1. law, govt and politics
2. legal issues
3. civil law
4. copyright

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General Aptitude
Q. No. 1 – 5 Carry One Mark Each
1.
Five teams have to compete in a league, with every team playing every other team exactly
once before going to the next round. How many matches will have to be held to complete
the league round of matches?
(A) 20
Answer: (B)
Exp:
(B) 10
(C) 8
(D) 5
For a match to be played, we need 2 teams
No. of matches = No. of ways of selections 2 teams out of 5  5 C2  10
2.
Fill in the blank with the correct idiom/phrase.
That boy from the town was a _____ in the sleepy village.
(A) dog out of herd
(C) fish out of water
(B) sheep from the heap
(D) bird from the flock
Answer:
(C)
Exp: From the statement, it appears that boy found it tough to adapt to a very different
situation.
3.
Ans:
4.
Ans:
5.
Choose the statement where underlined word is used correctly.
(A) When the teacher eludes to different authors, he is being elusive.
(B) When the thief keeps eluding the police, he is being elusive.
(C) Matters that are difficult to understand, indentify or remember are allusive.
(D) Mirages can be allusive, but a better way to express them is illusory.
(B)
Tanya is older than Eric. Cliff is older than Tanya. Eric is older than Cliff.
If the first two statements are true, then the third statement is:
(A) True
(B) False
(C) Uncertain
(D) Data insufficient
(B)
Choose the appropriate word/phrase, out of the four options given below, to complete the
following sentence:
Apparent lifelessness ___________ dormant life.
(A) harbours
Answer: (A)
(B) leads to
(C) supports
(D) affects
Exp: Apparent: looks like
dormant: hidden
Harbour: give shelter
Effect (verb): results in
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Q. No. 6 – 10 Carry Two Marks Each
6.
A coin tossed thrice. Let X be the event that head occurs in each of the first two tosses.
Let Y be the event that a tail occurs on the third toss. Let Z be the event that two tails
occurs in three tosses.
Based on the above information, which one of the following statements is TRUE?
(A) X and Y are not independent
(B) Y and Z are dependent
(C) Y and Z are independent
(D) X and Z are independent
Answer:
Exp:
(D)
X = {HHT, HHH}
Y depends on X
Z = {TTH, TTT}
∴ (D) is the correct choice.
7.
Given below are two statements followed by two conclusions. Assuming these statements
to be true, decide which one logically follows.
Statements:
I. No manager is a leader.
II. All leaders are executives.
Conclusions:
I. No manager is an executive.
II. No executive is a manager.
(A) Only conclusion I follows.
(B) Only conclusion II follows.
(C) Neither conclusion I nor II follows.
(D) Both conclusions I and II follow.
Answer:
Exp:
(D)
S 2:
S  1:
M
L
L
E
Therefore concluding diagram can be
E
E
L
E
or
L
E
M
or
Z
M
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In the given figure angle Q is a right angle, PS:QS=3:1, RT:QT=5:2 and PU:UR=1:1. If
area of triangle QTS is 20 cm2, then the area of triangle PQR in cm2 is ________.
R
U
P
Answer:
Exp:
T
S
Q
280
Let area of triangle PQR be „A‟
SQ
1
1


PQ 1  3 4
QT
2
2


QR 2  5 7
1
Area of  le QTS   SQ  QT
2
1 1
 2

   PQ    QR 
2 4
 7

1 2 1

     PQ  QR 
4 7 2


1
 Area of  le PQR
14
1
A
14
A  14  20  280cm 2
given 20cm 2 
9.
Select the appropriate option in place of underlined part of the sentence.
Increased productivity necessary reflects greater efforts made by the employees.
(A) Increase in productivity necessary
(B) Increase productivity is necessary
(C) Increase in productivity necessarily
(D) No improvement required
Answer:
10.
(C)
Right triangle PQR is to be constructed in the xy-plane so that the right angle is at P and
line PR is parallel to the x-axis. The x and y coordinates of P, Q, and R are to be integers
that satisfy the inequalities: 4  x  5 and 6  y  16. How many different triangles could
be constructed with these properties?
(A) 110
Answer:
(B) 1,100
(C) 9,900
(D) 10,000
(B)
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Instrumentation Engineering
Q. No. 1 – 25 Carry One Mark Each
1.
The capacitor shown in the figure is initially charged to +10 V. The switch closes at time
t=0. Then the value of Vc (t) in volts at time t=10 ms is _____ V.
t 0

