Quiz Electronics Engineering
Exam: NIC
Topic: Miscellaneous
Date: 17/11/2020
Each Question carries 1 Mark
Negative Marking: 1/4
Time: 10 Minutes
Q1. Which of this is not a guided media?
(a) Fiber optical cable
(b) Coaxial cable
(c) Wireless LAN
(d) Copper wire
Q2. Which of the following are electromagnetic in nature?
1. Alpha ray
2. X-ray
3. Gamma rays
4. Cathode rays
(a) 1 and 2
(b) 2 and 3
(c) 3 and 4
(d) 1cand 4
Q3. An unbiased coin is tossed an infinite number of times. The probability that the fourth head appears at the tenth toss is
(a) 0.082
(b) 0.073
(c) 0.062
(d) 0.091
Q4. The output of an Ex–NOR gate is 1. Which input combination is correct?
(a) A = 1, B = 0
(b) A = 0, B = 1
(c) A = 0, B = 0
(d) None of the above
Q5. The number of bits in a binary PCM system is increased from n to n+1. As a result, the signal quantization noise ratio will improve by a factor
(a) (?+1)/?
(b) 2(n+1)/n
(c) 22(n+1)/n
(d) independent of n
Q6. An instruction cycle refers to
(a) Fetching an instruction
(b) Fetching, decoding and executing an instruction
(c) Executing an instruction
(d) Clock speed
Q7. The impulse response of a system is h(t) = ?(t – 0.5). If two such systems are cascaded, the impulse response of the overall system will be
(a) 0.5?(t – 0.25)
(b) ?(t – 0.25)
(c) ?(t – 1)
(d) 0.5?(t – 1)
Q8. Piezo-electric crystal is generally employed for the measurement of which one of the following?
(a) Acceleration
(b) Velocity
(c) Flow
(d) Temperature
Q9. An operational amplifier possesses
(a) very large input resistance and very large output resistance
(b) very small input resistance and very small output resistance
(c) very large input resistance and very small output resistance
(d) very small input resistance and very large output resistance
Q10. The value of the current flowing in the 1 ohm register in the circuit shown in the given figure will be
(a) Zero
(b) 5 A
(c) 6 A
(d) 10 A
SOLUTIONS
S1. Ans.(c)
Sol. Wired medium is called guided media and wireless medium is called Unguided media. Example of guided media are Optical Fiber Cable, Coaxial Cable, Twisted Fiber Cable etc. Example of unguided media are wireless LAN.
S2. Ans.(b)
Sol. Alpha ray and cathode ray are not electromagnetic in nature.
S3. Ans.(a)
Sol. It means 3 – head appears in 1st 9 trials.
Probability of getting exactly 3 head in 1st 9 trials
= (_ ^9)C_3 ×(1/2)^3 (1/2)^6= (_ ^9)C_3 ×(1/2)^9
and in 10th trial head must appears.
So, required probability
= (_ ^9)C_3 (1/2)^9×1/2=84/1024=0.082
S4. Ans. (c)
Sol. Truth Table of Ex-NOR gate is
Therefore, The output of Ex-NOR gate is always 1 if both the inputs are same.
S5. Ans.(d)
Sol. It is given that the number of bits in a binary PCM system is increased from n to n+1.
We know that, SQNR = 3/2 2^2n
For n1 = n,
〖(SQNR)〗_1 = 3/2 2^2n
For n2 = n + 1,
〖(SQNR)〗_2 = 3/2 2^(2(n+1))
Now,
〖(SQNR)〗_2/(〖(SQNR)〗_1 ) = (3/2 2^(2(n+1)))/(3/2 2^2n ) = (3/2 2^2n.2^2)/(3/2 2^2n ) = 4
So, 〖(SQNR)〗_2 = 4〖(SQNR)〗_1
Therefore, Signal to quantization noise ratio increases by factor of 4. So this is improvement in SQNR is independent of n.
S6. Ans.(b)
Sol: An Instruction Cycle is the basic process of a computer. It is the process by which a computer retrieves a program instruction from its memory, determines what actions the instruction is asking to take and carries out those actions. It is also known as fetch decode-execute cycle.
S7. Ans.(c)
Sol. According to the property of Impulse response –
?(t – a)* ?(t – b) = ?(t – a – b)
Here, When the system is cascaded, the overall impulse response is –
⇒ ?(t – 0.5)* ?(t – 0.5) = ?(t – 0.5 – 0.5)
So, ?(t – 0.5)* ?(t – 0.5) = ?(t – 1)
S8. Ans.(a)
Sol. A piezoelectric crystal is an accelerometer that follows the piezoelectric effect of certain materials to measure dynamic changes in mechanical parameters like acceleration, vibration, and mechanical shock.
S9. Ans. (c)
Sol. In Ideal case an Op-Amp has infinite input resistance so that no current can flow across both terminals and zero output resistance so that there is no loss of signal. Hence, A practical Op-Amp has high input impedance and low output resistance.
S10. Ans.(b);
Sol. According to Superposition theorem,
CASE I: Only voltage source is present
⇒ I_1=5/1 = 5 Amp . . . . . . . . . . . . . . . . . . . . . . . . . (1)
CASE II: Only current source is present
⇒ I_2 = 0 . . . . . . . . . . . . . . . . . . . . . . . . . (2)
All 5 A current flow through the short circuit branch.
Adding 1 and 2, we get
⇒ I = I_1+I_2
So, I = 5 + 0 = 5A
