Grand Test - 1 - PDF Flipbook
Grand Test - 1
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GATE
EEE
GrandTests
Test-01Solutions
APTITUTE
One Mark Questions:
1. The question below consists of a pair of related words followed
by four pairs of words. Select the pair that best expresses the
relation in the original pair.
UNEMPLOYED: WORKER:
a) fallow: land
b) unaware: sleeper
c) wit: jester
d) renovated: house
Answer: (a)
Solution:
Analogy
A worker who is unemployed is un productive just as a land
which is fallow is un productive.
2. Choose the most appropriate word from the options given below
to complete the following sentence.
Under ethical guidelines recently adopted by the Indian Medical
Association, human genes are to be manipulated only to correct
diseases for which ______ treatments are unsatisfactory.
a) similar
b) most
c) uncommon
d) available
Answer: (d)
1
Solution:
The National Institute of Health has adopted ethical guidelines
regarding manipulation of human genes. The human genes are
to be manipulated only to correct diseases for which certain
kinds of treatments are unsatisfactory.
a) If treatments are similar to those based on manipulation of
human genes unsatisfactory then there is no point in trying
to correct diseases in this way.
b) If most treatments are unsatisfactory, then atleast some
treatments are satisfactory. Therefore, these satisfactory
treatments may be used to correct diseases.
c) To state that uncommon treatments are unsatisfactory
would mean common treatments are satisfactory, then
there is no need to manipulate human genes to correct
diseases.
d) Correct: If available or other types of treatments are
unsatisfactory, then manipulation of genes could be
allowed.
3. Choose the grammatically INCORRECT sentence:
a) They gave us the money back less the service charges of
Three Hundred rupees.
b) This country's expenditure is not less than that of Bangladesh.
c) The committee initially asked for a funding of Fifty Lakh
rupees, but later settled for a lesser sum.
2
d) This country's expenditure on educational reforms is very
less.
Answer: (d)
Solution:
The country’s expenditure on educational reforms is very low.
4. A number is as much as greater than 75 as it is smaller than 117.
The number is:
a) 91
b) 93
c) 89
d) 96
Answer: (d)
Solution:
From given options
→ option A ⟹ 91 – 75 = 16
→ option B ⟹ 93 – 75 = 18
→ option C ⟹ 89 – 75 = 14
→ option D ⟹ 96 – 75 = 21
The difference of given options with 75 is greatest in option (d)
so, it is much greater than 75 and it also smaller than 117.
Therefore option (d) is correct.
Alternative method:
From options, maximum value from 75 and nearer smaller than
117 is 96 only.
→ option A ⟹ 91 – 75 = 16
3
5. If the radius of a right circular cone is increased by 50% its
volume increases by
a) 75%
b) 100%
c) 125%
d) 237.5%
Answer: (c)
Solution:
Volume of a right circular one,
= 1 2
3
R = radius of a cone
H = height of the cone
∝ 2
1 = 2
2 = (15)2 2 with increasing
2 = 2.25 2 = 2.25 1
∴ Volume increases = 2− 1 × 100
1
= 2.25 1− 1 × 100
1
= 1(1.25) × 100
1
= 1.25 × 100 = 125%
4
Two Marks Questions:
6. After several defeats in wars, Robert Bruce went in exile and
wanted to commit suicide. Just before committing suicide, he
came across a spider attempting tirelessly to have its net. Time
and again, the spider failed but that did not deter it to refrain
from making attempts. Such attempts by the spider made Bruce
curious. Thus, Bruce started observing the near-impossible goal
of the spider to have the net. Ultimately, the spider succeeded in
having its net despite several failures. Such act of the spider
encouraged Bruce not to commit suicide. And then, Bruce went
back again and won many a battle, and the rest in history.
Which one of the following assertions is best supported by the
above information?
a) Failure is the pillar of success.
b) Honesty is the best policy.
c) Life begins and ends with- adventures.
d) No adversity justifies giving up hope.
Answer: (d)
Solution:
Option A
Failure is the pillar of success. It is probably true but it doesn’t
support completely the above information.
Option B
Honesty is the best policy. We cannot assert this statement from
the given passage as it doesn’t talk about “Honesty”.
5
Option C
Life begins and ends with adventures. This statement doesn’t go
with the passage as it doesn’t mention about adventures.
Option D
Correct: No adversity justifies giving up hope. The passage
talks about ‘Robert Bruce’ and spider’s not giving up hopes in
adverse situations. It can be inferred that no adversity justifies
giving up hope.
