(b) &
(c) %
(d) ::
Answer:
(d) ::

PART – 2
II. Answer to all the questions

Question 1.
Define Functions.
Answer:
A large program can typically be split into small sub – programs (blocks) called as functions where each sub-program can perform some specific functionality. Functions reduce the size and complexity of a program, makes it easier to understand, test and check for errors.

Question 2.
Write about strlen() function.
Answer:
The strlen() takes a null terminated byte string source as its argument and returns its length. The length does not include the null(\0) character.

Question 3.
What are importance of void data type? void type has two important purposes:
Answer:

1. To indicate the function does not return a value.
2. To declare a generic pointer.

Question 4.
What is Parameter and list its types?
Answer:
Arguments or parameters are the means to pass values from the calling function to the called function. The variables used in the function definition as parameters are known as formal parameters. The constants, variables or expressions used in the function call are known as actual parameters.

Types : Default arguments and Constant Arguments.

Question 5.
Write a note on Local Scope.
Answer:

1. A local variable is defined within a block. A block of code begins and ends with curly braces { }.
2. The scope of a local variable is the block in which it is defined.
3. A local variable cannot be accessed from outside the block of its declaration.
4. A local variable is created upon entry into its block and destroyed upon exit.

PART – 3
III. Answer to all the questions

Question 1.
What is Built – in functions?
Answer:
C++ provides a rich collection of functions ready to be used for various tasks. The tasks to be performed by each of these are already written, debugged and compiled, their definitions alone are grouped and stored in files called header files. Such ready – to – use sub programs are called pre – defined functions or built – in functions.

Question 2.
What is the difference between isupper() and toupper() functions?
Answer:
isupper():

• This function is used to check the given character is uppercase.
• This function will return 1 if true otherwise 0.

toupper():

• This function is used to convert the given character into its uppercase.
• This function will return the upper case equivalent of the given character. If the given character itself is in upper case, the output will be the same.

Question 3.
Write about strcmp() function.
Answer:
The strcmp() function takes two arguments: string1 and string2. It compares the contents of string1 and string2 lexicographically.
The strcmp() function returns a:

1. Positive value if the first differing character in string1 is greater than the corresponding character in string2. (ASCII values are compared)
2. Negative value if the first differing character in string1 is less than the corresponding character in string2.
3. 0 if string1 and string2 are equal.

Question 4.
Write short note on pow() function in C++.
Answer:
The pow() function returns base raised to the power of exponent. If any argument passed to pow() is long double, the return type is promoted to long double. If not, the return type is double. The pow() function takes two arguments:

1. base – the base value
2. exponent – exponent of the base

Example:
cout << pow(5, 2);

Output:
25

Question 5.
What are the information the prototype provides to the compiler?
Answer:
The prototype above provides the following information to the compiler:

1. The return value of the function is of type long.
2. Fact is the name of the function.
3. The function is called with two arguments:
   • The first argument is of int data type.
   • The second argument is of double data type, int display(int, int) // function prototype//.

The above function prototype provides details about the return data type, name of the function and a list of formal parameters or arguments.

Question 6.
What is default arguments? Give example.
Answer:
In C++, one can assign default values to the formal parameters of a function prototype. The Default arguments allows to omit some arguments when calling the function.
1. For any missing arguments, complier uses the values in default arguments for the called function.

2. The default value is given in the form of variable initialization.
Example : void defaultvalue(int n1 = 10, n2 = 100);

3. The default arguments facilitate the function call statement with partial or no arguments.
Example :

1. defaultvalue (x, y);
2. defaultvalue (200, 150);
3. defaultvalue (150);
4. defaultvalue (x, 150);

4. The default values can be included in the function prototype from right to left, i.e., we cannot have a default value for an argument in between the argument list.
Example:

1. void defaultvalue (int n1=10, n2);//invalid prototype.
2. void defaultvalue (int n1, n2 = 10);//valid prototype.

PART – 4
IV. Answers to all the questions

Question 1.
Explain Call by value method with suitable example.
Answer:
This method copies the value of an actual parameter into the formal parameter of the function. In this case, changes made to formal parameter within the function will have no effect on the actual parameter.

Output:
Example : Function call by value
Enter the Value for A : 5
The Value inside display function (a * a) : 25
The Value inside main function 5

Question 2.
What is Recursion? Write a program to find GCD using recursion.
Answer:
A function that calls itself is known as recursive function. And, this technique is known as recursion.

Output:
Enter two numbers: 81 153
gcd : 9

Question 3.
What are the different forms of function return? Explain with example.
Answer:
The return statement:
Returning from the function is done by using the return statement. The return statement stops execution and returns to the calling function. When a return statement is executed, the function is terminated immediately at that point. The return statement is used to return from a function. It is categorized as a jump statement because it terminates the execution of the function and transfer the control to the called statement.
Syntax:
return expression/variable;

Example : retum(a + b); retum(a);
return; // to terminate the function

The Returning values:
The functions that return no value is declared as void. The data type of a function is treated as int, if no data type is explicitly mentioned. For example,
For Example:
int add (int, int);
add (int, int);
In both prototypes, the return value is int, because by default the return value of a function in C++ is of type int.

Returning Non – integer values:
A string can also be returned to a calling statement.

Output:
Example: Function with Non Integer Return Chennai

The Returning by reference:
#include
using namespace std;
int main( )
{

int n 1 = 150;
int &n 1 ref = n1;
cout << “\nThe Value of N1 = “<< n1 << “and n 1 Reference =”<< n 1 ref;
n 1 ref++;
cout << “\n After nl increased the Value of N1 =”<< n1;
cout << “and n 1 Reference = ”<< n 1 ref;
retum(0);

}
Output:
The Value of N1 = 150 and nl Reference =150
After n1 increased the Value of N1 = 151 and n1 Reference =151

Question 4.
Explain scope of variable with example.
Answer:
Scope refers to the accessibility of a variable.
There are four types of scopes in C++

1. Local Scope
2. Function Scope
3. File Scope
4. Class Scope

1. Local Scope:

• A local variable is defined within a block. A block of code begins and ends with curly braces {}.
• The scope of a local variable is the block in which it is defined.
• A local variable cannot be accessed from outside the block of its declaration.
• A local variable is created upon entry into its block and destroyed upon exit;
  Example:
  int main( )
  {
  int a,b;   //Local variable
  }

2. Function Scope:

• The scope of variable within a function is extended to the function block and all sub-blocks therein.
• The lifetime of a function scope variable is the lifetime of the function block.
  Example:
  int. sum(intx, int y);  //x and y has function scope.

3. File Scope:

• A variable declared above all blocks and functions (including main()) has the scope of a file.
• The lifetime of a file scope variable is the lifetime of a program.
• The file scope variable is also called as global variable.
  Example:
  #include
  using namespace std;
  int x,y; //x and y are global variable
  void main()
  {
  ……..
  }

4. Class Scope:

• Data members declared in a class has the class scope.
• Data members declared in a class can be accessed by all member functions of the class.
  Example:
  Class example
  {

int x,y; //x and y can be accessed by print() and void():
void print();
Void total();

  };

Question 5.
Write a program to accept any integer number and reverse it.
Answer:

Output:
Enter number : 1 2 3
Reverse number: 3 2 1

Samacheer kalvi 11th Computer Science Functions Additional Questions and Answers

PART – 1
I. Choose the correct answer

Question 1.
………………. is the name of the function.
(a) Pre – defined
(b) Built – in
(c) Library
(d) All the above
Answer:
(d) All the above

Question 2.
………………. is used to check whether the given character is an alphabet or not.
(a) isalnum()
(b) isalpha()
(c) isalph()
(d) isal()
Answer:
(b) isalpha()

Question 3.
The strcpy() function takes two arguments of ……………….
(a) target and source
(b) upper and lower
(c) base and exponent
(d) none of these
Answer:
(a) target and source

Question 4.
………………. takes a null terminated byte string source as its argument and returns its length.
(a) strcpy()
(b) strlen()
(c) strcmp()
(d) strcat()
Answer:
(b) strlen()

Question 5.
The pow() function takes the two arguments of ……………….
(a) target and source
(b) upper and lower
(c) base and exponent
(d) source and exponent
Answer:
(c) base and exponent

Question 6.
………………. is the name of the function.
(a) fact
(b) task
(c) arguments
(d) none of these
Answer:
(d) none of these

Question 7.
The C++ program always have main() function to begin the program execution.
(a) 1
(b) 2
(c) 3
(d) null
Answer:
(a) 1

Question 8.
Arguments are also called as ……………….
(a) variable
(b) constant
(c) function
(d) parameters
Answer:
(d) parameters

Question 9.
In C++ the arguments can be passed to a function in ………………. ways.
(a) 2
(b) 1
(c) 3
(d) 7
Answer:
(a) 2

Question 10.
Inline functions execute faster but requires more ……………….
(a) variables
(b) pointers
(c) memory
(d) functions
Answer:
(c) memory

PART – 2
II. Very Short Answers

Question 1.
Write about reusability.
Answer:

1. Few lines of code may be repeatedly used in different contexts. Duplication of the same code can be eliminated by using functions which improves the maintenance and reduce program size.
2. Some functions can be called multiple times with different inputs.

Question 2.
What is user – defined functions?
Answer:
C++ also provides the facility to create new functions for specific task as per user requirement. The name of the task and data required (arguments) are decided by the user and hence they are known as User-defined functions.

Question 3.
What is constant arguments and write its syntax?
Answer:
The constant variable can be declared using const keyword. The const keyword makes variable , value stable. The constant variable should be initialized while declaring. The const modifier enables to assign an initial value to a variable that cannot be changed later inside the body of the function.
Syntax:
(const )

Question 4.
What are the advantages of inline functions?
Answer:
Advantages of inline functions:

1. Inline functions execute faster but requires more memory space.
2. Reduce the complexity of using STACKS.

Question 5.
What is function scope?
Answer:
Function Scope:

1. The scope of variables declared within a function is extended to the function block, and all sub – blocks therein.
2. The life time of a function scope variable, is the life time of the function block. The scope of.

PART – 3
III. Short Answers

Question 1.
What is divide and conquer?
Answer:
Divide and Conquer

1. Complicated programs can be divided into manageable sub programs called functions.
2. A programmer can focus on developing, debugging and testing individual functions.
3. Many programmers can work on different functions simultaneously.

Question 2.
Define library functions.
Answer:
C++ provides a rich collection of functions ready to be used for various tasks. The tasks to be performed by each of these are already written, debugged and compiled, their definitions alone are grouped and stored in files called header files. Such ready – to – use sub programs are called pre – defined functions or built – in functions or Library Functions.

Question 3.
What is isdigit()? Give example.
Answer:
This function is used to check whether a given character is a digit or not. This function will return 1 if the given character is a digit, and 0 otherwise.

Example:
using namespace std;
#include
#include int main( )
{

char ch;
cout << “\n Enter a Character:”; cin >> ch;
cout << “\n The Return Value of isdigit(ch) is << isdigit(ch);

}
Output – 1
Enter a Character: 3
The Return Value of isdigit(ch) is : 1

Output – 2
Enter a Character: A
The Return Value of isdigit(ch) is :0

Question 4.
Write a program using pow() and sin() function.
Answer:
The pow() function returns base raised to the power of exponent. The sin() function takes a single argument in radians.
#include
#include using namespace std;
int main ()
{

double base, exponent, result;
base = 5;
exponent = 4;
result = pow(base, exponent);
cout << “pow(“ << base << “A” << exponent <<”) =” << result;
double x = 25;
result = sin(x);
cout << “\nsin(“<< x <<”)=”<<result;
return 0;

}
Output:
pow (5^4) = 625
sin (25) = – 0.132352

Question 5.
What is return statement with example?
Answer:
The return statement stops execution and returns to the calling function. When a return statement is executed, the function is terminated immediately at that point. The return statement is used to return from a function. It is categorized as a jump statement because it terminates the execution of the function and transfer the control to the called statement.

Example:
return(a + b); return(a);
return; // to terminate the function

Question 6.
What is scope resolution operation?
Answer:

1. The scope operator reveals the hidden scope of a variable. The scope resolution operator (::) is used for the following purposes.
2. To access a Global variable when there is a Local variable with same name. An example using Scope Resolution Operator.

PART – 4
IV. Explain in Detail

Question 1.
Explain about generating random numbers with suitable program.
Answer:
The srand() function in C++ seeds the pseudo random number generator used by the rand() function. The seed for rand() function is 1 by default. It means that if no srand() is called before rand(), the rand() function behaves as if it was seeded with srand( 1). The srand() function takes an unsigned integer as its parameter which is used as seed by the rand() function. It is defined inor header file.
#include
#include using namespace std; int main()
{

int random = rand(); /* No srand() calls before rand(), so seed = 1*/
cout << “\nSeed = 1, Random number =” << random;
srand(10);
/* Seed= 10 */
random = rand();
cout << “\n\n Seed =10, Random number =” << random;
return 0;

}
Output:
Seed = 1, Random number = 41
Seed =10, Random number 71

Question 2.
Explain about Inline functions with a suitable program.
Answer:
An inline function looks like normal function in the source file but inserts the function’s code directly into the calling program. To make a function inline, one has to insert the keyword inline in the function header.

Output:
Enter the Principle Amount Rs. :60000
Enter the Number of Years : 10
Enter the Rate of Interest :5
The Simple Interest = Rs. 30000

Question 3.
Write about function prototype.
Answer:
C++ program can contain any number of functions. But, it must always have only one main() function to begin the program execution. We can write the definitions of functions in any order as we wish. We can define the main() function first and all other functions after that or we can define all the needed functions prior to main(). Like a variable declaration, a function must be declared before it is used in the program. The declaration statement may be given outside the main() function

The prototype above provides the following information to the compiler:

1. The return value of the function is of type long.
2. Fact is the name of the function.
3. The function is called with two arguments:
   • The first argument is of int data type.
   • The second argument is of double data type, int display(int, int)//function prototype//

The above function prototype provides details about the return data type, name of the function and a list of formal parameters or arguments.

Question 4.
Explain about address method.
Answer:
This method copies the address of the actual argument into the formal parameter. Since the address of the argument is passed, any change made in the formal parameter will be reflected back in the actual parameter.
#include
using namespace std;
void display(int & x) //passing address of a//
{

x = x*x;
cout << “\n\n The Value inside display function (n1 x n1) :”<< x ;

}
int main()
{
intn 1;
cout << “\n Enter the Value for N1 cin >> n1;
cout << “\n The Value of N1 is inside main function Before passing:” << n1;
display(n1);
cout << “\n The Value of N1 is inside main function After passing (n1 x n1):”<< n1; retum(O);
}

Output:
Enter the Value for N1 : 45
The Value of N1 is inside main function Before passing : 45
The Value inside display function (n1 x n1) : 2025
The Value ofNl is inside main function After passing (n1 x n1): 2025

Question 5.
Given data: Principal amount Rs. 50000

Number of years 5
Rate of interest 7

To find the simple interest of the above mentioned given data. Write a C++ program using inline functions.
Answer:
#include
using namespace std;
inline float simple interest(float p1, float n 1, float r 1)
{

float sil=(pl*nl*rl)/100;
retum(sil);

}
int main ()
{

float si,p,n,r;
cout << “\n Enter the Principle Amount Rs. :”; cin >> p;
cout << “\n Enter the Number of Years :”; cin >> n;
cout << “\n Enter the Rate of Interest :”; cin >> r;
si = simple interest(p, n, r);
cout << “\n The Simple Interest = Rs.” << si;
return 0;

}

Output:
Enter the Principle Amount Rs. : 50000
Enter the Number of Years : 5
Enter the Rate of Interest : 7
The Simple Interest = Rs. 17500

Students can Download Tamil Nadu 11th Chemistry Model Question Paper 3 English Medium Pdf, Tamil Nadu 11th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 11th Chemistry Model Question Paper 3 English Medium

Time: 2½ Hours
Total Score: 70 Marks

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I. II, III and IV are to be attempted separately.
4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
5. Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered in detail. Draw diagrams wherever necessary.

Part – I

Answer all the questions.
Choose the most suitable answer from the given four alternatives.

Question 1.
Choose the disproportionation reaction among the following redox reactions.

Answer:

Solution:
\(P_{4}(s)+3 N a O H+3 H_{2} \rightarrow P H_{3}(g)+3 N a H_{2} P O_{2}(a q)\)

Question 2.
Consider the following sets of quantum numbers:

Which of the following sets of quantum number is not possible?
(a) (i), (ii), (iii) and (iv)
(b) (ii), (iv) and (v)
(c) (i) and (ill) (d) (ii), (iii) and (iv)
Answer:
(b) (ii), (iv) and (v)

Solution:
(ii) l can have the values from O to n-1
n = 2; possible ‘l’ values are 0, 1 hence l = 2 is not possible.
(iv) for l = O; m = – 1 not possible
(v) for n = 3 l = 4 and m = 3 not possible

Question 3.
Which of the following is arranged in order of increasing radius?
\((a) \mathrm{K}_{(\mathrm{aq})}^{+}<\mathrm{Na}_{(\mathrm{aq})}^{+}<\mathrm{Li}^{+}_{(\mathrm{eq})}\)
\((b) \mathrm{K}_{(\mathrm{aq})}^{+}>\mathrm{Na}_{(\mathrm{aq})}^{+}>\mathrm{Zn}_{(\mathrm{eq})}^{2+}\)
\((c) \mathrm{K}_{(\mathrm{aq})}^{+}>\mathrm{Li}_{(\mathrm{aq})}^{+}>\mathrm{Na}_{(\mathrm{aq})}^{+}\)
\((d) \mathbf{L}_{(\mathrm{aq})}^{+}<\mathbf{N} \mathbf{a}_{(\mathbf{a} \mathbf{q})}^{+}<\mathbf{K}_{(\mathbf{a} \mathbf{q})}^{+}\)
Answer:
\(\text { (d) } \mathbf{L} \mathbf{i}_{(\mathbf{z} q)}^{+}<\mathbf{N} \mathbf{a}_{(\mathbf{a} \mathbf{q})}^{+}<\mathbf{K}_{(\mathbf{a} \mathbf{q})}^{+}\)

Question 4.
Statement – I: The magnetic moment of para-hydrogen is zero.
Statement – II: The spins of two hydrogen atoms in para H₂ molecule neutralise each other.
(a) Statements – I and lIare correct and Statement-Il is the correct explanation of statement – I.
(b) Statements – I and H are correct but Statement-Il is not the correct explanation of statement – I.
(e) Statement – I is correct lut Statement-Il is wrong.
(d) Statement – I is wrong but Statement-Il is correct.
Answer:
(a) Statements – I and lIare correct and Statement-Il is the correct explanation of statement – I.

Question 5.
Which of the following has the highest tendency to give the reaction,
\(\mathbf{M}^{+}(\mathrm{g}) \frac{\text { Aqueous }}{\text { Medium }} \rightarrow \mathrm{M}^{+}(\mathrm{aq})\)
(a) Na
(b) Li
(c) Rb
(d) K
Answer:
(b) Li

Question 6.
The value of universal gas constant depends upon
(a) Temperature of the gas
(b) Volume of the gas
(c) Number of moles of the gas
(d) units of Pressure and volume
Answer:
(d) units of Pressure and volume

Question 7.
The heat of formation of CO and CO, are —26.4 kcal and —94 kcal, respectively. Heat of combustion of carbon monoxide will be …………………… .
(a) +26.4 keal
(b) -67.6 kcal
(c) -120.6 kcal
(d) +52.8 kcal
Answer:
(b) -67.6 kcal

Solution:

Question 8.
In a chemical equilibrium, the rate constant for the forward reaction is 2.5 x 10² and the equilibrium constant is 50. The rate constant for the reverse reaction is
(a) 11.5
(b) 5
(c) 2 x 10²
(d) 2 x 10^-3
Answer:
(b) 5

Solution:

Question 9.
The Vant Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is ………………………………. . [NEET]
(a) O
(b) 1
(c) 2
(d) 3
Answer:
(b) 1

Solution: Ba(OH)₂ dissociates to form Ba2 and 20H- ion
\(\alpha=\frac{(i-1)}{(n-1)}\)
i = α (n – 1) + 1
∴ n = i = 3 (for Ba (OH)₂, α = 1)

Question 10.
Match the list I and list II using the correct code given below in the list:

Code: A B C D
(a) 4 3 2 1
(b) 1 2 3 4
(c) 3 1 4 2
(d) 2 4 1 3
Answer:
(a) 4 3 2 1

Question 11.
Nitrogen detection in an organic compound is carried out by Lassaigne’s test. The blue colour formed is due to the formation of
(a) Fe₃ [Fe(CN)₆]₂
(b) Fe₄ [Fe(CN)₆]₃
(c) Fe₄ [Fe(CN)₆]₂
(d) Fe₃ [Fe(CN)₆]₃
Answer:
(b) Fe₄ [Fe(CN)₆]₃

Question 12.
What is the hybridisation state of benzyl carbonium ion?
(a) sp²
(b) spd²
(c) sp²
(d) sp²d
Answer:
(a) sp²

Question 13.
Statement – I: Boiling point of methane is lower than that of butane.
Statement – II: The boiling point of continuous chain alkanes increases with increase in length of carbon chain.
(a) Statement -I and II are correct and statement – II is correct explanation of statement – I.
(b) Statement – I and II are correct but statement – II is not correct explanation of statement – I.
(c) Statement – I is correct but statement – II is wrong.
(d) Statement – II is wrong but statement – 1 is correct.
Answer:
(a) Statement -I and II are correct and statement – II is correct explanation of statement – I.

