UGC NET Electronic Science September 2022 Question Paper with Answer Key & Detailed Solutions

UGC NET Electronic Science September 2022 Question Paper with Answer Key and Full Explanations

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1. C

Transconductance g_m is the rate at which the drain current changes with respect to the gate-source voltage:

g_m = ∂I_D / ∂V_GS

Linear (Triode) Region

Condition:

V_DS < V_GS − V_T

Drain Current:

I_D = μ_nC_ox(W/L) [(V_GS − V_T)V_DS − V_DS²/2]

Taking derivative with respect to V_GS:

g_m = μ_nC_ox(W/L)V_DS

Let:

k_n = μ_nC_ox(W/L)

Then:

g_m = k_nV_DS

In the linear region, transconductance is directly proportional to V_DS.

Saturation Region

Condition:

V_DS ≥ V_GS − V_T

Drain Current:

I_D = (1/2)k_n(V_GS − V_T)²

Taking derivative with respect to V_GS:

g_m = k_n(V_GS − V_T)

Alternative forms:

g_m = 2I_D / (V_GS − V_T)

g_m = √(2k_nI_D)

2. (A)

3. (B)

Solution:

The total input impedance Z_in is given by the sum of the series components and the parallel combination of the diode's junction capacitance and negative resistance:

Z_in = R_s + jωL_s + [ (-R) || (1 / jωC) ]

Solve the parallel impedance part:

Z_parallel = -R / (1 - jωRC) = [ -R(1 + jωRC) ] / [ 1 + (ωRC)² ]

Separate the real part of the total impedance:

Re(Z_in) = R_s - R / [ 1 + (ωRC)² ]

The resistive cutoff frequency (f_r) is the frequency at which the real part of the input impedance becomes zero (the point where the device stops exhibiting negative resistance behavior):

R_s - R / [ 1 + (2πf_rRC)² ] = 0

Solve for f_r:

1 + (2πf_rRC)² = R / R_s

(2πf_rRC)² = (R / R_s) - 1

2πf_rRC = √[ (R / R_s) - 1 ]

f_r = [ 1 / 2πRC ] √[ (R / R_s) - 1 ]

Conclusion: Option (2) is the correct expression.

4. (B)

Solution:

From the band diagram, we observe the vertical relationships between energy levels:

The total energy difference between E_c and E_v is the bandgap, E_g.
On the p-side, the conduction band energy relative to the Fermi level is:
(E_c − E_F)_p = E_g − qV_p
On the n-side, the conduction band energy relative to the Fermi level is:
(E_c − E_F)_n = qV_n
The built-in potential energy qV_bi is the shift in E_c between the two sides:
qV_bi = (E_c − E_F)_p − (E_c − E_F)_n
qV_bi = (E_g − qV_p) − (qV_n)
qV_bi = E_g − (qV_n + qV_p)

Thus, option (2) is correct.

5. (B)

6. (A)

7. (B)

8. (A)

9. (C)

z-transform of a^n equals to

z-transform of aⁿ equals to

1/{1 - a z^(-1)} and |z| > |a|

So, z-transform of e^(-t) equals to

1/{1 - e^(-t) z^(-1)} and |e^(-t)| > |a|

Given z is sampled at 10 Hz, so t = 1/f = 1/10 = 0.1

Now, 1/{1 - e^(-t) z^(-1)} becomes

1/{1 - e^(-0.1) z^(-1)}

or, z/{z - e^(-0.1)}

or, z/{z - 0.9} as e^(-0.1) = 0.9

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:

an odd number of input variables are 1
an even number of input variables are 1
an odd number of input variables are 0
an even number of input variables are 0
an odd number of input variables are between 0 and 1

Choose the correct answer from the options given below:

(A), (C) only
(E), (D) only
(A), (D) only
(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).
R_b = Sampling Rate × Bits per frame
R_b = 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

32) The number of reference patterns in the code book (N), the number of samples in each block (K) and the coded transmission rate in bits per sample (r) are related as :

(1) r = N log₂(K)
(2) r = K log₂(N)
(3) r = K / log₂(N)
(4) r = log₂(N) / K

Answer: Option D

Step 1: Understand the Parameters

N = Number of reference patterns (code vectors) in the codebook
K = Number of samples in each block (vector dimension)
r = Coded transmission rate in bits per sample

Step 2: Determine the Number of Bits Required per Codeword

A codebook containing N reference patterns requires log₂(N) bits to uniquely identify each codeword.

Bits per Codeword = log₂(N)

Step 3: Calculate Bits per Sample

Each codeword represents a block containing K samples. Therefore, the transmission rate in bits per sample is:

r = (Bits per Block) / (Samples per Block)

r = log₂(N) / K

Step 4: Match with the Given Options

Correct Answer: Option (D)

r = log₂(N) / K

Bits per Vector = log₂(N)
Samples per Vector = K
Bits per Sample = log₂(N) / K

Therefore, the correct relation is: r = log₂(N) / K.

33) A graded indexed optical fiber has a parabolic refractive index profile (α = 2). If the fiber has a numerical aperture = 0.22 the total number of guided modes at a wavelength of 1310 nm is given by :

(1) 174
(2) 1740
(3) 119
(4) 274

Answer: Option A

Solution

Normalized frequency, V = 2*pi*a/λ * NA [a = core radius = 25 micrometer]

= 26.38

and, number of guided modes,

M = V^2 / 4 = 171

34) In optical Communication system the Zero-dispersion wave length is given by :

(1) 0 nm
(2) 1310 nm
(3) 1550 nm
(4) 800 nm

Answer: Option B

35) An R-L load is connected to a 230V, 400 Hz step down dc converter. The average load current is 100 A. The load resistance is 0.5 Ohm. The allowed ripple limit is 15%. The minimum value of inductor to limit the maximum ripple current is :

(1) 66.7 mH
(2) 6.67 mH
(3) 1.67 mH
(4) 16.7 mH

Answer: Option B

Solution

This is a standard single-phase step-down (buck) chopper design problem.