7. The number of people diagnosed with dengue fever (contracted
from the bite of a mosquito) in north India is twice the number
diagnosed last year. Municipal authorities have concluded that
measures to control the mosquito population have failed in this
region.
Which one of the following statements, if true, does not
contradict this conclusion?
a) A high proportion of the affected population has returned
from neighbouring countries where dengue is prevalent.
b) More cases of dengue are now reported because of an
increase in the Municipal Office's administrative efficiency.
c) Many more cases of dengue are being diagnosed this year
since the introduction of a new and effective diagnostic test.
d) The number of people with malarial fever (also contracted
from mosquito bites) has increased this year.
Answer: (d)
6
Solution:
It is a strengthen the argument question
a) Unwarranted (out of scope) it doesn’t support the conclusion.
The passage doesn’t mention ‘neighbouring countries’.
b) Contradicts the conclusion
c) Contradicts the conclusion
d) Correct: This is the right answer as it supports the
conclusion. Municipal authorities have concluded that
measures to control the mosquito population have failed in
this region.
8. Hari (H), Gita (G), Irfan (I) and Saira (S) are siblings (i.e.
brothers and sisters). All were born on 1st January. The age
difference between any two successive siblings (that is born one
after another is less than 3 years.
Given following facts:
(i) Hari's age + Gita's age > Irfan's age + Saira's age
(ii) The age difference between Gita and Saira is 1 year.
However, Gita is not the oldest and Saira is not the
youngest
(iii) There are no twins
In what order were they born (oldest first)?
a) HSIG
b) SGHI
c) IGSH
d) IHSG
7
Answer: (b)
Solution:
S – 5 years
G – 4 years
H – 3 years
I – 1 year
Assume, the age of Irfan is one year, Hari is 3 years, Gita is 4
years and Saira is 5 years.
∴ The age difference between all of them is not more than 3
years.
Hari’s age + Gita’s age = 3 + 4 = 7 years
Irfan’s age + Saira’s age = 1 + 5 = 6 years
∴ 7 years > 6 years
The age difference between Gita and Saira is 1 years
∴ The order of they born from oldest is SGHI
9. The sum of n terms of the series 4 + 44 + 444 +.... is
a) (4/81) [10n+1 – 9n – 1]
b) (4/81) [10n-1 – 9n – 1]
c) (4/81) [10n+1 – 9n – 10]
d) (4/81) [10n – 9n – 10]
Answer: (c)
Solution:
The sum of n terms of the series 4 + 44 + 444 + …. is in the
given alternatives if n = 1, which will give 4 and if n = 2 which
8
will give sum of first two numbers (i.e) 4 + 44 = 48 is our
answer.
Ex: If n = 1
a) 4 [10 +1 − 9 − 1]
81
= 4 [101+1 − 9(1) − 1]
81
= 4 [100 − 9 − 1]
81
= 4 × 90
81
∴ ‘A’ is not.
b) 4 [10 −1 − 9 − 1]
81
= 4 [101−1 − 9 − 1]
81
= 4 [1 − 9(1) − 1] = 4 (−9)
81 81
∴ It is not.
c) 4 [10 +1 − 9 − 10]
81
= 4 [101+1 − 9(1) − 10]
81
= 4 [100 − 9 − 10]
81
= 4 × 81 = 4
81
If n = 2
4 [102+1 − 9(2) − 10]
81
= 4 [1000 − 18 − 10]
81
= 4 [972]
81
∴ It is true
9
10. A transporter receives the same number of orders each day.
Currently, he has some pending orders (backlog) to be shipped.
If he uses 7 trucks, then at the end of the 4th day he can clear all
the orders. Alternatively, if he uses only 3 trucks, then all the
orders are cleared at the end of the 10th day. What is the
minimum number of trucks required so that there will be no
pending order at the end of the 5th day?
a) 4
b) 5
c) 6
d) 7
Answer: (c)
Solution:
7 trucks can clear all the orders at the end of 4th day.
3 trucks can clear all the orders at the end of 10th day.
Transporter receives every day same number orders currently,
he has some pending orders (backlog).