Question 14.
In the reaction


Answer:

Question 15.
Which of the following binds with haemoglobin and reduce the oxygen-carrying capacity of blood?
(a) CO₂
(b) NO₂
(c) SO₃
(d) CO
Answer:
(d) CO

Part – II
Answer any six questions in which question No. 21 is compulsory. [6 x 2 = 12]

Question 16.
State and explain Pauli’s exclusion principle.
Answer:
Pauli’s exclusion principle states that “No two electrons in an atom can have the same set of values of all four quantum numbers”.
Illustration: H(Z = 1) Is¹.

One electron is present in hydrogen atom, the four quantum numbers are n – 1, / = 0, m = 0 and s = + ½. For helium Z = 2. He: Is². In this one electron has the quantum number same as that of hydrogen. n = l, l = 0m = 0 and s = + ½. For other electron, fourth quantum number is different. i.e. n = 1, l = 0, m = 0 and s = – ½.

Question 17.
Give the general electronic configuration of lanthanides and actinides.
Answer:

• The electronic configuration of lanthanides is 4f^1-14 5d^10-1 6s².
• The electronic configuration of actinides is 5f^1-14 6d^10-1 7s².

Question 18.
Why alkaline earth metals are harder than alkali metals?
Answer:

1. The strength of metallic bond in alkaline earth metals is higher than alkali metals due to the presence of 2 electrons in its outermost shell as compared to alkali metals, which have only 1 electron in valence shell. Therefore, alkaline earth metals are harder than alkali metals.
2. The alkaline earth metals have greater nuclear charge and more valence electrons, thus metallic bonding is more effective. Due to this they are harder than alkali metals.

Question 19.
What are the limitations of the thermodynamics?
Answer:

1. Thermodynamics suggests feasibility of reaction but fails to suggest rate of reaction. It is concerned only with the initial and the final states of the system. It is not concerned with the path by which the change occurs.
2. It does not reveal the mechanism of a process.

Question 20.
Write the application of equilibrium constant?
Answer:

• predict the direction in which the net reaction will take place
• predict the extent of the reaction and
• calculate the equilibrium concentrations of the reactants and products.

Question 21.
How much volume of 6 M solution of NaOH is required to prepare 500 ml of 0.250 M NaOH solution?
Answer:
6% \(\frac{v}{v}\) aqueous solution contains 6g of methanol in 100 ml solution.

∴ To prepare 500 ml of 6% v/v solution of methanol 30g methanol is taken in a 500 ml standard flask and required quantity of water is added to make up the solution to 500 ml.

Question 22.
Define – Inductive effect.
Answer:

• It is defined as the change in the polarization of a covalent bond due to the presence of adjacent . bonded atoms or groups in the molecule. It is denoted as I-effect.
• Atoms or groups which lose electron towards a carbon atom are said to have a +1 effect.
  Example, CH₃ -, (CH₃)₂ CH (CH₃)₂ C- etc.
• Atoms or groups which draw electrons away from a carbon atom are said to have a -I effect.
  Example, -NO₂ -I, -Br, -OH, C₆H₅ etc.

Question 23.
How will you convert n-hexane into benzene? ,
Answer:
Alkanes with six to ten carbon atoms are converted into homologous of benzene at higher temperature and in the presence of a catalyst. This process is known as aromatisation.

For example, n -Hexane passed over Cr₂O₃ supported on alumina at 873 K gives benzene.

Question 24.
Why haloalkanes have higher boiling point than parent alkanes having the same number of carbons?
Answer:
Haloalkanes have higher boiling point than the parent alkane having the same number of carbon atoms because the intermolecular forces of attraction and dipole-dipole interactions are comparatively stronger in haloalkanes.

Part-III

Answer any six questions in which question No. 25 is compulsory. [6 x 3 = 18]

Question 25.
Balance the following equations by using oxidation number method.
K₂Cr₂O₇ + KI + H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + I₂ + H₂O
Answer:

Question 26.
Explain the periodic trend of ionization potential.
Answer:
(a) The energy required to remove the most loosely held electron from an isolated gaseous atom is called as ionization energy.
(b) Variation in a period.

Ionization energy is a periodic property.
On moving across a period from left to right, the ionization enthalpy value increases. This is due to the following reasons.

• Increase of nuclear charge in a period
• Decrease of atomic size in a period

Because of these reasons, the valence electrons are held more tightly by the nucleus. Therefore, ionization enthalpy increases.

(c) Variation in a group
As we move from top to bottom along a group, the ionization enthalpy decreases. This is due to the following reasons.

• A gradual increase in atomic size
• Increase of screening effect on the outermost electrons due to the increase of number of inner electrons.

Hence, ionization enthalpy is a periodic property.

Question 27.
Distinguish between hard water and soft water.
Answer:

Question 28.
Explain how the equilibrium constant K_c predict the extent of a reaction.
Answer:
(i) The value of equilibrium constant K_c tells us the extent of the reaction i.e., it indicates how far the reaction has proceeded towards product formation at a given temperature.
(ii) A large value of K_c indicates that the reaction reaches equilibrium with high product yield on the other hand, lower value of K_c indicates that the reaction reaches equilibrium with low product yield.
(iii) If K_c > 10³, the reaction proceeds nearly to completion.
(iv) If K_c < 10^-3 the reaction rarely proceeds.
(v) It the K_c is in the range 10^-3 to 10³, a significant amount of both reactants and products are present at equilibrium.

Question 29.
How many moles of solute particles are present in one litre of 10^-4 M K₂SO₄?
Answer:
In 10^-4 M K₂ SC₄), solution, there are 10^-4 moles of potassium sulphate.
K₂SO₄ molecule contains 3 ions (2K⁺ and 1 SO_4^2-)
1 m;ole of I^SO,, contains 3 x 6.023 x 10²³ ions
10″4 mole of K₂SO₄ contains 3 x 6.023 x 10²³ x 10^-4 ions = 18.069 x 10¹⁹

Question 30.
Draw the Lewis structures for the following.
(i) SO_4^2-
(ii) O₃
Answer:

Question 31.
Write a short notes on hyperconjugation.
Answer:

• The delocalization of electrons of a bond is called as hyperconjugation. It is a special stabilizing effect that results due to the interaction of electrons of a cr bond with the adjacent empty non-bonding P-orbitals resulting in an extended molecular orbital.
• Hyperconjugation is a permanent effect.
• For example, in propene, the a- electrons of C—H bond of methyl group can be delocalised into the n- orbital of doubly bonded carbon as represented below.
  

Question 32.
How will you prepare propyne using alkyene dihalide?
Answer:

Question 33.
Explain how the oxides of sulphur pollute the atomospheric air? Give its harmful effects.
Answer:

• Sulphur dioxide and sulphur trioxide are produced by burning sulphur-containing fossil fuels and by roasting of sulphide ores.
• SO₂ is a poisonous gas for both animals and plants. SO₂ causes eye irritation, coughing and respiratory diseases like asthma, bronchitis.
• SO₂ is oxidised to more harmful SO₃ gas in the presence of particulate matter present in the polluted air:
  SO₃ combines with atmospheric water vapour to form H₂SO₄ which comes down along with rain in the form of acid rain:
• Acid rain causes stone leprosy, affect aquatic ecosystem, corrode water pipes and causes respiratory ailment in humans and animals.

Part – IV

Answer all the questions.

Question 34.
(a) Urea is prepared by the reaction between ammonia and arbondioxide.
2NH_3(g) + CO_2(g) → (NH₂)₂ CO_(aq) + H₂O_(l)
In one process, 637.2 g of NH₃ are allowed to react with 1142 g of CO₂.
(i) Which of the two reactants is the limiting reagent? (2)
(ii) Calculate the mass of (NH₂)₂ CO formed. (2)
(iii) How much Of the excess reagent in grams is left at the end of the reaction? (1)
[OR]
(b) (i) A student reported the ionic radii of isoelectronic species X³⁺, Y²⁺ and Z^– as 136 pm, 64 pm and 49 pm respectively. Is that order correct? Comment. (2)
(ii) Explain any three factors which influence the ionization energy. (3)
Answer:
(a) (i) 2NH_3(g) + CO_2(g) → (NH₄)₂ CO_(aq) + H₂O_(l)
2 moles 1 mole 1 mole
No. of moles of ammonia =\(\frac{637.2}{17}\) = 37.45 mole
No. of moles of CO₂ = \(\frac{1142}{44}\) = 25.95 mole

As per the balanced equation, one mole of C02 requires 2 moles of ammonia.
∴ No. of moles of NH3 required to react with 25.95 moles of CO₂ is
= \(\frac{2}{1}\) x 25.95 = 51.90 moles.
∴ 37.45 moles of NH₃ is not enough to completely react with CO₂ (25.95 moles).
Hence, NH₃ must be the limiting reagent, and CO₂ is excess reagent.
(ii) 2 moles of ammonia produce 1 mole of urea.
∴ Limiting reagent 37.45 moles of NH₃ can produce \(\frac{1}{2}\) x 37.45 moles of urea.
= 18.725 moles of urea.
∴ The mass of 18.725 moles of urea = No. of moles x Molar mass
= 18.725 x 60
= 1123.5 g of urea.

(iii) 2 moles of ammonia requires 1 mole of CO₂.
∴ Limiting reagent 37.45 moles of NH₃ will require \(\frac{1}{2}\) x 37.45 moles of CO₂.
= 18.725 moles of CO₂.
∴ No. of moles of the excess reagent (CO₂) left = 25.95 – 18.725 = 7.225
The mass of the excess reagent (C02) left = 7.225 x 44
= 317.9 g of CO₂.
[OR]

(b) (i) X³⁺, Y²⁺, Z^– are isoelectronic.
∴ Effective nuclear charge is in the order
\(\left(Z_{\text {eff }}\right)_{Z}<\left(Z_{\text {eff }}\right)_{Y^{2+}}<\left(Z_{\text {eff }}\right)_{X^{3+}}\) and hence, ionic radii should be in the order \(r_{z}>r_{y^{2}}>r_{x^{3+}}\)
∴ The correct values are:

(ii) 1. Size of the atom: If the size of an atom is larger, the outermost electron shell from the nucleus is also larger and hence the outermost electrons experience lesser force of attraction. Hence it would be more easy to remove an electron from the outermost shell. Thus, ionization energy decreases with increasing atomic sizes.
\(Ionization enthalpy \propto \frac{1}{\text { Atomic size }}\)

2. Magnitude of nuclear charge: As the nuclear charge increases, the force of attraction . between the nucleus and valence electrons also increases. So, more energy is required to remove a valence electron. Hence I.E increases with increase in nuclear charge.
Ionization enthalpy α nuclear charge

3. Screening or shielding effect of the inner electrons: The electrons of inner shells form a cloud of negative charge and this shields the outer electron from the nucleus. This screen reduces the coulombic attraction between the positive nucleus and the negative outer electrons. If screening effect increases, ionization energy decreases.
\(Ionization enthalpy \propto \frac{1}{\text { Screening effects }}\)

4. Penetrating power of subshells s, p, d and f: The s-orbital penetrate more closely to the nucleus as compared to p-orbitals. Thus, electrons in s-orbitals are more tightly held by the nucleus than electrons in p-orbitals. Due to this, more energy is required to remove a electron from an s-orbital as compared to a p-orbital. For the same value of ‘n’, the penetration power decreases in a given shell in the order, s > p > d > f.

5. Electronic configuration: If the atoms of elements have either completely filled or exactly half filled electronic configuration, then the ionization energy increases.

Question 35.
(a) (i) Write the important common features of group 2 elements? (3)
(ii) What is meant by retrograde solubility? (2)
[OR]
(b) (i) Derive a general expression for the equilibrium constant K_p and K_c for the reaction. (3)
(ii) Write the K_p and K_c for NH₃ formation reaction. (2)
Answer:
(a) (i)

• Group 2 elements except beryllium are commonly known as alkaline earth metals because their oxides and hydroxides are alkaline in nature and these metal oxides are found in the Earth’s crust.
• Many alkaline earth metals are used in creating colours and used in fireworks.
• Their general electronic configuration is ns².
• Atomic and ionic radii of alkaline earth metals are smaller than alkali metals, on moving down the group, the radii increases.
• These elements exhibit +2 oxidation state in their compounds.
• Alkaline earth metals have higher ionizatoin enthalpy values than alkali metals and they are less electropositive than alkali metals.
• Hydration enthalpies of alkaline earth metals decreases as we go down the group.
• Electronegativity values of alkaline earth metals decrease down the group.
• Alkaline earth metal salts moistened with concentrated hydrochloric acid gave a characteristic coloured flame, when heated on a platinum wire in a flame.

(ii) Gypsum is a soft mineral and it is less soluble in water as the temperature increases. This is known as retrograde solubility, which is a distinguishing characteristic of gypsum.

[OR]

(b) (i) Consider a general reaction in which all reactants and products are ideal gases.

The equilibrium constant k_c is

The ideal gas equation is PV = nRT or P = \(\frac{n}{V}\) RT Since, Active mass = molar concentration = \(\frac{n}{V}\)
p = Active mass x RT

Based on the above expression, the partial pressure of the reacants and products can be expressed as,

By comparing equation (1) and (4), we get K_p = K_c (RT)^(Δng) ………………………. (5)

where Δn_g is the difference between the sum of number of moles of products and the sum of number of moles of reactants in the gas phase.

Question 36.
(a) (i) 0.24 g of a gas dissolves in 1 L of water at 1.5 atm pressure. Calculate the amount of dissolved gas when the pressure is raised to 6.0 atm at constant temperature. (3) (ii) What is a vapour pressure of liquid? (2)

[OR]

(b) (i) Explain the ionic bond formation in MgO and CaF₂. (3)
(ii) What is bond angle?
Answer:
(a) (i) \(P_{\text {solute }}=\mathrm{K}_{\mathrm{H}} x_{\text {solute in solution }}\)
At pressure 1.5 atm, \(p_{1}=\mathbf{K}_{\mathrm{H}} x_{1}\)  ……………. (1)
At pressure 6.0 atm, \(p_{2}=\mathbf{K}_{\mathbf{H}} \boldsymbol{x}_{\mathbf{2}}\) ………………… (2)
Dividing equation (1) by (2)
We get \(\frac{p_{1}}{p_{2}}=\frac{x_{1}}{x_{2}} \Rightarrow \frac{1.5}{6.0}=\frac{0.24}{x_{2}}\)
Therefore \(x_{2}=\frac{0.24 \times 6.0}{1.5}=0.96 \mathrm{g} / \mathrm{L}\)

(ii) 1. The pressure of the vapour in equilibrium with its liquid is called vapour pressure of the liquid at the given temperature.
2. The relative lowering of vapour pressure is defined as the ratio of lowering of vapour pressure to vapour pressure of pure solvent.
Relative lowering of vapour pressure = \(\frac{p_{\text {solvent }}^{\circ}-p_{\text {solution }}}{p_{\text {solvent }}^{\circ}}\)

[OR]

(b) (i) Magnesium oxide (MgO) :
Electronic configuration of Mg – ls² 2s² 2p⁵ 3s²
Electronic configuration of O – ls² 2s² 2p⁶ 3s² 3p⁴

• Magnesium has two electrons in its valence shell and oxygen has six electrons in its valence shell.
• By losing two electrons, Mg acquires the inert gas configuration of Neon and becomes a dipositive cation, Mg²⁺:
  \(\mathrm{Mg} \longrightarrow \mathrm{Mg}^{2+}+2 e^{-}\)
• Oxygen accepts the two electrons to become a dinegative oxide anion, O^2- thereby attaining the inert gas configuration of Neon:
  \(0+2 e^{-} \longrightarrow 0^{2-}\)
• These two ions, Mg²⁺ and O^2- combine to form an ionic crystal in which they are held together by electrostatic attractive forces.
• During the formation of magnesium oxide crystal 601.6 kJ mol^-1 energy is released. This favours the formation of magnesium oxide (MgO) and its stabilisation.

CaF₂, Calcium fluoride:

• Calcium, Ca : [Ar] 4s² Fluorine, F : [He] 2s² 2p⁵
• Calcium has two electrons in its valence shell and fluorine has seven electrons in its valence shell.
• By losing two electrons, calcium attains the inert gas configuration of Argon and becomes a dipositive cation, Ca²⁺.
• Two fluorine atoms, each one accepts one electron to become two uninegative fluoride ions (F-) thereby attaining the stable configuration of Neon.
• These three ions combine to form an ionic crystal in which they are held together by electrostatic attractive force.
• During the formation of calcium fluoride crystal 1225.91 kJ mol^-1 of energy is released. This favours the formation of calcium fluoride, CaF₂ and its stabilisation.

(ii) Covalent bonds are directional in nature and are oriented in specific direction in space. This directional nature creates a fixed angle between two covalent bonds in a molecule and this angle is termed as bond angle.

Question 37.
(a) (i) 0.33 g of an organic compound containing phosphorous gave 0.397 g of Mg₂P₂O₇ by the analysis. Calculate the percentage of P in the compound. (3)
(ii) Give the IUPAC names of the following compounds. (2)

[OR]

(b) (i) Describe the mechanism of nitration of benzene. (3)
(ii) How will you prepare m-dinitro benzene. (2)
Answer:
(a) (i) Weight of organic substance (w) = 0.33g
Weight of Mg₂P₂O₇ (x) = 0.397g
Percentage of phosphorous = \(\frac{62}{222} \times \frac{x}{w} \times 100=\frac{62}{222} \times \frac{0.397}{0.33} \times 100=33.59 \%\)
(ii) (a) 4-chloropent-2-ype .
(b) Acetophenone

(b) (i) Step-1: Generation of NO₂ electrophile.
HNO₃ + H₂SO₄ → NO₂ + HSO₄ + H₂O

Step-2: Attack of the electrophile on benzene ring to form arenium ion.

Question 38.
(a) Complete the following reaction, identify the products.

[OR]
(b) (i) What is green chemistry?
(ii) Differentiate – BOD and COD.
Answer:

[OR]
(b) (i)

• Green chemistry is a chemical philosophy encouraging the design of products and processes that reduces or eliminates the use and generation of hazardous substances.
• Efforts to control environmental pollution resulted in the development of science for the synthesis of chemicals favourable to the environment.
• Green chemistry means the science of environmentally favourable chemical synthesis.

(ii)

Students can Download Tamil Nadu 11th Chemistry Previous Year Question Paper June 2019 English Medium Pdf, Tamil Nadu 11th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 11th Chemistry Previous Year Question Paper June 2019 English Medium

Time: 2½ Hours
Total Score: 90 Marks

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I. II, III and IV are to be attempted separately.
4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
5. Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered in detail. Draw diagrams wherever necessary.

Part-I

Answer all the Questions. [15 x 1 = 15]
Choose the most appropriate answers from the given alternatives and write the option code and corresponding answer.

Question 1.
The oxidation number of Carbon in CH₂F₂ is …………………………. .
(a) + 4
(b) – 4
(c) 0
(d) + 2
Answer:
(c) 0

Question 2.
The energy of an electron in the third orbit of hydrogen atom is – E. The energy of an electron in the first orbit will be.
(a) – 3E
(b) \(-\frac{E}{3}\)
(c) \(-\frac{E}{9}\)
(d) – 9E
Answer:
(d) – 9E

Question 3.
The effective nuclear charge experienced by d1 electron in the given electronic configuration, (Is)² (2s, 2p)⁸ (3s, 3p)⁸ (3d)¹ (4s)² is ……………………. .
(a) 4
(b) 3
(c) 2.1
(d) 6.9
Answer:
(b) 3

Question 4.
The type of H-bonding present in orthonitro phenol and p-nitro phenol are respectively.
(a) Inter molecular H-bonding and intra molecular H-bonding
(b) Intra molecular H-bonding and inter molecular H-bonding
(c) Intra molecular H-bonding and no H-bonding
(d) Intra molecular H-bonding and intra molecular H-bonding
Answer:
(b) Intra molecular H-bonding and inter molecular H-bonding

Question 5.
When CaC₂ is heated in atmospheric nitrogen in an electric furnace, the compound formed is
(a) Ca(CN)₂
(b) CaNCN
(c) CaC₂N₂
(d) CaNC₂
Answer:

Question 6.
When an ideal gas undergoes unrestrained expansion, on cooling occurs because the molecules …………………………. .
(a) are above the inversion temperature
(b) exert on attractive force on each other
(c) do work equal to the loss in kinetic energy
(d) collide with loss of energy
Answer:
(b) exert on attractive force on each other

Question 7.
Among the following statements, which on is/are correct?
(i) During cyclic process the amount of heat absorbed by the surrounding is equal to work done on the surrounding.
(ii) Refractive index is an example for intensive property.
(iii) If the enthalpy change of a process is positive then the process is spontaneous.
(iv) The entropy of an isolated system increases during spontaneous process.
(a) (i), (ii), (iii)
(b) (i), (iv)
(c) (ii), (iv)
(d) (ii) only
Answer:
(d) (ii) only

Question 8.
If Kb and Kf for reversible reaction are 0.8 x 10 5 and 1.6 x 10^ respectively, the value of equilibrium constant is.
(a) 20
(b) 0.2 x 10^-1
(c) 0.05
(d) 0.2
Answer:
(a) 20

Question 9.
Assertion: Mixture of carbon tetrachloride and chloroform show positive deviation from Raoult’s law.
Reason: In the mixture, the inter molecular force of attraction between chloroform and carbon tetrachloride is weaker than those between molecules of carbon tetrachloride and chloroform molecules.
(a) Both assertion and reason are correct and reason is the correct explanation of assertion.
(b) Both assertion and reason are correct and reason is not the correct explanation for assertion.
(c) Both assertion and reason are false.
(d) Assertion is true, but reason is false.
Answer:
(a) Both assertion and reason are correct and reason is the correct explanation of assertion.

