Quiz Electronics Engineering

Exam: NIC

Topic: Miscellaneous

Date: 19/11/2020

Each Question carries 1 Mark

Negative Marking: 1/4

Time: 10 Minutes

Q1. A series R-L-C circuit will have unity power factor if operated at a frequency of

(a) 1/(2π√LC)

(b) LC

(c) 1/(LC)

(d) 1/√LC

Q2. A device with input x(t) and output y(t) is characterized by: y(t) = x2(t). An FM signal with frequency deviation of 90 kHz and modulating signal bandwidth of 5 kHz is applied to this device. The bandwidth of the output signal is

(a) 380 kHz

(b)190 kHz

(c) 370 kHz

(d) 95 kHz

Q3. Which machine instruction is used for comparing two operands in microprocessor?

(a) CLC

(b) CMP

(c) CMC

(d) None of the Above

Q4. The Poynting vector P ⃗ = E ⃗ × H ⃗

(a) Power/unit area

(b) Volts

(c) Power

(d) Volt/unit length

Q5. A fair coin is tossed 10 times. What is the probability that ONLY the first two tosses will yield heads?

(a) (1/2)^2

(b) (_ ^10)C_2 (1/2)^2

(c) (1/2)^10

(d) 〖(_ ^10)C_2 (1/2)〗^10

Q6. After firing an SCR, the gate pulse is removed. The current in the SCR will

(a) rise up.

(b) remain same.

(c) rise a little and then fall to zero.

(d) immediately fall to zero.

Q7. Which network topology supports bidirectional links between each possible node?

(a) Mesh topology

(b) Bus topology

(c) Ring topology

(d) Star topology

Q8. In a vacuum tube, diode current flows due to

(a) electrons only

(b) holes only

(c) positive ions only

(d) none of the above

Q9. When signed numbers are used in binary arithmetic, then which of the following notations would have unique representation for zero?

(a) Sign-magnitude

(b) 1’s complement

(c) 2’s complement

(d) 9,s complement

Q10. The 1’s complement of the number 11011 is

(a) 00101

(b) 11011

(c) 10101

(d) 00100

SOLUTIONS

S1. Ans.(a)

Sol. We know that power factor is

cosϕ = R/Z

At resonance condition, Z = R

⇒ cosϕ = R/Z = 1

Hence, A series RLC circuit will have unity power factor at resonance condition only and fr =1/(2π√LC)

S2. Ans.(c)

Sol. Given:

Frequency deviation(Δf) = 90 kHz

Message signal frequency(fm) = 5 kHz

We know that,

BW = 2(Δf + fm)

As y(t) = x2(t), that means it is a doubler circuit. There will be no change in the frequency of the message signal but there will be change in Frequency deviation. It will be doubled.

Now,

BW = 2(2 x 90 + 5) = 2(180 + 5)

So, BW = 370 kHz

Therefore, The bandwidth of the output signal is 370 kHz.

S3. Ans.(b)

Sol.

CMP: compares the data byte in the register or memory with the contents of accumulator.

CMC: complement carry flag

CLC: clear carry flag

S4. Ans.(a)

Sol. We know that,

Dimension of E ⃗ = V/m

Dimension of H ⃗ = A/m

Dimension of P ⃗ = V.A/m^2 = Watt/m^2

Hence, Dimension of P ⃗ is Power/unit area.

S5. Ans.(c)

Sol. p(only first two tosses are heads) = p(H,H,T,T,T,….T)

Now, each toss is independent.

So, required probability

= p(H)× p(H)× [p(T)]^8 ….

= (1/2)^2 (1/2)^8=(1/2)^10

S6. Ans.(b)

Sol: After the SCR is turned ON, anode current goes on increasing and when becomes more than the latching current the gate loses control over the Anode current. Thus, gate pulse is removed. So, The current in the SCR will remain same.

S7. Ans.(a)

Sol. Mesh topology routes data sent through any of several possible nodes to the receiving node. It is expensive for setup as it connects every node to all nodes on the network. However, it is considered the most reliable because it is most redundant of all topology i.e. if one node fails, the network can route traffic to another node.

In a mesh topology every node is connected to every other node i.e., bidirectional links between each possible nodes. So, Total number of cable links with n nodes = n(n – 1)/2

S8. Ans.(a)

Sol. After removal of air from the tube, vacuum is created in the tube and electrons get a clear path to flow from the cathode to anode, and a current flowing from anode to cathode is created. The current starts to flow when cathode is heated and a positive voltage is applied to the anode.

S9. Ans.(c)

Sol. In 1’s complement, 9’s complement and sign-magnitude zero is represented in more than one notation but in 2’s complement it is represented in only 1 way.

S10. Ans.(d)

Sol. In Complement, we convert binary ‘1’ to ‘0’ and ‘0’ to ‘1’.

So, 1’s complement of 11011 = 00100