10
TECHNICAL
One Mark Questions:
11. If A and B are real symmetric matrices of order n then which
of the following is true.
a) AAT = I
b) A = A-1
c) AB = BA
d) (AB)T = BTAT
Answer: (d)
Solution:
By the properties of transpose of matrices option (d) is correct.
i.e. By the reversal law of the transpose of the product matrices,
we have
(AB)T = BTAT
12. Among the following, the pair of vectors orthogonal to each
other is
a) [3, 4, 7], [3, 4, 7]
b) [1, 0, 0], [1, 1, 0]
c) [1, 0, 2], [0, 5, 0]
d) [1, 1, 1], [−1, −1, −1]
Answer: (c)
Solution:
The two vectors 1× and 1 × are said to be orthogonal if
1× × 1 = 0 (or) YXT = 0
Let X = [1 0 2] and [0 5 0]
11
0
Then XYT = [1 0 2]�5� = 0 + 0 + 0 = 0
0
∴X, Y are orthogonal to each other
3 4 45
13. Perform the following operations on the matrix � 7 9 105�
13 2 195
i. Add the third row to the second row.
ii. Subtract the third column from the first column.
The determinant of the resultant matrix is ______.
Answer: 0
Solution:
3 4 45
= � 7 9 105�
13 2 195
The value of the determinant remains same after performing the
operations Ri + K Rj and Ci + K Cj.
∴The determinant of resultant matrix = | |
We can see that, C3 = 15 C1
∴ | | = 0
14. The directional derivative of f (x, y, z) = 2x2 + 3y2 + z2 at the
point p(2, 1, 3) in the direction of the vector � = ̅ − 2 � is
_____.
a) -2.785
b) -2.145
c) -1.789
d) 1.000
12
Answer: (c)
Solution:
∇ ](2,1,3) = 8 + 6 + 6 �
Directional derivative = ∇ ∙ � = −4
| � | √5
15. The v-i characteristic of an element is shown in the figure
given below. The element is
a) non-linear, active, non-bilateral
b) linear, active, non-bilateral
c) non-linear, passive, non-bilateral
d) non-linear, active, bilateral
Answer: (a)
16. The value of resistance 'R' shown in the given figure is ____Ω.
a) 3.5 Ω
b) 2.5 Ω
c) 1 Ω
d) 4.5 Ω
Answer: (a)
13
Solution:
By applying KVL in 1st loop
50 = 6i + 7(i – 4)
⟹ 13i = 78
⟹i=6A
Now, by applying KVL in 2nd loop
7×2=4×R
R = 3.5 Ω
17. In the circuit shown in the given figure,
The current through the inductor L is _____A.
a) 0 A
b) 3 A
c) 4 A
d) 8 A
Answer: (d)
14
Solution:
= �52 − 42 = 3
( − 4)2 + 32 = 52
= 8 Amp
18. Assertion (A): The intrinsic carrier concentration of Si at
room temperature is more than that of GaAs.
Reason (R): Si is an indirect band-gap semiconductor while
GaAs is a direct band-gap semiconductor.
a) Both A and R are true and R is the correct explanation of A.
b) Both A and R are true but R is NOT the correct explanation
of A.
c) A is true but R is false.
d) A is false but R is true.
Answer: (b)
Solution:
Both the statements are true but R is not the correct explanation
of A because there is no relation in intrinsic concentration with
direct band gap or indirect band gap.
15
19. Which one of the following statements is correct?
In a transistor,
a) ICBO is greater than ICEO, and does not depend upon
temperature.
b) ICBO is greater than ICO, and doubles for every ten degrees rise
in temperature.
c) ICBO is equal to ICO and double for every ten degrees rise in
temperature.
d) ICEO is equal to ICO, and doubles for every ten degrees rise in
temperature.
Answer: (b)
Solution:
We know that ICEO (β + 1) ICEO and ICBO are slightly greater
than ICO due to some current induced and flow at the surface and
it also doubles for every 100 rise in temperature.
20. Which logical operation is performed by ALU of 8085 to
complement a number?
a) AND
b) NOT
c) OR
d) EXCLUSIVE OR
Answer: (d)
16
Solution:
When we XOR of a number
A = a7 …… a0
with B = 11111111
we get output = ̅
∵ XOR of any digit with 1 is ⨁1 = �
21. The discrete -time equation Y (n + 1) + 0.5n y(n) = 0.5x (n + 1)
is not attributable to a
a) memory less system
b) time-varying system
c) linear system
d) causal system
Answer: (a)
Solution:
Since the output for each value of the independent variable at a
given time is not dependent on the input at the same time, hence
the system is having memory. Therefore, the system is not
attributable to a memory less system.
NOTE: A system is said to be memory less if its output for each
value of the independent variable at a given time is dependent
only on the input at the same time.
(i) Memory less system is also called static system.
(ii) A system having memory is also called dynamic system.
17
22. The sum of two or more arbitrary sinusoids is
a) Always periodic
b) Periodic under certain conditions
c) Never periodic
d) Periodic only if all the sinusoids are identical in frequency
and phase
Answer: (b)
Solution:
The sum of two sinusoids is periodic if the reatio of the time
periods of the sinusoids is a rational number.