Question 10.
Shape and hybridisation of IF₅ are
(a) Trigonal bipyramidal, sp³d²
(b) Trigonal bipyramidal, sp³d
(c) Square pyramidal, sp³d²
(d) Octahedral, sp³d²
Answer:
(c) Square pyramidal, sp³d²

Question 11.
Which of the following is optically active?
(a) 3-chloro pentane
(b) 2-chloro propane
(c) meso-tartaric acid
(d) glucose
Answer:
(d) glucose

Question 12.
Which of the following species is not electrophile in nature? ‘
(a) Cl⁺
(b) BH₃
(c) H₃O⁺
(d) ⁺NO₂
Answer:
(c) H₃O⁺

Question 13.
group is ortho para directing and deactivating group.
(a) amino
(b) methyl
(c) halogen
(d) aldehyde
Answer:
(c) halogen

Question 14.
The raw material for Rasching process is
(a) chloro benzene
(b) phenol
(c) benzene
(d) anisole
Answer:
(c) benzene

Question 15.
cause kidney damage.
(a) Cadmium, Mercury
(b) Lead, Cadmium
(c) Freon, Fluoride
(d) Copper, Cadmium
Answer:
(a) Cadmium, Mercury

Part-II

Answer any six from the following questions. Question No. 24 is compulsory. [6 x 2 = 12]

Question 16.
What is syngas? How it is prepared?
Answer:
The mixture of CO and H₂ gases are called syngas.
Preparation: C + H₂O → CO + H₂

Question 17.
Write any two similarities between beryllium and aluminum.
(i) Beryllium and aluminium have same electronegativity values.
(ii) BeCl₂ and A1C1₃ forms dimeric structure. Both are soluble in organic solvents and are strong Lewis acids.

Question 18.
What is inversion temperature?
Answer:
The temperature below which a gas obey Joule-Thomson effect is called inversion temperature
\(\text { (T) } . \quad \mathrm{Ti}=\frac{2 \mathrm{a}}{\mathrm{Rb}}\)

Question 19.
What is the effect of added inert gas on the reaction of equilibrium?
Answer:
When an inert gas (i.e., a gas which does not react with any 6ther species involved in equilibrium) is added to an equilibrium system at constant volume, the total number of moles of gases present in the container increases, that is, the total pressure of gases increases, the partial pressure of the reactants and the products are unchanged. Hence at constant volume, addition of inert gas has no effect on equilibrium.

Question 20.
Linear form of carbon dioxide molecule has two polar bonds. Yet the molecule has zero dipole moment. Why?
Answer:

• The linear form of carbon dioxide has zero dipole moment, even though it has two polar bonds.
• In C02, there are two polar bonds [C = O], which have dipole moments that are equal in magnitude but have opposite direction.
• Hence the net dipole moment of the CO₂ is μ = μ₁ + μ₂ = μ₁ + (- μ₁) = 0
  • 
• In this case, \(\mu=\vec{\mu}_{1}+\vec{\mu}_{2}=\vec{\mu}_{1}+\left(-\vec{\mu}_{1}\right)=0\)

Question 21.
How do you detect the presence of nitrogen and sulphur together in an organic compound?
Answer:
If both N and S are present, a blood-red colour is obtained due to the following reactions

Question 22.
What happens when acetylene undergoes ozonolysis?
Answer:
When acetylene undergoes ozonolysis,

Question 23.
What is Green chemistry?
Answer:

• Green chemistry is a chemical philosophy encouraging the design of products and processes that reduces or eliminates the use and generation of hazardous substances.
• Efforts to control environmental pollution resulted in development of science for the synthesis of chemicals favourable to environment.
• Green chemistry means science of environmentally favourable chemical synthesis.

Question 24.
Calculate the orbital angular momentum for d and f orbital.
Answer:
Orbital angular momentum = √1(1 (1+1) h/2π.
For d orbital = √(2 x 3) h/2π = √6 h/2π
For f orbital =  √(3(3 + 1)) h/2π = √12 h/2π

Part-III

Answer any six froita the following questions. Q. No. 33 is compulsory. [6 x 3 = 18]

Question 25.
What do you understand by the term mole?
Answer:
The mole is defined as the amount of a substance which contains 6.023 x 10²³ particles such as atoms, molecules or ions. It is represented by the symbol “n”.

Question 26.
Ionisation potential of nitrogen is greater than that of oxygen. Explain by giving appropriate reason.
Answer:
N (Z = 7) Is² 2s² 2p_x¹ 2p_y¹ 2p_z¹. It has exactly half filled electronic configuration and it is more stable. Due to stability, ionization energy of nitrogen is high.

O (Z = 8) Is² 2s² 2p_x¹ 2p_y¹ 2p_z¹ It has incomplete electronic configuration and it requires less ionization energy. I.E₁ N > I.E₁ O

Question 27.
Among the alkali metal halides, which is covalent? Explain with reason.
Answer:
All alkali metal halides are ionic crystals. However lithium halides shows covalent character, as it is the smallest cation that exerts high polarising power on the halides.

Question 28.
Derive ideal gas equation.
Answer:
The gaseous state is described completely using the following four variables T, P, V and n. Each gas law relates one variable of a gaseous sample to another while the other two variables are held constant.

Therefore, combining all equations into a single equation will enable to account for the change in any or all of the variables.

• Boyle’s law: V α \(\frac{1}{P}\)
• Charles’law: V α T
• Avogadro’s law: V α n

We can combine these equations into the following general equation that describes the physical behaviour of all gases.
\(\mathrm{V} \propto \frac{\mathrm{nT}}{\mathrm{P}}\)
\(\mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{P}}\), where R = Proportionately constant.

The above equation can be rearranged to give PV = nRT – Ideal gas equation. Where, R is also known as Universal gas constant.

Question 29.
Define molar heat capacity. Give its unit.
Answer:
Molar heat capacity is defined as “the amount of heat absorbed by one mole of a substance in raising the temperature by 1 Kelvin”. It is denoted as C_m.

Unit of Molar heat capacity: SI unit of C_m is JK^-1 mol^-1.

Question 30.
What is vapour pressure of a liquid? What is relative lowering of vapour pressure?

1. The pressure of the vapour in equilibrium with its liquid is called vapour pressure of the liquid at the given temperature.
2. The relative lowering of vapour pressure is defined as the ratio of lowering of vapour pressure to vapour pressure of pure solvent.
3. Relative lowering of vapour pressure = \(\frac{p_{\text {solvent }}-p_{\text {solution }}}{p_{\text {solvent }}}^{\circ}\)

Question 31.
Explain a suitable method for purifying and separating liquids present in a mixture having very close boiling point.
Answer:
Fractional distillation: This method is used to purify and separate liquids present in the mixture having their boiling-point close to each other. The process of separation of the components in liquid mixture at their respective boiling points in the form of vapours and the subsequent condensation of those vapours is called fractional distillation. This method is applied in distillation of petroleum, coal tar and crude oil.

Question 32.
What is polymerisation? Explain the two types of polymerisation reaction of acetylene.
ANswer:
A polymer is a large molecule formed by the combination of large number of small molecules (monomers). This process is known as polymerisation.

Acetylene undergoes two types of polymerisation reactions, they are:

• Linear polymerisation
• Cyclic polymerisation

(i) Linear polymerisation: Acetylene forms linear polymer, when passed into a solution of cuprous chloride and ammonium chloride.

(ii) Cyclic polymerisation: Acetylene undergoes cyclic polymerisation on passing through red hot iron tube. Three molecules of acetylene polymerises to form benzene.

Question 33.
The bond length between all the four carbon atoms is same in 1,3 – butadine. Explain with reason.
Answer:

1,3-butadiene is a conjugated molecule with four overlapping p-orbital on adjacent atoms. And π-electrons are delocalised over four atoms. This shortens the bond length of central C – C bond. Thus, the bond length between all the four C-atoms are same in 1,3-butadiene.

Part – IV

Answer all the following questions. [5 x 5 = 25]

Question 34.
(a) (i) What are auto redox reactions? Give example.
(ii) Define orbital. What are the n and 1 values of 3p_x and 4d_x²-y² electron?
[OR]
(b) (i) Why hydrogen peroxide is stored in plastic containers, not in glass container?
(ii) Give the general electronic configuration of lanthanides and actinides.
Answer:
(a) (i) Auto redox reactions: It is a redox reaction in which a substance act as both the oxidizing agent and the reducing agent. (Reaction occur in the same substance)

(ii) 1. Orbital is a three dimensional space which the probability of finding the electron is maximum.
2. For 3p_x electron – n value =3, l value =1
3. For 4d_x²-y² electron n value = 4, l value = 2

[OR]

(b) (i) The aqueous solution of hydrogen peroxide is spontaneously disproportionate to give oxygen. The reaction is slow but it is explosive when it is catalyzed by metal or alkali dissolved from glass. For this reason, its solution are stored in plastic bottles.
\(\mathrm{H}_{2} \mathrm{O}_{2(\mathrm{aq})} \rightarrow \mathrm{H}_{2} \mathrm{O}_{(\mathrm{i})}+1 / 2 \mathrm{O}_{2(\mathrm{g})}\)

(ii) 1. The electronic configuration of lanthanides is 4f^1-14 5d^0-1 6s².
2. The electronic configuration of actinides is 5f^1-14 6d^0-1 7s².

Question 35.
(a) (i) Why blue colour appears during the dissolution of alkali metals in liquid ammonia?
(ii) What is Boyle’s temperature? What happens to real gases above and below the Boyle’s temperature?

[OR]

(b) (i) Derive the relation between KP and Kc for general homogeneous gaseous reaction.
(ii) How do you measure heat changes of a constant pressure?
Answer:
(a) (i) M + (x + y)NH₃ → [M(NH₃)_x]⁺ + [e(NH₃)_y]^–
The blue colour of the solution is due to the ammoniated electron which absorbs energy in the visible region of light and thus imparts blue colour to the solution. The solutions are paramagnetic and on standing slowly liberate hydrogen resulting in the formation of amide. In concentrated solution, the blue colour changes to bronze colour and become diamagnetic.

(ii)

1. Over a range of low pressures, the real gases can behave ideally at a particular temperature called as Boyle temperature or Boyle point.
2. The Boyle point varies with the nature of the gas.
3. Above the Boyle point, the compression point Z > 1 for real gases i.e. real gases show positive deviation.
4. Below the Boyle point, the real gases first show a-decrease for Z, reaches a minimum and then increase with increase in pressure. So, it is clear that at low pressure and at high temperature, the real gases behave as ideal gases.
5. 

[OR]

(b) (i) Consider a generaL reaction in which all reactants and products arc ideal gases.

P = Active mass x RT
Based on the above expression, the partial pressure of the reactants and products can be expressed as.
\(\begin{array}{ll}
p_{A}^{x}=[\mathrm{A}]^{x} \cdot[\mathrm{RT}]^{x} & p_{\mathrm{B}}^{y}=[\mathrm{B}]^{y} \cdot[\mathrm{RT}]^{y} \\
p_{\mathrm{C}}^{\prime}=[\mathrm{C}]^{\prime} \cdot[\mathrm{RT}]^{\prime} & p_{\mathrm{D}}^{m}=[\mathrm{D}]^{m} \cdot[\mathrm{RT}]^{m}
\end{array}\)

On substitution in equation (2),

where Δn_g is the difference between the sum of number of moles of products and the sum of number of moles of reactants in the gas phase.

(ii) 1. Measurement of heat change at constant pressure can be done in a coffee cup calorimeter.
2. We know that ΔH = q_p (at constant P) and therefore, heat absorbed or evolved, q_p at constant pressure is also called the heat of reaction or enthalpy of reaction, ΔH_r.
3. In an exothermic reaction, heat is evolved, and system loses heat to the surroundings. Therefore, q_p will be negative and Reaction AH. will also be negative. Mixture
4. Similarly in an endothermic reaction, heat is absorbed, q_p is positive and ΔH_r will also be positive.

Question 36.
(a) (i) Draw the M.O. diagram of oxygen molecule. Calculate its bond order and magnetic character.
(ii) Draw and explain the graph obtained by plotting solubility versus temperature for calcium chloride.

[OR]

(b) (i) Write the IUPAC names of the following compunds:

(ii) Calculate the formal charge on carbon and oxygen for the following structure

(a)

1. Electronic configuration of O atom is Is² 2s² 2p⁴
2. Electronic configuration of O₂ molecule is

3. Bond order \(\frac{N_{b}-N_{a}}{2}=\frac{10-6}{2}=2\)
4. Molecule has two unpaired electrons, hence it is paramagnetic.

(ii) Even though the dissolution Of calcium chloride is exothermic, the solubility increases moderately with increase in temperature. Here, the entropy factor also plays a significant role in deciding the position of the equilibrium.

[OR]

(b) (i) (x) 5-oxohexanoic acid (y) 4-chloropent-2-yne (z) 2-ethoxypropane
(ii)

Question 37.
(a) (i) Explain about the inductive effect.
(ii) What do you mean by confirmation? Explain about staggered conformation in ethane.

[OR]

(b) (i) Among the following compounds, o-dichloro benzene and p-dichloro benzene, which has higher melting point? Explain with reason.
(ii) Write notes on the adverse effect caused by ozone depletion.
Answer:
(a) (i) Inductive effect:
1. It is defined as the change in the polarization of a covalent bond due to the presence of adjacent bonded atoms or groups in the molecule. It is denoted as I-effect.
2. Atoms or groups which lose electron towards a carbon atom are said to have a +1 effect.
Example, CH₃ —, (CH₃)₂ CH —, (CH₃)₂ C— etc.
3. Atoms or groups which draw electrons away from a carbon atom are said to have a -I effect.
Example, —NO₂. -I, Br, -OH, C₆H₅ etc.
4. For example, consider ethane and ethyl chloride. The C—C bond in ethane is non¬polar while the C-C bond in ethyl chloride is polar. We know that chlorine is more electronegative than carbon and hence it attracts the shared pair of electrons between C—Cl in ethyl chloride towards itself.

This develops a slight negative charge on chlorine and a slight positive charge on carbon to which chlorine is attached. To compensate it, the C₁ draws the shared pair of electron between itself and C₂. This polarization effect is called inductive effect.

(ii) The rotation about C—C single bond axis yielding several arrangements of a hydrocarbon called conformers.

Staggered conformation: In this conformation, the hydrogens of both the carbon atoms are far apart from each other. The repulsion between the atoms is minimum and it is the most stable conformer.

[OR]

(b) (i) The melting point of p-dichloro benzene is higher than the melting point of the corresponding ortho and meta isomers. The higher melting which leads to more close packing of its moleules in the crystal lattice and consequently strong intermolecular attractive forces which requires more energy for melting.
p-dichlorobenzene > o-dichlorobenzene > m-dichlorobenZene

(ii) 1. The formation and destruction of ozone is a regular natural process, which never disturbs the equilibrium level of ozone in the stratosphere. Any change in the equilibrium level of ozone in the atmosphere will adversely affect the life in biosphere in the following ways.
2. Depletion of ozone layer will allow more UV rays to reach the earth surface and would cause skin cancer and also decreases the immunity level in human beings.
3. UV radiations affects plant proteins which lead to harmful mutation in plant cells.
4. UV radiations affect the growth of phytoplankton and as a result ocean food chain is disturbed and it even damages the fish productivity.

Question 38.
(a) (i) Calculate the uncertainty in the position of an electron, if the uncertainty in its velocity is 5.7 x 10⁵ ms^-1.
(ii) What is the mass of glucose (C₆ H₁₂ O₆) in it one litre solution which is isotonic with 6 gl^-1 of urea (NH₂ CONH₂)?

[OR]

(b) (i) An organic compound (A) of molecular formula C₂H₆O, on heating with conc.H₂SO₄ gives compound (B). (B) on treating with cold dilute alkaline KMnO₄ gives compound (C). Identify (A), (B) and (C) and explain the reactions.
(ii) A simple aromatic hydrocarbon (A) reacts with chlorine to give compound (B). Compound (B) reacts with ammonia to give compound (C) which undergoes carbylamine reaction. Identify (A), (B) and (C) arid explain the reactions.

(a) (i) Uncertainty in velocity = Δv = 5.7 x 10⁵ ms^-1.
Mass of an electron = m = 9.1 x 10^-31 kg.
Uncertainty in position = Δx = ?

(ii) Osmotic pressure of urea solution (π₁) = CRT

(b)

Students can Download Tamil Nadu 11th Chemistry Model Question Paper 2 English Medium Pdf, Tamil Nadu 11th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 11th Chemistry Model Question Paper 2 English Medium

Time: 2½ Hours
Total Score: 70 Marks

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I. II, III and IV are to be attempted separately.
4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
5. Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered in detail. Draw diagrams wherever necessary.

PART-I

Answer all the Questions. [15 x 1 = 1]
Choose the most suitable answer from the given four alternatives.

Question 1.
Match the List-I with List-Il using the correct code given below the list.

Code :
A B C D
(a) 4 3 1 2
(b) 3 4 1 2
(c) 2 1 4 3
(d) 3 1 2 4
Answer:
(b) 3 4 1 2

Question 2.
The energies E₁ and E₂ of two radiations are 25 eV and 50 eV respectively. The relation between their wavelengths i.e. Xl and X2 will be ……………… .
(a) \(\frac{\lambda_{1}}{\lambda_{2}}=1\)
(b) λ₁ = 2λ₂
(c) \(\lambda_{1}=\sqrt{25 \times 50} \lambda_{2}\)
(d) 2λ₁ = λ₂
Answer:
(b) λ₁ = 2λ₂

Solution:

Question 3.
In which of the following options the order of arrangement does not agree with the variation of property indicated against it? (NEET 2016 Phase-l)
(a) 1 < Br < Cl < F (increasing electron gain enthalpy)
(b) Li < Na < K < Rb (increasing metallic radius)
(c) AI³⁺ < Mg²⁺ < Na⁺ < F- (increasing ionic size)
(d) B < C < O < N (increasing first ionization enthalpy)
Answer:
(a) 1 < Br < Cl < F (increasing electron gain enthalpy)

Question 4.
For decolourisation of I mole of acidified KMnO4, the moles of H202 required is
(a) 1/2
(b) 3/2
(c) 5/2
(d) 7/2
Answer:
(c) 5/2

Question 5.
The compound (X) on heating gives a colourless gas and a residue that is dissolved in water to obtain (B). Excess of CO₂ is bubbled through aqueous solution of B, C is formed. Solid (C) on heating gives back X. (B) is ………………………. .
(a) CaCO₃
(b) Ca(OH)₂
(c) Na₂CO₂
(d) NaHCO₃
Answer:
(b) Ca(OH)₂

Solution:

Question 6.
Which law is used in the process of enriching the isotope of U²³⁵ from other isotopes?
(a) Boyle’s law
(b) Dalton’s law of partial pressure
(c) Graham’s law of diffusion
(d) Charles’ law
Answer:
(c) Graham’s law of diffusion

Question 7.
The enthalpies of formation of Al₂O₃ and Cr₂O₃ are -1596 kJ and -1134 kJ, respectively. ΔH for reaction 2A1 + Cr₂O₃ → 2Cr + Al₂O₃ is …………………… .
(a) -1365 kJ
(b) 2730 kJ
(c) -2730 kJ
(d) -462 kJ
Answer:
(d) -462 kJ

Solution:

Question 8.
Solubility of carbon dioxide gas in cold water can be increased by
(a) increase in pressure
(b) decrease in pressure
(c) increase in volume
(d) none of these
Answer:
(a) increase in pressure

Solution:

increase in pressure, favours the forward reaction.

Question 9.
Two liquids X and Y on mixing gives a warm solution. The solution is
(a) ideal
(b) non-ideal and shows positive deviation from Raoults law
(c) ideal and shows negative deviation from Raoults Law
(d) non-ideal and shows negative deviation from Raoults Law
Answer:
(d) non-ideal and shows negative deviation from Raoults Law

Solution: ΔH_mix is negative and show negative deviation from Raoults law.

Question 10.
Which of the following molecules have bond order equal to 1?
(a) NO, HF, HC1, Li₂, CO
(b) H₂, Li₂, HF, Br₂, HC1
(c) Li₂, B₂, CO, NO, He₂⁺
(d) B₂, CO, He₂⁺, NO, HF
Answer:
(b) H₂, Li₂, HF, Br₂, HC₁

Question 11.
How many cyclic and acyclic isomers are possible for the molecular formula C3H60?
(a) 4
(b) 5
(c) 9
(d) 10
Answer:
(c) 9

Question 12.
Statement-I: Fluoro acetic acid is stronger acid than acetic acid
Statement-II: Fluorine has high electronegativity and it is facilitate to dissociate the O-H bond easily.
(а) Statement-I and II are correct and statement-II is correct explanation of statement-I.
(b) Statement-I and II are correct but statement-II is not correct explanation of statement-I.
(c) Statement-I is correct but statement-II is wrong.
(d) Statement-I is wrong but statement-II is correct
Answer:
(а) Statement-I and II are correct and statement-II is correct explanation of statement-I.

Question 13.
Consider the following statements.
(i) Alkanes are saturated hydrocarbons in which all the bonds between the carbon atoms are single bond.
(ii) Alkenes are saturated hydrocarbons in which atleast one carbon – carbon double bond is present.
(iii) Alkynes are unsaturated hydrocarbons in which atleast one carbon – carbon triple bond is present.
Which of the above statement is/are correct?
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iii)
(d) only (ii)
Answer:
(b) (i) and (iii)

Question 14.
Product X is ……………….. .
(a) CH₄
(b) C₂H₆
(c) HCHO
(d) CH₃OH
Answer:
(a) CH₄

Question 15.
Eutrophication causes reduction in
(a) dissolved oxygen
(b) dissolved nitrogen
(c) dissolved salts
(d) all of the above
Answer:
(a) dissolved oxygen

Part-II

Answer any six questions in which question No. 22 is compulsory. [6 x 2 = 12]

Question 16.
Explain about the factors that affect electronegativity?
Answer:

• Effective nuclear charge: As the nuclear charge increases, electronegativity also increases along the periods.
• Atomic radius: The atoms in smaller size will have larger electronegativity.