C  1F
R  10k
Vc (t)

Answer: 3.67
Exp: It is a source free network where capacitor voltage
Vc  t   V0 e  t  ; t  0
  RC 
 10e 100t
Vc 10  103   10e
2.
 10010103 


1
100
 3.67V
The voltage (E 0 ) developed across a glass electrode for pH measurement is related to the
temperature (T) by the relation
(A) E 0 
Answer:
3.
1
T2
(B) E o 
(C) E0  T
(D) E0  T 2
(C)
Let 3+4j be a zero of a fourth order linear –phase FIR filter. The complex number which
is NOT a zero of this filter is
(A) 3  4j
Answer:
Exp:
1
T
(B)
3
4

j
25 25
(C)
3
4

j
25 25
(D)
1 1
 j
3 4
(D)
The property of a FIR filter is that if Zo is a zero then the remaining zeros are
1
,
Z0
 Z0 *,


4
3
 25  j 25  ,


3  4 j,
1 
 *
 Z0 

4
3
 25  j 25 


So option (D) is not matching with any of this.
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4.
The bridge most suited for measurement of a four-terminal resistance in the range of
0.001 to 0.1 is
(A) Wien‟s bridge
(C) Maxwell‟s bridge
Answer: (B)
5.
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(B) Kelvin double bridge
(D) Schering bridge
A load resistor RL is connected to a battery of voltage E with internal resistance Ri
through a resistance Rs as shown in the figure. For fixed values of R L and R i , the value
of R S   0  for maximum power transfer to RL is
Rs
Ri
E
RL
(B) R L  R i
(A) 0
(D) R L  R i
(C) R L
Answer: (A)
Exp: When load is constant, we should see for what value of resistance current will be
maximum and PR L  max .
6.
A mass-spring-damper system with force as input and displacement of the mass as output
has a transfer function G(s)  1/  s2  24s  900 . A force input F(t)=10 sin(70t) Newtons
is applied at time t = 0s. A beam from an optical stroboscope is focused on the mass. In
steady state, the strobe frequency in hertz at which the mass appears to be stationary is
(A) 5 / 
(B) 15 / 
(C) 35 / 
(D) 50 / 
Answer: (C)
7.
A light detector circuit using an ideal photo-diode is shown in the figure. The sensitivity
of the photo-diode is 0.5A / W. With Vr= 6V, the output voltage V0  1.0 for 10W
of incident light. If Vr is changed to 3V, keeping all other parameters the same, the value
of Vo in volts is __________ V.
R
Incident

light
V0

Vr
Answer: -1
Exp: Here Vo is independent of the value of Vr
It depends on the intensity of incident light.
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The logic evaluated by the circuit at the output is
X
Output
Y
(A) XY  YX
(B)
(C) XY  XY
Answer: (A)
Exp:
 X  Y  XY
(D) XY  XY  X  Y
Output of upper AND gate is XY
Output of lower AND gate is XY
Output of OR gate is XY  XY .
9.
The value of
1
 z
2
dz, where the contour is the unit circle traversed clockwise, is
(A) 2i
Answer:
Exp:
(C) 2i
(B) 0
(D) 4i
(B)
Given
1
 z
2
dz where c is unit circle.
1
z2
f(z) is not analytic at z = 0
Let f (z) 
By Laurent series, f(z) can be expressed as
1
 1
f (z)  (zero tens)  0.  1. 2 
z 
z




Analytic part
Principal part
Principal part has only two terms
 z  0 is pole of order 2
 Res f (z)z0  coefficient of
1
z
 b1  0
 By Cauchy‟s residue theorem
1
 z
2
dz  2i  (sum of residues)
 2i  0  0
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Consider the ammeter-voltmeter method of determining the value of the resistance R
using the circuit shown in the figure. The maximum possible errors of the voltmeter and
ammeter are known to be 1% and 2% of their readings, respectively. Neglecting the
effects of meter resistances, the maximum possible percentage error in the value of R
determined from the measurements, is ___ %.
Source
Vi
Answer:
R