23. Which one of the following statements is INCORRECT with
reference to pneumatic system?
a) Operating pressure is low compared to hydraulic system
b) Leaks can create problems as well as fire hazards
c) They are insensitive to temperature changes
d) High compressibility of air results in longer time delays
Answer: (b)
24. Consider the following three block diagram A, B and C shown
below:
18
Which one of the following statements is correct in respect of
the above block diagrams?
a) Only A and B are equivalent
b) Only A and C are equivalent
c) Only B and C are equivalent
d) A, B and C are equivalent
Answer: (d)
Solution:
Block diagram 'B' can be obtained from 'A' and 'C' can be
obtained from 'B'.
25. The breakaway point in the root loci plot for the loop transfer
function ( ) = is ________.
( +3)2
a) −2.5
b) −2.0
c) −1.0
d) −0.5
19
Answer: (c)
Solution:
Characteristic equation = 1 + G(s) H(s)
= 1 +
( +3)2
=0
= − ( + 3)2
= −( 3 + 6 2 + 9 )
for breakaway point,
= 0
⟹ ( 3 + 6 2 + 9 ) = 3 2 + 12 + 9
1,2 = −1, −3
But -3 does not lie on root locus.
Hence, option (c) is correct
26. A d.c shunt generator when driven without connecting field
winding shows an open circuit terminal voltage of 12 V. When
field winding is connected and excited, the terminal voltage
drops to zero because
a) field resistance is higher than critical resistance
b) there is no residual magnetism in the field circuit
c) field winding has got wrongly connected
d) there is a fault in armature circuit
Answer: (c)
20
Solution:
If shunt field resistance is more than the critical field resistance,
there will be no voltage buildup beyond residual voltage but
voltage would not be zero.
27. The current drawn by a 220 V dc motor of armature resistance
0.5 Ω and back emf 200 V is ________ A.
a) 40 A
b) 44 A
c) 400 A
d) 440 A
Answer: (a)
Solution:
Ea = V – IaRa
Ea = 200 = 200 – Ia × 0.5 ⟹ = 40
28. The good effect of corona on overhead lines is to
a) increase the line carrying capacity due to conducting ionized
air envelop around the conductor.
b) increase the power factor due to corona loss.
c) reduce the radio interference from the conductor.
d) reduce the steepness of surge fronts.
Answer: (d)
Solution:
Corona, is helpful in one respect, namely, it reduces the effect of
surges and acts as a relief valve for them. This is so because the
surges are partially dissipated as corona.
21
29. What is the approximate value of the surge impedance loading
of a 400 kV, 3-phase 50 Hz overhead single circuit transmission
line _____ MW?
a) 230 MW
b) 400 MW
c) 1000 MW
d) 1600 MW
Answer: (b)
Solution:
= ( )2
0
For single-circuit line Z0 = 400 Ω/phase
∴ = 400×400 = 400
400
30. The snubber circuit is used in thyristor circuits for
a) triggering
b) dv/dt protection
c) di/dt protection
d) phase shifting
Answer: (b)
Solution:
22
The capacitor Cs in parallel with thyristor is sufficient to prevent
unwanted dv/dt triggering of thyristor.
When switch is closed, capacitor acts as short-circuit, therefore
voltage across thyristor is zero. With the passage of time,
voltage across capacitor builds up a slow rate such that dv/dt
across thyristor is less than specified maximum dv/dt rating of
the device.
31. In a switched-mode power supply (SMPS), after conversion of
a.c supply to a highly filtered d.c voltage, a switching transistor
is switched ON and OFF at a very high speed by a pulse width
modulator (PWM) which generates very-high frequency square
pulses. The frequency of the pulses is typically in the range of
________ Hz.
a) 100 Hz-200 Hz
b) 500 Hz-1 kHz
c) 2 kHz-5 kHz
d) 20 kHz-50 kHz
Answer: (d)
Solution:
The frequency of the pulses is kept as high as possible to reduce
the size of the pulse transformer.
23
32. A thyristor has a PIV of 650 V. The voltage safety factor is 2.
Then the voltage upto which the device can be operated is given
by ______ V.
a) 1300 V
b) 650 V
c) 325 V
d) 230 V
Answer: (d)
33. Which of the following are the characteristics of a
thermocouple type of indicating instrument?
1. Its accuracy is very high, as high as about 1 percent.
2. It has linear scale because a d’Arsonval movement is used for
measuring the output.
3. It is an RF instrument and can be used for frequency up to
about 50 MHz.
