UGC-NET Electronic Science Question Paper With Answer Key and Full Explanation [Sept 2022]

 

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UGC-NET Electronic Science Sept. 2022 Answers with Explanations

1. (C)

2. (A)

3. (B)

4. (B)

5. (B)

6. (A)

7. (B)

8. (A)

9. (C)

10. (B)

11. A, B, C. D

12. (B)

13. (D)

14. (A)

15. B

Solution

min slew rate = 2*pi*f*Vpeak    (Unit V/s)

= 2*pi* 20*10^3*10

= 12.56*10^5 V/s

= 1.25 V/microsec

16. (B)

 

17) Consider the expression Y = P ⊕ Q ⊕ R where P, Q, R are the input variables and Y is the output variable. Y will be logic 0 if:

1. an odd number of input variables are 1
2. an even number of input variables are 1
3. an odd number of input variables are 0
4. an even number of input variables are 0
5. an odd number of input variables are between 0 and 1

Choose the correct answer from the options given below:

1. (A), (C) only
2. (E), (D) only
3. (A), (D) only
4. (B) only

Answer: Option D

18. (C)

19. (D)

20. (A)

21.(C)

22. (B)

Solution

number of clock cycle *  (1 / operating frequency )

= 12 * (1/12*10^6) = 1 millisecond

23.(C)

24. (A)

25. (A)

Solution

Q/r * 1/4*pi*ϵ
={5*10^(-6)/2} * {9*10^9}
= 22.5 kV

26. (A)

A sphere of radius r₁ = 30 cm has a charge density variation ρ(r) = ρ₀ (r / r₁) where ρ₀ = 200 pC/m³. Find the total charge on the sphere.

Solution

The charge element is:

dQ = ρ dV

For a spherical shell:

dV = 4πr² dr

Substitute ρ(r):

dQ = ρ₀ (r / r₁) × 4πr² dr

Total charge:

Q = ∫₀ʳ¹ ρ₀ (r / r₁) 4πr² dr

Q = (4πρ₀ / r₁) ∫₀ʳ¹ r³ dr

Q = (4πρ₀ / r₁) [r⁴ / 4]₀ʳ¹

Q = πρ₀ r₁³

Now substitute values:

r₁ = 0.3 m

Q = π × 200 × (0.3)³ pC

Q = π × 200 × 0.027 pC

Q = 5.4π pC

Q ≈ 17 pC

Final Answer: 17 pC

27. (A)

28. (B)

29(B)

30(C)

31. (D)
Solution

Step 1: Calculate bits per sample (n)
The number of quantization levels (L) is 64.
Formula: n = log2(L)
n = log2(64) = 6 bits per sample

Step 2: Calculate total bits per TDM frame
A frame contains one sample from each of the 10 signals plus the sync bits.
Total bits = (Number of signals × bits per sample) + synchronization bits
Total bits = (10 × 6) + 5 = 65 bits per frame

Step 3: Calculate the bit rate (R_b)
The sampling rate is given as 8 kHz (8,000 samples per second).
Rb = Sampling Rate × Bits per frame
Rb = 8,000 × 65 = 520,000 bits/sec (520 kbps)

Step 4: Determine the minimum channel bandwidth
For PCM transmission, the minimum bandwidth is typically taken as equal to the bit rate (R_b).
Minimum Bandwidth = 520 kHz

Correct Option: (4) 520 kHz

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