Question 17.
How do you convert para-hydrogen into ortho-hydrogen?
Answer:
Para hydrogen can be converted into ortho hydrogen by the following ways:

• By treating with catalysts platinum or iron.
• By passing an electric discharge.
• By heating > 800°C.
• By mixing with paramagnetic molecules such as O₂, NO, NO₂.
• By treating with nascent/atomic hydrogen.

Question 18.
Gases don’t settle at the bottom of a container?
Answer:
Gases by definition are the least dense state of matter. They have negligible intermolecular forces of attraction. So they are all free to roam separately. So the least dense gas particles will not sink at the bottom of a container.

Question 19.
Define – Normality.
Answer:
Normality (N): It is defined as the number of gram equivalents of solute in 1 litre of the solution.
\(\mathrm{N}=\frac{\text { Number of gram equivalent of solute }}{ \text { Volume of solution (in }\mathrm{L})}\)

Question 20.
If there is no change in concentration why is the equilibrium state considered dynamic?
Answer:
At chemical equilibrium the rate of two opposing reactions are equal and the concentration of reactants and products do not change with time. This condition is not static and is dynamic, because both the forward and reverse reactions are still occurring with the same rate and no macroscopic change is observed. So chemical equilibrium is in a state of dynamic equilibrium.

Question 21.
In CH4, NH3 and H20 the central atom undergoes sp3 hybridization – yet their bond angles are different. Why?
Answer:

1. In CH₄, NH₂ and H₂O the central atom undergoes sp³ hybridisation. But their bond angles are different due to the presence of lone pair of electrons.
2. It can be explained by VSEPR theory. According to this theory, even though the hybridisation . is same, the repulsive force between the bond pairs and lone pairs are not same.
3. Bond pair-Bond pair < Bond pair — Lone pair < Lone pair -Lone pair
   So due to the varying repulsive force the. bond pairs and lone pairs are distorted from regular geometry and organise themselves in such a way that repulsion will be minimum and stability will be maximum.
4. In case of CH₄, there are 4 bond pairs and no lone pair of electrons. So it remains in its regular geometry, i.e., tetrahedral with bond angle = 109° 28’.
5. H₂O has 2 bond pairs and 2 lone pairs. There is large repulsion between Ip-lp. Again repulsion between Ip-bp is more than that of 2 bond pairs. So 2 bonds are more restricted to form inverted V shape (or) bent shape molecule with a bond angle of 104° 35’.
6. NH₃ has 3 bond pairs and 1 lone pair. There is repulsion between Ip-bp. So 3 bonds are more restricted to form pyramidal shape with bond angle equal to 107° 18’.

Question 22.
Identify the compound in the following reaction,

Answer:

Question 23.
ClassIfy the following compounds in the form of alkyl, allylic, vinyl, benzyllc halides,
(a) CH₂ = CH – CH₂ CI
(b) C₆H₅CH₂I
(c) CH₃ – CH – CH₃
(d) CH₂ = CH – Cl
Answer:

Question 24.
Define — Smog.
Answer:

1. Smog is a combination of’ smoke and fog which form droplets that remains suspended in the air.
2. Smog is a chemical misture of gases that forms a brownish yellow haze. It mainly consists of ground level ozone, oxideš of nitrogen, volatile organic compounds, SO2, acidic aerosols and some other gases.

Part – III

Answer any six questions in which question No. 30 is compulsory. [6 x 3 = 18]

Question 25.
Explain how effective nuclear charge is related with stability of the orbital?
Answer:

1. In a multi-electron atom, in addition to the electrostatic attractive force between the electron and nucleus, there exists a repulsive force among the electrons.
2. These two forces are operating in the opposite direction. This results in the decrease in the nuclear force of attraction on electron.
3. The net charge experienced by the electron is called effective nuclear charge.
4. The effective nuclear charge depends on the shape of the orbitals and it decreases with increase in azimuthal quantum number.
5. The order of the effective nuclear charge felt by a electron in an orbital within the given shell is s > p > d > f.
6. Greater the effective nuclear charge, greater is the stability of the orbital. Hence, within a given energy level, the energy of the orbitals are in the following order s < p < d < f.

Question 26.
Write a notes on hydrogen sponge.
Answer:
(i) Hydrogen sponge (or) Metal hydride e.g., palladium-hydrogen system is a binary hydride (Pdll).
(ii) Upon heating, H atoms diffuse through the metal to the surface and recombine to form molecular hydrogen. Since no other gas behaves this way with palladium, this process has been used to separate hydrogen gas from other gases:
\(2 \mathrm{Pd}_{(\mathrm{s})}+\mathrm{H}_{2(\mathrm{g})} \rightleftharpoons 2 \mathrm{PdH}_{(\mathrm{s})}\)
(iii) The hydrogen molecule readily adsorb on the palladium surface, where it dissociates into atomic hydrogen. The dissociated atoms, dissolve into the interstices or voids (octahedral or tetrahedral) of the crystal lattice.

(iv) Technically, the formation of metal hydride is by chemical reaction but it behaves like a physical storage method, i. e., it is absorbed and released like a water sponge. Such a reversible uptake of hydrogen in metals and alloys is also attractive for hydrogen storage and for rechargeable metal hydride battery applications.

Question 27.
In what way lithium differs from other metals of the same group?
Answer:

Lithium	Other elements of the family
Very hard.	Very Soft.
High melting and boiling point.	Low melting and boiling point.
Least reactive.	More reactive.
Reacts with nitrogen to get Li3N.	No reaction.
Reacts with bromine slowly.	Reacts violently.
Burnt in air gives monoxide only.	Burnt in air gives peroxides also, apart from monoxides. K, Rb and Cs gave superoxides.
Compounds are partially soluble in water.	Highly soluble in water.
Lithium nitrate decomposes to form an oxide.	Other metals on heating gives nitrite.
Extremely small in size.	Comparatively large in size.
Li+ has greater polarizing power.	Other M+ ions have comparatively larger polarizing power.

Question 28.
A 0.25 M glucose solution at 370.28 K has approximately the pressure as blood does what is the osmotic pressure of blood?
Sol.
C = 0.25 M
T = 370.28 K
(π)_glucose = CRT
(π) = 0.25 mol L^-1 x 0.082 L atm K^-1 mol^-1 x 370.28K
= 7.59 atm

Question 29.
Write the important principles of VSEPR theory to predict the shape of molecules.
Answer:

1. The shape of a molecule depends upon the no. of electron pairs around the central atom.
2. There is a repulsive force between the electron pairs, which tend to repel one another.
3. The electron pairs in space tend to occupy such positions that they are at maximum distance, so that the repulsive force will be minimum.
4. A multiple bond is treated as if it is a single bond and the remaining electron pairs which constitute the bond may be regarded as single super pair.

Question 30.
0.26 g of an organic compound gave 0.039 g of water and 0.245 g of CO₂ on combustion. Calculate the percentage of carbon and hydrogen.
Answer:
Weight of organic compound = 0.26 g
Weight of water = 0.039 g
Weight of CO₂ = 0.245 g

Percentage of hydrogen
18 g of water contains 2 g of hydrogen
0.039 g of water contains = \(\frac{2}{18} \times \frac{0.039}{0.26} \text { of } \mathrm{H}\)
∴ % of hydrogen = \(\frac{0.039}{0.26} \times \frac{2}{18}\) x 100 = 1.66%

Percentage of carbon
44 g of CO₂ contains 12 g of C
0. 245 g of CO₂ contains = \(\frac{12}{44} \times \frac{0.245}{0.26}\) g of C
% of Carbon = \(\frac{12}{44} \times \frac{0.245}{0.26}\) x 100 = 25.69%

Question 31.
How will you distinguish l-butyne and 2-butyne?

Answer:
In l-butyne, terminal carbon atom contains atom one acidic hydrogen, therefore it will react with silver nitrate in the presence of ammonium hydroxide to give silver butynide. Whereas 2-butyne does not undergo such type of the reaction, because of the absence of acidic hydrogen.

Question 32.
Write the structure of the following compounds,
(a) l-Bromo-4 -ethyl cyclohexane
(b) 1, 4-Dichlorobut-2-ene
(c) 2-Chloro-3-methyl pentane
Answer:
(a) l-bromo-4 ethylcyclo hexane
(b) 1,4-dichloro but-2-ene
(c) 2-chloro-3-methyl pentane Br

Question 33.
Explain how does green house effect cause global warming.
Answer:

1. The earth’s atmosphere allows most of the visible light from the sun to pass through and reach the earth’s surface. As earth’s surface is heated by sunlight, it radiates a part of this energy back towards the space as longer IR wavelengths.
2. Some of the heat is trapped by CH₄, CO₂, CFCs and water vapour present in the atmosphere. They absorb IR radiations and block a large portion of earth’s emitted radiations.
3. The radiations thus absorbed is partly remitted to the earth’s surface. Therefore the earth’s surface gets heated up by a phenomenon called greenhouse effect.
4. Thus greenhouse effect is defined as the heating up of the earth surface due to trapping of infrared radiations reflected by earth’s surface by CO₂ layer in the atmosphere. The heating up of the earth through the greenhouse effect is called global warming.

Part-IV

Answer all the questions. [5 x 5 = 25]

Question 34.
(a) (i) Write the steps to be followed for writing empirical formula. (3)
(ii) What do you understand by the terms empirical formula and molecular formula? (2)
[OR]
(b) (i) Explain the shape of s and p-orbital. (2)
(ii) The mass of an electron is 9.1 x 10^-31 kg. If its kinetic energy is 3.0 x 10^-25 J. Calculate its wavelength. (3)
Answer:
(a) (i) Empirical formula shows the ratio of numher of atoms of different elements in one molecule of the compound.

Steps for finding the Empirical formula:
The percentage of the elements in the compound is determined by suitable methods and from the data collected; the empirical formula is determined by the following steps.

1. Divide the percentage of each element by its atomic mass. This will give the relative number of atoms of various elements present in the compound.
2. Divide the atom value obtained in the above step by the smallest of them so as to get a simple ratio of atoms of various elements.
3. Multiply the figures so obtained, by a suitable integer if necessary in order to obtain whole number ratio.
4. Finally write down the symbols of the various elements side by side and put the above numbers as the subscripts to the lower right hand of each symbol. This will represent the empirical formula of the compound.
5. Percentage of Oxygen = 100 – Sum of the percentage masses of all the given elements.

(ii)

(b) (i) s-orbital: For is orbital, l = 0, m = 0, f(θ) = 1/√2 and g((φ) = 1√2π. Therefore, the angular distribution function is equal to 1/2√π. i.e. it is independent of the angle θ and φ. Hence, the probability of finding the electron is independent of the direction from the nucleus. So, the shape of the s orbital is spherical.

p-orbital: For p orbitals l = 1 and the corresponding m values are -1, 0 and +1. The three different m values indicates that there are three different orientations possible for p orbitals. These orbitals are designated as p_x, p_y and p_z. The shape of p orbitals are dumb bell shape.

(ii) Step I. Calculation of the velocity of electron
Kinetic energy =1/2 mu² = 3.0 x 10^-25 J = 3.0 x 10^-25 kg m² s^-2

Step II. Calculation of wavelength of the electron According to de Broglie’s equation,

Question 35.
(a) (i) Explain the exchange reactions of deuterium, (2)
(ii) Explain the action of soap with hard water. (3)
[OR]
(b) (i) Derive ideal gas equation. (3)
(ii) CO₂ gas cannot be liquified at room temperature. Give the reason. (2)
Answer:
(a) (i) Deuterium can replace reversibly hydrogen in compounds either partially or completely depending upon the reaction conditions. These reactions occur in the presence of deuterium.

(ii) Action of soap with hard water:

• The cleaning capacity of soap is reduced when used in hard water.
• Soaps are sodium or potassium salts of long chain fatty acids.
• When soap is added to hard water, the divalent magnesium and calcium cations present in hard water relict with soap.
• The sodium salts present in soaps are converted to their corresponding magnesium and calcium.salts which are precipitated as scum or precipitate.
  

(b) (i) Ideal gas equation:
The gaseous state is described completely using the following four variables T, P, V and n. Each gas law relates one variable of a gaseous sample to another while the other two variables are held constant.

Therefore, combining all equations into a single equation will enable to account for the change in any or all of the variables.

Boyle’s law: V α \(\frac{1}{P}\)
Charles’ law: V α T
Avogadro’s law: V α n

We can combine these equations into the following general equation that describes the physical behaviour of all gases.
\(\mathrm{V} \propto \frac{\mathrm{nT}}{\mathrm{P}}\)

The above equation can be rearranged to give PV = nRT – Ideal gas equation.

Where, R is also known as Universal gas constant.
(ii) Only below the critical temperature, by the application of pressure, a gas can be liquefied. CO₂ has critical temperature as 303.98 K. Room temperature means (30 + 273 K) 303 K. At room temperature, (critical temperature) even by applying large amount of pressure CO₂ cannot be liquefied. Only below the critical temperature, it can be liquefied. At room temperature, CO₂ remains as gas.

Question 36.
(a) (i) Why deep-sea divers use air diluted with helium gas in their air tanks? (3)
(ii) What is molal depression constant? Does it depend on nature of the solute. (2)

[OR]

(b) (i) Write the resonance structures for ozone molecule and N20? (2)
(ii) Draw MO diagram of CO and calculate its bond order. (3)
Answer:
(a) (i) 1. Deep-sea divers carry a compressed air tank for breathing at high pressure under water. This air tank contains nitrogen and oxygen which are not very soluble in blood and other body fluids at normal pressure.
2. As the pressure at the depth is far greater than the surface atmospheric pressure, more nitrogen dissolves in the blood when the diver breathes from tank.
3. When the divers ascends to the surface, the pressure decreases, the dissolved nitrogen comes out of the blood quickly forming bubbles in the blood stream. These bubbles restrict blood flow, affect the transmission of nerve impulses and can even burst the capillaries or block them. This condition is called “the bends” which are painful and dangerous to life.
4. To avoid such dangerous condition they use air diluted with helium gas (11.7 % helium, 56.2% nitrogen and 32.1% oxygen) of lower solubility of helium in the blood than nitrogen.

(ii) K_f = molar freezing point depression constant (or) cryoscdpic constant. ΔT_f = K_f. m, where ΔT_f = depression in freezing point, m = molality of the solution K_f= cryoscopic constant If m = 1, ΔT_f = k_f

i. e., cryoscopic constant is equal to the depression in freezing point for 1 molal solution cryoscopic constant depends on the molar concentration of the solute particles. K_f is directly proportional to the molal concentration of the solute particles.
\(\Delta \mathrm{T}_{f}=\mathrm{K}_{f} \times \frac{\mathrm{W}_{\mathrm{B}} \times 1000}{\mathrm{M}_{\mathrm{B}} \times \mathrm{W}_{\mathrm{A}}}\)

W_B = mass of the solute, W_A = mass of solvent, M_B = molecular mass of the solute.

[OR]

(b)

• Electronic configuration of C atom: Is², 2s² 2p²
  Electronic configuration of O atom: ls² 2s² 2p⁴
• Electronic configuration of CO molecule is: \(\sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \pi 2 p_{y}^{2} \pi 2 p_{z}^{2} \sigma 2 p_{x}^{2}\)
• Bond order = \(\frac{N_{b}-N_{a}}{2}=\frac{10-4}{2}=3\)
• Molecule has no unpaired electron, hence it is diamagnetic.

Question 37.
(a) (i) What are cis-trans isomerism? Explain with example.
(ii) Give the IUPAC names of the following compounds,

[OR]

(b) (i) What is enzymatic browning?
(ii) How will you convert nitrile into primary amine?
Answer:
(a) (i)

• Geometrical isomers are the stereoisomers which have different arrangement of groups or atoms around a rigid framework of double bonds. This type of isomerism occurs due to restricted rotation of double bonds or about single bonds in cyclic compounds.
• In 2-butene, the carbon-carbon double bond is sp² hybridised. The carbon-carbon double bond consists of a a bond and a n bond. The presence of π bond lock the molecule in one position. Hence, rotation around C = C bond is not possible.
• 
• These two compounds are termed as geometrical isomers and are termed as cis and trans form.
• The cis isomer is the one in which two similar groups are on the same side of the double bond. The_trans isomer is that in which two similar groups are on the opposite side of the double bond. Hence, this type of isomerism is called cis-trans isomerism.

(ii)

[OR]

(b) (i) 1. Apples contains an enzyme called polyphenol oxidase (PPO) also known as tyrosinase.
2. Cutting an apple exposes its cells to the atmospheric oxygen and oxidizes the phenolic compounds present in apples. This is called the “enzymatic browning” that turns a cut apple brown.
3. In addition to apples, enzymatic browning is also evident in bananas, pears, avocados and even potatoes.

(ii) Converting nitrile into primary amine:

Question 38.
(a) An organic compound (A) of a molecular formula C₂H₄, decolourises bromine water. (A) on reacts with Cl₂ gives (B). (A) reacts with HBr to give (C). (A) reacts with hydrogen in the presence of Ni to give D. Identify A, B, C and D. Explain the reactions. (5)

[OR]

(b) (i) Explain the harmful effects of acid rain. (3)
(ii) What is Eutrophication? (2)
Answer:
(a) 1. C₂H₄ (A), decolorises the bromine water. Therefore it contains double bond. Hence (A) is ethylene.

[OR]

(b) (i)

1. Acid rain causes damage to buildings made us of marbles. This attack on marble is termed as stone leprosy.
   CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂↑
2. Acid rain affects plant and animal life in aquatic ecosystem.
3. It is harmful for agriculture, as it dissolves in the earth and removes the nutrients needed for the growth of plants.
4. It corrodes water pipes resulting in the leaching of heavy metals such as iron, lead and copper into drinking water which have toxic effects.
5. It causes respiratory ailment in humans and animals.

(ii) Eutrophication: When the growth of algae increases in the surface of water, dissolved oxygen in water is greatly reduced. This phenomenon is known as eutrophication. Due to this growth of fishes gets inhibited.

Students can Download Tamil Nadu 11th Physics Model Question Paper 1 English Medium Pdf, Tamil Nadu 11th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 11th Physics Model Question Paper 1 English Medium

Instructions:

1. The question paper comprises of four parts
2. You are to attempt all the parts. An internal choice of questions is provided wherever: applicable
3. All questions of Part I, II, III, and IV are to be attempted separately
4. Question numbers 1 to 15 in Part I are Multiple choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are lo be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are lo be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are lo be answered in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 70

Part-I

Answer all the questions. [15 x 1 = 15]

Question 1.
A dimensionless quantity
(a) never has a unit
(b) always has a unit
(c) may have or have not a unit
(d) none of the above
Answer:
(c) may have or have not a unit

Question 2.
The magnitude of average velocity is equal to average speed when the particle moving with
(a) variable speed
(b) constant velocity
(c) variable velocity
(d) constant acceleration
Answer:
(b) constant velocity

Question 3.
Two blocks of masses m and 2 m are placed on a smooth horizontal surface as shown. In the first case only a force F₁ is applied from the left. Later only a force F₂ is applied from the-right. If the force acting at the interface of the two blocks in the two cases is same, then

F₁ : F₂ is ……………. .
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1 : 3
Answer:
(c) 2 : 1

Question 4.
The instantaneous angular position of a point on a rotating wheel is given by the equation θ(t) = 2t³ – 6t². The torque on the wheel becomes zero at …………… .
(a) t = 1s
(b) t = 0.5s
(c) t = 0.25 s
(d) t = 2s
Answer:
(a) t = 1s

Hint: According to question, torque τ = 0 it means that \(\propto=\frac{d^{2} \theta}{d t^{2}}\)

Question 5.
The workdone by the conservative force for a closed path is
(a) always negative
(b) zero
(c) always positive
(d) not defined
Answer:
(b) zero

Question 6.
The rain drops are spherical in shape due to
(a) gravity
(b) due to contraction
(c) surface tension
(d) viscosity
Answer:
(c) surface tension

Question 7.
Which of the following diagrams does not represent a streamline flow?

Answer:

Question 8.
When a spiral spring is stretched by a force, the resultant strain is
(a) volume
(b) shear
(c) tensile
(d) all these
Answer:
(b) shear

Question 9.
In a simple harmonic oscillation, the acceleration against displacement for one complete oscillation will be
(a) an ellipse
(b) a circle
(c) a parabola
(d) a straight line
Answer:
(d) a straight line

Question 10.
Beats are produced by two progressive waves. Maximum loudness at the waxing is x times the loudness of each wave. The value of x is
(a) 1
(b) \(\sqrt{2}\)
(c) 2
(d) 4
Answer:
(d) 4

Hint: The maximum amplitude is the sum of two amplitude, i.e. a + a = 2a Hence, maximum intensity is proportional to 4a² Loudness of one wave is given by \(x=\frac{4 a^{2}}{a^{2}}=4\)

Question 11.
Water does not freeze at the bottom of the lakes in winter because ………… .
(a) ice is a good conductor of heat
(b) ice reflects heat and light
(c) of anomalous expansion of water between 4°C to 0°C
(d) nothing can be said ‘
Answer:
(c) of anomalous expansion of water between 4°C to 0°C

Question 12.
According to Stefan Boltzmann law, the heat radiated by a black body is directly proportional to
(a) T²
(b) T³
(c) T⁴
(d) T⁸
Answer:
(c) T⁴

Question 13.
Two springs of spring constant 1500 Nm^-1 and 300 Nm^-1 respectively are stretched with the same force. They will have potential energy in the ratio
(a) 1 : 2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1
Answer:
(b) 2 : 1

Hint:

Question 14.
When the axis of rotation passes through its centre of gravity, then the moment of inertia of a rigid body is ………………. .
(a) reduced to its minimum value
(b) zero
(c) increased to its maximum value
(d) infinity
Answer:
(a) reduced to its minimum value

Question 15.
Power is given by
\((a) \mathrm{P}=\frac{\overrightarrow{\mathrm{F}}}{\overline{\mathrm{V}}}
(b) \frac{\vec{F}^{2}}{\vec{V}}
(c) \frac{\overrightarrow{\mathrm{F}}}{\overrightarrow{\mathrm{V}}^{2}}
(d) \overrightarrow{\mathbf{F}} \cdot \overrightarrow{\mathbf{V}}\)
Answer:
\((d)\overrightarrow{\mathbf{F}} \cdot \overrightarrow{\mathbf{V}}\)

Part-II

Answer any six questions in which Q. No 23 is compulsory. [6 x 2 = 12]

Question 16.
In solids both longitudinal and transverse waves are possible, but transverse waves are not produced in gases. Why?
Answer:
It is because transverse waves travel in the form of crests and troughs which involves the change in shape. For the propagation of transverse waves, medium should posses elasticity of shape. Therefore transverse waves are not produced in gases.

