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# Electrical Engineering (EE) 2011 GATE Paper with solution (solved) GATE Notes | EduRev

## GATE : Electrical Engineering (EE) 2011 GATE Paper with solution (solved) GATE Notes | EduRev

``` Page 1

?EE-Paper Code-A? GATE 2011                          www.gateforum.com

© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the
written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                                        1
Q. No. 1 – 25 Carry One Mark Each

1. Roots of the algebraic equation
3 2
x x x 1 0 + + + = are
(A) ( ) 1, j, j + + - (B) ( ) 1, 1, 1 + - + (C) (0,0,0) (D) ( ) 1, j, j - + -
Exp: -
( )( )
3 2 2
x x x 1 0 x 1 x 1 0 + + + = ? + + =

2
x 1 0; x 1 0 ? + = + =

x 1 x j ? = - = ±

2. With K as a constant, the possible solution for the first order differential equation
3x
dy
e
dx
-
= is
(A)
3x
1
e K
3
-
- + (B)
3x
1
e K
3
- + (C)
3x
1
e K
3
-
- + (D)
x
3e K
-
- +
Exp: -
3x 3x
dy
e dy e dx
dx
- -
= ? =
Integrate on both sides

3x
3x
e 1
y K e K
3 3
-
-
= + = - +
-

3. The r.m.s value of the current i(t) in the circuit shown below is
(A)
1
A
2

(B)
1
A
2

(C) 1A
(D) 2A
Exp: - 1 rad /sec ? =
L C
X 1 ; X 1 = O = O

( )
Sin t
I t Sin t;
1
= =
O
rms
1
I A
2
=
1F
1H
~
1O
( ) i t
( ) 1.0sint V
+

1O
1Sint
1O

+

-

Page 2

?EE-Paper Code-A? GATE 2011                          www.gateforum.com

© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the
written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                                        1
Q. No. 1 – 25 Carry One Mark Each

1. Roots of the algebraic equation
3 2
x x x 1 0 + + + = are
(A) ( ) 1, j, j + + - (B) ( ) 1, 1, 1 + - + (C) (0,0,0) (D) ( ) 1, j, j - + -
Exp: -
( )( )
3 2 2
x x x 1 0 x 1 x 1 0 + + + = ? + + =

2
x 1 0; x 1 0 ? + = + =

x 1 x j ? = - = ±

2. With K as a constant, the possible solution for the first order differential equation
3x
dy
e
dx
-
= is
(A)
3x
1
e K
3
-
- + (B)
3x
1
e K
3
- + (C)
3x
1
e K
3
-
- + (D)
x
3e K
-
- +
Exp: -
3x 3x
dy
e dy e dx
dx
- -
= ? =
Integrate on both sides

3x
3x
e 1
y K e K
3 3
-
-
= + = - +
-

3. The r.m.s value of the current i(t) in the circuit shown below is
(A)
1
A
2

(B)
1
A
2

(C) 1A
(D) 2A
Exp: - 1 rad /sec ? =
L C
X 1 ; X 1 = O = O

( )
Sin t
I t Sin t;
1
= =
O
rms
1
I A
2
=
1F
1H
~
1O
( ) i t
( ) 1.0sint V
+

1O
1Sint
1O

+

-

?EE-Paper Code-A? GATE 2011                          www.gateforum.com

© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the
written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                                        2
4. The fourier series expansion ( )
0 n n
n 1
f t a a cosn t b sin n t
8
=
= + ? + ?
?
of the periodic
signal shown below will contain the following nonzero terms
( )
0 n
A a and b , n 1,3,5,... = 8
( )
0 n
B a and a , n 1,2,3,... = 8
( )
0 n n
C a a and b , n 1,2,3,... = 8
( )
0 n
D a anda n 1,3,5,... = 8
Exp: -  ? it satisfies the half wave symmetry, so that it contains only odd harmonics.