Voltmeter


V


Ammeter
3
Vx Vx I2%

 R x  3%
I x I x  1%
Exp:
Rx 
11.
The highest frequency present in the signal x(t) is fmax. The highest frequency present in
the signal y(t)  x 2 (t) is
1
(B) f max
(C) 2f max
(D) 4f max
f max
2
Answer: (C)
Exp: Multiplication in time domain corresponds to convolution in frequency domain
(A)
x2  t   x f  * x f 
Using limit property of convolution x2(t) have maximum frequency 2fm.
12.
Let A be an n×n matrix with the rank r  0  r  n  . Then Ax=0 has p independent
solutions, where p is
(A) r
(B) n
Answer: (C)
Exp: Given AX=0
  A n n   r
0  r  n 
(C) n-r
(D) n+r
p = no. of independent solution
i.e nullity = p
We know that rank + nullity = n
rpn
pnr
13.
The double integral
(A)

x
y
0
0
(C)

a
a
0
x
Answer:
a
y
0
0

f (x, y) dxdy is equivalent to
f (x, y) dxdy
(B)

a
y
0
x
f (x, y) dxdy
(D)

a
a
0
0
f (x, y) dxdy
f (x, y) dxdy
(C)
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Exp:
Given double integral
a
y
0
0

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f (x, y)dx. dy
and x=0 to x=y, y=0 to y=a
the diagram is
yx
ya
(0,a)
(a,a)
x0
(0,0)
(a,0)
By applying change of order of integration
a
 
a
x 0 y  x
14.
f (x, y)dydx
In the circuit shown, the switch is momentarily closed and then opened. Assuming the
logic gates to have non-zero delay, at steady state, the logic states of X and Y are
X
Y
1
(A) X is latched, Y toggles continuously
(C) Y is latched, X toggles continuously
(B) X and Y are both latched
(D) X and Y both toggle continuously
Answer: (D)
Exp: The above circuit is a stable multivibrator circuit, where odd numbers of inverter are there
in the loop. In such a circuit, irrespective of the position of output, it always toggles.
 Latching means X and Y will be fixed to same value, in this case it is not possible.
15.
Consider the logic circuit with input signal TEST shown in the figure. All gates in the
figure shown have identical non-zero delay. The signal TEST which was at logic LOW is
switched to logic HIGH and maintained at logic HIGH. The output
TEST
(A) stays HIGH throughout
(C) pulses from LOW to HIGH to LOW
Answer: (D)
Output
(B) stays LOW throughout
(D) pulses from HIGH to LOW to HIGH
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Exp:
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For analysis point of view, assume delay of each gate is 10 msec. However we can take
any value.
 By referring the circuit the upper input to the NAND gate is direct test signal.
The lower input to NAND gate is TEST but with a delay of 30 nsec.
 Assuming the delay of NAND gate is 0. First draw output waveform (ideal case) then
shift that by 10 msec. i.e. introduce the delay.
Test
0
Test
 with delay  30n sec 
0
Output with
delay  0
30 n sec
0
Output with
NAND gate
delay  10 n sec
0
10 nsec
40 n sec
So we can clearly say that initial output change from high to low, then it changes from
low to high and then finally at steady state output is 1.
Note: Saying output is high (option A) will be wrong here. We are not interested to find
steady state output.
16.
A p-type semiconductor strain gauge has a nominal resistance of 1000  and a gauge
factor of +200 at 25o C. The resistance of the strain gauge in ohms when subjected to a
strain of 104 m m at the same temperature is _________ .
Answer:
1020
1
Exp:
R R
 R  200  10 4  1000

 20
GF 
p-type semiconductor have a positive gauge factor
i.e., Resistance  with  in strain
∴ Rstrained = 1000+20 = 1020
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17.
1
has zero initial conditions. The percentage
s 1
overshoot in its step response is _________ %.
A system with transfer function G  s  
Answer:
Exp:
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2
100
Comparing the denominator,   0

% Overshoot  e
18.
100  100
Liquid flow rate is measured using
(A) a Pirani gauge
(B) a pyrometer
(C) an orifice plate
(D) a Bourdon tube
Answer:
19.
12
(C)
The magnitude of the directional derivative of the function f  x, y   x 2  3y2 in
a
direction normal to the circle x 2  y2  2, at the point (1, 1) , is
(A) 4 2
Answer:
Exp:
(B) 5 2
(C) 7 2
(D)
9 2
(A)
Let f (x, y)  x 2  3y2
Let   x 2  y2  2 and P is (1,1)
Normal to the surface   


j
x
y
 2xi  2yj
i
 at (1,1)  2i  2j

Le a    2i  2j

We need to calculate magnitude of directional derivatives of f along a at (1,1)
Magnitude of directional deviations  f.aˆ
f
f
f  i  j  2xi  6yj
x
y
f (1,1)  2i  6 j

a  44 2 2
aˆ 
2i  2 j
2 2

i j
2
 i  j
Magnitude of direction derivative   2i  6j .