Question 17.
It is possible that the brakes of a bus are so perfect that the bus stops instantaneously? If not, why?
Answer:
No. If the car stops instantaneously, this means that the velocity is reduced to zero is an infinitesimally small interval of time. This would further mean that the car has infinite deceleration. This is not possible. Thus, we cannot have a car which can stop instantaneously.

Question 18.
Why does a rubber ball bounce higher on hills than in plains?
Answer:
The maximum height attained by the projectile is inversely proportional to acceleration due to gravity. At greater height, acceleration due to gravity will be lesser than plains. So ball can bounce higher in hills than in plains.

Question 19.
State conservation of angular momentum.
Answer:
The law of conservation of angular momentum states that when no external torque acts on the body the net angular momentum of a rotating rigid body remains constant.

Question 20.
Mountain roads rarely go straight up but wind up gradually. Why?
Answer:
Frictional force f is given by f = μ mg cos θ. If the roads go straight up the angle of slope θ would be large and frictional force will be less and vehicles may slip.

Question 21.
State Newton’s second law.
Answer:
The force acting on an object is equal to the rate of change of its momentum
\(\overrightarrow{\mathrm{F}}=\frac{d \vec{p}}{d t}\)

Question 22.
Why the specific heat capacity of constant pressure is greater than the specific heat capacity at constant volume?
Answer:
It implies that to increase the temperature of the gas at constant volume requires less heat than increasing the temperature of the gas at constant pressure. In other words s_p is always greater than s_v.

Question 23.
Define angle of contact.
Answer:
The angle between tangents drawn at the point of contact to the liquid surface and solid surface inside the liquid is called the angle of contact for a pair of solid and liquid.

Question 24.
Define forced oscillation. Give an example.
Answer:
The body executing vibration initially vibrates with its natural frequency and due to the presence of external periodic force, the body later vibrates with the frequency of the applied periodic force. Such vibrations are known as forced vibrations.
Example: Sound boards of stringed instruments.

Part – III

Answer any six questions in which Q.No. 29 is compulsory. [6 x 3 = 18]

Question 25.
Convert G = 6.67 x 10^-11 Nm² kg^-2 to cm³ g^-1 s^-2
Answer:
G = 6.67 x 10^-11 N m² kg^-2
= 6.67 x 10^-11 (kg m s^-2) (m² kg^-2)
= 6.67 x 10^-11 kg^-1 m³ s^-2
= 6.67 x 10^-11 (1000 g)^-1 (100 cm)³ (s^-2) .
= 6.67 x 10^-11 x \(\frac{1}{1000}\) x 100 x 100 x 100 g^-1 cm³ s^-2
= 6.67 x 10^-8 g^-1 cm³ s^-2

Question 26.
A object is thrown with initial speed 5 ms^-1 with an angle of projection 30°. What is the height and range reached by the particle?
Answer:
Data: Initial speed = 5 ms^-1.
Angle of projection = 30°

Question 27.
Three point masses m₁, m₂, m₃ are located at the vertices of equilateral triangle of side “a”. What is the moment of inertia of system about an axis along the altitude of triangle passing through m₁.
Answer:

Question 28.
A box is pulled with a force of 25 N to produce a displacement of 15 m. If the angle between the force and displacement is 30°, find the work done by the force?
Answer:

Force, F = 25 N
Displacement, dr = 15 m
Angle between F and dr, θ = 30°
Work done, W = F dr cos θ
W = 25 x 15 x cos 30 = 25 x 15 x \(\frac{\sqrt{3}}{2}\)
W = 324.76 J

Question 29.
The time period of a mass suspended by a spring (force constant K) is T. If the spring is cut into three equal pieces, what will be the force constant of each part? If the same mass be suspended from one piece what will be the periodic time?
Answer:
Consider the spring be made of a combination of three springs in series each of spring constant k. The effective spring constant K is given by
\(\frac{1}{K}=\frac{1}{k}+\frac{1}{k}+\frac{1}{k}=\frac{3}{k}\)
or K = \(\frac{k}{3}\) or k = 3K

∴ Time period of vibration of a body attached to the end of this spring,

When the spring is cut into three pieces, the spring constant k, time period of vibration of a body attached to the end of this spring,

T₁ = 2π\(\sqrt{\frac{m}{k}}\) ………………………………. (2)

From eqns. (1) and (2)
\(\frac{T_{1}}{T}=\frac{1}{\sqrt{3}} \text { or } T_{1}=\frac{T}{\sqrt{3}}\)

Question 30.
Give various applications of viscosity.
Answer:

1. The oil used as a lubricant for heavy machinery parts should have a high viscous coefficient. To select a suitable lubricant, we should know its viscosity and how it varies with temperature. [Note: As temperature increases, the viscosity of the liquid decreases].
2. Also, it helps to choose oils with low viscosity used in car engines (light machinery).
3. The highly viscous liquid is used to damp the motion of some instruments and is used as brake oil in hydraulic brakes.
4. Blood circulation through arteries and veins depends upon the viscosity of fluids.
5. Millikan conducted the oil drop experiment to determine the charge of an electron. He used the knowledge of viscosity to determine the charge.

Question 31.
Write a note on quasi-static process.
Answer:
A quasi-static process is an infinitely slow process in which the system changes its variables (P,V,T) so slowly such that it remains in thermal, mechanical and chemical equilibrium with its surroundings throughout.

Question 32.
A body cools from 60°C to 50°C in 10 min of room after 10 more minute.
Answer:
According to Newton’s law of cooling

Question 33.
Consider two springs with force constants 1 Nm^-1 and 2 Nm^-1 connected in parallel. Calculate the effective spring constant (k_p) and comment on k_p.
Answer:
k₁ = 1 Nm^-1, k₂ = 2 Nm^-1
k_p = k₁ + k₂ Nm^-1
k_p = 1 + 2 = 3 Nm^-1
k_p > k₁ and k_p > k₂

Therefore, the effective spring constant is greater than both k₁ and k₂.

Part-IV

Answer all the questions. [5 x 5 = 25]

Question 34.
(a) Describe the construction and working of venturimeter and obtain an equation for the volume of liquid flowing per second through a wider entry of the tube.
Answer:

Venturimeter: This device is used to measure the rate of flow (or say flow speed) of the incompressible fluid flowing through a pipe. It works on the principle of Bernoulli’s theorem. It consists of two wider tubes A and A’ (with cross sectional area A) connected by a narrow tube B (with cross sectional area a). A manometer in the form of U-tube is also attached between the wide and narrow tubes. The manometer contains a liquid of density ‘p_m’.

Let P₁ be the pressure of the fluid at the wider region of the tube A. Let us assume that the fluid of density ‘p’ flows from the pipe with speed ‘v₁’ and into the narrow region, its speed increases to ‘v₂‘ According to the Bernoulli’s equation, this increase in speed is accompanied by a decrease in the fluid pressure P₂ at the narrow region of the tube B. Therefore, the pressure difference between the tubes A and B is noted by measuring the height difference (ΔP = P₁ – P₂) between the surfaces of the manometer liquid.

From the equation of continuity, we can say that Av₁ = av₂ which means that
\(v_{2}=\frac{\mathbf{A}}{a} v_{1}\)

Using Bernoulli’s equation,

From the above equation, the pressure difference

Thus, the speed of flow of fluid at the wide end of the tube

The volume of the liquid flowing out per second is

[OR]

(b) Derive the time period of satellite orbiting the Earth.
Answer:
Time period of the satellite: The distance covered by the satellite during one rotation in its orbit is equal to 2π(R_E + h) and time taken for it is the time period T. Then,

Squaring both sides of the equation (2) we get,

Equation (3) implies that a satellite orbiting the Earth has the same relation between time and distance as that of Kepler’s law of planetary motion. For a satellite orbiting near the surface of the Earth, h is negligible compared to the radius of the Earth R_E. Then,

By substituting the values of R_E = 6.4 x 10⁶ m and g = 9.8 ms^-2, the orbital time period is obtained as T = 85 minutes.

Question 35.
(a) State and explain work energy principle.
(b) Mention any three examples for it.
Answer:
(a) Work and energy are equivalents. This is true in the case of kinetic energy also. To prove this, let us consider a body of mass m at rest on a ffictionless horizontal surface. The work (W) done by the constant force (F) for a displacement (s) in the same direction is,
W = Fs ………………………….. (1)

The constant force is given by the equation,
F = ma ………………………….. (2)

The third equation of motion can be written as,
v² = u² + 2as
\(a=\frac{v^{2}-u^{2}}{2 s}\)

Substituting for a in equation (2),

Substituting equation (3) in (1),

(b) (i) If the work done by the force on the body is positive then its kinetic energy increases.
(ii) If the work done by the force on the body is negative then its kinetic energy decreases.
(iii) If there is no work done by the force on the body then there is no change in its kinetic energy, which means that the body has moved at constant speed provided its mass remains constant.
(iv) When a particle moves with constant speed in a circle, there is no change in the kinetic energy of the particle. So according to work energy principle, the work done by centripetal force is zero.

[OR]

(b) Derive an expression for time taken by the round object to reach the ground on inclined plane.
Answer:
Let us assume a round object of mass m and radius R is rolling down an inclined plane without slipping as shown in figure. There are two forces acting on the object along the inclined plane. One is the component of gravitational force (mg sin θ) and the other is the static frictional force (f). The other component of gravitation force (mg cos θ) is cancelled by the normal force (N) exerted by the plane. As the motion is happening along the incline, we shall write the equation for motion from the free body diagram (FBP) of the object.

For translational motion, mg sin0 is the supporting force and f is the opposing force,
mg sinθ f = ma

For rotational motion, let us take the torque with respect to the center of the object. Then mg sin θ cannot cause torque as it passes through it but the frictional force/can set torque of Rf.
Rf = Iα
By using the relation, a = rα, and moment of inertia I = mK², we get,

We can also find the expression for final velocity of the rolling object by using third equation of motion for the inclined plane.

v² = u² + las. If the body starts rolling from rest, u = 0. When h is the vertical height of the incline, the length of the incline s is, s = \(\frac{h}{\sin \theta}\)

The time taken for rolling down the incline could also be written from first equation of motion as, v = u + at. For the object which starts rolling from rest, u = 0. Then,

The equation suggests that for a given incline, the object with the least value of radius of’ gyration K will reach the bottom of the incline first.

Question 36.
(a) Define heat engine. Derive the expression for carnot engine efficiency.
Answer:
Head engine: Heat engine is a device which takes heat as input and converts this heat in to work by undergoing a cyclic process.

Efficiency of a Carnot engine: Efficiency is defined as the ratio of work done by the working substance in one cycle to the amount of heat extracted from the source.

From the first law of thermodynamics, W = Q_H – Q_L
Applying isothermal conditions, we get,

Here we omit the negative sign. Since we are interested in only the amount of heat (QL) ejected into the sink, we have

By applying adiabatic conditions, we get,

By dividing the above two equations, we get

Which implies that \(\frac{V_{2}}{V_{1}}=\frac{V_{3}}{V_{4}}\) ………………………… (5)
Substituting equation (5) in (4), we get

Note: T_L and T_H should be expressed in Kelvin scale.

Important results:

1. η is always less than 1 because T_L is less than T_H. This implies the efficiency cannot be 100%.
2. The efficiency of the Carnot’s engine is independent of the working substance. It depends only on the temperatures of the source and the sink. The greater the difference between the two temperatures, higher the efficiency.
3. When T_H = T_L the efficiency η = 0. No engine can work having source and sink at the same temperature.

[OR]

(b) Derive the expression for moment of inertia of a rod about its center and perpendicular to the rod.
Answer:
Let us consider a uniform rod of mass (M) and length (1) as shown in figure. Let us find an expression for moment of inertia of this rod about an axis that passes through the center of mass and perpendicular to the rod. First an origin is to be fixed for the coordinate system so that it coincides with the center of mass; which is also the geometric center of the rod. The rod is now along the x axis. We take an infinitesimally small mass (dm) at a distance (x) from the origin. The moment of inertia (dI) of this mass (dm) about the axis is,

As the mass is uniformly distributed, the mass per unit length (λ) of the rod is, λ = \(\frac{M}{l}\)

The (dm) mass of the infinitesimally small length as, dm = λdx = \(\frac{M}{l}\) dx
The moment of inertia (I) of the entire rod can be found by integrating dl,

As the mass is distributed on either side of the origin, the limits for integration are taken from to – l/2 to l/2.

Question 37.
(a) Show that the resultant intensity at any point depends on the phase difference at that point in interference of waves.
Answer:
Consider two harmonic waves having identical frequencies, constant phase difference φ and same wave form (can be treated as coherent source), but having amplitudes A₁ and A₂, then
y₁ = A₁ sin (kx – ωt) …………………………. (1)
y₂ = A₂ sin (kx – ωt + φ) ……………………. (2)

Suppose they move simultaneously in a particular direction, then interference occurs (i.e., overlap of these two waves). Mathematically
y = y₁ + y₂ ………………………………. (3)

Therefore, substituting equation (1) and equation (3) in equation (3), we get
y = A₁ sin (kx – ωt) + A₂ sin (kx – ωt + φ)

Using trigonometric identity sin (α + β) = (sin α cos β + cos α sin β), we get
y = A₁ sin (kx – ωt) +A₂ [sin (kx – ωt) cos φ + cos (kx – ωt) sin φ]
y = sin (kx – ωt) (A₁ + A₂ cos φ) + A₂ sin φ cos (kx – ωt) …………………………. (4)

Let us re-define A cos θ = (A₁ + A₂ cos φ) ……………………………….. (5)
and A sin θ = A₂ sin φ …………………………………. (6)

then equation (4) can be rewritten as y = A sin (kx – ωt) cos θ + A cos (kx – ωt) sin θ
y = A (sin (kx – ωt) cos θ + sin θ cos (kx – ωt))
y = A sin (kx – ωt + 9) …………………………….. (7)

By squaring and adding equation (5) and equation (6), we get
A₂ = A₁2 + A₂2 + 2A₁ A₂ cos φ …………………………… (8)

Since, intensity is square of the amplitude (I = A₂), we have
I = I₁ + I₂ + 2\(2 \sqrt{\mathrm{I}_{1} \mathrm{I}_{2}}\) cos φ …………………………….. (9)

This means the resultant intensity at any point depends on the phase difference at that point.

(i) For constructive interference: When crests of one wave overlap with crests of another wave, their amplitudes will add up and we get constructive interference. The resultant wave has a larger amplitude than the individual waves as shown in figure (a).

The constructive interference at a point occurs if there is maximum intensity at that point, which means that
cos φ = + 1 ⇒ φ = 0, 2π, 4π,… = 2nπ,
where n = 0, 1, 2,…

This is the phase difference in which two waves overlap to give constructive interference. Therefore, for this resultant wave,
\(I_{\text {maximum }}=(\sqrt{I_{1}}+\sqrt{I_{2}})^{2}=\left(A_{1}+A_{2}\right)^{2}\)

Hence, the resultant amplitude A = A₁ + A₂

(ii) For destructive interference: When the trough of one wave overlaps with the crest of another wave, their amplitudes “cancel” each other and we get destructive interference as shown in figure (b). The resultant amplitude is nearly zero. The destructive interference occurs if there is minimum intensity at that point, which means cos φ = – 1 ⇒ φ = π, 3π, 5π,… = (2 n-1) π, where n = 0,1,2,…. i.e. This is the phase difference in which two waves overlap to give destructive interference. Therefore,
\(I_{\text {maximum }}=(\sqrt{I_{1}}+\sqrt{I_{2}})^{2}=\left(A_{1}+A_{2}\right)^{2}\)

Hence, the resultant amplitude
A= |A₁ – A₂|

(b) Define adiabatic process. Derive an expression for work done in an adiabatic process.
Answer:
Adiabatic process: This is a process in which no heat flows into or out of the system (Q = 0). But the gas can expand by spending its internal energy or gas can be compressed through some external work. So the pressure, volume and temperature of the system may change in an adiabatic process.

For an adiabatic process, the first law becomes ΔU = W.

This implies that the work is done by the gas at the expense of internal energy or work is done on the system which increases its internal energy.

Work done in an adiabatic process: Consider μ moles of an ideal gas enclosed in a cylinder having perfectly non conducting walls and base. A frictionless and insulating piston of cross sectional area A is fitted in the cylinder.

Let W be the work done when the system goes from the initial state (P_i, V_i, T_i) to the final state (P_f, V_f, T_f) adiabatically.
\(\mathrm{W}=\int_{V_{1}}^{V_{f}} \mathrm{P} d \mathrm{V}\)

By assuming that the adiabatic process occurs quasi-statically, at every stage the ideal gas law is valid. Under this condition, the adiabatic equation of state is P_Vr = Constant (or)

\(P=\frac{\text { Constant }}{\mathrm{V} r}\)
can be substituted in the equation (1), we get

From ideal gas law, P_f V_f = μRT_f  and P_i V_i = μRT_i
Substituting in equation (2), we get
∴ W_adia = \(\frac{\mu \mathrm{R}}{\gamma-1}\) [T_f – T_i].

In adiabatic expansion, work is done by the gas. i.e., Watfoj is positive. As T_i > T_f, the gas cools during adiabatic expansion.

In adiabatic compression, work is done on the gas. i.e., Wadia is negative. As T_i < T_f, the temperature of the gas increases during adiabatic compression.

PV diagram -Work done in the adiabatic process

To differentiate between isothermal and adiabatic curves in PV diagram, the adiabatic curve is drawn along with isothennal curve for and T_f and T_i Note that adiabatic curve is steeper than isothermal curve. This is because ϒ > 1 always.

Question 38.
(a) Describe the construction and working of venturimeter and obtain an equation for the volume of liquid flowing per second though a wider entry of the tube.
Answer:
Venturimeter : This device is used to measure the rate of flow (or say flow speed) of the incompressible fluid flowing through a pipe. It works on the principle of Bernoulli’s theorem. It consists of two wider tubes A and A’ (with cross sectional area A) connected by a narrow tube B (with cross sectional area a). A manometer in the form of U-tube is also attached between the wide and narrow tubes. The manometer contains a liquid of density ‘p_m‘

Let P₁ be the pressure of the fluid at the wider region of the tube A. Let us assume that the fluid of density ‘p’ flows from the pipe with speed ‘v₁’ and into the narrow region, its speed increases to ‘v₂’. According to the Bernoulli’s equation, this increase in speed is accompanied by a decrease in the fluid pressure P₂ at the narrow region of the tube B. Therefore, the pressure difference between the tubes A and B is noted by measuring the height difference (ΔP = P₁ – P₂) between the surfaces of the manometer liquid.

From the equation of continuity, we can say that Av₁ = av₂ which means that
\(v_{2}=\frac{\mathrm{A}}{a} v_{1}\)

Using Bernoulli’s equation,

From the above equation, the pressure difference

Thus, the speed of flow of fluid at the wide end of the tube A

The volume of the liquid flowing out per second is

[OR]

(b) Define gravitational potential energy and derive the expression for it.
Answer:
Gravitational potential energy: Potential energy of a body at a point in a gravitational field is the work done by an external agent in moving the body from infinity to that point.

Expression for Gravitational potential energy: The gravitational force is a conservative force and hence we can define a gravitational potential energy associated with this conservative force field.

Two masses m₁ and m₂ are initially separated by a distance r’. Assuming m₁, to be fixed in its position, work must be done on m₂ to move the distance from r’ to r.

To move the mass m₂ through an infinitesimal displacement \(d \vec{r}\) from \(\vec{r}\) to \(\vec{r}\) + \(d \vec{r}\), work has to be done externally. This infinitesimal work is given by
\(d \mathrm{W}=\overrightarrow{\mathrm{F}}_{e x t} \cdot d \vec{r}\) ……………(1)

The work is done against the gravitational force, therefore,
\(\overrightarrow{\mathbf{F}}_{e x t}=\frac{\mathrm{G} m_{1} m_{2}}{r^{2}} \hat{r}\) ………………………. (2)

Substituting equation (2) in (1), we get

Thus the total work done for displacing the particle from r’ to r is f’

This work done W gives the gravitational potential energy difference of the system of masses m₁ and m₂ when the separation between them are r and r’ respectively.

Case 1: If r < r’ : Since gravitational force is attractive, m₂ is attracted by m₁. Then m₂ can move from r to r’ without any external work. Here work is done by the system spending its internal energy and hence the work done is said to be negative.

Case 2: If r > r’ : Work has to be done against gravity to move the object from r’ to r. Therefore work is done on the body by external force and hence work done is positive.