? It satisfies the even symmetry. So
n
b 0 =

5. A 4 – point starter is used to start and control the speed of a
(A)  dc shunt motor with armature resistance control
(B)  dc shunt motor with field weakening control
(C)  dc series motor
(D)  dc compound motor

6. A three-phase, salient pole synchronous motor is connected to an infinite bus. Ig
is operated at no load a normal excitation. The field excitation of the motor is
first reduced to zero and then increased in reverse direction gradually. Then the
armature current
(A) Increases continuously
(B) First increases and then decreases steeply
(C) First decreases and then increases steeply
(D) Remains constant

7. A nuclear power station of 500 MW capacity  is located at 300 km away from a
load center. Select the most suitable power evacuation transmission configuration
among the following options

(A)

(B)

(C)

(D)
( ) f t
0
t
~

132kV,300km doublecircuit
~

132kv,300 km single circuitwith40%seriescapacitor compensation
~

400kV,300kmsinglecircuit
~

400kV,300km doublecircuit
Page 3

?EE-Paper Code-A? GATE 2011                          www.gateforum.com

© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the
written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                                        1
Q. No. 1 – 25 Carry One Mark Each

1. Roots of the algebraic equation
3 2
x x x 1 0 + + + = are
(A) ( ) 1, j, j + + - (B) ( ) 1, 1, 1 + - + (C) (0,0,0) (D) ( ) 1, j, j - + -
Exp: -
( )( )
3 2 2
x x x 1 0 x 1 x 1 0 + + + = ? + + =

2
x 1 0; x 1 0 ? + = + =

x 1 x j ? = - = ±

2. With K as a constant, the possible solution for the first order differential equation
3x
dy
e
dx
-
= is
(A)
3x
1
e K
3
-
- + (B)
3x
1
e K
3
- + (C)
3x
1
e K
3
-
- + (D)
x
3e K
-
- +
Exp: -
3x 3x
dy
e dy e dx
dx
- -
= ? =
Integrate on both sides

3x
3x
e 1
y K e K
3 3
-
-
= + = - +
-

3. The r.m.s value of the current i(t) in the circuit shown below is
(A)
1
A
2

(B)
1
A
2

(C) 1A
(D) 2A
Exp: - 1 rad /sec ? =
L C
X 1 ; X 1 = O = O

( )
Sin t
I t Sin t;
1
= =
O
rms
1
I A
2
=
1F
1H
~
1O
( ) i t
( ) 1.0sint V
+

1O
1Sint
1O

+

-

?EE-Paper Code-A? GATE 2011                          www.gateforum.com

© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the
written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                                        2
4. The fourier series expansion ( )
0 n n
n 1
f t a a cosn t b sin n t
8
=
= + ? + ?
?
of the periodic
signal shown below will contain the following nonzero terms
( )
0 n
A a and b , n 1,3,5,... = 8
( )
0 n
B a and a , n 1,2,3,... = 8
( )
0 n n
C a a and b , n 1,2,3,... = 8
( )
0 n
D a anda n 1,3,5,... = 8
Exp: -  ? it satisfies the half wave symmetry, so that it contains only odd harmonics.
? It satisfies the even symmetry. So
n
b 0 =

5. A 4 – point starter is used to start and control the speed of a
(A)  dc shunt motor with armature resistance control
(B)  dc shunt motor with field weakening control
(C)  dc series motor
(D)  dc compound motor

6. A three-phase, salient pole synchronous motor is connected to an infinite bus. Ig
is operated at no load a normal excitation. The field excitation of the motor is
first reduced to zero and then increased in reverse direction gradually. Then the
armature current
(A) Increases continuously
(B) First increases and then decreases steeply
(C) First decreases and then increases steeply
(D) Remains constant

7. A nuclear power station of 500 MW capacity  is located at 300 km away from a
load center. Select the most suitable power evacuation transmission configuration
among the following options

(A)

(B)

(C)

(D)
( ) f t
0
t
~

132kV,300km doublecircuit
~

132kv,300 km single circuitwith40%seriescapacitor compensation
~

400kV,300kmsinglecircuit
~

400kV,300km doublecircuit
?EE-Paper Code-A? GATE 2011                          www.gateforum.com

© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the
written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                                        3
8. The frequency response of a linear system ( ) G j? is provided in the tubular form
below
( ) G j?
1.3 1.2 1.0 0.8 0.5 0.3
( ) G j ? ?
O
130 -
O
140 -
O
150 -
O
160 -
O
180 -
O
200 -
( )
O
A 6 dB and 30   ( )
O
B 6 dB and 30 -
( )
O
C 6 dB and 30 -   ( )
O
D 6 dB and 30 - -
Exp: - At ( ) G jw 180 ? = - magnitude M=0.5
So G.M = 20 log
1
6dB
0.5
? ?
=
? ?
? ?

At ( ) G jw 1 = phase angle ( ) G jw 150 ? = -
So P.M = ( )
0
180 150 30 + - =

9. The steady state error of a unity feedback linear system for a unit step input is
0.1. The steady state error of the same system, for a pulse input r(t) having a
magnitude of 10 and a duration of one second, as shown in the figure is

(A) 0 (B) 0.1 (C) 1 (D) 10
Exp: - For step input
ss
1
e 0.1 k 9
1 k
= = ? =
+

( )
9
G S
S 1
=
+

Now the input is pulse ( ) ( ) ( ) r t 10 u t u t 1 ? ? = - -
? ?