Given

Supply voltage, Vs = 230 V
Chopper frequency, f = 400 Hz
Average load current, Io = 100 A
Load resistance, R = 0.5 Ω
Ripple current limit = 15%

First find the average output voltage:

Vo = Io × R = 100 × 0.5 = 50 V

For a buck converter,

Vo = DVs

So the duty ratio is

D = 50 / 230 = 0.2174

Ripple Current

Allowed peak-to-peak ripple current:

ΔI = 0.15 × 100 = 15 A

For a buck converter operating in continuous conduction mode (CCM),

ΔI = Vo(1 − D) / (Lf)

Substituting the values:

15 = [50 × (1 − 0.2174)] / (L × 400)

15 = 39.13 / (400L)

L = 39.13 / 6000

L = 0.00652 H

L ≈ 6.5 mH

The nearest option is:

Answer: (B) 6.67 mH

The key formula used is:

ΔI = Vo(1 − D) / (Lf)

This is the inductor ripple-current expression for a buck converter operating in continuous conduction mode.

36) A thyristor can be fired with dv/dt of 100 V/µs. If the forward breakover current of the device is 5 mA. The value of diffusion capacitance is given by :

(1) 25 pF
(2) 2.5 pF
(3) 50 pF
(4) 5.0 pF

Answer: Option C

i = C*dV/dt

or, 5 * 10^(-3) = C * 100/10^(-6)

or, C = 5 * 10^(-3) / 100 * 10^(-6)

= 0.05 * 10^(9)

= 50 pF

37) Which of the following statements is not correct about Electromagnetic Flow meter?

(1) This technique can be used for electrical non-conducting fluids like gases.
(2) The principle of operation is directly analogue to Faraday's law of electromagnetic induction for solid conductors.
(3) Electromagnetic flow meter is suitable for fluids containing solid matter.
(4) The use of these meters is limited to fluids having conductivity atleast of the order of 5 x 10^-6 Ω^-1/cm (mho/cm).

Answer: Option A

38) For an Ionization chamber, If 'n' is ionization density, 'Ni' represents ionizing particles arriving each second 'L' is active length of Chamber 'P' is pressure of gas in Chamber and 'e' is the charge of an electron. Then ionization current is calculated using equation.

(1) Is = nL / (Ni P e)
(2) Is = (Ni n L P) / e
(3) Is = (Ni n L e) / P
(4) Is = Ni n L P e

Answer: Option D

39) Which of the following is not a desirable characteristic of manometer fluid?

(1) It should have high viscosity.
(2) It should be non-corrosive.
(3) It should be free from capillary effects.
(4) It should have negligible surface tension.

Answer : Option A

40) The Q of an unknown coil using parallel connection measurement method can be calculated using which of the following equations.

(1) Q_p =

C₁Q₁ (C₁ - C₂)(Q₁Q₂)

(2) Q_p =

(C₁ - C₂) C₁Q₂

(3) Q_p =

(C₁ - C₂)Q₁Q₂ /

ΔQ

(4) Q_p =

C₁Q₁ (C₁ + C₂)(Q₁Q₂)

Answer: Option C

41) The pull up to pull down ratio for an n mos inverter driven by another n mos inverter can be evaluated from:

(A) V_inv = V_t -

V_td √(Z_pu/Z_pd)

(B) V_inv =

-V_td √(Z_pu/Z_pd)

(C)

W_pd L_pd

(V_inv - V_t)² =

W_pu L_pu

(-V_td)²

(D)

W_pd L_pd

(V_inv - V_t) =

W_pu L_pu

(-V_td)²

Choose the correct answer from the options given below:

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

Answer: Option D

42) The layers of nmos involves a subset of layers. For nmos we require :

(A) Contact, (Black or Brown)

(B) Polysilicon (Green)

(D) Implant (Yellow)

Choose the correct answer from the options given below:

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

Answer: Option C

43) The layout of nmos involves a subset of the layers and features set out. For nmos we require

(A) N-diffusion (Green)

(B) Implant (Yellow)

(D) Metal (Blue)

Choose the correct answer from the options given below:

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

Answer: Option A

44) The electron current density in semiconductor devices is given as

(A) qμ_nnε + qD_n

dn /dx

(B) qμ_ppε + qD_p

dp dx

kT/ q

dn /dx

)

(D) qμ_n (nε -

kT q

dn dx

)

Choose the correct answer from the options given below:

(1) (A) and (B) only
(2) (A) and (C) only
(3) (A) and (D) only
(4) (C) and (D) only

Answer: option B

45) The number of occupied conduction band levels in intrinsic semiconductor is

(A) N_c exp (−

E_c - E_f kT

)

(B) ∫

^E_top
_{E_c}

N(E) f(E) dE

E_f - E_c kT

)

(D) ∫

^E_top
_{E_c}

exp(N(E)) dE

Choose the correct answer from the options given below:

(1) (A) and (B) only
(2) (B) and (C) only
(3) (C) and (D) only
(4) (A) and (D) only

Answer: Option A

46. (B)

47. (A)

48. (A)

49. (C)

50) Which of the following statements are correct?

(A)In a boost regulator, the output voltage is filtered with choke-input filter.
(B)A capacitor may be needed in a discrete voltage regulator to prevent oscillations.
(C)Thermal shutdown occurs in an IC regulator if internal temperature is too high.
(D)An increase of line voltage into a power supply produces an increase in load voltage.
(E)A series regulator is an example of a switching regulator.

Choose the correct answer from the options given below :

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

Answer; option B

Solution

Statement Analysis:

(A) In a boost regulator, the output voltage is filtered with choke-input filter.

Status: Incorrect. In a Boost converter (regulator), the inductor (choke) is located at the input side to store energy. The output stage consists of a diode and a capacitor. The output filter is therefore a capacitor-input filter, not a choke-input filter.