 2

26
2

8
2
4 2
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20.
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A power line is coupled capacitively through various parasitic capacitances to a shielded
signal line as shown in the figure. The conductive shield is grounded solidly at one end.
Assume that the length of the signal wire extending beyond the shield, and the shield
resistance are negligible. The magnitude of the noise voltage coupled to the signal line is
1
(A) directly proportional to C1G
2
C2S
C1S
Power
line
Signal
line
(B) inversely proportional to
the power line frequency
Conductive
shield
(C) inversely proportional to C15
V1 sin  1t 
(D) Zero
Answer:
21.
CSG
C1G
~
(D)
The torque transmitted by a cylindrical shaft is to be measured by using two strain gauges.
The angles for mounting the strain gauges relative to the axis of the shaft for maximum
sensitivity are
(A) 45o
Answer:
(B) 60o
(D) 180o
(C) 90
(A)
Exp:
45
45
22.
The output voltage of the ideal transformer with the polarities and dots shown in the
figure is given by
1: N


Vi sin t


V0

(A) NVi sin t
Answer:
(B)  NVi sin t
(C)
1
Vi sin t
N
(D) 
1
Vi sin t
N
(B)
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Exp:
23.
First mark the mutual voltage polarity using dot convention (when a reference current
enters at dot of one coil, it generates positive polarity on dot terminal of other coil)
1: N
 Using Transformer ratio

V2 N 2


Vi N1

V0
V2
V
sin

t
V
N
i
 V2  V1    NVi sin t

1


 Vo  V2   NVi sin t
The filter whose transfer function is of the form G  s  
s 2  bs  c
is
s 2  bs  c
(A) a high-pass filter
(B) a low-pass filter
(C) an all-pass filter
(D) a band-reject filter
Answer:
24.
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(C)
An apparatus to capture ECG signals has a filter followed by a data acquisition system.
The filter best suited for this application is
(A) low pass with cutoff frequency 200 Hz
(B) high pass with cutoff frequency 200 Hz
(C) band pass with lower and upper cutoff frequencies 100 Hz and 200 Hz for its pass band
(D) band reject with lower and upper cutoff frequencies 1 Hz and 200 Hz for its stop band
Answer:
(A)
Exp:
More appropriate will be a Band pass filter from 1Hz to 200Hz
25.
The figure shows a half-wave rectifier circuit with input voltage V  t   10sin 100t 
volts. Assuming ideal diode characteristics with zero forward voltage drop and zero reverse
current, the average power consumed in watts by the load resistance RL is ____ W.


Vt
R L  100

Answer:
Exp:
0.25
The voltage waveform at R  100 is as shown below.
Pavg
10 2 
V2
 rms 
R
100
 0.25 Watt
2
10
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Q. No. 26 – 55 Carry Two Marks Each
26.
Consider a low-pass filter module with a pass-band ripple of  in the gain magnitude. If
M such identical modules are cascaded, ignoring the loading effects, the pass-band ripple
of the cascade is
(A) 1  1   
Answer:
27.
M
(B) M

(C) 1 M

(D) 1  
M
*
The probability density function of a random variable X is pX  x   e x x  0 and 0
otherwise. The expected value of the function g x  x   e3x 4 is ________.
Answer:
Exp:
4
Probability density function of X is
given as Px (x)  e  x ; x  0
 0 ; otherwise
3x
and g x (x)  e 4

E  g x (x)    f x (x)g x (x)dx
0
3x

  e  x e 4 dx
0

3x
x
  e 4 dx
0


x
  e 4 dx
0



 e x / 4 

1 
 
 4 0
 4 e   1
 4[0  1]  4
  AX  Bu, where A   1 2  and B  1 .
A system is represented in state-space as X
1
 6 



The value of  for which the system is not controllable is _________.
Answer: -3
Exp: For a system to be uncontrollable, its controllability determinant should be equal to zero.
28.
Qc  B AB  0
 3 
 1 2 1 
AB  





 6  22  1 21   6
Qc  B AB 
1
3
 0    6  3  0    3
1 6
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29.
The Seebeck coefficients, in V
o
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C, for copper, constantan and iron, with respect to
platinum, are 1.9, 38.3 and 13.3 respectively. The magnitude of the thermo emf E
developed in the circuit shown in the figure, in millivolts is __________mV.
copper
T2  50o C
T1  0o C
E
constantan
T3  15o C
Answer:
Exp:
T1  0o C
iron
1.236
Cu Constantan  1.9   38.3

 VCu Constantan  40.2  50  2010 V  2.01mV
 40.2 V C 

FeConstantan  13.3   38.3  51.6 V C VFe constantan  51.6   15  0.774mV
Vout  2.01  0.774  1.236mV
30.
The current in amperes through the resistor R in the circuit shown in the figure is
__________.