( )
s
1 e
r s 10
s
-
? ? -
=
? ?
? ?

( )
( ) ( )
s
ss
S 0 S 0
S 10 1 e
S.R s
S
e Lt Lt
S 10 1 G S H S
S 1
-
? ?
? ? -
? ?
= =
+ +
+

0
0
10
= =

10. Consider the following statement
(i)  The compensating coil of a low power factor wattmeter compensates the
effect of the impedance of the current coil.
(ii) The compensating coil of a low power factor wattmeter compensates the
effect 0of the impedance of the voltage coil circuit.
( ) r t
10
1s
t
Page 4

?EE-Paper Code-A? GATE 2011                          www.gateforum.com

© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the
written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                                        1
Q. No. 1 – 25 Carry One Mark Each

1. Roots of the algebraic equation
3 2
x x x 1 0 + + + = are
(A) ( ) 1, j, j + + - (B) ( ) 1, 1, 1 + - + (C) (0,0,0) (D) ( ) 1, j, j - + -
Exp: -
( )( )
3 2 2
x x x 1 0 x 1 x 1 0 + + + = ? + + =

2
x 1 0; x 1 0 ? + = + =

x 1 x j ? = - = ±

2. With K as a constant, the possible solution for the first order differential equation
3x
dy
e
dx
-
= is
(A)
3x
1
e K
3
-
- + (B)
3x
1
e K
3
- + (C)
3x
1
e K
3
-
- + (D)
x
3e K
-
- +
Exp: -
3x 3x
dy
e dy e dx
dx
- -
= ? =
Integrate on both sides

3x
3x
e 1
y K e K
3 3
-
-
= + = - +
-

3. The r.m.s value of the current i(t) in the circuit shown below is
(A)
1
A
2

(B)
1
A
2

(C) 1A
(D) 2A
Exp: - 1 rad /sec ? =
L C
X 1 ; X 1 = O = O

( )
Sin t
I t Sin t;
1
= =
O
rms
1
I A
2
=
1F
1H
~
1O
( ) i t
( ) 1.0sint V
+

1O
1Sint
1O

+

-

?EE-Paper Code-A? GATE 2011                          www.gateforum.com

© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the
written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                                        2
4. The fourier series expansion ( )
0 n n
n 1
f t a a cosn t b sin n t
8
=
= + ? + ?
?
of the periodic
signal shown below will contain the following nonzero terms
( )
0 n
A a and b , n 1,3,5,... = 8
( )
0 n
B a and a , n 1,2,3,... = 8
( )
0 n n
C a a and b , n 1,2,3,... = 8
( )
0 n
D a anda n 1,3,5,... = 8
Exp: -  ? it satisfies the half wave symmetry, so that it contains only odd harmonics.
? It satisfies the even symmetry. So
n
b 0 =

5. A 4 – point starter is used to start and control the speed of a
(A)  dc shunt motor with armature resistance control
(B)  dc shunt motor with field weakening control
(C)  dc series motor
(D)  dc compound motor

6. A three-phase, salient pole synchronous motor is connected to an infinite bus. Ig
is operated at no load a normal excitation. The field excitation of the motor is
first reduced to zero and then increased in reverse direction gradually. Then the
armature current
(A) Increases continuously
(B) First increases and then decreases steeply
(C) First decreases and then increases steeply
(D) Remains constant

7. A nuclear power station of 500 MW capacity  is located at 300 km away from a
load center. Select the most suitable power evacuation transmission configuration
among the following options

(A)

(B)

(C)

(D)
( ) f t
0
t
~

132kV,300km doublecircuit
~

132kv,300 km single circuitwith40%seriescapacitor compensation
~

400kV,300kmsinglecircuit
~

400kV,300km doublecircuit
?EE-Paper Code-A? GATE 2011                          www.gateforum.com

© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the
written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                                        3
8. The frequency response of a linear system ( ) G j? is provided in the tubular form
below
( ) G j?
1.3 1.2 1.0 0.8 0.5 0.3
( ) G j ? ?
O
130 -
O
140 -
O
150 -
O
160 -
O
180 -
O
200 -
( )
O
A 6 dB and 30   ( )
O
B 6 dB and 30 -
( )
O
C 6 dB and 30 -   ( )
O
D 6 dB and 30 - -
Exp: - At ( ) G jw 180 ? = - magnitude M=0.5
So G.M = 20 log
1
6dB
0.5
? ?
=
? ?
? ?