(B) A capacitor may be needed in a discrete voltage regulator to prevent oscillations.

Status: Correct. Voltage regulators use feedback loops. Without proper "frequency compensation" (usually provided by input and output capacitors), these loops can become unstable and cause the circuit to vibrate or "oscillate," which can ruin the output voltage or damage components.

(C) Thermal shutdown occurs in an IC regulator if internal temperature is too high.

Status: Correct. This is a fundamental safety feature in Integrated Circuit (IC) regulators like the 7805 or LM317. If the chip pulls too much current and gets too hot, an internal sensor shuts it down to prevent the silicon from melting.

(D) An increase of line voltage into a power supply produces an increase in load voltage.

Status: Correct. This refers to Line Regulation. While a high-quality regulator tries to keep the output (load) voltage perfectly steady, in real-world power supplies, a rise in the input (line) voltage will always cause a small, measurable increase in the output voltage before the regulator can fully compensate.

(E) A series regulator is an example of a switching regulator.

Status: Incorrect. A series regulator is a type of Linear Regulator. It works by constantly "burning off" excess voltage as heat. A Switching Regulator (like a Buck or Boost) works by rapidly turning a switch on and off to transfer energy efficiently. They are two completely different categories of regulators.

51) The steps for subtracting two positive numbers (M-N) using (r-1)'s complement are:

(A)Add the minuend M to the (r-1)'s complement of the subtrahend N.
(B)If an end carry occurs, add 1 to the least significant bit.
(C)If end carry does not occur, take the (r-1)'s complement of the number obtained in step A and place negative sign in front.
(D)Take (r-1)'s complement of the number obtained in step B on step C.

Choose the correct answer from the options given below :

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

Answer; option B

Solution

Step 1: The Complement (Statement A)
To subtract

$N$

from

$M$

, you don't actually "subtract." Instead, you find the

 $(r - 1)$

's complement of

$N$

and add it to

$M$

Example in 1's complement: To do

 $5 - 3$

, you take the 1's complement of

$3$

and add it to

$5$

Statement (A) is correct.

Step 2: Check for a Carry (Statements B and C)
After the addition in Step 1, you look at the "end carry" (the extra bit at the very left). Two things can happen:

Case 1: There IS an end carry.
This means the result is positive (

 $M \geq N$

). In

 $(r - 1)$

's complement systems, you must perform an "End-Around Carry." This means you remove that extra carry from the left and add 1 to the least significant bit (LSB) to get the final answer.

Statement (B) is correct.

Case 2: There is NO end carry.
This means the result is negative (

 $M < N$

). The number you have is currently "hidden" in its complement form. To see the real answer, you must take the and put a minus sign in front of it.

Statement (C) is correct.

Why Statement (D) is Wrong:

Statement (D) says: "Take (r-1)'s complement of the number obtained in step B on step C."
This doesn't make logical sense in the algorithm. Steps B and C are mutually exclusive—you either do one or the other depending on if a carry exists. You never perform a "Step D" on top of them.

52) For an R S flip flop :

(A) R = 0, S = 0, Q_n+1 = 1
(B) R = 0, S = 0, Q_n+1 = Q_n
(C) R = 0, S = 0, Q_n+1 = Q_n
(D) R = 1, S = 0, Q_n+1 = 0
(E) R = 0, S = 1, Q_n+1 = 1

Choose the correct answer from the options given below :

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

Answer: Option A

Solution

To determine the correct statements, we must refer to the standard characteristic behavior (truth table) of an RS (Reset-Set) Flip-Flop:

S (Set)	R (Reset)	Q_n+1 (Next State)	Operation
0	0	Q_n	Hold (No Change)
0	1	0	Reset
1	0	1	Set
1	Indeterminate	Invalid	Forbidden State

Evaluation of Statements:

(A) R=0, S=0, Q_n+1=1: Incorrect. (Should be Q_n).
(B) R=0, S=0, Q_n+1=Q_n: Incorrect. (This is toggle behavior, not RS).
(C) R=0, S=0, Q_n+1=Q_n: Correct. (Hold state).
(D) R=1, S=0, Q_n+1=0: Correct. (Reset state).
(E) R=0, S=1, Q_n+1=1: Correct. (Set state).

Conclusion: Statements (C), (D), and (E) are correct.

Correct Option: (1) (C), (D), (E) only

53) The methods of passing parameters to a procedure are:

(A) Using interrupt routine
(B) Using general memory and named memory locations
(C) Using register to pass values of the variables
(D) Using the stack to pass values of variables

Choose the correct answer from the options given below :

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

Answr:option D

Solution

Analysis of the Statements:

(A) Using interrupt routine:

Incorrect. Interrupt routines are mechanisms used to handle asynchronous events (like hardware signals) or system calls. While an interrupt might involve data transfer, it is not a standard "method" for passing parameters to a general procedure.

(B) Using general memory and named memory locations:

Correct. One of the oldest methods is to store parameters in fixed memory locations (like global variables) that both the caller and the procedure can access.

(C) Using register to pass values of the variables:

Correct. This is a very common and fast method, especially in assembly language and modern RISC architectures. The caller loads values into specific CPU registers before the procedure call, and the procedure reads them from those same registers.

(D) Using the stack to pass values of variables:

Correct. This is the standard method used by most high-level programming languages (like C or Java). Parameters are "pushed" onto the stack by the caller and "popped" or accessed via stack offsets by the procedure. This method is essential for supporting recursion.

Conclusion:

Statements (B), (C), and (D) are the standard methods of passing parameters.

54) Which of the following steps are necessary for developing an assembly language program?