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Tamilnadu Samacheer Kalvi 11th Computer Science Solutions Chapter 8 Iteration and Recursion

Samacheer Kalvi 11th Computer Science Iteration and Recursion Text Book Back Questions and Answers

PART – 1
I. Choose The Correct Answer

Question 1.
A loop invariant need not be true ………………….
(a) at the start of the loop
(b) at the start of each iteration
(c) at the end of each iteration
(d) at the start of the algorithm
Answer:
(d) at the start of the algorithm

Question 2.
We wish to cover a chessboard with dominoes, the number of black squares and the number of white squares covered by dominoes, respectively, placing a domino can be modeled by ………………….
(a) b : = b + 2
(b) w : = w + 2
(c) b, w : = b + 1, w + 1
(d) b : = w
Answer:
(c) b, w : = b + 1, w + 1

Question 3.
If m x a + n x b is an invariant for the assignment a, b : = a + 8, b + 7, the values of m and n are ………………….
(a) m = 8, n = 7
(b) m = 7, n = – 8
(c) m = 7, n = 8
(d) m = 8, n = – 7
Answer:
(b) m = 7, n = – 8

Question 4.
Which of the following is not an invariant of the assignment?
m, n : = m + 2, n + 3
(a) m mod 2
(b) n mod 3
(c) 3 x m – 2 x n
(d) 2 x m – 3 x n
Answer:
(c) 3 x m – 2 x n

Question 5.
If Fibonacci number is defined recursively as

to evaluate F(4), how many times F( ) is applied?
(a) 3
(b) 4
(c) 8
(d) 9
Answer:
(d) 9

Question 6.
Using this recursive definition

how many multiplications are needed to calculate a¹⁰?
(a) 11
(b) 10
(c) 9
(d) 8
Answer:
(b) 10

PART – 2
II. Short Answers

Question 1.
What is an invariant?
Answer:
An expression involving variables, which remains unchanged by an assignment to one of these variables is called as an invariant of the assignment.

Question 2.
Define a loop invariant.
Answer:
In iteration, the loop body is repeatedly executed as long as the loop condition is true. Each time the loop body is executed, the variables are updated. However, there is also a property of the variables which remains unchanged by the execution of the loop body. This unchanging property is called the loop invariant.

Question 3.
Does testing the loop condition affect the loop invariant? Why?
Answer:
No, the loop condition do not affect the loop invariant. Because the loop invariant is true at four points.

1. At the start of loop.
2. At the start of each iteration.
3. At the end of each iteration.
4. At the end of the loop.

Question 4.
What is the relationship between loop invariant, loop condition and the input – output recursively?
Answer:
A loop invariant is a condition [among program variables] that is necessarily true immediately before and immediately after, each iteration of a loop. A loop invariant is some condition that holds for every iteration of the loop.

Question 5.
What is recursive problem solving?
Answer:
Recursion is a method of solving problems that involves breaking a problem down into smaller and smaller sub problems until user gets in to a small problem that it can be solved trivially. Usually recursion involves a function calling itself. While it may not seem like much on the surface, recursion allows us to write elegant solutions to problems that may otherwise be very difficult to program.

Question 6.
Define factorial of a natural number recursively.
Answer:

Recursive Algorithm:
Fact (n)
– – inputs : n
– – outputs : Fact = n!
if (n = 0) – – base case
1
else
n * fact (n – 1) – – recursive step

PART – 3
III. Explain in Brief

Question 1.
There are 7 tumblers on a table, all standing upside down. You are allowed to turn any 2 tumblers simultaneously in one move. Is it possible to reach a situation when all the tumblers are right side up? (Hint: The parity of the number of upside down tumblers is invariant.)
Answer:
Let u – No. of tumblers right side up
v – No. of tumblers up side down
Initial stage : u = 0, v = 7 (All tumblers upside down)
Final stage output: u = 7, v = 0 (All tumblers right side up)

Possible Iterations:
(i) Turning both up side down tumblers to right side up
u = u + 2, v = v – 2 [u is even]

(ii) Turning both right side up tumblers to upside down.
u = u – 2, v = v + 2 [u is even]

(iii) Turning one right side up tumblers to upside down and other tumbler from upside down to right side up.
u = u + 1 – 1 = u, v = v + 1 – 1 = v [u is even]

Initially u = 0 and continuous to be even in all the three cases. Therefore u is always even. Invariant: u is even (i. e. No. of right side up tumblers are always even)
But in the final stage (Goal), u = 7 and v = 0 i. e. u is odd.
Therefore it is not possible to reach a situation where all the tumblers are right side up.

Question 2.
A knockout tournament is a series of games. Two players complete in each game; the loser is knocked out (i.e. does not play any more), the winner carries on. The winner of the tournament is the player that is left after all other players have been knocked out. Suppose there are 1234 players in a tournament. How many games are played before the tournament winner is decided?
Answer:

On the other hand let n be the number of games, played and r be the number of players remaining in the tournament.
After every game, r will be reduced by 1.
r → no. of players remaining
n → no. of games played
If r = 2 then n = 1
As n increases, r decreases
n, r : = n + 1, r – 1
n + r = (n + 1) + (r – 1)
= n + 1 + r – 1
= n + r
Therefore n + r is invariant. n + r = 1234 (No. of players initially)
The winner of the tournament is the player that is left after all other players have been knocked out.
After all the games, only one player (winner) is left out.
i. e. n = 1
Put n = 1 in (1)
n + r = 1234 …. (1)
1 + r = 1234
r = 1234 – 1 = 1233
No. of games played = 1233

Question 3.
King Vikramaditya has two magic swords. With one, he can cut off 19 heads of a dragon, but after that the dragon grows 13 heads. With the other sword, he can cut off 7 heads, but 22 new heads grow. If all heads are cut off, the dragon dies. If the dragon has originally 1000 heads, can it ever die? (Hint: The number of heads mod 3 is invariant.)
Answer:
No. of heads of dragon = 1000
sword 1 : cuts 19 heads but 13 heads grow back.
sword 2 : cuts 7 heads but 22 heads grow back.
Let n be the number of heads of the dragon at initial state.

Case 1 : King uses Sword 1
Sword 1 cuts off 19 heads but 13 heads grow back.
n : = n – 19 + 13 = n – 6 No. of heads are reduced by 6.

Case 2 : King uses Sword 2
Sword 2 cuts 7 heads but 22 heads grow back.
n : = n – 7 + 22 = n + 15
No. of heads are increased by 15.

Note:
In the above two cases either 6 heads are removed or 15 heads added. Both 6 and 15 are multiples of 3.
Therefore repeating case 1 and case 2 recursively will either reduce or increase dragon heads in multiples of 3.
That is the invariant is n mod 3.
If n mod 3 = 0 then there is a possibility that the dragon dies.
But 1000 is not a multiple of 3 1000 mod 3 = 1 ≠ 0
It is not possible to kill the dragon. The dragon never dies.

PART – 4
IV. Explain in Detail

Question 1.
Assume an 8 x 8 chessboard with the usual coloring. “Recoloring” operation changes the color of all squares of a row or a column. You can recolor repeatedly. The goal is to attain just one black square. Show that you cannot achieve the goal. (Hint: If a row or column has b black squares, it changes by (|(8 – b) – b|).
Answer:
In a chess board no. of squares in any row or column = 8
Total No. of squares = 8 x 8 = 64
No. of black squares = 32
No. of white squares = 32
Let No. of blacks in a selected recoloring row or column = b
No. of black squares after recoloring operation = 8 – b
Initial state b = 32
Desired final state b = 1
Let us tabulate all the possible outcomes after recoloring a row or column.

Difference is the no. of black squares left out in a row or column after recoloring operation. Difference is even. Difference is invariant.
But our goal is one ( 1 is odd) black square, so, it is not possible to attain the goal.

Question 2.
Power can also be defined recursively as

Construct a recursive algorithm using this definition. How many multiplications are needed to calculate a¹⁰?
Answer:
Power:
power (5, 2) = 5 x 5 = 25
power (x, n) raise x to the power n

Algorithm:

To find a¹⁰:

Question 3.
A single – square – covered board is a board of 2ⁿ x 2ⁿ squares in which one square is covered with a single square tile. Show that it is possible to cover this board with triominoes without overlap.
Answer:
size of the board is 2nⁿ x 2ⁿ
Number of squares = 2ⁿ x 2ⁿ = 4ⁿ
Number of squares covered = 1
Number of squares to be covered = 4ⁿ – 1
4ⁿ – 1 is a multiple of 3

Case 1 : n = 1
The size of the board 2 x 2
one triominoe can cover 3 squares without overlap.

We can cover it with one triominoe and solve the problem.

Case 2 : n ≥ 2
1. place a triominoe at the center of the entire board so as to not cover the covered sub – board.

2. One square in the board is covered by a tile. The board has 4 sub – boards of size 2^{2n – 1} x 2^{2n – 1}.

Out of 4 sub – boards one sub – board is a single square covered sub – board.

One triominoe can cover remaining three sub – boards into single square covered sub – board. The problem of size n is divided into 4 sub – problems of size (n – 1). Each sub – board has 2^{2n – 1} x 2^{2n – 1} – 1 = 2^{2n – 2} – 1 = 4^{n – 1} – 1 squares to be covered.

4^{n – 1} – 1 is also a multiple of 3
In this, the 2ⁿ x 2ⁿ board is reduced to boards of size 2×2 having are square covered. A triominoe can be placed in each of these boards and hence the whole original 2ⁿ x 2ⁿ . board is covered with triominoe with out overlap.

Samacheer kalvi 11th Computer Science Iteration and Recursion Additional Questions and Answers

PART – 1
I. Choose the correct answer

Question 1.
………………… is an algorithm design technique, closely related to induction.
(a) Iteration
(b) Invariant
(c) Loop invariant
(d) Recursion
Answer:
(d) Recursion

Question 2.
In which year E W Dijkstra was awarded ACM Turing Award?
(a) 1972
(b) 1974
(c) 1970
(d) 1911
Answer:
(a) 1972

Question 3.
In a loop, if L is an invariant of the loop body B, then L is known as a …………………
(a) recursion
(b) variant
(c) loop invariant
(d) algorithm
Answer:
(c) loop invariant

Question 4.
Recursion must have at least ………………… base case.
(a) one
(b) two
(c) three
(d) four
Answer:
(a) one

Question 5.
The unchanged variables of the loop body is …………………
(a) loop invariant
(b) loop variant
(c) condition
(d) loop variable
Answer:
(a) loop invariant

Question 6.
………………… is the algorithm design techniques to execute the same action repeatedly.
(a) Iteration
(b) Recursion
(c) Both a & b
(d) none of these
Answer:
(c) Both a & b

Question 7.
If L is a loop variant, then it should be true at ………………… important points in the algorithm.
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(c) 4

Question 8.
The loop invariant need not be true at the …………………
(a) Start of the loop
(b) end of the loop
(c) end of each iteration
(d) middle of algorithm
Answer:
(d) middle of algorithm

Question 9.
In an expression if the variables has the same value before and after an assignment, then it is of an assignment.
(a) variant
(b) Invariant
(c) iteration
(d) variable
Answer:
(b) Invariant

Question 10.
The input size to a sub problem is than the input size to the original problem.
(a) equal
(b) smaller
(c) greater
(d) no criteria
Answer:
(b) smaller

Question 11.
When the solver calls a sub solver, then it is called …………………
(a) Iterative call
(b) solver call
(c) recursive call
(d) conditional call
Answer:
(c) recursive call

Question 12.
How many cases are needed for a recursive solvers?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Question 13.
Which of the following is updated when each time the loop body is executed?
(a) data type
(b) subprogram
(c) function
(d) variable
Answer:
(d) variable

Question 14.
Which is the key to constract iterative algorithm …………………
(a) loop invariant
(b) Variable
(c) loop
(d) Recursive
Answer:
(a) loop invariant

PART – 3
II. Short Answers

Question 1.
When the loop variant will be true?
Answer:
The loop invariant is true before the loop body and after the loop body, each time.

Question 2.
What is an invariant?
Answer:
An expression involving variables, which remains unchanged by an assignment to one of these variables, is called an invariant of the assignment.

PART – 3
III. Explain in Brief

Question 1.
Write a note on Recursion.
Answer:
Recursion:
Recursion is another algorithm design technique, closely related to iteration, but more powerful. Using recursion, we solve a problem with a given input, by solving the same problem with a part of the input, and constructing a solution to the original problem from the solution to the partial input.

Question 2.
Write down the steps to be taken to construct a loop.
Answer:

1. Establish the loop invariant at the start of the loop.
2. The loop body should so update the variables as to progress toward the end, and maintain the loop invariant, at the same time.
3. When the loop ends, the termination condition and the loop invariant should establish the input – output relation.

Question 3.
Write a note on Iteration.
Answer:
Iteration:
In iteration, the loop body is repeatedly executed as long as the loop condition is true. Each time the loop body is executed, the variables are updated. However, there is also a property of the variables which remains unchanged by the execution of the loop body. This unchanging property is called the loop invariant. Loop invariant is the key to construct and to reason about iterative algorithms.

PART – 4
IV. Explain in Detail

Question 1.
Explain Loop invariant with a near diagram.
Answer:
In a loop, if L is an invariant of the loop body B, then L is known as a loop invariant, while C
– – L
B
– – L
The loop invariant is true before the loop body and after the loop body, each time. Since L is true at the start of the first iteration, L is true at the start of the loop also (just before the loop). Since L is true at the end of the last iteration, L is true when the loop ends also (just after the loop). Thus, if L is a loop variant, then it is true at four important points in the algorithm, as annotated in the algorithm.

1. At the start of the loop (just before the loop)
2. at the start of each iteration (before loop body)
3. at the end of each iteration (after loop body)
4. at the end of the loop (just after the loop)

Question 2.
Explain recursive problem solving.
Answer:
To solve a problem recursively, the solver reduces the problem to sub – problems, and calls another instance of the solver, known as sub – solver, to solve the sub – problem. The input size to a sub – problem is smaller than the input size to the original problem. When the solver calls a sub – solver, it is known as recursive call. The magic of recursion allows the solver to assume that the sub – solver (recursive call) outputs the solution to the sub – problem. Then, from the solution to the sub – problem, the solver constructs the solution to the given problem. As the sub – solvers go on reducing the problem into sub – problems of smaller sizes, eventually the sub-problem becomes small enough to be solved directly, without recursion. Therefore, a recursive solver has two cases:
1. Base case:
The problem size is small enough to be solved directly. Output the solution. here must be at least one base case.

2. Recursion step:
The problem size is not small enough. Deconstruct the problem into a sub – problem, strictly smaller in size than the given problem. Call a sub – solver to solve the sub problem. Assume that the sub – solver outputs the solution to the sub problem. Construct the solution to the given problem. This outline of recursive problem solving technique is shown below.

Whenever we solve a problem using recursion, we have to ensure these two cases: In the recursion step, the size of the input to the recursive call is strictly smaller than the size of the given input, and there is at least one base case.

Question 3.
Give an example for loop invariant.
Answer:
The loop invariant is true in four crucial points in a loop. Using the loop invariant, we can construct the loop and reason about the properties of the variables at these points.
Example:
Design an iterative algorithm to compute an. Let us name the algorithm power(a, n). For example,
power(10, 4) = 10000
power (5, 3) = 125
power (2, 5) = 32
Algorithm power(a, n) computes aⁿ by multiplying a cumulatively n times,

The specification and the loop invariant are shown as comments.
power (a, n)
– – inputs: n is a positive integer
– – outputs: p = aⁿ
p, i : = 1, 0
while i ≠ n
– – loop invariant: p = aⁱ
p, i : = p x a, i + 1
The step by step execution of power (2, 5) is shown in Table. Each row shows the values of the two variables p and i at the end of an iteration, and how they are calculated. We see that p = aⁱ is true at the start of the loop, and remains true in each row. Therefore, it is a loop invariant.

When the loop ends, p = aⁱ is still true, but i = 5. Therefore, p = a⁵. In general, when the loop ends, p = aⁿ . Thus, we have verified that power(a, n) satisfies its specification.

Question 4.
There are 6 equally spaced trees and 6 sparrows sitting on these trees,one sparrow on each tree. If a sparrow flies from one tree to another, then at the same time, another sparrow flies from its tree to some other tree the same distance away, but in the opposite direction. Is it possible for all the sparrows to gather on one tree?
Answer:
Let us index the trees from 1 to 6. The index of a sparrow is the index of the tree it is currently sitting on. A pair of sparrows flying can be modeled as an iterative step of a loop. When a sparrow at tree i flies to tree i + d, another sparrow at tree j flies to tree j – d. Thus, after each iterative step, the sum S of the indices of the sparrows remains invariant. Moreover, a loop invariant is true at the start and at the end of the loop.
At the start of the loop, the value of the invariant is
S = 1 + 2 + 3 + 4 + 5 + 6 = 21
When the loop ends, the loop invariant has the same value. However, when the loop ends, if all the sparrows were on the same tree, say k, then S = 6k

It is not possible – 21 is not a multiple of 6. The desired final values of the sparrow indices is not possible with the loop invariant. Therefore, all the sparrows cannot gather on one tree.

Question 5.
Customers are waiting in a line at a counter. The man at the counter wants to know how many customers are waiting in the line.
Answer:
Customers are waiting in a line at a counter. The man at the counter wants to know how many customers are waiting in the line.

Instead of counting the length himself, he asks customer A for the length of the line with him at the head, customer A asks customer B for the length of the line with customer B at the head, and so on. When the query reaches the last customer in the line, E, since there is no one behind him, he replies 1 to D who asked him. D replies 1 + 1 = 2 to C, C replies 1 + 2 = 3 to B, B replies 1 + 3 = 4 to A, and A replies 1 + 4 = 5 to the man in the counter.

Question 6.
Design a recursive algorithm to compute aⁿ. We constructed an iterative algorithm to compute an in Example 8.5. aⁿ can be defined recursively as

Answer:
The recursive definition can be expressed as a recursive solver for computing power(a, n).

The recursive process resulting from power(2, 5)

power (2, 5)
= 2 x power (2, 4)
= 2 x 2 x power (2, 3)
= 2 x 2 x 2 x power (2, 2)
= 2 x 2 x 2 x 2 x power (2, 1)
= 2 x 2 x 2 x 2 x 2 x power (2, 0) = 2 x 2 x 2 x 2 x 2 x 1
= 2 x 2 x 2 x 2 x 2 = 2 x 2 x 2 x 4 = 2 x 2 x 8 = 2 x 16 = 32

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Tamilnadu Samacheer Kalvi 11th Accountancy Solutions Chapter 12 Final Accounts of Sole Proprietors – I

Samacheer Kalvi 11th Accountancy Final Accounts of Sole Proprietors – I Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the Correct Answer

Question 1.
Closing stock is an item of ……………..
(a) Fixed asset
(b) Current asset
(c) Fictitious asset
(d) Intangible asset
Answer:
(b) Current asset

Question 2.
Balance sheet is ……………..
(a) An account
(b) A statement
(c) Neither a statement nor an account
(d) None of the above
Answer:
(b) A statement

Question 3.
Net profit of the business increases the ……………..
(a) Drawings
(b) Receivables
(c) Debts
(d) Capital
Answer:
(d) Capital

Question 4.
Carriage inwards will be shown ……………..
(a) In the trading account
(b) In the profit and loss account
(c) On the liabilities side
(d) On the assets side
Answer:
(a) In the trading account

Question 5.
Bank overdraft should be shown ……………..
(a) In the trading account
(b) Profit and loss account
(c) On the liabilities side
(d) On the assets side
Answer:
(c) On the liabilities side

Question 6.
Balance sheet shows the …………….. of the business.
(a) Profitability
(b) Financial position
(c) Sales
(d) Purchases
Answer:
(b) Financial position

Question 7.
Drawings appearing in the trial balance is ……………..
(a) Added to the purchases
(b) Subtracted from the purchases
(c) Added to the capital
(d) Subtracted from the capital
Answer:
(d) Subtracted from the capital

Question 8.
Salaries appearing ill the trial balance is shown on the ……………..
(a) Debit side of trading account
(b) Debit side of profit and loss account
(c) Liabilities side of the balance sheet
(d) Assets side of the balance sheet
Answer:
(b) Debit side of profit and loss account

Question 9.
Current assets does not include ……………..
(a) Cash
(b) Stock
(c) Furniture
(d) Prepaid expenses
Answer:
(c) Furniture

Question 10.
Goodwill is classified as ……………..
(a) A current asset
(b) A liquid asset
(c) A tangible asset
(d) An intangible asset
Answer:
(d) An intangible asset

II. Very Short Answer Questions

Question 1.
Write a note on trading account.
Answer:
Trading refers to buying and selling of goods with the intention of making profit. The trading account is a nominal account which shows the result of buying and selling of goods for an accounting period. Trading account is prepared to find out the difference between the revenue from sales and cost of goods sold.

Question 2.
What are wasting assets?
Answer:
These are the assets which get exhausted gradually in the process of excavation. Examples: mines and quarry.

Question 3.
What are fixed assets?
Answer:
Fixed assets are those assets which are acquired or constructed for continued use in the business and last for many years such as land and building, plant and machinery, motor vehicles, furniture, etc.

Question 4.
What is meant by purchases returns?
Answer:
Goods purchased which are returned to suppliers are termed as purchases returns or returns outward.

Question 5.
Name any two direct expenses and indirect expenses.
Answer:
Direct expenses:

• Carriage inwards or freight inwards
• Wages Indirect expenses:

Indirect expenses:

• Office and administrative expenses
• Selling and distribution expenses

Question 6.
Mention any two differences between trial balance and balance sheet.
Answer:

Question 7.
What are the objectives of preparing trading account?
Answer:

1. Provides information about gross profit or gross loss.
2. Provides an opportunity to safeguard against possible losses.

Question 8.
What is the need for preparing profit and loss account?
Answer:

1. Ascertainment of net profit or net loss
2. Comparison of profit
3. Control on expenses
4. Helpful in the preparation of balance sheet.