4. It cannot be damaged by overloads.
a) 1 and 2
b) 2 and 3
c) 3 and 4
d) 1 and 3
Answer: (d)
Solution:
• Thermo-couple instrument has a “square-law response” so it
does not have a linear scale.
24
• Their accuracy can be as high as l% for frequencies upto 50
MHz and can be classified as RF instruments. Above this
frequency the effective resistance of heater wire is increased
on account of skin effect, thereby reducing the accuracy.
Tubular designs are used for heater wire above 3 A to reduce
errors on account of skin effect produced by high frequencies.
For small currents upto 3 A, the heater wire is solid and very
thin.
• At normal rated current, the heater attains a temperature of
3000C. If we pass twice the rated current, the heater would
give a temperature of nearly 4 times the normal temperature
i.e., 12000 C. It is obvious that square law rate will bring the
heater to nearly its burn out temperature. Thus it is imperative
that the heater be protected against damaging over loads. The
fuses do not provide any protection as due to overload the
heater wire may burn out before fuse blows out. Thus, the
overload capacity of thermo-couple instruments is small as
compared with other instruments and is about 150 percent of
the full scale current.
34. A 1 cm piezoelectric transducer having a g-coefficient of 58
V/kg/m2 is subjected to a constant pressure of 10-3 kg/m2 for
about 15 minutes. The Piezo voltage developed by the
transducer will be _______ mV.
a) 116 mV
b) 58 mV
25
c) 29 mV
d) 0 mV
Answer: (d)
Solution:
It can't measure static pressure.
35. The maximum space rate of change of the function which is in
increasing direction of the function is known as
a) curl of the vector function
b) gradient of the scalar function
c) divergence of the vector function
d) Stokes theorem
Answer: (b)
Solution:
Gradient of a scalar;
∇ = maximum rate of change of scalar A with respect to given
coordinates system.
26
Two Marks Questions:
36. The differential 2 + + sin = 0 is
2
a) Linear
b) Non-linear
c) Homogeneous
d) Of degree two
Answer: (b)
Solution:
Given equation is a non-linear differential equation
37. If L{f(t)} = F(s) then L{f ( t – T)} is equal to
a) esTF(s)
b) e-sT F(s)
c) ( )
1−
d) ( )
1− −
Answer: (b)
Solution:
By second shifting Theorem,
L{f(t – T)} = e-sT F(s)
38. The bilinear transformation w = −1
+1
a) maps the inside of the unit circle in the z-plane to the left
half of the w-plane
b) maps the outside the unit circle in the z-plane to the left half
of the w-plane
27
c) maps the inside of the unit circle in the z-plane to right half of
the w-plane
d) maps the outside of the unit circle in the z-plane to the right
half of the w-plane
Answer: (a)
Solution:
Given w = −1
+1
w = + = ( )
+
⟺ = −1( ) = − +
−
∴ = − −1
−1
Unit circle | | = 1
Consider | | < 1
which represents inside of this unit circle | | = 1
⟹ �− − −11� < 1
⟹ | + + 1| < | + − 1|
⟹ |( + 1) + | < |( − 1) + |
⟹ ( + 1)2 + 2 < ( − 1)2 + 2
⟹ 2 + 1 + 2 + 2 < 2 + 1 − 2 + 2
⟹ 4 < 0
⟹ < 0
⟹ The function w = −1 maps the inside of unit circle in the z
+1
plane to the left half of the w plane.
28
39. The function f(x) = ex – 1 is to be solved using Newton-
Raphson method. If the initial value of x0 is taken 1.0, then the
absolute error observed at 2nd iteration is ______.
Answer: 0.06
Solution:
′( ) =
1 = 0 − ( 0) = 1 − � − 1� = 1
′( 0)
2 = 1 − ( 1)
′( 1)
1 1
� −1 1�
= −
= 0.06
40. Let g: [0, ∞) → [0, ∞) be a function defined by g(x) = x - [x],
where [x] represents the integer part of x. (That is, it is the
largest integer which is less than or equal to x). The value of the
constant term in the Fourier series expansion of g(x) is _____.
Answer: 0.5
Solution:
g(x) = x – [x] is a periodic function with period ‘1’.
∴ 0 = 1 ∫02 ( )
2
(if ‘2L’ is the periodic of g(x)
= ∫01 �∵ − [ ] = (0, 1)�
= 1/2
29
41. The effective inductance of the circuit across the terminals
A, B in the Figure shown below is
a) 9H
b) 21H
c) 11H
d) 6H
Answer: (c)
Solution:
The relevant circuit is shown in Fig.