(A) Write down in general terms the algorithm for the task program has to solve using flow chart
(B) Write an initialization chek-list for your program
(C) Determine instruction statements required to do each part of the program
(D) Start writing assembly language code for the program
(E) Execute the program on the microprocessor

Choose the correct answer from the options given below :

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

Answer: Option D

Analysis of the Steps:

(A) Write down in general terms the algorithm... using flow chart:
Necessary. Developing an algorithm or flowchart is the first crucial design step. It allows the programmer to map out the logic before dealing with low-level hardware details.

(B) Write an initialization check-list for your program:
Necessary. In assembly language (microprocessor programming), you must manually manage hardware resources. This involves deciding which registers to use, setting initial memory values, and configuring stack pointers.

(C) Determine instruction statements required to do each part...:
Necessary. This is the bridge between the logic (flowchart) and the actual code. You decide which specific assembly instructions (e.g., MOV, ADD, JMP) will implement each block of your flowchart.

(D) Start writing assembly language code for the program:
Necessary. This is the core implementation phase where the source code is actually produced.

(E) Execute the program on the microprocessor:
While execution is part of the testing and debugging phase, technically, "developing" a program refers to the creative and constructive process of building the source code. Execution happens after the program has been developed and assembled into machine code.

Conclusion:

Steps (A), (B), (C), and (D) represent the design and implementation phases of development. If we look at the provided choices, the most comprehensive list of "development" steps that does not include the post-development execution phase is:

56) For sky waves, following statements are given :

(A) n > 1, this shows
81N/f²
positive
(B) n < 1, shows
81N/f²
Negative
(C) n < 1 shows
81N/f²
< 0
(D) v_g · v_p = c²
(E) n = 0 shows
81N/f²
= 1, f = f_c

Choose the correct answer from the options given below :

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

Answer: option C

To find the solution for question 56, we need to analyze the properties of sky waves propagating through the ionosphere.

Analysis of the Statements:

Refractive Index (

$n$

) and Curvature:

The refractive index of an ionospheric layer is given by:

 $n = \sqrt{1 - \frac{81 N}{f^{2}}}$

, where

$N$

is electron density and

$f$

is frequency.

Since

 $\frac{81 N}{f^{2}}$

is always positive, is always less than 1 (

 $n < 1$

). Statement (A) is incorrect.

As a sky wave travels upward into the ionosphere, the electron density (

$N$

) increases, which causes the refractive index (

$n$

) to decrease. Mathematically,

 $\frac{d n}{d z}$

is negative.

The path of a wave is related to the derivative of

$n$

. For a wave to bend back toward Earth, its second derivative (

 $\frac{d^{2} z}{d x^{2}}$

) must be negative. Since

 $n < 1$

in the ionosphere and it bends downward, statements (B) and (C) are both correct. They represent the same physical condition (negative curvature).

Velocity Relation (Statement D):
In a dispersive medium like the ionosphere, the relationship between phase velocity (

 $v_{p} = \frac{c}{n}$

) and group velocity (

 $v_{g} = c \cdot n$

) is:

 $v_{g} \cdot v_{p} = (c \cdot n) \cdot (\frac{c}{n}) = c^{2}$

Thus, statement (D) is correct.

Critical Frequency (Statement E):
The critical frequency (

 $f_{c}$

) is the highest frequency returned from a layer at vertical incidence. At this point, the refractive index

$n$

becomes zero at the peak of the layer.

 $n = 0 ⟹ 1 - \frac{81 N}{f^{2}} = 0 ⟹ \frac{81 N}{f^{2}} = 1 and f = f_{c}$

Thus, statement (E) is correct.

Conclusion:

Statements (B), (C), (D), and (E) are all correct based on ionospheric propagation theory.

57) For Poynting vector, following are given :

(A) Poynting Vector around a point source is radial.
(B) There are only E_θ & H_φ present and are transverse.
(C) There is only E_φ component is present
(D) There is only E_θ is present
(E) Poynting vector does not exist around a point source

Choose the correct answer from the options given below :

(1) (A) and (B) only
(2) (B) and (E) only
(3) (B) and (D) only
(4) (A) and (E) only

Answer: A

Analysis of the Statements:

Definition: The Poynting Vector represents the directional energy flux of an electromagnetic field and is given by the cross product:

 $S = E \times H$

(A) Poynting Vector around a point source is radial:
Correct. A point source radiates energy outward in all directions. In spherical coordinates

 $(r, θ, ϕ)$

, the energy flows away from the center along the radius. Therefore, the Poynting vector has only a radial component (

 $S_{r}$

(B) There are only

 $E_{θ}$

and
$H_{ϕ}$
present and are transverse:

Correct. In the far-field (radiation zone) of a typical point source model like a Hertzian dipole, the electromagnetic waves become transverse. The electric field (

$E$

) typically has a polar component (

 $E_{θ}$

) and the magnetic field (

$H$

) has an azimuthal component (

 $H_{ϕ}$

). Since

 $S = E \times H = (E_{θ} \hat{θ}) \times (H_{ϕ} \hat{ϕ}) = (E_{θ} H_{ϕ}) \hat{r}$

, this confirms that the fields are transverse to the direction of propagation (radial) and to each other.

(C) & (D) Only one field component is present:
Incorrect. You need both an electric field (

$E$

) and a magnetic field (

$H$

) to have a non-zero Poynting vector. One alone does not result in energy propagation.

(E) Poynting vector does not exist around a point source:
Incorrect. Any source that radiates electromagnetic energy has an associated Poynting vector.

Conclusion:

Statements (A) and (B) are correct.

58) For electrical drives

(A) For T_m(t) < T_L(t),
dω/dt
> 0
(B) For T_m(t) > T_L(t),
dω/dt
> 0, speed increases
(C) For T_m(t) = T_L(t),
dω/dt
= 0, drive attains steady state
(D) The Equation J
dωdt
+ B
dω(t)dt
= T_m(t) - T_L(t)
(E) Equation J
dωdt
+ Bω = T_m(t) - T_L(t)

Choose the correct answer from the options given below :

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

Answer: A

Evaluating the Statements:

(A) For

 $T_{m} < T_{L}, \frac{d ω}{d t} > 0$

Incorrect. If the "drag" (

 $T_{L}$

) is stronger than the "push" (

 $T_{m}$

), the motor must slow down. Acceleration (

 $\frac{d ω}{d t}$

) would be negative (

 $< 0$

), not positive.