At ( ) G jw 1 = phase angle ( ) G jw 150 ? = -
So P.M = ( )
0
180 150 30 + - =

9. The steady state error of a unity feedback linear system for a unit step input is
0.1. The steady state error of the same system, for a pulse input r(t) having a
magnitude of 10 and a duration of one second, as shown in the figure is

(A) 0 (B) 0.1 (C) 1 (D) 10
Exp: - For step input
ss
1
e 0.1 k 9
1 k
= = ? =
+

( )
9
G S
S 1
=
+

Now the input is pulse ( ) ( ) ( ) r t 10 u t u t 1 ? ? = - -
? ?

( )
s
1 e
r s 10
s
-
? ? -
=
? ?
? ?

( )
( ) ( )
s
ss
S 0 S 0
S 10 1 e
S.R s
S
e Lt Lt
S 10 1 G S H S
S 1
-
? ?
? ? -
? ?
= =
+ +
+

0
0
10
= =

10. Consider the following statement
(i)  The compensating coil of a low power factor wattmeter compensates the
effect of the impedance of the current coil.
(ii) The compensating coil of a low power factor wattmeter compensates the
effect 0of the impedance of the voltage coil circuit.
( ) r t
10
1s
t
?EE-Paper Code-A? GATE 2011                          www.gateforum.com

© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the
written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                                        4
(A)  (i) is true but (ii) is false (B) (i) is false but (ii) is true
(C)  both (i) and (ii) are true (D) both (i) and (ii) are false

11. A low – pass filter with a cut-off frequency of 30Hz is cascaded with a high-pass
filter with a cut-off frequency of 20Hz. The resultant system of filters will function
as
(A)  an all-pass filter  (B)  an all-stop filter
(B)  an band stop (band-reject) filter (D)  a band – pass filter
Exp: -

So it is a band pass filter

12.

The CORRECT transfer characteristic is

(A)    (B)

(C)    (D)

R
12V +
12V +
O
V
12V -
12V -
R R
R
R
vi
-
+
-
+
Vo
6v -
12v -
Vi
12v + 12V +
6v -
12v -
Vi
O
V
6v +
Vo
Vi
6v + 6v -
12v +
Vo
Vi
6v + 6v -
12v +
20 30
Page 5

?EE-Paper Code-A? GATE 2011                          www.gateforum.com

© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the
written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                                        1
Q. No. 1 – 25 Carry One Mark Each

1. Roots of the algebraic equation
3 2
x x x 1 0 + + + = are
(A) ( ) 1, j, j + + - (B) ( ) 1, 1, 1 + - + (C) (0,0,0) (D) ( ) 1, j, j - + -
Exp: -
( )( )
3 2 2
x x x 1 0 x 1 x 1 0 + + + = ? + + =

2
x 1 0; x 1 0 ? + = + =

x 1 x j ? = - = ±

2. With K as a constant, the possible solution for the first order differential equation
3x
dy
e
dx
-
= is
(A)
3x
1
e K
3
-
- + (B)
3x
1
e K
3
- + (C)
3x
1
e K
3
-
- + (D)
x
3e K
-
- +
Exp: -
3x 3x
dy
e dy e dx
dx
- -
= ? =
Integrate on both sides

3x
3x
e 1
y K e K
3 3
-
-
= + = - +
-

3. The r.m.s value of the current i(t) in the circuit shown below is
(A)
1
A
2

(B)
1
A
2

(C) 1A
(D) 2A
Exp: - 1 rad /sec ? =
L C
X 1 ; X 1 = O = O

( )
Sin t
I t Sin t;
1
= =
O
rms
1
I A
2
=
1F
1H
~
1O
( ) i t
( ) 1.0sint V
+

1O
1Sint
1O

+

-

?EE-Paper Code-A? GATE 2011                          www.gateforum.com

© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the
written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                                        2
4. The fourier series expansion ( )
0 n n
n 1
f t a a cosn t b sin n t
8
=
= + ? + ?
?
of the periodic
signal shown below will contain the following nonzero terms
( )
0 n
A a and b , n 1,3,5,... = 8
( )
0 n
B a and a , n 1,2,3,... = 8
( )
0 n n
C a a and b , n 1,2,3,... = 8
( )
0 n
D a anda n 1,3,5,... = 8
Exp: -  ? it satisfies the half wave symmetry, so that it contains only odd harmonics.