According to the dot convention, the coupling
between
L1 and L2 is opposing (M = 1 H)
L2 and L3 is aiding (M = 2 H)
L3 and L1 is opposing (M = 3 H)
LAB = Lequ = 4 + 5 + 6 – 2 + 4 – 6 = 11 H
30
42. A 3V dc supply with an internal resistance of 2Ω supplies a
passive non-linear resistance characterized by the relation VNL =
I2NL. The power dissipated in the non-linear resistance is
a) 1.0W
b) 1.5W
c) 2.5W
d) 3.0W
Answer: (a)
Solution:
The relevant circuit is shown in Fig.
The load is a nonlinear resistance described by VNL = I2NL
3 − = 2
3 − 2 = 2
2 + 2 − 3 = 0 ⇒ ( + 3)( − 1) = 0
= −3 1
(Since the load is given as a passive element, it cannot deliver
power to the source. Hence INL has to be 1 A.)
Taking = 1 , = 1
∴ Power dissipated = = 1
31
43. A 10V battery with an internal resistance of 1Ω is connected
across a nonlinear load whose V-I characteristic is given by
7I = V2 + 2V. The current delivered by the battery is
______ A.
Answer: 5A
Solution:
The relevant circuit is shown in Fig.
Non-linear load characteristics:
7I = V2 + 2V ……….. (i)
Loop equation:
I + V = 10 ……….. (ii)
Solving (i) and (ii)
V2 + 2V = 7 (10 – V)
V2 + 2V = 70 – 7V
V2 + 9V – 70 = 0
(V – 5) (V + 14) = 0
V = 5V, –14V
As V cannot be negative, V is taken as 5V
∴ I = 10 – V = 10 – 5 = 5A
32
44. In the voltage double circuit shown in figure, the switch 'S' is
closed at t = 0. Assuming diodes D1 & D2 to be ideal, load
resistance to be infinite and initial capacitor voltages to be zero,
the steady state voltage across capacitors C1 & C2 will be
a) VC1 = 10V, VC2 = 5V
b) VC1 = 10V, VC2 = -5V
c) VC1 = 5V, VC2 = 10V
d) VC1 = 5V, VC2 = -10V
Answer: (d)
Solution:
At t = 0, S-Closed and during the +ve cycle of i/p,
D1 – Forward biased
D2 – Reverse biased
‘C1’, Charges towards the more value and
‘C1’, will charge Upto + 5V
∴ 1 = + 5
33
During the –ve half cycle of i/p voltage
D1 – off
D2 – ON
Apply KVL
+ 2 + 1 = 0
2 = − − 1
= −5 − 5 = −10
∴ 2 will charge upto -10V
45. Figure shows as electronic voltage regulator the zener diode
may be assumed to require a minimum current of 25 mA for
satisfactory operations. The value of R required for satisfactory
voltage regulation of the circuit is ____ ohms.
Answer: 80
34
Solution:
From the given problem
= = 10
= 25
= �20− 10� = 10
But =
100
=
100
= 10
100
= 100
∴ = + = 125
= 10 = 10 = 80
1 125×10−3
46. The Boolean expression + + + +
can be simplified to
a) + +
b) + +
c) + +
d) + +
Answer: (b)
35
Solution:
= ∑ (1, 2, 5, 6, 7)
= � + ̅ +
(or)
= � + ̅ +
47. In the following sequential circuit, the initial state (before the
first clock pulse) of the circuit is Q1Q0 = 00. The state (Q1Q0),
immediately after the 333rd clock pulse is
a) 00
b) 01
c) 10
d) 11
Answer: (b)
36
Solution:
It is a synchronous counter
Clock Flip Flop Inputs Outputs
0 1
0 = � � 1� 0 = 1 0 = 0 1 = � � � 0� 00
01 00 1
11 01 01 0
20 11 01 1
30 10 10 1
41 00 10 0
Every four cycles the output repeated
∴ 333 = 83 → ①
4
After 332 clockpulse the output is Q1 Q0 = 0 0
After 333 clockpulse the output is Q1 Q0 = 0 1
48. The function shown in the figure can be represented as
a) ( ) − ( − ) + ( − ) ( − ) − � − 2 � ( − 2 )
b) ( ) + ( − ) − ( − 2 )
c) ( ) − ( − ) + ( − ) ( ) − � − 2 � ( )
d) ( ) + ( − ) ( − ) − 2 ( −2 ) ( − 2 )
37
Answer: (a)
Solution:
(T, 0) (2T, 1)
x(t) – 0 = 1−0 ( − )
2 −
x(t) = 1 ( − ) = − 1 < < 2
( ) = ( ) − ( − ) + 1 [ ( − ) − ( − 2 )]
( ) = ( ) − ( − ) + 1 [( ( − ) − ( − 2 )]
( ) = . ( )
( ) = ( ) − ( − ) + 1 [( − ) ( − ) −
( − 2 ) ( − 2 )]
( ) = ( ) − ( − ) + ( − ) ( − ) − ( −2 ) ( − 2 )
38
49. The transfer function of a linear time invariant system is given
as ( ) = 1 ∙ The steady state value of the output of this
2+3 +2
system for a unit impulse input applied at time instant t = 1 will
be
a) 0
b) 0.5
c) 1
d) 2
Answer: (a)
Solution:
For the given transfer function of LTI system
( ) = 1 = 1
2+3 +2 ( +1)( +2)
For the input, x(t) = 1 δ(t – 1)
X(s) = 1 e-s
∴ output Y(s) = X(s) G(s) = −
( +1)( +2)
Steady state value of y(t) = ( ) = 0
→ 0
50. For a periodic signal v (t) = 30 sin100t + 10 cos300t + 6 sin
(500t + π/4), the fundamental frequency in rad/s is _____.