1V

1
1
1A
1
Answer:
Exp:
R  1
1
1V
Using supermesh analysis

Mesh 1, 3 form super mesh
2I1  I2  2I3  I2  0
 2I1  2I2  2I3  0
1
1
… (1)
Writing KCL at Q
I1  I3  1
I2
… (2)
1
I1
R  1
1A
I3
Writing KCL on mesh 2
2I2  I1  I3  1
… (3)
Q
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Solution Equation 1, 2, 3
I1  0, I2  I3  1A
Current through R is I3  1A
Note: Strictly saying it is an ambiguous question as the direction of current is not
mentioned, so it could be –1A as well.
31.
In the circuit shown, the voltage source V  t   15  0.1 sin (100t) volts. The PMOS
transistor is biased such that it is in saturation with its gate-source capacitance being 4nF
and its transconductance at the operating part being 1 mA/V. Other parasitic impedances
of the MOSFET may be ignored. An external capacitor of capacitance 2 nF is connected
across the PMOS transistor as shown. The input impedance in mega ohm as seen by the
voltage source is ___________ M.
VS
PMOS
2nF
Vt
Answer:
32.
2k
1
The probability that a thermistor randomly picked up from a production unit is defective
is 0.1. The probability that out of 10 thermistors randomly picked up, 3 are defective is
(A) 0.001
Answer:
Exp:
~
(B) 0.057
(C) 0.0107
(D) 0.3
(B)
p = Probability of a thermister is defective
 0.1
q  1  p  1  0.1  0.9
n  10
Let X be random variable which is no. of defective pieces.
Required probability = P(x  3)
 10 C3 p3q 7
 10 C3 (0.1)3 (0.9)7
 0.057
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33.
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A liquid level measurement system employing a radio-isotopic is mounted on a tank as
shown in the figure. The absorption coefficient of water for the radiation is 7.7 m1. If the
height of water in the tank is reduced from 100 mm to 90 mm, the percentage change in
the radiation intensity received by the detector, neglecting abosroption of the radiation by
air, is ___________%.
Detector
Water
Source
Answer:
Exp:
8
General formula for the amount of intensity received , I  Io ex
I0  Intensity when absorbing material is nill or empty
  absorption coefficient
  density of the absorption material
x  thickness (or) height of the absorbing, material
By considering the given parameters and neglecting others
Intensity received,
(a) When xheight = 100 mm = 0.1 m
I1  e7.70.1  e0.77
(b) When x = 90 mm = 0.9 m
I2  e7.70.09  e0.693
So, %change 
I I
change
 100  2 1  100  8%
Initial
I1
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A beam of monochromatic light passes through two glass slabs of the same geometrical
thickness at normal incidence. The refractive index of the first slab is 1.5 and that of the
second, 2.0. The ratio of the time of passage of the beam through the first to the second
slab is _________.
Answer:
Exp:
0.75
We know, time 
distance d

velocity v
∴ The ratio of the time for 2 slabs 
t1
t2

d1 v1 d1 v 2  d1  d 2 becauseof same 
 


d 2 v 2 v1 d 2 geometrical thickness


v2
v1

v2 2
v1 1

1 1.5

 2 2.0
 0.75
NOTE: If a light of velocity “v” is passing through a matter of R.I = 0.7 (say)
Then the final output velocity will be 
35.
v
0.7
In the circuit shown in the figure, it is found that VBE  0.7 V amnd VE  0V. If dc  99
for the transistor, then the value of RB in kilo ohms is __________ k .
10V
RB
VE  0V
10k
IE
10V
Answer:
93
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Exp:
10V
RB
VB  0.7V
VE  0V
FE
1k
10V
10  0.7
IB 
RB
     (1)
I E     1 I B 
 IB 
0  (10V)
 10mA
1k
10
mA    99 
100
 100 A
Put the volume of IB in equation (1), we get
RB 
36.
9.3 9.3