III. Short Answer Questions

Question 1.
What are final accounts? What are its constituents?
Answer:
Businessmen want to know the profitability and the financial position of the business. These can be ascertained by preparing the final accounts or financial statements. The final accounts or financial statements include the following:

1. Income statement or trading and profit and loss account; and
2. Position statement or Balance sheet.

Question 2.
What is meant by closing entries? Why are they passed?
Answer:
Balances of all the nominal accounts are required to be closed on the last day of the accounting year to facilitate the preparation of trading and profit and loss account. It is done by passing necessary closing entries in the journal proper. Purchases has debit balance and purchases returns has credit balance. At the end of the accounting year, the balance in purchases returns account is closed by transferring to purchase account.

Question 3.
What is meant by gross profit and net profit?
Answer:

1. If the amount of sales exceeds the cost of goods sold, the difference is gross profit.
   Sales – Cost of goods sold = Gross profit.
2. If the total of the credit side of the profit and loss account exceeds the debit side, the difference is termed as net profit.

Question 4.
“Balance sheet is not an account” – Explain.
Answer:
A balance sheet is a part of the final accounts. However, the balance sheet is a statement and not an account. It has no debit or credit sides and as such the words ‘To’ and ‘By’ are not used before the names of the accounts shown therein.

Question 5.
What are the advantages of preparing a balance sheet?
Answer:
Balance sheet discloses the financial position of a business on a particular date, it gives
the balances only for the date on which it is prepared. It shows the financial position of the business according to the going concern concept.

Question 6.
What is meant by grouping and marshalling of assets and liabilities?
Answer:
1. The term ‘grouping’ means showing the items of similar nature under a common heading. For example, the amount due from various customers will be shown under the head‘sundry debtors’.

2. ‘Marshalling’ is the arrangement of various assets and liabilities in a proper order. Marshalling can be made in one of the following two ways:

• In the order of liquidity
• In the order of permanence

IV. Exercises

Question 1.
Prepare trading account in the books of Sivashankar from the following figures:

Answer:
Trading account of Sivashankar

Question 2.
Prepare trading account in the books of Mr. Sanj ay for the year ended 31^st December 2017:

Answer:
Trading account of Mr. Sanjay for the year ended 31^st December, 2017

Question 3.
Prepare trading account in the books of Mr. Sanj ay for the year ended 31^st December 2017:

Answer:
Trading account of Saravanan for the year ended 31^st December, 2017

Question 4.
From the following details for the year ended 31^st March, 2018, prepare trading account

Answer:
Trading account for the year ended 31^st March, 2018

Question 5.
Ascertain gross profit or gross loss from the following:

Answer:
Profit and Loss Account

Question 6.
From the following balances taken from the books of Victor, prepare trading account for the year ended December 31,2017:

Answer:
Trading Account of Victor for the year ended 31^st December, 2017

Hint : Closing stock will not appear in trading account because of adjusted purchases have been given.

Question 7.
Compute cost of goods sold from the following information:

Answer:
Cost of goods sold = Opening stock + Net purchases + Direct expenses – Closing stock
= 10,000 + 80,000 + 7,000 – 15,000
= ₹ 82,000
Note: Indirect expenses do not form part of cost of goods sold.

Question 8.
Find out the amount of sales from the following information:

Answer:
Cost of goods sold = Opening stock + Net purchases + Direct expenses – Closing stock
= 30,000 + 2,00,000 + 0 – 20,000
= ₹ 2,10,000

Therefore, percentage of gross profit on cost of goods sold is
\(\frac { 30 }{ 70 }\) x 100 = 42.85% (42.857142 ……….)
Gross profit = 42.85% on 2,10,000 i.e., \(\frac { 42.85 }{ 100 }\) x 2,10,000 = ₹ 90,000
Sales = Cost of goods sold + Gross Profit
= 2,10,000 + 90,000 (Fractions to be rounded)
= ₹ 3,00,000

Question 9.
Prepare profit and loss account in the books of Kirubavathi for the year ended 31^st December, 2016 from the following information:

Answer:
Profit and loss account of Kirubavathi for the year ended 31^st Dec, 2016

Question 10.
Ascertain net profit or net loss from the following:

Answer:
Profit and loss account

Question 11.
From the following details, prepare profit and loss account.

Answer:
Profit and loss account

(Hint: Freight inwards will not appear in profit and loss account as it is a direct expense)

Question 12.
From the following information, prepare profit and loss account for the year ending 31^st December, 2016.

Answer:
Profit and loss account for the year ended 31^st December, 2016

Question 13.
From the following balances obtained from the books of Mr. Ganesh, prepare trading and profit and loss account:

Answer:
Trading and Profit & loss account of Mr. Ganesh

Question 14.
From the following balances extracted from the books of a trader, ascertain gross profit and net profit for the year ended March 31,2017:

Answer:
Trading and Profit & Loss account for the year ended 31^st March, 2017

Question 15.
From the following particulars, prepare balance sheet in the books of Bragathish as on 31^st December, 2017:

Answer:
Balance Sheet of Bragathish as on 31^st December, 2017

Question 16.
Prepare trading and profit and loss account in the books of Ramasundari for the year ended 31^st December, 2017 and balance sheet as on that date from the following information:

Answer:
Trading and Profit & Loss a/c of Ramasundari for the year ended 31 Dec, 2017 Cr.

Balance Sheet of Ramasundari as on 31^st March, 2018

Question 17.
From the Trial balance, given by Saif, prepare final accounts for the year ended 31^st March, 2018 in his books:

Answer:
Trading and Profit & Loss a/c of Saif for the year ended 31 March, 2018

Balance Sheet of Saif as on 31^st March, 2018

Question 18.
Prepare trading and profit and loss account and balance sheet in the books of Deri, a trader, from the following balances as on March 31, 2018.

Answer:
Trading and Profit & Loss a/c of Deri for the year ended 31^st March, 2018

Balance Sheet of Deri as on 31^st March, 2018

Textbook Case Study Solved

Question 1.
Mr. Abhinav started a small shop of selling dairy products. He wanted to maintain proper books of accounts. But, he had very little knowledge of accounting. He maintained only three books – purchases, sales and cash book by himself. He bought some dairy products and a refrigerator to store the milk products for which the payment was made by cheque but recorded the same in the purchases book. He also spent for the transportation charges and paid some money to the person who unloaded the stock. He recorded the same in the cash book.

He made both cash and credit sale for the next few weeks. He entered the entire sales in the sales book. In the middle of the month, he was in need of some money for his personal use. So he took some money, but did not record in the books.
Now, discuss on the following points:

Question 1.
Do you think Mr. Abhinav needs an accountant? Why do you think so?
Answer:
Yes, Mr. Abhinav needs an accountant because he records all cash and credit transactions.

Question 2.
Does he maintain enough books of accounts?
Answer:
Yes, he maintains enough books of accounts.

Question 3.
What other books do you think that he needs to maintain?
Answer:
He needs to maintain a petty cashbook.

Question 4.
What will be the impact on the profit, if he records the purchase of refrigerator in the purchases book?
Answer:
The purchase book will be overcast because this transaction will be recorded in proper journal.

Question 5.
Is it important to record the money taken for personal use? Will it affect the final accounts?
Answer:
Yes, then only the actual profit or loss can be found out in the business.

Question 6.
Identify some of the accounting principles relevant to this situation.
Answer:
Some of the accounting principles relevant to this situation are: matching principles, business entity concept, money measurement concept, dual output concept, periodicity concept and going concern concept.

Samacheer Kalvi 11th Accountancy Final Accounts of Sole Proprietors – I Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer

Question 1.
Income statement is divided into ……………….. parts.
(a) one
(b) two
(c) three
(d) four
Answer:
(b) two

Question 2.
The first part of the income statement is ………………..
(a) Final account
(b) Trading account
(c) Profit and Loss account
(d) Balance Sheet
Answer:
(b) Trading account

Question 3.
Balances of all ……………….. accounts are required to be closed on the last day of the accounting year.
(a) Nominal
(b) Personal
(c) Real
(d) Representative personal
Answer:
(a) Nominal

Question 4.
Trading account is a ……………….. account.
(a) Personal
(b) Nominal
(c) Real
(d) Representative personal
Answer:
(b) Nominal

Question 5.
Sales – Gross Profit = ………………..
(a) Sales
(b) Cost of goods sold
(c) Gross profit
(d) Gross loss
Answer:
(b) Cost of goods sold

Question 6.
……………….. account is the second part of income statement.
(a) Trading
(b) Profit and Loss
(c) Balance sheet
(d) Final
Answer:
(b) Profit and Loss

Question 7.
Which one is correctly matched?
(a) Bad debts – Indirect expense
(b) Wages – Asset
(c) Salary – Trading account
(d) Net Profit – Asset
Answer:
(a) Bad debts – Indirect expense

Question 8.
Balances of all the personal and real account are shown in ………………..
(a) Trading account
(b) Profit and loss account
(c) Income statement
(d) Balance sheet
Answer:
(d) Balance sheet

Question 9.
A balance sheet is a part of the ……………….. account.
(a) Trading
(b) Profit and Loss
(c) Income statement
(d) Final
Answer:
(d) Final

Question 10.
……………….. is a summary of the personal and real accounts.
(a) Balance sheet
(b) Final account
(c) Trading account
(d) Profit and loss account
Answer:
(a) Balance sheet

Question 11.
The balance sheet of business concern can be presented in the ……………….. forms.
(a) two
(b) three
(c) four
(d) six
Answer:
(a) two

Question 12.
Marshalling can be made in one of the ……………….. ways.
(a) three
(b) two
(c) four
(d) five
Answer:
(b) two

Question 13.
These are the assets which get exhausted gradually in the process of excavation.
(a) Wasting assets
(b) Nominal assets
(c) Liquid assets
(d) Current assets
Answer:
(a) Wasting assets

Question 14.
……………….. liabilities are not shown in the balance sheet.
(a) Contingent
(b) Current
(c) Liquid
(d) Fixed
Answer:
(a) Contingent

Students can Download Accountancy Chapter 14 Computerised Accounting Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Accountancy Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Accountancy Solutions Chapter 14 Computerised Accounting

Samacheer Kalvi 11th Accountancy Computerised Accounting Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the Correct Answer

Question 1.
In accounting, computer is commonly used in the following areas:
(a) Recording of business transactions
(b) Payroll accounting
(c) Stores accounting
(d) All the above
Answer:
(d) All the above

Question 2.
Customised accounting software is suitable for ………………
(a) Small, conventional business
(b) Large, medium business
(c) Large, typical business
(d) None of the above
Answer:
(b) Large, medium business

Question 3.
Which one is not a component of computer system?
(a) Input unit
(b) Output unit
(c) Data
(d) Central Processing Unit
Answer:
(c) Data

Question 4.
An example of output device is ………………
(a) Mouse
(b) Printer
(c) Scanner
(d) Keyboard
Answer:
(b) Printer

Question 5.
One of the limitations of computerised accounting system is ………………
(a) System failure
(b) Accuracy
(c) Versatility
(d) Storage
Answer:
(a) System failure

Question 6.
Expand CAS ………………
(a) Common Application Software
(b) Computerised Accounting System
(c) Centralised Accounting System
(d) Certified Accounting System
Answer:
(b) Computerised Accounting System

Question 7.
Which one of the following is not a method of codification of accounts?
(a) Access codes
(b) Sequential codes
(c) Block codes
(d) Mnemonic codes
Answer:
(a) Access codes

Question 8.
TALLY is an example of ………………
(a) Tailor – made accounting software
(b) Ready – made accounting software
(c) In – built accounting software
(d) Customised accounting software
Answer:
(b) Ready – made accounting software

Question 9.
People who write codes and programs are called as ………………
(a) System analysts
(b) System designers
(c) System operators
(d) System programmers
Answer:
(d) System programmers

Question 10.
Accounting software is an example of ………………
(a) System software
(b) Application software
(c) Utility software
(d) Operating software
Answer:
(b) Application software

II. Very Short Answer Questions

Question 1.
What is a computer?
Answer:
A computer can be described as an electronic device designed to accept raw data as input, processes them and produces meaningful information as output. It has the ability to perform arithmetic and logical operations as per given set of instructions called program. Today, computers are used all over the world in several areas for different purposes.

Question 2.
What is CAS?
Answer:
Computerised accounting system (CAS) refers to the system of maintaining accounts using computers. It involves the processing of accounting transactions through the use of hardware and software in order to keep and produce accounting records and reports.

Question 3.
What is hardware?
Answer:
The physical components of a computer constitute its hardware. Hardware consists of input devices and output devices that make a complete computer system.

Question 4.
What is meant by software?
Answer:
A set of programs that form an interface between the hardware and the user of a computer system are referred to as software.

Question 5.
What is accounting software?
Answer:
The main function of CAS is to perform the accounting activities in an organisation and generate reports as per the requirements of the users. To obtain the desired results optimally, need based software or packages are to be installed in the organisation.

Question 6.
Name any two accounting packages.
Answer:

1. Readymade software
2. Customised software

Question 7.
Give any two examples of readymade software.
Answer:

1. Tally
2. Busy

Question 8.
What is coding?
Answer:
Code is an identification mark, generally, computerised accounting involves codification of accounts.

Question 9.
What is grouping of accounts?
Answer:
In any organisation, the main unit of classification is the major head which is further divided, into minor heads. Each minor head may have number of sub – heads. After classification of accounts into various groups.

Question 10.
What are mnemonic codes?
Answer:
A mnemonic code consists of alphabets or abbreviations as symbols to codify a piece of information.

III. Short Answer Questions

Question 1.
What are the various types of accounting software?
Answer:

1. Readymade software
2. Customised software and
3. Tailormade software

Question 2.
Mention any three limitations of computerised accounting system.
Answer:
Heavy cost of installation, Cost of training and fear of unemployment

Question 3.
State the various types of coding methods.
Answer:
Sequential codes, block codes and mnemonic codes.

Question 4.
List out the various reports generated by computerised accounting system.
Answer:

1. Liabilities and capital
2. Assets
3. Revenues and
4. Expenses.

Under Liabilities and Capital:
Capital, Non – current liabilities and current liabilities.

Under Assets:
Non – current assets and Current assets.

Question 5.
State the input and output devices of a computer system.
Answer:
Input devices: keyboard, optical scanner, mouse, joystick, touch screen and stylus.
Output devices: Monitor and printer.

Textbook Case Study Solved

Question 1.
The manager of a medium – sized business is considering the introduction of computerised accounting system. Some staff feels that it is an opportunity to learn new skill. The manager has promised free framing for their staff, So, the staff realise that their own skill can be enhanced. Also, there is a demand for highly skilled staff. But, some staff feels threatened by these changes. They feel that they may not be able to leam new skill. Moreover, some of them are nearing their retiring age. So they think that it is not needed for them. But the manager expects the cooperation from all the staff.
Now, discuss on the following points:

Question 1.
Will it be expensive for the business to introduce computerised accounting system?
Answer:
No, it will not be expensive. It may be integrated with enhanced MIS, multi – lingual and data organisation capabilities to simplify all the business processes easily and cost – effectively.

Question 2.
Will everyone get the access to use the computers? In such a case, how to protect data?
Answer:
It is an opportunity to leam new skill at free of cost. Retrieval of data is easier as the records are kept in soft copy in data base. By giving instructions, data can be retrieved quickly.

Question 3.
“People at the retirement age are not required to leam new skill” – Do you think so?
Answer:
No, the computerised accounting system is easy to leam by everyone. It is not a difficult one to retiring people also.

Question 4.
What are the factors to be considered by the managers before introducing CAS?
Answer:
The manager has to give an opportunity to leam new skill to his employees. It will take time but he has to face the employees problems:

1. Heavy cost of installation.
2. Cost of training.
3. Fear of unemployment.
4. Disruption of work and so on.

Samacheer Kalvi 11th Accountancy Computerised Accounting Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer

Question 1.
Which one is output device?
(a) Monitor
(b) Keyboard
(c) Mouse
(d) Optical scanner
Answer:
(a) Monitor

Question 2.
Components of CAS can be classified into …………….. categories.
(a) Six
(b) Seven
(c) Five
(d) Three
Answer:
(a) Six

Question 3.
Which one is operating system?
(a) File manager
(b) COBOL
(c) Windows
(d) PASCAL
Answer:
(c) Windows

Question 4.
Which one is matched correctly?
(a) Land & building – current assets
(b) Goodwill – non – current liabilities
(c) Patents – intangible assets
(d) Sales – expenses
Answer:
(c) Patents – intangible assets

Question 5.
There are …………….. methods of codification.
(a) Two
(b) Three
(c) Four
(d) Five
Answer:
(b) Three

II. Very Short Answer Questions

Question 1.
Expands of MIS and CPU.
Answer:

1. MIS – Management Information System
2. CPU – Central Processing Unit

Question 2.
Mention any two features of CAS.
Answer:

1. Simple and integrated
2. Speed

Question 3.
What is utility software?
Answer:
These are designed specifically for managing the computer device and its resources.

Question 4.
What is system operators?
Answer:
People who operate the systems and use it for different purposes. They are also called as end users.

Question 5.
What do you mean by DATA?
Answer:
The facts and figures that are fed into a computer for further processing are called data.

III. Short Answer Questions

Question 1.
What are the types of people interacting with a computer system?
Answer:

1. System analysts
2. System programmers
3. System operators

Question 2.
What is opearating system? Give two examples.
Answer:
A set of tools and programs to manage the overall working of a computer using a defined set of hardware components is called an operating system.
Example:

1. DOS
2. Windows.

Question 3.
Mention any three advantages of CAS.
Answer:

1. Faster processing
2. Accurate information and
3. Reliability

Question 4.
What are the three types of procedures in a computer system?
Answer:

1. Hardware oriented procedure
2. Software oriented procedure and
3. Internal procedure

Question 5.
Can you explain data?
Answer:
The facts and figures that are fed into a computer for further processing are called data. Data are raw input until the computer system interprets them using machine language, stores them in memory, classifies them for processing and produces results in conformance with the instructions given to it. Processed and useful data are called information which is used for decision making.

Students can Download Accountancy Chapter 13 Final Accounts of Sole Proprietors – II Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Accountancy Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Samacheer Kalvi 11th Accountancy Final Accounts of Sole Proprietors – II Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the Correct Answer

Question 1.
A prepayment of insurance premium will appear in ………………
(a) The trading account on the debit side
(b) The profit and loss account on the credit side
(c) The balance sheet on the assets side
(d) The balance sheet on the liabilities side
Answer:
(c) The balance sheet on the assets side

Question 2.
Net profit is ………………
(a) Debited to capital account
(b) Credited to capital account
(c) Debited to drawings account
(d) Credited to drawings account
Answer:
(b) Credited to capital account

Question 3.
Closing stock is valued at ………………
(a) Cost price
(b) Market price
(c) Cost price or market price whichever is higher
(d) Cost price or net realisable value whichever is lower
Answer:
(d) Cost price or net realisable value whichever is lower

Question 4.
Accrued interest on investment will be shown ………………
(a) On the credit side of profit and loss account
(b) On the assets side of balance sheet
(c) Both (a) and (b)
(d) None of these
Answer:
(c) Both (a) and (b)

Question 5.
If there is no existing provision for doubtful debts, provision created for doubtful debts is ………………
(a) Debited to bad debts account
(b) Debited to sundry debtors account
(c) Credited to bad debts account
(d) Debited to profit and loss account
Answer:
(d) Debited to profit and loss account

II. Very Short Answer Questions

Question 1.
What are adjusting entries?
Answer:
Adjustment entries are the journal entries made at the end of the accounting period to account for items which are omitted in trial balance and to make adjustments for outstanding and prepaid expenses and revenues accrued and received in advance.

Question 2.
What is outstanding expense?
Answer:
Expenses which have been incurred in the accounting period but not paid till the end of the accounting period are called outstanding expenses.

Question 3.
What is prepaid expense?
Answer:
Prepaid expenses refer to any expense or portion of expense paid in the current accounting year but the benefit or services of which will be received in the next accounting period. They are also called as unexpired expenses.

Question 4.
What are accrued incomes?
Answer:
Accrued income is an income or portion of income which has been earned during the current accounting year but not received till the end of that accounting year.

Question 5.
What is provision for discount on debtors?
Answer:
Cash discount is allowed by the suppliers to customers for prompt payment of amount due either on or before the due date. A provision created on sundry debtors for allowing such discount is called provision for discount on debtors.

III. Short Answer Questions

Question 1.
What is the need for preparing final accounts?
Answer:

1. To record omissions in trial balance such as closing stock, interest on captial, interest on drawings, etc.
2. To bring into account outstanding and prepaid expenses.
3. To bring into account income accrued and received in advance.
4. To create reserves and provisions.

Question 2.
What is meant by provision for doubtful debts? Why is it created?
Answer:
Provision for bad and doubtful debts refers to amount set aside as a charge against profit to meet any loss arising due to bad debt in future. The amount of doubtful debts is calculated on the basis of some percentage on debtors at the end of the accounting period after deducting further bad debts (if any). A provision for doubtful debts is created and is charged to profit and loss account.

Question 3.
Explain how closing stock is treated in final accounts?
Answer:
The unsold goods in the business at the end of the accounting period are termed as closing stock. As per As-2 (Revised), the stock is valued at cost price or net realisable value, whichever is lower.

Presentation in final accounts:

1. In the trading account: Shown on the credit side.
2. In the balance sheet: Shown on the assets side under current assets.

Question 4.
Give the adjusting entries for interest on capital and interest on drawings.
Answer:
Adjusting Entry: Interest on Capital

Adjusting Entry: Interest on Drawings

Question 5.
Explain the accounting treatment of bad debts, provision for doubtful debts and provision for discount on debtors.
Answer:

1. Bad Debts: When it is definitely known that amount due from a customer (debtor) to whom goods were sold on credit, cannot be realised at all, it is treated as bad debts.
2. Provision for bad and doubtful debts refers to amount set aside as a charge against profit to meet any loss arising due to bad debt in future.
3. Cash discount is allowed by the suppliers to customers for prompt payment of amount due either on or before the due date.