a) 100
b) 300
c) 500
d) 1500
Answer: (a)
39
Solution:
Given,
v (t) = 30 sin100t + 10 cos 300t + 6 sin (500t + π/4)
ω1 = 100, ω2 = 300, ω3 = 500
⟹ 2 = 100, 2 = 300, 2 = 500
1 2 3
⟹ 1 = 2 , 2 = 2 , 3 = 2
100 300 500
⟹ 1 = 3 , 1 = 5
2 1 3 1
Fundamental period, T = x × 1
x = LCM (1, 1) = 1
1 = 2 ⟹ = 1 × 2 =
100 100 50
0 = 1 = 50
⟹ 0 = 2 0 = 100 /
51. Let a causal LTI system be characterized by the following
differential equation, with initial rest condition
2 + 7 + 10 ( ) = 4 ( ) + 5 ( )
2
Where, x (t) and y (t) are the input and output respectively. The
impulse response of the system is (u (t) is the unit step function).
a) 2e-2t u(t) – 7e-5t u(t)
b) -2e-2t u(t) + 7e-5t u(t)
c) 7e-2t u(t) – 2e-5t u(t)
d) -7e-2t u(t) + 2e-5t u(t)
Answer: (b)
40
Solution:
Taking the Laplace transform,
TF = ( ) = 5 +4
( ) 2+7 +10
= 5 +4 = −2 + 7
( +2)( +5) +2 +5
= −1[ ] = −2 −2 ( ) + 7 −5 ( )
52. The block diagram shown in fig given is a unity feedback
closed loop control system. The steady state error in the
response of the above system to unit step input is
a) 25%
b) 0.75%
c) 6%
d) 33%
Answer: (a)
Solution:
( ) = 3 . 15 & ( ) = 1
+15 +1
= l i→m0 ( ) = (3)(15) = 3
(15)
Steady state error,
1 = 1 = 1 = 0.25 = 25%
1+ 1+3 4
41
53. For a system having transfer function ( ) = − + +11, a unit step
input is applied at time t = 0. The value of the response of the
system at t = 1.5 sec (rounded off to three decimal places) is
________.
Answer: 0.554
Solution:
= 1−
1+
= 1 ( )
Output = (TF) (input)
= �11−+ � 1
Output = −1 �1 × 11+− �
= 1 − 2 −
Output at (t = 1.5) = 1 – 2e-1.5
= 0.554
54. A 240V d.c series motor takes 40A when giving its rated
output at 1500 rpm. Its resistance is 0.3 ohms. The value of
resistance which must be added to obtain rated torque at 1000
rpm is
a) 6 ohms
b) 5.7 ohms
c) 2.2 ohms
d) 1.9 ohms
Answer: (d)
42
Solution:
V = 240 V, Ia = 40 A,
Ra + Rse = 0.3 Ω
N = 1500 rpm
Eb1 = V – Ia (Ra + Rse)
Eb1 = 240 – 40 (0.3)
Eb1 = 228 V
∝ ⟹ ∝ 2 [∵ ∝ ]
To have same torque even at 1000 rpm armature current must be
same in both cases
∝
⟹ ∝
⟹ ∝
[∵ ]
2 = 2
1 1
1000 = 240−40(0.3+ )
1500 228
The resistance to be inserted (r) = 1.9 Ω
43
55. A 15 kW, 230 V dc shunt motor has armature circuit resistance
of 0.4 Ω and field circuit resistance of 230 Ω. At no load and
rated voltage, the motor runs at 1400 rpm and the line current
draws by the motor is 5 A. At full load, the motor draws a line
current of 70 A. Neglect armature reaction the full load speed of
the motor in rpm is ______.