 106  93k
IB 100
The output frequency of an LC tank oscillator employing a capacitive
sensor acting as
the capacitor of the tank is 100 kHz. If the sensor capacitance increases by 10%, the
output frequency in kilohertz becomes ________ kHz.
Answer:
Exp:
f1 
95.35
1
2 LC
Now C'  C  0.1C  1.1C
f2 
1

1

f
2 LC 2 1.1LC
1.1
100kHz
f2 
 95.346 kHz
1.1
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The resolving power of a spectrometer consisting of a collimator, a grating and a
telescope can be increased by
(A) increasing the angular magnification of the telescope
(B) increasing the period of the grating
(C) decreasing the period of the grating
(D) decreasing the slit-width of the collimator
Answer:
Exp:
(C)
Resolving power of a grating plate can be increased by increasing the number of slits.
i.e., by decreasing the grating period.
38.
A transfer function G(s) with the degree of its numerator polynomial zero and the degree
of its denominator polynomial two has a Nyquist plot shown in the figure. The transfer
function represents
lm G j
  
0
5j
4j
3j
2j
1j
 0
0.3
1j
Re  G  j 
0.1
(A) a stable, type-0 system
(B) a stable, type-1 system
(C) an unstable, type-0 system
(D) an unstable, type-1 system
Answer:
Exp:
0.1
0.2
(D)
Consider the Transfer function to be
G s 
1
s  s  1
Magnitude M 
0
1


2 2  1

1
 2  1
Phase   270  tan 1  
  0 M  ,
   M  0,
  270
  270  90  180

∴ our assumption of G(s) is true
The system is type-1 and is unstable.
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For the circuit shown in the figure, the rising edge triggered D-flip flop with asynchronous
reset has a clock frequency of 1 Hz. The NMOS transistor has an ON resistance of 1000
and an OFF resistance of infinity. The nature of the output waveform is
VS
HIGH
Q
D
1A
Output
CLK
R
Q

2 F
0.1V

NMOS
A
 B
C
 D
Answer:
Exp:
(A)
Before solving the problem, consider the following fact.
 clock is +ve edge triggered and input to D flip-flop is constant i.e logic 1.
 Reset pin of Flip flip is active high and it can function at any moment of time
independent to clock. Also note that reset will be triggered when V  terminal voltage just
crosses 0.1 V.
 N-MOS behave as open switch (off state, R   ) when Q  0 and it will behave as a
resistance of 1000 when Q  1 (as ON resistance is specified).
 Now for analysis, assume capacitor is initially uncharged and a +ve edge is triggered
to D-flipflop, as its input is high. Then Q=1 and Q  0 .
* When Q  0 , N-MOS is off, assuming the input resistance of comparator infinite. Then
the source current can flow only through the capacitor. Now we have to see how much
time the capacitor will take to reach 0.1 V. So that V  V and R will change from 0 to 1
and as the reset terminal will be triggered the state of flip-flop changes i.e.  Q  0, Q  1.
and N-MOS gets ON. As the integration of step is ramp if we integrate the constant
current voltage will be ramp.
 For this use the capacitor voltage equation.
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V0
1A
1 t
i c d
C 0
t
1
0.1V 
1  106 d
6 0
2  10
t
0.1   t  0.2sec  200 millisec
2
Vc (t) 
V0
2 F

So at t = 0.2sec, N-MOS is ON. Now the circuit is equivalent to a current source
connected to a parallel combination of RC with non zero initial voltage of capacitor
(0.1V).

Now the capacitor will discharge from its initial voltage with time constant   RC  2ms
and at 10 it will react steady state i.e., at 20 m.sec. The nature of discharge is
exponential decay.
Vs
1A
2 F
1000 

So we can conclude from the above analysis that capacitor takes 200 msec to charge upto
0.1V and 20 msec to discharge to the steady state value.

So if the +ve edge of clock appeared at t = 0 then at 200 msec it completed its charging
and at 220 msec it completed its discharging. This continues in each clock cycles.