(B) For

 $T_{m} > T_{L}, \frac{d ω}{d t} > 0$

, speed increases:

Correct. If the "push" is stronger than the "drag," the motor accelerates, and the speed goes up.

(C) For

 $T_{m} = T_{L}, \frac{d ω}{d t} = 0$

, drive attains steady state:

Correct. When the forces are balanced, there is no change in speed. It keeps spinning at a constant rate, which is called "steady state."

(D) The Equation

 $J \frac{d ω}{d t} + B \frac{d ω (t)}{d t} = T_{m} - T_{L}$

Incorrect. This equation is mathematically weird because it has acceleration in both terms on the left. The friction term (

$B$

) should depend on speed (

 $ω$

), not acceleration.

(E) Equation

 $J \frac{d ω}{d t} + B ω = T_{m} - T_{L}$

Correct. This is the fundamental equation for motor dynamics. It says: (Net Torque) = (Force to overcome inertia) + (Force to overcome friction).

Conclusion:

Statements (B), (C), and (E) are correct.

59. (C)

60) For Induction motors :

(A) In regenerative breaking the synchronous speed is greater than rotor speed
(B) The slip is Negative in regenerative breaking
(C) In regenerative breaking rotor speed is greater than synchronous speed
(D) The slip is positive in regenerative breaking

Choose the correct answer from the options given below :

(1) (A) and (B) only
(2) (A) and (D) only
(3) (B) and (C) only
(4) (C) and (D) only

Answer: C

Evaluating the Statements:

(A) In regenerative braking the synchronous speed is greater than rotor speed (

 $N_{s} > N_{r}$

Incorrect. This is for normal driving, not braking.

(B) The slip is Negative in regenerative braking:
Correct. As we saw above, when

 $N_{r} > N_{s}$

, the math results in a negative slip.

(C) In regenerative braking rotor speed is greater than synchronous speed (

 $N_{r} > N_{s}$

Correct. This is the physical requirement for regeneration to happen.

(D) The slip is positive in regenerative braking:
Incorrect. Positive slip happens during normal driving.

Conclusion:

Statements (B) and (C) are correct.

61) Read the following statements :

(A) A selective filtering method is most commonly used for the generation of SSB signals
(B) A selective filtering method is most commonly used for the generation of DSB signals
(C) A VSB modulator is also known as asymmetric side band system
(D) A VSB modulator is also known as symmetric side band system

Choose the correct answer from the options given below :

(1) (A) and (B) only
(2) (A) and (C) only
(3) (B) and (D) only
(4) (B) and (C) only

Answer: B

62) In digital data transmission, a line code should have the following properties:

(A) Transmission bandwidth should be as small as possible
(B) Transmitted power should be as high as possible
(C) Transmission bandwidth should be as high as possible
(D) Transmitted power should be as low as possible

Choose the correct answer from the options given below :

(1) (A) and (B) only
(2) (B) and (C) only
(3) (C) and (D) only
(4) (A) and (D) only

Answer: D

63) Which of the following statements are correct about an IR photodiode?

(A) IR photodiodes cause no risk of contamination or physical damage to the object surface
(B) For higher temperatures, IR photodiode devices have shorter life
(C) IR photodiodes can not be used to measure temperature of moving objects (conveyor belt)
(D) IR photodiodes are capable of measuring hazardous and physically in accessible process

Choose the correct answer from the options given below :

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

Answer: C

64) Which of the following statements are correct about AC Bridges?

(A) The Maxwell bridge is suitable for measurement of Medium-Q Coils (1 < Q < 10)
(B) For measurements at low frequencies, an oscillator supplies excitation voltage while at higher frequencies, power line may serve as source of excitation
(C) The Hay bridge is convenient for measuring very low-Q coils (Q < 1)
(D) The Schering bridge is particularly useful for the measure of insulating properties i.e. for phase angles nearly 90°

Choose the correct answer from the options given below :

(1) (A) and (B) only
(2) (A) and (D) only
(3) (B) and (C) only
(4) (B) and (D) only

Answer: B

65) For transducers, following statements are given :

(A) An optical interferometer is useful for measuring extremely small motions
(B) The damping ratio of a seismic instrument should be low for good dynamic performance
(C) A rate gyro is a relative motion measuring device
(D) A pneumatic motion transducer is non linear over a wide range of motion
(E) A piezoelectric accelerometer can only be used for static motion measurement

Choose the correct answer from the options given below :

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

Answer: C

Solution

(A) An optical interferometer is useful for measuring extremely small displacements.

(B) A seismic instrument is used to measure vibration. For good dynamic performance, its damping ratio should be in the range of 0.7 to 1.0 (approximately critically damped).

(C) A rate gyroscope (rate gyro) is used to measure angular velocity (angular motion).

(D) A pneumatic transducer converts an electrical signal (voltage/current) into a proportional pneumatic pressure signal.

(E) A piezoelectric accelerometer is commonly used for vibration and shock measurements.

