Quiz Electronics Engineering 16 Oct 2020

Quiz Electronics Engineering
Exam: NIC
Topic: Miscellaneous
Date: 16/10/2020

Each Question carries 1 Mark
Negative Marking: 1/4
Time: 10 Minutes

Q1. What is the electric field when the voltage applied is 5V and the length is 100cm?
(a) 0.5V/m
(b) 5V/m
(c) 50V/m
(d) None of the above

Q2. What is the direction of data bus?
(a) Bi-directional
(b) Uni-directional out microprocessor
(c) Uni-directional into microprocessor
(d) Mixed direction i.e. some lines into microprocessor and some other out of microprocessor

Q3. As compared to BJT, a power MOSFET has
(a) higher switching losses and lower conduction loss
(b) lower switching losses and lower conduction loss
(c) lower switching losses and higher conduction loss
(d) higher switching losses and higher conduction loss

Q4. In the IPv4 addressing format the number of networks allowed under Class B addresses is
(a) 220
(b) 234
(c) 221
(d) 214

Q5. The dominant mode in a rectangular waveguide is TE10, because this mode has
(a) no attenuation
(b) the highest cut-off wavelength
(c) no cut-off
(d) no magnetic field component

Q6. Ampere-second is the unit of
(a) emf
(b) power
(c) electric charge
(d) energy

Q7. Modulation is the process of
(a) reducing distortion in RF amplifiers
(b) generating constant-frequency radio waves
(c) improving thermal stability of a transistor
(d) combining audio and radio frequency waves at the transmitting end of a communication system

Q8. The response of a linear, time-invariant, discrete-time system to a unit step input u(n) is the unit impulse 𝛿(n). The system response to a ramp input nu(n) would be
(a) u(n)
(b) u(n – 1)
(c) n𝛿(n)
(d) ■(∞@Σ@k=0) k𝛿(n – k)

Q9. The decimal equivalent of BCD 8421 number (10000000. 1001) is
(a) 128.56
(b) 128.9
(c) 80.25
(d) 128.25

Q10. Microprogramming refers to
(a) the use of storage to implement the control unit
(b) programming at micro level
(c) emulation
(d) array processing

SOLUTIONS

S1. Ans.(b)
Sol. Given:
Applied Voltage(V) =5V
length(L) = 100cm = 1m
Now,
E = V/L
⇒ E = 5/1
So, E = 5V/m.

S2. Ans.(a)
Sol. Data Bus is bi-directional and indicated by arrow in both direction. Data Bus is used to transfer data between microprocessor and memory and also between microprocessor and I/O devices. Same pins are used for data transfer.

S3. Ans.(c)
Sol. Power MOSFETS have lower switching losses but its ON-resistance and conduction losses are more. A BJT has higher switching loss but lower conduction loss. At higher frequency power MOSFET is better but at lower frequency BJT is superior.

S4. Ans.(d)
Sol. Class C has 16 network bits and 16 host bits.
Now, out of 16 network bits first 2-bits are reserved for Network ID i.e. “10” and remaining are 14 bits used for different networks. Therefore total number of networks possible are 214.

S5. Ans.(b)
Sol. TE10 is considered dominant mode because it has lowest cut off frequency which is equal to c/2a. We know that frequency is the inverse of wavelength i.e. (λ) = 1/f . As cut off frequency is minimum so it will have maximum wavelength.

S6. Ans.(c)
Sol. We know that,
⇒ I = q/t
⇒ q = I.t
So, Unit of electric-charge = Ampere–second.

S7. Ans (d)
Sol. Modulation is a process of varying amplitude, frequency or phase of carrier signal w.r.t modulating signal. So, It is the process of combining audio and radio frequency waves at the transmitting end of a communication system.

S8. Ans.(a)
Sol. The response of a linear, time-invariant, discrete-time system to a unit step input u(n) is the unit impulse 𝛿(n) that means the system acts as a Differentiator.
So, The system response to a ramp input nu(n) would be u(n).

S9. Ans(a)
Sol. Converting binary to decimal, we get
1 0 0 0 0 0 0 0 . 1 0 0 1 = 2^7 × 1 + 2^(-1) × 1 + 2^(-4) × 1
= 2^7 + 1/2 + 1/2^4
= 128 + 0.5 + 0.0625
= 128.5625
Therefore, (10000000. 1001)2 = 128.56

S10. Ans.(c)
Sol. Microprogramming consists of microcode with the help of which large number of architectures are achieved. It consists of large number of execution units and wider word lengths.

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