The following waveform gives further clarity as fCLK  1Hz, TCLK  1000ms
CLOCK
500 msec
Output
500 msec
0.1V
Ch arging (Ramp)
Discharging
(exp decay)
200 msec
20 msec
Q
Reset
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40.
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Consider the circuits shown in the figure. The magnitude of the ratio of the currents, i.e.,
|I1/I2|, is _____________.
5
3
3
5
4
I1
2V
4
I2
2V
Answer:
Exp:
By reciprocity theorem, I1 = I2
So
41.
1
I1
1
I2
The linear I-V characteristics of 2-terminal non-ideal dc sources X and Y are shown in the
figure. If the sources are connected to a 1 resistor as shown, the current through the
resistor in amperes is _________A.
SourceY


SourceY
3
Current  A  2

SourceX
1
SourceX
1

0
0
1
2
3
4
5
Voltage  V 
Answer:
Exp:
1.75
The transfer characteristic curve and the circuit of a non ideal source is given below.
I
Vs
Rs
Rs
Slope  
1
Rs
Vs
V
Vs


VY  3V

Comparing the curve of X and Y
we can conclude that
X:
Vx  4V, R x  2
Y:
Vy  3V, R y  1
R x  2
R r  1
1
 V  4V
x

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When we connect these the circuit becomes,
I1 
42.
43
 1.75A
2 11
The figure shows a spot of light of uniform intensity 50 W/m2 and size 10mm×10 mm
incident at the exact centre of a photo-detector, comprising two identical photo-diodes D1
and D2. Each diode has a sensitivity of 0.4 A/W and is operated in the photoconductive
mode. If the spot of light is displaced upwards by 100μm, the resulting difference
between the photocurrents generated by D1 and D2 in micro amperes, is _________ μA.
10mm
D1
Photo-
10mm
detector
D2
Spot of
light
Answer:
Exp:
40
Total illuminated Area = 10×10×10-6m2 = 10-4m2
Light power = 50×10-4W = 5mW
Output from both diodes = 0.4×5mA = 2mA
So output from each diode = 1mA
In second case,
Area of one diode (D1) = 10×5.1×10-6
= 0.51×10-4m2
Light output (D1) = 0.51×10-4×50W
= 2.55 mW
Current in D1 = 2.55×0.4 = 1.02 mA
Current in D2 = Total current – ID1
= 2-1.02 = 0.98 mA
 ID1  ID2  1.02  0.98 mA  0.04mA  40A
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An ADC is interfaced with a microprocessor as shown in the figure. All signals have been
indicated with typical notations. Acquisition of one new sample of the analog input signal
by the microprocessor involves
Analog
Data
Bus
D7
D7
Input
Data linesD0toD7
D0
D0
RD
Address
Bus
Enablebuffers
Address
A15
Decode
A0
Logic
Start conversion
WR
INT
Endof conversion
Microprocessor
ADC
(A) One READ cycle only
(B) One WRITE cycle only
(C) One WRITE cycle followed by one READ cycle
(D) One READ cycle followed by one WRITE cycle
Answer:
44.
(C)
In the figure shown, RT represents a resistance temperature device (RTD), whose
characteristic is given by RT = Ro (1+αT), where R0 = 100, α = 0.0039℃-1 and T denotes
the temperature in ℃. Assuming the op-amp to be ideal, the value of V0 in volts when T =
100℃, is _________V.
V1  1V



V0
RT

100
Answer:
Exp:
1.39
Using the above information
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R T  R o 1  T 
 100 1  0.0039  100  139
 Since it is a non inverting amplifier its output voltage w.r.t ground is
 Rf 
 139 
1 
 Vi  1 
  2.39V
Ri 
 100 

 But here Vo is not w.r.t ground, it is across RT
So Vo  2.39  1  1.39V
45.
The signal x[n] = sin(πn/6)/(πn) is processed through a linear filter with the impulse
response h[n] = sin(cn)/(πn) where c>π/6. The output of the filter is
(A) sin(2cn)/(πn)
(2) sin(πn/3)/(πn)
(C) [sin(πn/6)/(πn)]2
(D) [sin(πn/6)/(πn)]
Answer:
Exp:
(D)
 sin n   sin n   sin n 
 n    n    n 

 
 

where   min  ,  
Output g  n   x  n  * h  n 
In frequency domain they will be multiplied.
46.
If the deflection of the galvanometer in the bridge circuit shown in the figure is zero, then
the value of Rx in ohms is ________ . .
100 
200
100  100
50 
G
100
Rx
2V
Answer:
33.33
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When the current through galvanometer is 0 then its 2 end voltage are same or we can say
the open circuit voltage across galvanometer is 0.
200
100
100 
100
Va
Vb
100
200
100
100
100
Rx
Va 100
Vb
Rx
2V
2V
By voltage division, Va  0.5V

2
Vb  
 100  R x

Rx

Va = Vb
2R x
1
100

 100  R x  4R x  R x 
 33.33
2 100  R x
3
47.
In the circuit shown in the figure, both the NMOS transistors are identical with their
threshold voltages being 5V. Ignoring channel length modulation, the output voltage V out
in volt is ___________V.
30V
NMOS