66) Match List I with List II :

List I

(A) JFET
(B) MOSFET
(C) Tunnel diode
(D) HEMT

List II

(I) Two-dimensional electron gas
(II) Negative resistance device
(III) Enhancement mode
(IV) Voltage controlled resistor

Choose the correct answer from the options given below :

(1) (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
(2) (A)-(III), (B)-(II), (C)-(I), (D)-(IV)
(3) (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
(4) (A)-(I), (B)-(IV), (C)-(III), (D)-(II)

Answer: A
67. (A)

68) Match List I with List II :

List I
Coefficients of s² + a₁s + a₀ = 0

(A) a₁² > 4a₀
(B) a₁² = 4a₀
(C) a₁² < 4a₀
(D) a₁ = 0
a₀ ≠ 0

List II
Nature of Roots

(I) Negative real and equal
(II) Conjugate Imaginary
(III) Negative Real and Unequal
(IV) Conjugate Complex (Real part negative)

Choose the correct answer from the options given below :

(1) (A)-(IV), (B)-(II), (C)-(I), (D)-(III)
(2) (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
(3) (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
(4) (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

Answer: C

69) Match List I with List II :

List I
Bias Configuration of BJT

(A) Fixed Bias Configuration
(B) Emitter Bias Configuration
(C) Voltage Divider Configuration
(D) Feedback Bias Configuration

List II
Stability factor equation

(I) S(V_BE) =
-β / R_B1 + β(R_E / R_B)
(II) S(V_BE) = -β / R_B
(III) S(V_BE) =
-β / R_B1 + β(R_C / R_B)
(IV) S(V_BE) =
-β / R_th1 + β(R_E / R_th)

Choose the correct answer from the options given below :

(1) (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
(2) (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
(3) (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
(4) (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

Answer: C

70) Match List I with List II :

List I (Decimal)

(A) +43
(B) +123
(C) -54
(D) -107

List II (8-bit sign magnitude number)

(I) 10110110
(II) 01111011
(III) 00101011
(IV) 11101011

Choose the correct answer from the options given below :

(1) (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
(2) (A)-(III), (B)-(II), (C)-(I), (D)-(IV)
(3) (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
(4) (A)-(II), (B)-(III), (C)-(IV), (D)-(I)

Answer: B

Solution

Option III --> 0 0101011 = +43

left MSB 0 for sign '+' and 0101011 is equal to 43 in decimal

71) Match List I with List II :

List I (Communication Mode)

(A) Mode 0
(B) Mode 1
(C) Mode 2
(D) Mode 3

List II (Features)

(I) High speed, 8-bit shift register; one
baud rate of f/12
(II) Standard 8-bit UART; variable baud rate using timer 1 overflow
(III) Multiprocessor 9-bit UART; variable baud rate using timer 1 overflow
(IV) Multiprocessor 9-bit UART; two
baud rates of f/32 or f/64

Choose the correct answer from the options given below :

(1) (A)-(I), (B)-(II), (C)-(IV), (D)-(III)
(2) (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
(3) (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
(4) (A)-(III), (B)-(II), (C)-(I), (D)-(IV)

Answer: A

72) Match List I with List II :

List I (Optical Amplifiers)

(A) PDFA
(B) TDFA
(C) EDFA
(D) Gain shifted EDFA

List II (Operating band)

(I) S-band
(II) C-band
(III) L-band
(IV) O-band

Choose the correct answer from the options given below :

(1) (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
(2) (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
(3) (A)-(IV), (B)-(I), (C)-(III), (D)-(II)
(4) (A)-(II), (B)-(I), (C)-(IV), (D)-(III)

Answer: C

73) Match List I with List II :

List I

(A) Shannon's theorem
(B) Shannon-Hartley theorem
(C) Bayes's theorem
(D) Parseval's theorem

List II

(I) Capacity of Gaussian Noise channel
(II) Rate of Information
(III) Energy of a signal
(IV) Conditional probabilities

Choose the correct answer from the options given below :

(1) (A)-(II), (B)-(I), (C)-(III), (D)-(IV)
(2) (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
(3) (A)-(III), (B)-(II), (C)-(IV), (D)-(I)
(4) (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

Answer: B

74) Match List I with List II :

List I
Effect of ζ

(A) 0 < ζ < 1
(B) ζ = 1
(C) ζ = 0
(D) ζ = -1

List II
Condition of System

(I) Over damped
(II) Undamped
(III) Unstable
(IV) Under damped

Choose the correct answer from the options given below :

(1) (A)-(III), (B)-(II), (C)-(I), (D)-(IV)
(2) (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
(3) (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
(4) (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

Answer: D

75) Match List I with List II :

List I (Transducer)

(A) Pirani gage
(B) Dielectric gage
(C) Magnetostriction gage
(D) Variable capacitance pressure gage

List II (Application)

(I) Liquid level
(II) Displacement
(III) Gas flow
(IV) Sound

Choose the correct answer from the options given below :

(1) (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
(2) (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
(3) (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
(4) (A)-(III), (B)-(I), (C)-(II), (D)-(IV)

Answer: D

76) Arrange the following in ascending order of their dielectric constant :

(A) Si₃N₄

(B) Ge

(D) SiO₂

Choose the correct answer from the options given below :

(1) (A), (B), (D), (C)

(2) (B), (C), (A), (D)

(3) (C), (D), (B), (A)

(4) (D), (A), (C), (B)

Answer: D

77) Arrange the following in descending order of their switching times :

(A) Schottky diode

(B) Power transistor (Darlington)

(D) Triac

Choose the correct answer from the options given below :

(1) (C), (D), (B), (A)

(2) (B), (C), (D), (A)

(3) (C), (B), (D), (A)

(4) (D), (B), (C), (A)

Answer: D

Breakdown of Switching Speeds:

(D) Triac: These are thyristor-based devices used for AC power control. They are generally the slowest because they often rely on the AC line current passing through zero to turn off. Their switching times are typically in the range of tens of microseconds to milliseconds.

(B) Power transistor (Darlington): Darlington transistors are bipolar devices. They suffer from "storage time" effects due to minority carrier injection, which makes them relatively slow to turn off. Their switching times are typically in the range of several microseconds.

(C) IGBT (Insulated Gate Bipolar Transistor): IGBTs combine the high-speed characteristics of a MOSFET with the high-voltage/current capabilities of a BJT. They are faster than standard power transistors but slower than pure majority-carrier devices. Their switching times are typically in the range of hundreds of nanoseconds to a few microseconds.