IV. Exercises

Question 1.
Pass adjusting entries for the following:
(a) The closing stock was valued at ₹ 5,000
(b) Outstanding salaries ₹ 150
(c) Insurance prepaid ₹ 450
(d) ₹ 20,000 was received in advance for commission.
(e) Accrued interest on investments is ₹ 1,000.
Answer:
Adjusting Entries

Question 2.
For the fol owing adjustments, pass adjusting entries:
(a) Outstanding wages ₹ 5,000.
(b) Depreciate machinery by ₹ 1,000.
(c) Interest on capital @ 5% (Capital: ₹ 20,000)
(d) Interest on drawings ₹ 50
(e) Write off bad debts ₹ 500
Answer:
Adjusting Entries

Question 3.
On preparing final accounts of Suresh, bad debt account has a balance of ₹ 800 and sundry debtors account has a balance of ₹ 16,000 of which ₹ 1,200 is to be written off as further bad debts. Pass adjusting entry for bad debts. And also show how it would appear in profit and loss account and balance sheet.
Answer:
Adjusting Entry

Profit and Loss Account

Balance Sheet

Question 4.
The trial balance on March 31, 2016 shows the following:
Sundry debtors ₹ 30,000; Bad debts ₹ 1,200
It is found that 3% of sundry debtors is doubtful of recovery and is to be provided for. Pass journal entry for the amount of provision and also show how it would appear in the profit and loss account and balance sheet.
Answer:

Profit and Loss Account for the year ended 31.03.2016

Balance Sheet as on 31.03.2016

Question 5.
The trial balance of a trader on 31^st December, 2016 shows debtors as ₹ 50,000.
Adjustments:
(a) Write off ₹ 1,000 as bad debts
(b) Provide 5% for doubtful debts
(c) Provide 2% for discount on debtors
Show how these items will appear in the profit and loss A/c and balance sheet of the trader.
Answer:
Profit and Loss Account for the year ended 31^st December, 2016

Balance Sheet as on 31^st December, 2016

Question 6.
On 1^st January, 2016, provision for doubtful debts account had a balance of ₹ 3,000. On December 31, 2016, sundry debtors amounted to ₹ 80,000. During the year, bad debts to be written off were ₹ 2,000. A provision for 5% was required for next year. Pass journal entries and show how these items would appear in the final accounts.
Answer:
Adjusting Entries

Profit and Loss Account for the year ended 31^st December, 2016

Balance Sheet as on 31.12.2016

Question 7.
The following are the extracts from the trial balance.
Sundry debtors ₹ 30,000; Bad debts ₹ 5,000 Additional information:
(a) Write off further bad debts ₹ 3,000.
(b) Create 10% provision for bad and doubtful debts.
You are required to pass necessary adjusting entries and show how these items will appear in profit and loss account and balance sheet.
Answer:
Adjusting Entries

Profit and Loss Account

Balance Sheet

Question 8.
The following are the extracts from the trial balance.

Additional information:
(a) Additional bad debts ₹ 3,000.
(b) Keep a provision for bad and doubtful debts @ 10% on sundry debtors.
You are required to pass necessary adjusting entries and show how these items will appear in profit and loss account and balance sheet.
Answer:
Adjusting Entries

Profit and Loss Account

Balance Sheet

Question 9.
The accounts of Lakshmi traders showed the following balance on 31^st March, 2016.
Sundry debtors ₹ 60,000; Bad debts ₹ 2,000
Provision for doubtful debts ₹ 4,200
At the time of preparation of final accounts on 31^st March, it was found that out of sundry debtors, ₹ 1,000 will be irrecoverable. It was decided to create a provision of 2% on debtors to meet any future possible bad debts.
Pass necessary journal entries and show how these items would appear in the final accounts.
Answer:
Adjusting Entries

Profit and Loss Account for the year ended 31.03.2016

Balance Sheet as on 31.03.2016

Question 10.

The following are the extracts from the trial balance.
Additional information:
(a) Create a provision for doubtful debts @ 10% on sundry debtors.
(b) Create a provision for discount on debtors @ 5% on sundry debtors.
You are required to pass necessary adjusting entries and show how these items will appear in the final accounts.
Answer:
Adjusting Entries

Profit and Loss Account

Balance Sheet

Question 11.
Following are the extracts from the trial balance.

Additional information:
(a) Additional bad debts 1,000
(b) Create a provision for doubtful debts @ 5% on sundry debtors.
(c) Create a provision for discount on debtors @ 2% on sundry debtors.
You are required to pass necessary journal entries and show how these items will appear in the final accounts.
Answer:
Adjusting Entries

Profit and Loss Account

Balance Sheet

Question 12.
The following are the extracts from the trial balance.

Answer:
Profit and Loss Account

Balance Sheet

Question 13.
Prepare trading account of Archana for the year ending 31^st December, 2106 from the following information.

Adjustments:
(a) Closing stock ₹ 1,00,000
(b) Wages outstanding ₹ 12,000
(c) Freight inwards paid in advance ₹ 5,000
Answer:
Trading A/c of Archana for the year ended 31.12.2016

Question 14.
Prepare profit and loss account of Manoj for the year ending on 31^st March, 2016

Adjustments:
(a) Salary outstanding ₹ 400
(b) Rent paid in advance ₹ 50
(c) Commission receivable ₹ 100
Answer:
Profit and Loss A/c of Manoj for the year ended 31.03.2016

Question 15.
From the trial balance of Sumathi and the adjustments prepare the trading and profit and loss account for the year ended 31^st March, 2016, and a balance sheet as on that date.

Adjustments
(a) Six months interest on loan is outstanding.
(b) Two months rent is due from tenant, the monthly rent being ₹ 25.
(c) Salary for the month of March 2016, ₹ 75 is unpaid.
(d) Stock in hand on March 31, 2016 was valued at ₹ 1,030.
Answer:
Trading and Profit & Loss A/c of Sumathi
for the year ended 31^st March, 2016

Balance Sheet of Sumathi as on 31.03.2016

Question 16.
The following trial balance was extracted from the books of Arun Traders as on 31^st March, 2018.

Answer:
Prepare trading and profit and loss account for the year ending 31^st March, 2018 and balance sheet as on that date after considering the following:
(a) Depreciate Plant and machinery @ 20%
(b) Wages outstanding amounts to ₹ 750.
(c) Half of repairs and maintenance paid is for the next year.
(d) Closing stock was valued at ₹ 15,000.
Answer:
Trading and Profit & Loss A/c of Arun Traders for the year ended 31.03.2018

Balance Sheet of Arun Traders as on 31.03.2018

Question 17.
The following is the trial balance of Babu as on 31^st December, 2016.

Prepare trading and profit and loss account for the year ended 31^st December, 2016 and a balance sheet as on that date after the following adjustments.
(a) Salaries outstanding ₹ 500
(b) Interest on investments receivable at 10%.
(c) Provision required for bad debts is 5%.
(d) Closing stock is valued at ₹ 9,900.
Answer:
Trading and Profit & Loss A/c for the year ended 31.12.2016

Balance Sheet as on 31.12.2016

Question 18.
From the following trial balance of Ramesh as on 31^st March, 2017, prepare the trading and profit and loss account and the balance sheet as on that date.

Answer:
Adjustments:
Closing stock was valued at ₹ 35,000
(b) Unexpired advertising ₹ 250
(c) Provision for bad and doubtful debts is to be increased to ₹ 3,000
(d) Provide 2% for discount on debtors.
Answer:
Trading and Profit & Loss A/c of Ramesh for the year ended 31.03.2017

Balance Sheet of Ramesh as on 31.03.2017

Question 19.
Following are the ledger balances of Devi as on 31^st December, 2016.

Prepare trading and profit and loss account for the year ended 31^st December, 2016 and balance sheet as on that date.
(a) Stock on 31^st December, 2016 ₹ 5,800.
(b) Write off bad debts ₹ 500.
(c) Make a provision for bad debts @ 5%.
(d) Provide for discount on debtors @ 2%.
Answer:
Trading and Profit & Loss A/c of Devi for the year ended 31.12.2016

Balance Sheet of Devi as on 31.12.2016

Question 20.
From the following trial balance of Mohan for the year ended 31^st March, 2017 and additional information, prepare trading and profit and loss account and balance sheet.

Additional information:
(a) Closing stock is valued at ₹ 15,500
(b) Write off ₹ 500 as bad debts and create a provision for bad debts @ 10% on debtors.
(c) Depreciation @ 10% required
Answer:
Trading and Profit and Loss A/c of Mohan for the year ended 31.03.2017

Balance Sheet of Mohan as on 31.03.2017

Question 21.
From the following trial balance of Subramaniam, prepare his trading and profit and loss account and balance sheet as on 31^st December, 2016.

Take into account the following adjustments:
(a) Charge interest on drawings at 8%.
(b) Outstanding salaries ₹ 3,000
(c) Closing stock was valued at ₹ 48,000
(d) Provide for 5% interest on capital.
Answer:
Trading and Profit & Loss A/c of Subramaniam for the year ended 31.12.2016

Balance Sheet of Subramaniam as on 31.12.2016

Question 22.
Prepare trading and profit and loss account and balance sheet from the following trial balance of Madan as on 31^st March, 2018.

Adjustments:
(a) The closing stock was ₹ 80,000
(b) Provide depreciation on plant and machinery @ 20%
(c) Write off ₹ 800 as further bad debts
(d) Provide the doubtful debts @ 5% on sundry debtors
Answer:
Trading and Profit & Loss A/c for the year ended 31.03.2018

Balance Sheet as on 31.03.2018

Question 23.
From the following information prepare trading and profit and loss account and balance sheet of Kumar for the year ending 31^st December, 2017.

Adjustments:
(a) The closing stock on 31st December, 2017 was valued at ₹ 3,900.
(b) Carriage inwards prepaid ₹ 250
(c) Rent received in advance ₹ 100
(d) Manager is entitled to receive commission @ 5% of net profit after providing such commission.
Answer:
Trading and Profit & Loss Account of Kumar for the year ended 31.12.2017

Balance Sheet of Kumar as on 31.12.2017

Question 24.
From the following information, prepare trading and profit and loss account and balance sheet in the books of Sangeetha for the year ending 31^st March, 2018.

Adjustments:
(a) Stock on 31st March, 2018 ₹ 14,200
(b) Income tax of Sangeetha paid ₹ 800
(c) Charge interest on drawings @ 12% p.a.
(d) Provide managerial remuneration @ 10% of net profit before charging such commission.
Answer:
Trading and Profit & Loss Account of Sangeetha for the year ended 31.03.2018

Balance Sheet of Sangeetha as on 31.03.2018

Textbook Case Study Solved

Question 1.
James is a trader who sells washing machines on credit. But, he does not remember the due date to collect the money from his debtors. Some of his customers do not pay on time. His cash inflow is becoming worse. As a result, he could not pay his telephone bill and rent at the end of the accounting period. Hence, he showed only the amount paid as expense. He has many washing machines unsold at the year end. He is worried about the performance of his business. So, he is planning to appoint a manager to take care of his business. The new manager insists James to apply the accounting principle of prudence and matching and also to allow cash discount.
Now, discuss on the following points:

Question 1.
Why does James sell on credit?
Answer:
James sells goods on credit to increase the sales volume and reduce the stock.

Question 2.
Are there any ways to encourage his debtors to make the payment on time?
Answer:
Yes, there are many ways to encourage his debtors to make the payment on time by way of cash discount and trade discount.

Question 3.
What might happen if the debtors do not pay?
Answer:
If the debtors do not pay, the bad debts will be increased in the business.

Question 4.
In what ways prudence and matching principles can be applied for the business of James?
Answer:
Prudence principle can be applied for the business here closing stock was valued on cost price or market price whichever is lower under the prudence principle. Matching principle can be applied here for revenue and expense.

Question 5.
What will be the impact on income statement and the balance sheet, if the outstanding expenses are not adjusted?
Answer:
Outstanding expenses to be added with the concerned expenditure in the income Statement and the outstanding expenses will be recorded in liabilities side.

Question 6.
On what basis the unsold washing machines should be valued?
Answer:
The unsold washing machine should be valued at cost price or market price, whichever is lower under prudence principle. Managerial commission can be given to motivate the new manager to retain him in the business of James.

Samacheer Kalvi 11th Accountancy Final Accounts of Sole Proprietors – II Additional Questions and Answers

I. Multiple Choice Questions (Other important questions)
Choose the correct answer

Question 1.
If closing stock is already adjusted, adjusted purchase account and ………………. stock will appear in trial balance.
(a) Opening
(b) Closing
(c) Average
(d) None of these
Answer:
(b) Closing

Question 2.
Outstanding expense account is a ………………. account.
(a) Nominal
(b) Real
(c) Representative personal
(d) Personal
Answer:
(c) Representative personal

Question 3.
When bad debts already appears in the trial balance, it is taken only to debit side of ………………. account.
(a) Profit and Loss
(b) Balance sheet
(c) Asset side
(d) None of these
Answer:
(a) Profit and Loss

Question 4.
Income tax paid by the business for the proprietor is treated as ……………….
(a) Expense
(b) Profit and Loss A/c
(c) Drawings
(d) None of these
Answer:
(c) Drawings

Question 5.
Commission on net profit after charging such commission:

Answer:

Students can Download Accountancy Chapter 11 Capital and Revenue Transactions Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Accountancy Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Accountancy Solutions Chapter 11 Capital and Revenue Transactions

Samacheer Kalvi 11th Accountancy Capital and Revenue Transactions Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the Correct Answer

Question 1.
Amount spent on increasing the seating capacity in a cinema hall is ……………….
(a) Capital expenditure
(b) Revenue expenditure
(c) Deferred revenue expenditure
(d) None of the above
Answer:
(a) Capital expenditure

Question 2.
Expenditure incurred ₹ 20,000 for trial run of a newly installed machinery will be ……………….
(a) Preliminary expense
(b) Revenue expenditure
(c) Capital expenditure
(d) Deferred revenue expenditure
Answer:
(c) Capital expenditure

Question 5.
Revenue expenditure is intended to benefit ……………….
(a) Past period
(b) Future period
(c) Current period
(d) Any period
Answer:
(c) Current period

Question 6.
Pre – operative expenses are ……………….
(a) Revenue expenditure
(b) Prepaid revenue expenditure
(c) Deferred revenue expenditure
(d) Capital expenditure
Answer:
(d) Capital expenditure

II. Very Short Answer Questions

Question 1.
What is meant by revenue expenditure?
Answer:
The expenditure incurred for day to day running of the business or for maintaining the earning capacity of the business is known as revenue expenditure. It is recurring in nature. It is incurred to generate revenue for a particular accounting period. The revenue expenditure may be incurred in relation with revenue or in relation with a particular accounting period.

Question 2.
What is capital expenditure?
Answer:
It is an expenditure incurred during an accounting period, the benefits of which will be available for more than one accounting period. It includes any expenditure resulting in the acquisition of any fixed asset or contributes to the revenue earning capacity of the business. It is non-recurring in nature.

Question 3.
What is capital profit?
Answer:
Capital profit is the profit which arises not from the normal course of the business. Profit on sale of fixed asset is an example for capital profit.

Question 4.
Write a short note on revenue receipt.
Answer:
Receipts which are obtained in the normal course of business are called revenue receipts. It is recurring in nature. The amount received is generally small.

Question 5.
What is meant by deferred revenue expenditure?
Answer:
An expenditure which is revenue expenditure in nature, the benefits of which is to be derived over a subsequent period or periods is known as deferred
revenue expenditure.

III. Short Answer Questions

Question 1.
Distinguish between capital expenditure and revenue expenditure.
Answer:

Question 2.
Distinguish between capital receipt and revenue receipt.
Answer:

Question 3.
What is deferred revenue expenditure? Give two examples.
Answer:
An expenditure, which is revenue expenditure in nature, the benefits of which is to be derived over a subsequent period or periods is known as deferred revenue expenditure. The benefit usually accrues for a period of two or more years. It is for the time being, deferred from being charged against income. It is charged against income over a period of certain years. Examples: Considerable amount spent on advertising and major repairs to plant and machinery.

IV. Exercises

Question 1.
State whether the following expenditures are capital, revenue or deferred revenue.
(a) Advertising expenditure, the benefits of which will last for three years.
(b) Registration fees paid at the time of registration of a building.
(c) Expenditure incurred on repairs and whitewashing at the time of purchase of an old building in order to make it usable.
Answer:
(a) Deferred revenue expenditure
(b) Capital expenditure
(c) Capital expenditure

Question 2.
Classify the following items into capital and revenue.
(a) Registration expenses incurred for the purchase of land.
(b) Repairing charges paid for remodelling the old building purchased.
(c) Carriage paid on goods purchased.
(d) Legal expenses paid for raising of loans.
Answer:
(a) Capital expenditure
(b) Capital expenditure
(c) Revenue expenditure
(d) Capital expenditure

Question 3.
State whether they are capital and revenue.
(a) Construction of building ₹ 10,00,000.
(b) Repairs to furniture ₹ 50,000.
(c) White-washing the building ₹ 80,000.
(d) Pulling down the old building and rebuilding ₹ 4,00,000.
Answer:
(a) Capital expenditure
(b) Revenue expenditure
(c) Revenue expenditure
(d) Capital expenditure

Question 4.
Classify the following items into capital and revenue.
(a) ₹ 50,000 spent for railway siding.
(b) Loss on sale of old furniture.
(c) Carriage paid on goods sold.
Answer:
(a) Capital expenditure
(b) Capital loss
(c) Revenue expenditure

Question 5.
State whether the following are capital, revenue and deferred revenue.
(a) Legal fees paid to the lawyer for acquiring a land ₹ 20,000.
(b) Heavy advertising cost of ₹ 12,00,000 spent on introducing a new product.
(c) Renewal of factory licence ₹ 12,000.
(d) A sum of ₹ 4,000 was spent on painting the factory.
Answer:
(a) Capital expenditure
(b) Deferred revenue expenditure
(c) Revenue expenditure
(d) Revenue expenditure

Question 6.
Classify the following receipts into capital and revenue.
(a) Sale proceeds of goods ₹ 75,000.
(b) Loan borrowed from bank ₹ 2,50,000.
(c) Sale of investment ₹ 1,20,000.
(d) Commission received ₹ 30,000.
(e) ₹ 1,400 wages paid in connection with the erection of new machinery.
Answer:
(a) Revenue receipts
(b) Capital receipts
(c) Capital receipts
(d) Revenue receipts
(e) Capital expenditure

Question 7.
Identify the following items into capital or revenue.
(a) Audit fees paid ₹ 10,000.
(b) Labour welfare expenses ₹ 5,000.
(c) ₹ 2,000 paid for servicing the company vehicle.
(d) Repair to furniture purchased second hand ₹ 3,000.
(e) Rent paid for the factory ₹ 12,000.
Answer:
(a) Revenue expenditure
(b) Revenue expenditure
(c) Revenue expenditure
(d) Capital expenditure
(e) Revenue expenditure

Textbook Case Study Solved

Question 1.
Sadhana decides to start a business of selling air – conditioners. She buys different brands of air – conditioners. She also buys a delivery van, some furniture and some tools to fix air- conditioners. She buys some stationery items and cleaning liquid. She spends some amount on advertising her shop. She records the entire amount spent in the trading account.
Now, discuss on the following points:

Question 1.
Is it correct to record the entire amount spent in the first year of trading in the trading account? What impact will it have on the profit for the year?
Answer:
It is not correct to record the entire amount spent in the first year of trading in the trading account. It will be shown as a capital expenditure. For example: furniture and delivery van.

Question 2.
What are her fixed assets?
Answer:

1. A delivery van.
2. Some furniture.
3. Some tools.

Question 3.
Does she apply accounting concepts? If not which is the concept she does not apply?
Answer:
No, she does not apply accounting concepts. She does not apply business entity concepts.

Question 4.
Can you help Sadhana to classify the expenditure?
Answer:
Yes, I can help Sadhana to classify the expenditure:

Question 5.
What other capital, revenue and deferred revenue expenditure her business may incur in the future?
Answer:
In the future:

Samacheer Kalvi 11th Accountancy Capital and Revenue Transactions Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer

Question 1.
Cost of acquisition of land and building is an example of ………………
(a) Capital expenditure
(b) Revenue expenditure
(c) Capital receipts
(d) Revenue receipts
Answer:
(a) Capital expenditure

Question 2.
There are types of expenditure ………………
(a) One
(b) Two
(c) Three
(d) Four
Answer:
(c) Three

Question 3.
……………… expenditure is recurring in nature.
(a) Capital expenditure
(b) Revenue expenditure
(c) Capital loss
(d) Capital profit
Answer:
(b) Revenue expenditure

Question 4.
Considerable amount spent on advertising is an example of ……………… expenditure.
(a) capital
(b) revenue
(c) deferred
(d) none of these
Answer:
(c) deferred

II. Very Short Answer Questions

Question 1.
What is capital receipt?
Answer:
Receipt which is not revenue in nature is called capital receipt. It is non-recurring in nature. The amount received is normally substantial. It is shown on the liabilities side of the balance sheet.

Question 2.
Write any two features of revenue expenditure.
Answer:

1. It is recurring in nature.
2. It is incurred for maintaining the earning capacity of the business.

III. Short Answer Questions

Question 1.
Write three features of deferred revenue expenditure.
Answer:

1. It is a revenue expenditure, the benefits of which is to be derived over a subsequent period or periods.
2. It is not fully written off in the year of actual expenditure. It is written off over a period of certain years.
3. The balance available after writing off (i.e., Actual expenditure – Amount written off) is shown on the assets side balance sheet.

Question 2.
What are the classifications of expenditures?
Answer:

1. Capital expenditure
2. Revenue expenditure
3. Deferred revenue expenditure

Question 3.
Write any three examples of revenue expenditure.
Answer:

1. Purchase of goods for sale
2. Administrative, selling and distribution expenses
3. Manufacturing expenses