Vout
10V
Answer:
Exp:
NMOS

20
VGS1  VGS2 (Both are identical transistors)
30  Vout  10  0
Vout  20V
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In the potentiometer circuit shown in the figure, the expression for Vx is
R
V
V
R

Vx
(A) (1-2α)V
(C) (α-1)V
(B) (1-α)V
(D) αV
Answer: (A)
Exp: Writing KVL as per the current direction
Vx   R 
2V
V0
R
R
R
V
 
V


2V
R
Vx
49.
An opamp has ideal characteristics except that its open loop gain is given by the
expression AV(s) = 104/(1+10-3s). This op-amp is used in the circuit shown in the figure.
The 3-dB bandwidth of the circuit, in rad/s, is


V1

AV
V0
9k

1k
(A) 102
(B) 103
(C) 104
(D) 106
Answer: (D)
Exp: Closed loop transfer function ACL
ACL 
AOL
1  AOL
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Here AOL 
A CL
Ri
104
1
, 

 0.1
3
1  10 s
Ri  Rf 9 1
104
3
104
s 
 1  10
104  0.1 1  103 s  103
1
1  103 s
10
10
 3

6
s
10  10 s  1
1 6
10
 In a first order transfer function
k
1  s
the bandwidth of system is given by 1
50.

B.W  106
A signal is band-limited to 0 to 12 kHz. The signal spectrum is corrupted by additive
noise which is band-limited to 10 to 12 kHz. Theoretically, the minimum rate in kilohertz
at which the noisy signal must be sampled so that the UNCORRUPTED PART of the
signal spectrum can be recovered, is _________ kHz.
Answer:
20
uncorrupted part
Exp:
noise
noise
12
51.
10
10
10 12
10
kHz.
The circuit shown in the figure is in series resonance of frequency f c Hz. The value of Vc
in volts is _____________V.
10
1V
f c Hz
Answer:
Exp:
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0.1H
0.1F
VC
100
Under series resonance condition, voltage across L, C is Q times of the supply voltage.
VC  QV
1 L 1
0.1

 100
R C 10 0.1  106
Vc  QV  100 1  100V
Q
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52.
 2t 
The fundamental period of the signal x  t   2cos 
  cos  t  , in seconds, is
 3 
__________s.
Answer:
Exp:
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6
0 
H.C.F  2,  
L.C.M  3,1


3
2 
  T  6 sec
T 3
53.
The open loop transfer function of a system is G  s  
s 2  6s  10
. The angles of arrival
s 2  2s  2
of its root loci are
(A) 
Answer:
Exp:

4
(B) 

3
(C) 

2
(D) 
5
6
(A)
Angle of arrival is calculated on a complex zero and it is given by,
a  180  GH
 at a  ve imaginary zero 
 s  3  i  s  3  i 
 s  1  i  s  1  i 
 3  i  3  i 3  i  3  i  
 2i 
G  3  i  
 3  i  1  i 3  i  1  i  2  2i 2
G s 
2

G  3  i   90o  180o  tan 1   180o 
2

o
o
o
 90  180  45  180o  135o
a  180  135  45  
4
Other angle will be same with opposite sign  
54.
4
The z-transform of x[n] = a|n|, 0<|α|<1, is X(z). The region of convergence of X(z) is
(A)   z 
(C) z 
Answer:
1

1

(B) |z|>a

1
(D) z  min   , 
 

(A)
x 
x
1

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n
Exp:
55.
1
x  n      u  n     u  n  1

1
z  ; z 

n
n
The number of clock cycles for the duration of an input pulse is counted using a cascade
of N decade counters (DC 1 to DC N) as shown in the figure. If the clock frequency in
mega hertz is f, the resolution and range of measurement of input width, both in μs, are
respectively
N  digit display
N  digit register
LATCH
Clock
DC1
DC 2
DC N
Generator
E R
E R
E R
E R :Enable Reset
Pulse
shaping
Input pulse
(A)


2N  1
1
and
f
f

(B)

10 N  1
10 N
(C)
and
f
f
Answer:


10 N  1
1
and
f
f


2N  1
2N
(D)
and
f
f
(B)
Exp:
T
Input pulse
Clock pulse
TCLK
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The Resolution (R) is the smallest change that is detectable.
R
1
 TCLK
f  MHz 
Range of measurement of input width = T

10
T

 1
T  10 N  1 TCLK
N
f
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