# Online Electronics and Communication Engineering Test - Previous Exam Papers Test 3

Instruction:

• This is a FREE online test. DO NOT pay money to anyone to attend this test.
• Total number of questions : 20.
• Time alloted : 30 minutes.
• Each question carry 1 mark, no negative marks.
• DO NOT refresh the page.
• All the best :-).

1.

A point charge of 6 μC is located at the origin, uniform line charge density of 180 nC/m of 8 m length lies along the x axis and a uniform sheet of charge equal to 25 nC/m2 of area p x 42 lies in the z = 0 plane. Calculate total electric flux leaving the surface of sphere of 4 m radius centered at origin.

A.
 86.97 μC
B.
 8.697 nC
C.
 8.697 μC
D.
 86.97 nC

Explanation:

According to Gauss's law total flux leaving the closed surface is equal to the charge enclosed by the closed surface

Qencl = (6 x 10-6) + (8 x 180 x 10-6) + (p x 42 x 25 x 10-9)

= (6 x 10-6) + (1.44 x 10-6) + (1.257 x 10-6)

= 8.697 μ Coulombs.

2.

Z = __________

A.
 C[A(B + C) + D]
B.
 (AB + AC + D)
C.
 (CA + AC + D)
D.
 none of these

Explanation:

Output of first MUX = A(B + C)

Output of second MUX = C D + A B).

3.

The following circuit is implementation of:

3. difference of full subtractor

A.
 1 only
B.
 3 only
C.
 2 only
D.
 1 and 2 both

Explanation:

Let the two bits be A and B. Then the truth table be:

Now it is evident from the truth tables that Sum and Difference are same.

4.

The probability density function of a random variable x is as shown.

The mean of the distribution is:

A.
B.
C.
D.

Explanation:

Mean of the distribution =

A = 1/5 then find mean.

5.

With respect to the following 555 timer circuit what will be the value of RA and R such that the LED turns on for a duration of 10 ms every time it receives negative trigger pulse. LED operating current = 20 mA, Vf = 1.4 V, Vo on = 13.4 V and C = 0.22 nf

A.
 41.4 KΩ, 600 Ω
B.
 45 KΩ, 630 Ω
C.
 43.5 KΩ, 580 Ω
D.
 44.9 KΩ, 600 Ω

Explanation:

Timer operated in mono, mode

RAC =

RA = 41.4 KΩ

R = = 600 Ω.

6.

Given f = GEF + KHIJ + LMON + TUVWXYZ and

1. E + F + G = LMN
2. V + W = U . Z . Y . X . T
3. NAND B
4. (UVW ⊕ WVU) = KJ(XY ⊕ XY)

Then f is equivalent to

A.
 0
B.
 1
C.
 EF + UZ + HI
D.
 None of these

Explanation:

(UVW ⊕ WVU) = 1 = KJ(XY ⊕ XY)

KJ = 1

NAND B

HI = 1

Since f = GEF + 1 + LMON + TUVWXYZ = 1.

7.

The circuit shown in the figure has 4 boxes each described by inputs P, Q, R and outputs Y, Z with
Y = P ⊕ Q ⊕ R, Z = RQ + P R + Q P

The circuit is a

A.
 4 bit adder giving P + Q
B.
 4 bit subtractor giving P - Q
C.
 4 bit subtractor giving Q - P
D.
 4 bit adder giving P + Q + R

Explanation:

Let P = 1101 Q = 1101

Yn = Pn ⊕ Qn ⊕ Rn

Z = Rn Qn + Pn Rn + Qn Pn

Constructing truth table

So that, Rn + 1 = Zn 1 ≥ n ≥ 3

Z4 = R5(MSB)

Hence, output is 00010 which show that it is a 4 bit subtractor giving P - Q.

8.

The value of the integral of the function g(x, y) = 4x3 + 10y4 along the straight line segment from the point (0, 0) to the point (1, 2) in the x-y plane is

A.
 33
B.
 35
C.
 40
D.
 56

Explanation:

(4x3 + 10y4)

y = 2x

[4x3 + 10(2x)4]dx = 33

9.

The eigen values of a skew symmetric matrix are

A.
 always zero
B.
 always pure imaginary
C.
 either zero or pure imaginary
D.
 always real

Explanation:

For a real skew symmetric matrix the non-zero eigen values are all pure imaginary and thus occurs in complex conjugate pair.

10.

A switch tail ring counter is made by using a single D FF. The resulting circuit is a

A.
 SR FF
B.
 JK FF
C.
 D FF
D.
 T FF

Explanation:

In both case input will always opposite to each other hence it is D FF.

11.

A transmitter radiates a power 20 KW and its base current is 18A. The radiation resistance of the antenna is

A.
 1111 Ω
B.
 900 Ω
C.
 75 Ω
D.
 62 Ω

Explanation:

.

12.

The Thevenin resistance of the circuit is

A.
B.
C.
 3 K
D.

Explanation:

Diode is the non-linear element

Find RTH across diode short voltage supplies.

RAB= RTH = 3K + (2K || 1K)

.

13.

Assertion (A): The small signal analysis of a transistor amplifier is done to obtain the current gain, voltage gain and the conversion efficiency of an amplifier.

Reason (R): The small signal analysis of a transistor amplifier uses the small signal parameters of the transistor.

A.
 Both A and R are individually true and R is the correct explanation of A
B.
 Both A and R are individually 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

14.

Consider the following, logic diagram :

The count sequence of the above logic diagram is :

A.
 0000 to 1111
B.
 0000 to 1001
C.
 0000 to 1010
D.
 1001 to 0000

15.

Which one of the following driving point functions does not represent an LC network?

A.
B.
C.
D.

16.

The pre-emphasis in FM aids in providing extra noise immunity by

A.
 emphasizing only the higher frequency components
B.
 emphasizing the entire frequency band
C.
 emphasizing only the lower frequency band
D.
 cutting down the low frequency noise

17.

Which of the following circuits come under the class of sequential logic circuits?

2. Full subtractor
4. J-K flip-flop
5. Counter
Select the correct answer from the codes given below:

A.
 1 and 2
B.
 1, 2 and 3
C.
 3 and 4
D.
 4 and 5

18.

Which of the following factors are responsible to design IC logic gates to operated at a fixed supply voltage of 5 volts?

1. Low heating of IC logic gates
2. Compatibility with other logic gates
3. Satisfactory and safe operation
4. Standardization from IC manufacturing point of view.
Select the correct answer from the codes given below:

A.
 1 only
B.
 2 only
C.
 2 and 3
D.
 3 and 4

19.

For the circuit shown in the given figure, assuming ideal op-amps, the output corresponding to the given input will be :

A.
B.
C.
D.

20.

Step responses of a set of three second-order underdamped systems all have the same percentage overshoot. Which of the following diagrams represents the pole of the three systems?

A.
B.
C.
D.