Answer: 1240.63
Solution:
No-load condition,
N1 = 1400 rpm
Eb1 = V – IaRa
= 230 – 4 × 0.4 = 228.4 V
Full-load condition,
Speed = N2
Eb2 = 230 – 69 × 0.4
= 202.4 V
44
∴ 2 = 2 × 1 [∴ 1 = 2]
1 1 2
2 = 202.4 ⟹ 2 = 1240.63
1400 228.4
56. A separately excited DC generator supplies 150 A to a 145 V
DC grid. The generator is running at 800 rpm. The armature
resistance of the generator is 0.1 Ω. If the speed of the generator
is increased to 1000 rpm, the current in amperes supplied by the
generator to the DC grid is _____ A.
Answer: 550
Solution:
Vt = 145 V (Grid)
Ia1 = 150 A; N1 = 800 rpm
Ra = 0.1 Ω
N2 = 1000 rpm; Vt2 = 145 V (Grid)
Ia2 =?
⟹ Eg1 = Vt + Ia1ra
= 145 + (150) (0.1) = 160 V
1 = 1 [∵ = = ]
2 2
2 = �1800000� (160)
= 200 V
⟹ 2 = 200−145
0.1
= 550
2 = 550
45
57. The load shown in the figure absorbs 4 kW at a power factor of
0.89 lagging.
Assuming the transformer to value of the reactance X to be
ideal, improve the input power factor to unity is ____.
Answer: 23-24
Solution:
= 1
2
2 = 4000 = 40.858
110×0.89
2 = cos−1(0.89) = 27.128
2 = 40.858∠ − 27.128
21 = 2 = 1 [40.858∠ − 27.128]
2
= 20.429∠ − 27.128
= 18.18 − 9.31
1 = 21 = 2 = 110∠00 = 220∠00
�12�
46
1 = 21 = 220∠00
21 20.429∠−27.128
= 10.769∠27.128
= 9.58 + 4.91
Equivalent circuit when referred to primary side
Primary power factor is unity when resonance occurs i.e.,
Reactive currents in two branch is same.
21 sin 2 = 9.31
= 21
= 21 = 220
9.31 9.31
= 23.63
58. A shunt reactor of 100 MVAR is operated at 98% of its rated
voltage and at 96% of its rated frequency. The reactive power
absorbed by the reactor is;
a) 98 MVAR
b) 104.02 MVAR
c) 96.04 MVAR
d) 100.04 MVAR
Answer: (d)
47
Solution:
Reactive power absorbed by reactor = 2
1 = 12
2 1
= 100
If V2 = 0.98 V1 & f2 = 0.96 f1
Then reactive power absorbed = 22
2 2
2 = (0.98 1)2
2 (0.96 1 )
2 = 1.000416 �2 1 2 1 �
2 = 1.000416 × 100
2 = 100.04
59. The generalized circuit constants of a 3-phase, 220 kV rated
voltage, medium length transmission line are A = D = 0.936 +
j0.016 = 0.936 ∠0.980, B = 33.5 + j138 = 142.0∠76.40 Ω, C = (-
5.18 + j914) ×10-6 Ω. If the load at the receiving end is 50 MW
at 220 kV with a power factor of 0.9 lagging, then magnitude of
line to line sending end voltage should be
a) 133.23 kV
b) 220.00 kV
c) 230.78 kV
d) 246.30 kV
Answer: (c)
48
Solution:
Power received by load = 50 MW
Current at receiving end = √3 cos ∅
∴ = 50×106
√3×220×103×0.9
= 145.79
= 145.79∠25.840
[∵ cos = 0.9 ⇒ = 25.840]
= +
= 0.936∠0.980 × �220�√3� × 103 + 14.2∠76.40 ×
145.79∠ − 25.840
= 133.246∠7.770
Line to line sending end voltage = √3 × 133.246∠7.770
= 230.78∠7.770
Magnitude of line to line sending end voltage = 230.78 kV
60. For a fault at the terminals of a synchronous generator, the
fault current is maximum for a
a) 3-phase fault
b) 3-phase to ground fault
c) line-to ground fault
d) line-to-line fault
Answer: (c)
Solution:
In case of line to ground fault IR0 = IR1 = IR2 and fault current is
If = 3IR1.
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