Quiz Electronics Engineering
Exam: NIC
Topic: Miscellaneous
Date: 09/10/2020
Each Question carries 1 Mark
Negative Marking: 1/4
Time: 10 Minutes
Q1. Assertion(A): The expression E ⃗ = -∇V, where E ⃗ is the electric field and V is the potential is not valid for time varying fields.
Reason(R): The curl of a gradient is identically zero.
(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.
Q2. What is the period of the sinusoidal signal x(n) = 5cos[0.2πn]?
(a) 0
(b) 5
(c) 1
(d) 10
Q3. The essential blocks of a Phase Locked Loop(PLL) are phase detector, amplifier
(a) High Pass Filter and Crystal Controlled Oscillator
(b) Low Pass Filter and Crystal Controlled Oscillator
(c) Low Pass Filter and Voltage Controlled Oscillator
(d) High Pass Filter and Voltage Controlled Oscillator
Q4. (1111.11)_2 is
(a) (1.01)_10
(b) -(0.75)_10
(c) (15.3)_10
(d) (15.75)_10
Q5. The property of a material to permit electric current to flow through it
(a) inductance
(b) conductance
(c) capacitance
(d) resistance
Q6. The total resistance between B and C will be
(a) 160/3 Ω
(b) 180/3 Ω
(c) 190/6 Ω
(d) 195/3 Ω
Q7. A local telephone network is an example of a _______ network.
(a) Packet switched
(b) Bit switched
(c) Circuit switched
(d) Line switched
Q8. The minimum number of flip-flops required for a mod- 14 counter is
(a) 6
(b) 4
(c) 8
(d) 16
Q9. What is the expression for capacitance of an infinitely conducting solid sphere of radius ‘R’ in free space?
(a) 2πε_oR
(b) 0.5πε_oR
(c) 8πε_oR
(d) 4πε_oR
Q10. Null character needs a space of
(a) 1 Bytes
(b) 4 Bytes
(c) 2 Bytes
(d) 0 Bytes
SOLUTIONS
S1. Ans.(a)
S2. Ans.(d)
Sol. Given:
x(n) = 5cos[0.2πn]
We know that, Time period of a discrete time signal is –
⇒ N = 2π/ω
Here,
ω = 0.2π
So, N = 2π/0.2π = 10
S3. Ans.(c)
Sol. The essential blocks of a Phase Locked Loop(PLL) are phase detector, amplifier, Low Pass Filter and Voltage Controlled Oscillator. When PLL is used as FM demodulator multiplier is replaced by phase detector.
Basic Block diagram of PLL detector :
S4. Ans.(d)
Sol. Converting binary to decimal, we get
(1111.11)_2 = 1×23 + 1×22 + 1×21 + 1×20 + 1×2-1 + 1×2-2
= 8 + 4 + 2 + 1 + 0.5 + 0.25
= 15.75
Therefore, (1111.11)_2 =〖 (15.75)〗_10
S5. Ans.(b)
Sol. With respect to the current flow, material has two basic properties one is to oppose the current flow called Resistance of the material and another one is to permit or pass the electric current, called Conductance of the material.
S6. Ans.(a)
Sol.
Since, 40Ω is connected in parallel with 20 Ω.
Here,
⇒ 1/R=1/40+1/20=3/40
So, R = 40/3 Ω
Total resistance across BC is
⇒ R_AB = 20+40/3+20
⇒ R_AB = (20×3+40+20×3)/3
⇒ R_AB = (60+40+60)/3
So, R_AB = 160/3 Ω
S7. Ans.(c)
Sol. In Packet Switching connectionless switching technique is used but in Circuit switching there is connection oriented switching technique. A Local Telephone network spreads in a local area to connect with the people with the help of wired medium. So, it is guided or connection oriented media.
S8. Ans (b)
Sol. Given:
Number of valid state(N) = 14
The number of flip-flops(n) required for a desired MOD number-N is found out using the equation:
Number of flip flop(n) = log_2N
Now,
n = log_2N
⇒ n = log_214
⇒ n = 3.80
But number of flip flop is always in natural number.
So, n = 4
Therefore, number of flip-flops is 4.
S9. Ans.(d)
Sol. We know that Q = CV
and Voltage of an infinitely conducting solid sphere of radius ‘R’ in free space is
V = Q/(4πε_0 R)
Now,
C = Q/V
⇒ C = Q/(Q/(4πε_0 R))
So, C = 4πε_0 R
S10. Ans.(a)
Sol. Null character doesn’t means empty. It means there is some character like “/0” or “/00” etc., which indicate that any data terminates there. To store it 1 byte space is required.