(A) Schottky diode: These are majority-carrier devices and have essentially zero reverse-recovery time. They are extremely fast, with switching times in the nanosecond range. They are the fastest on this list.

Final Arrangement (Slowest to Fastest):

(D) Triac > (B) Power transistor (Darlington) > (C) IGBT > (A) Schottky diode

78) Consider the network given in the following figure and arrange the following in their increasing order of numbers:

(A) Total numbers of branches

(B) Total numbers of non touching loops

(D) Number of nodes

Choose the correct answer from the options given below :

(1) (B), (C), (D), (A)

(2) (B), (C), (A), (D)

(3) (C), (B), (A), (D)

(4) (C), (B), (D), (A)

Answer: A

Solution

Total numbers of branches 5

Total numbers of non touching loops 0

Number of nodes 3

Number of independent loops = #branches - #nodes + 1

= 5-3+1 =3

79) The components of class D amplifier are given below. Arrange them in order of their use from input towards output :

(A) Amplifier

(B) Comparator

(D) Low pass filter

Choose the correct answer from the options given below :

(1) (C), (D), (B), (A)

(2) (D), (A), (B), (C)

(3) (C), (B), (A), (D)

(4) (B), (C), (D), (A)

Answer: C

Solution

Analysis:
A Class D amplifier operates using Pulse Width Modulation (PWM).

Input stage: A (C) Sawtooth generator provides a reference wave.

Modulation: This sawtooth wave and the audio input are fed into a (B) Comparator to create a PWM signal.

Power Stage: The PWM signal drives a switching stage, which acts as the power (A) Amplifier.

Output stage: Finally, a (D) Low pass filter is used to remove high-frequency switching noise and reconstruct the original audio signal for the speaker.

Order from Input to Output: (C), (B), (A), (D).

Correct Option: (C)

80) Arrange the following activities to properly execute TYPE 4 Interrupt in 8086 micro processor:

(A) Pushes the CS and IP value on stack for next instruction

(B) Pushes the Flag register on the stack

(D) Gets the CS value for start of interrupt service procedure

Choose the correct answer from the options given below :

(1) (A), (B), (D), (C)

(2) (C), (B), (A), (D)

(3) (B), (C), (A), (D)

(4) (D), (C), (B), (A)

Answer: C

Solution

When an 8086 microprocessor receives an interrupt, it performs the following hardware sequence:

(B) Pushes the Flag register onto the stack.

(C) Resets (clears) TF and IF to disable single-stepping and further maskable interrupts during ISR entry.

(A) Pushes the CS and IP values of the next instruction onto the stack (so it knows where to return).

(D) Gets the new CS and IP values from the interrupt vector table (IVT) to jump to the start of the interrupt service procedure.

Sequence of activities: (B), (C), (A), (D).

Correct Option: (3)

81) Compare the IC logic families based on their rise time in ascending order :

(A) MECL

(B) CMOS

(D) PECL

Choose the correct answer from the options given below :

(1) (D), (A), (C), (B)

(2) (B), (C), (A), (D)

(3) (C), (B), (D), (A)

(4) (B), (D), (A), (C)

Answer: A

Solution

PECL (Positive Emitter Coupled Logic) / MECL (Motorola Emitter Coupled Logic): ECL families are the fastest logic families because they operate in the non-saturation region. They have the smallest rise times (typically < 1-2 ns). PECL is generally slightly newer/refined compared to standard MECL.
Low power Schottky TTL (LS-TTL): These use Schottky diodes to prevent saturation, making them faster than standard TTL but slower than ECL. Rise times are typically around 6-10 ns.
CMOS: Standard CMOS families (like the 4000 series) are traditionally slower due to high input capacitance, though high-speed versions (HCMOS) exist. In general comparisons, they are placed at the slower end for rise time.

Correct Ascending Order: (D) PECL < (A) MECL < (C) Low power Schottky TTL < (B) CMOS.
Answer: (1)

82) Arrange the following in ascending order of Molecular density in various regions of ionosphere :

(A) F₁ Layer (B) F₂ Layer

Choose the correct answer from the options given below :

(1) (A), (D), (B), (C)

(2) (B), (C), (D), (A)

(3) (B), (A), (D), (C)

(4) (A), (B), (D), (C)

Answer: C

Solution

Atmospheric and molecular density decreases as altitude increases. Therefore, the highest layer has the lowest density, and the lowest layer has the highest density.

The layers of the ionosphere in order of increasing altitude are:
D Layer (Lowest) < E Layer < F

Therefore, the order of ascending molecular density (lowest to highest) is:
(B) F

Correct Order: (B), (A), (D), (C).
Answer: (3)

83) Arrange the following in decreasing order of the noise generated by them :

(A) Diode

(B) Transistor

(D) FET

Choose the correct answer from the options given below :

(1) (C), (B), (D), (A)

(2) (C), (B), (A), (D)

(3) (B), (C), (A), (D)

(4) (B), (C), (D), (A)

Answer: B

Solution

(C) Avalanche photo diode (APD): Highest noise due to the statistical nature of its internal gain (avalanche) process.
(B) Transistor (BJT): Higher noise than FETs because of shot noise from base current.
(A) Diode: Standard PN junction diodes generate shot noise and thermal noise.
(D) FET: Known for being low-noise devices, specifically in high-impedance applications.

Order: (C) > (B) > (A) > (D).
Correct Answer: (2)

84) Arrange the following in terms of their dielectric strength in ascending order :

(A) Bakelite

(B) Mica

(D) Teflon

Choose the correct answer from the options given below :

(1) (D), (A), (B), (C)

(2) (A), (C), (D), (B)

(3) (A), (B), (C), (D)

(4) (B), (A), (C), (D)

Answer: B

Solution

(A) Bakelite: ~15 kV/mm
(C) Paper (impregnated): ~20-30 kV/mm
(D) Teflon: ~60 kV/mm
(B) Mica: ~100+ kV/mm (highest dielectric strength among these).