? It satisfies the even symmetry. So
n
b 0 =

5. A 4 – point starter is used to start and control the speed of a
(A)  dc shunt motor with armature resistance control
(B)  dc shunt motor with field weakening control
(C)  dc series motor
(D)  dc compound motor

6. A three-phase, salient pole synchronous motor is connected to an infinite bus. Ig
is operated at no load a normal excitation. The field excitation of the motor is
first reduced to zero and then increased in reverse direction gradually. Then the
armature current
(A) Increases continuously
(B) First increases and then decreases steeply
(C) First decreases and then increases steeply
(D) Remains constant

7. A nuclear power station of 500 MW capacity  is located at 300 km away from a
load center. Select the most suitable power evacuation transmission configuration
among the following options

(A)

(B)

(C)

(D)
( ) f t
0
t
~

132kV,300km doublecircuit
~

132kv,300 km single circuitwith40%seriescapacitor compensation
~

400kV,300kmsinglecircuit
~

400kV,300km doublecircuit
?EE-Paper Code-A? GATE 2011                          www.gateforum.com

© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the
written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                                        3
8. The frequency response of a linear system ( ) G j? is provided in the tubular form
below
( ) G j?
1.3 1.2 1.0 0.8 0.5 0.3
( ) G j ? ?
O
130 -
O
140 -
O
150 -
O
160 -
O
180 -
O
200 -
( )
O
A 6 dB and 30   ( )
O
B 6 dB and 30 -
( )
O
C 6 dB and 30 -   ( )
O
D 6 dB and 30 - -
Exp: - At ( ) G jw 180 ? = - magnitude M=0.5
So G.M = 20 log
1
6dB
0.5
? ?
=
? ?
? ?

At ( ) G jw 1 = phase angle ( ) G jw 150 ? = -
So P.M = ( )
0
180 150 30 + - =

9. The steady state error of a unity feedback linear system for a unit step input is
0.1. The steady state error of the same system, for a pulse input r(t) having a
magnitude of 10 and a duration of one second, as shown in the figure is

(A) 0 (B) 0.1 (C) 1 (D) 10
Exp: - For step input
ss
1
e 0.1 k 9
1 k
= = ? =
+

( )
9
G S
S 1
=
+

Now the input is pulse ( ) ( ) ( ) r t 10 u t u t 1 ? ? = - -
? ?

( )
s
1 e
r s 10
s
-
? ? -
=
? ?
? ?

( )
( ) ( )
s
ss
S 0 S 0
S 10 1 e
S.R s
S
e Lt Lt
S 10 1 G S H S
S 1
-
? ?
? ? -
? ?
= =
+ +
+

0
0
10
= =

10. Consider the following statement
(i)  The compensating coil of a low power factor wattmeter compensates the
effect of the impedance of the current coil.
(ii) The compensating coil of a low power factor wattmeter compensates the
effect 0of the impedance of the voltage coil circuit.
( ) r t
10
1s
t
?EE-Paper Code-A? GATE 2011                          www.gateforum.com

© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the
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(A)  (i) is true but (ii) is false (B) (i) is false but (ii) is true
(C)  both (i) and (ii) are true (D) both (i) and (ii) are false

11. A low – pass filter with a cut-off frequency of 30Hz is cascaded with a high-pass
filter with a cut-off frequency of 20Hz. The resultant system of filters will function
as
(A)  an all-pass filter  (B)  an all-stop filter
(B)  an band stop (band-reject) filter (D)  a band – pass filter
Exp: -

So it is a band pass filter

12.

The CORRECT transfer characteristic is

(A)    (B)

(C)    (D)

R
12V +
12V +
O
V
12V -
12V -
R R
R
R
vi
-
+
-
+
Vo
6v -
12v -
Vi
12v + 12V +
6v -
12v -
Vi
O
V
6v +
Vo
Vi
6v + 6v -
12v +
Vo
Vi
6v + 6v -
12v +
20 30
?EE-Paper Code-A? GATE 2011                          www.gateforum.com

© All rights reserved by Gateforum Educational Services Pvt. Ltd.  No part of this document may be reproduced or utilized in any form without the
written permission. Discuss GATE 2011 question paper at www.gatementor.com.                                                                                        5
Exp: -  It is a Schmitt trigger and phase shift is zero.

13. A three-phase current source inverter used for the speed control of an induction
motor is to be realized using MOSFET switches as shown below. Switches S
1
to S
6

are identical switches.

The proper configuration for realizing switches S
1
to S
6
is

(A)  (B)  (C)  (D)

14. A point Z has been plotted in the complex plane, as shown in figure below.

The plot of the complex number
1
y is
z
=

d
I
A
1
S
B
2
S
3
S
A
B
4
S
6
S
2
S
l.M.
A
B
A
B
A
B
A
B
Im unit circle
o
Z
Re
Im unit circle
y
•

Re
( ) A Im unit circle
y
•

Re
( ) B
Im unit circle
y
•

Re
( ) C Im unit circle
y
•

Re
( ) D
```
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