Quiz Electronics Engineering 20 Oct 2020

Quiz Electronics Engineering
Exam: NIC
Topic: Miscellaneous
Date: 20/10/2020

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

Q1. Which of the following has the highest priority?
(a) INTR
(b) RST 6.5
(c) RST 7.5
(d) TRAP

Q2. In the case of a 70–MHz IF carrier for a transponder bandwidth of 36 MHz, energy must lie between(in MHz):
(a) 34 and 106
(b) 52 and 88
(c) 106 and 142
(d) 34 and 142

Q3. Assertion(A): Capacitance between two parallel plates area ‘A’ each and distance of separation ‘d’ is εA/d for large A/d ratio.
Reason(R): Fringing electric field can be neglected for large A/d ratio.
(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.

Q4. A continuous time system will be BIBO stable if all the eigen values are
(a) one
(b) distinct and their real parts negative
(c) negative
(d) zero

Q5. The phase controlled rectifiers used in speed control of DC motors converts fixed AC supply voltage into ___________ output voltage.
(a) variable frequency AC
(b) full rectified DC
(c) variable AC
(d) variable DC

Q6. Find out the integrating type analog to digital converter?
(a) Flash type converter
(b) Dual slope ADC
(c) Counter type converter
(d) Tracking converter

Q7. The underlying Transport layer protocol used by SMTP is ________
(a) Either TCP or UDP
(b) UDP
(c) TCP
(d) IMAP

Q8. The resistance of insulations, in general, _____________ with temperature rise.
(a) Decreases
(b) Increases rapidly
(c) Increases slowly
(d) Does not change

Q9. Which instruction is one byte instruction?
(a) JMP
(b) SHLD
(c) SPHL
(d) XRI

Q10. Kirchhoff’s current law is applicable to only
(a) closed loops in a network
(b) junction in a network
(c) electric circuits
(d) electronic circuits

SOLUTIONS

S1. Ans.(d)
Sol. Priority in decreasing order is –
⇒ TRAP > RST 7.5 > RST 6.5 > RST 5.5 > INTR

S2. Ans. (b)
Sol. Given:
• IF = 70 MHz
• BW = 36 MHz
Now,
fmin = 70 -18 = 52 MHz
So, fmax = 70 + 18 = 88 MHz

S3. Ans.(a)

S4. Ans.(c)
Sol. For a continuous time system to be BIBO the roots must lie to the left side of s-plane. Hence, roots are negative and so is the eigen values.

S5. Ans.(d)

S6. Ans.(b)
Sol. Other than dual slope ADC the rest belongs to direct type ADCs.

S7. Ans.(c)
Sol. Simple Mail Transfer Protocol (SMTP) as internet standard uses only reliable Transport protocol. TCP is a reliable protocol and is more appropriate because it ensures that every packet is delivered.

S8. Ans.(a)
Sol. The insulators have Negative temperature co-efficient so the resistances of insulations in general decreases with temperature rise. Insulators starts to conduct if temperature is increased more and more and conductance & resistance are inversely related.

S9. Ans.(c)
Sol. Number of bytes required by each instruction is –
• SPHL is 1 byte.
• SHLD is 3 bytes.
• JMP is 3 bytes
• XRI is 2 byte.

S10. Ans.(b)
Sol. Kirchhoff’s Current law state that at any node of an electrical circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node. Node is also termed as Junction.