Order: (A) < (C) < (D) < (B).
Correct Answer: (2)

85) Arrange the following thermal sensors in ascending order of their cost for an application :

(A) Thermocouple

(B) Thermistor

(D) IR Photodiode

Choose the correct answer from the options given below :

(1) (C), (B), (A), (D)

(2) (A), (B), (C), (D)

(3) (D), (A), (B), (C)

(4) (B), (A), (C), (D)

Answer: C

Solution

(B) Thermistor: Very inexpensive, mass-produced semiconductor devices.
(A) Thermocouple: Relatively cheap, consists essentially of two metal wires joined.
(C) RTD: More expensive due to high-purity materials (like platinum) and precise manufacturing.

Order: (B) < (A) < (D) < (C).
Correct Answer: (4)

86) For JFET and MOSFET, consider the following statements :

(A) MOSFET is a voltage controlled capacitor

(B) BJT is slower than FET

(D) Leakage currents are zero in FET

(E) MOSFET are slower than JFET

Choose the correct answer from the options given below :

(1) (A), (B) and (C) only

(2) (A), (B) and (D) only

(3) (B), (C) and (D) only

(4) (B) and (E) only

Answer: A

Solution

(A) MOSFET is a voltage controlled capacitor: True. The capacitance of the MOS structure depends on the applied gate voltage, which controls the depletion/inversion regions.
(B) BJT is slower than FET: True. BJTs are bipolar (minority carrier) devices and suffer from carrier storage effects, making their switching speed slower than unipolar (majority carrier) FETs.
(C) Input impedance of MOSFET is greater than JFET: True. A MOSFET has an insulated gate (using
```
 $S i O_{2}$ 
```
), leading to extremely high input impedance (
```
 $10^{10}$ 
```
to
```
 $10^{15} Ω$ 
```
), whereas a JFET uses a reverse-biased PN junction (
```
 $10^{9} Ω$ 
```
).
(D) Leakage currents are zero in FET: False. While they are very small (picoamps to nanoamps), they are not zero.
(E) MOSFET are slower than JFET: False. Modern MOSFETs are designed for very high-speed switching and generally outperform standard JFETs in this regard.

Statements (A), (B), and (C) are correct.
Answer: (1)

87) Given below are two statements :

Statement I : Turbo codes exhibit performance, in terms of bit error probability, that is very close to the Shannon limit and can be efficiently implemented for high speed use

Statement II : Convolutional codes provide worst performance in noisy channels where a high proportion of the bits are in error

In the light of the above statements, choose the correct answer from the options given below :

(1) Both Statement I and Statement II are true

(2) Both Statement I and Statement II are false

(3) Statement I is true but Statement II is false

(4) Statement I is false but Statement II is true

Answer: C

Statement I: True. Turbo codes were the first practical codes to closely approach the Shannon capacity limit. They can be implemented efficiently, though their iterative decoding adds some latency.
Statement II: False. Convolutional codes actually provide excellent performance in noisy channels for correcting random errors. They are widely used precisely because they perform well in noisy environments. While they may be outperformed by more modern Turbo or LDPC codes at very high noise levels, calling their performance "worst" is technically incorrect.

Statement I is true, but Statement II is false.
Answer: (3)

88) Given below are two statements :

Statement I : In proportional control, the actuating signal for the control action in a control system is proportional to the error signal

Statement II : It is desirable that control system be over damped for the point of view of quick response

In the light of the above statements, choose the answer answer from the options given below :

(1) Both statement I and Statement II are true

(2) Both statement I and Statement II are false

(3) Statement I is true but Statement II is false

(4) Statement I is false but Statement II is true

Answer: C

Solution

Statement I: True. By definition, in a proportional control system, the control output is directly proportional to the current error signal (
```
 $u (t) = K_{p} \cdot e (t)$ 
```
).
Statement II: False. Overdamped systems are sluggish and take a long time to reach the steady state. For a quick response, an underdamped system (typically with a damping ratio
```
 $ζ \approx 0.7$ 
```
) is preferred because it reaches the target value faster, albeit with some initial oscillation.

Since Statement I is true and Statement II is false, the correct answer is (3).

89) Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R :

Assertion (A) : The biggest advantage of using a PROM as a programmable logic device is that its design can be changed and modified rapidly.

Reason (R) : The PROM can generate any possible logic function of input variables because it generates every possible AND product term.

In the light of the above statements, choose the correct answer from the options given below :

(1) Both (A) and (R) are true and (R) is the correct explanation of (A)

(2) Both (A) and (R) are true but (R) is not the correct explanation of (A)

(3) (A) is true but (R) is false

(4) (A) is not true but (R) is true

Answer: D

Solution

Assertion (A): False. Standard PROMs are "One-Time Programmable" (OTP). Once programmed, the fuses are blown and the design cannot be changed or modified. Reprogrammability is a feature of EPROMs, EEPROMs, or Flash memory, but not basic PROMs.
Reason (R): True. A PROM architecture consists of a fixed "AND" array (which decodes inputs into all possible minterms) and a programmable "OR" array. This allows it to implement any Boolean function of its input variables.

Since Assertion (A) is false (not true) and Reason (R) is true, the correct answer is (4).

90) Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R :

Assertion (A) : Power MOSFETs have a positive temperature coefficient, which combats the possibility of thermal runaway

Reason (R) : Negative temperature coefficient results in decreased level of resistance with increase in temperature and causes thermal runaway

In the light of the above statements, choose the correct answer from the options given below :

(1) Both (A) and (R) are true and (R) is the correct explanation of (A)

(2) Both (A) and (R) are true but (R) is not the correct explanation of (A)

(3) (A) is true but (R) is false

(4) (A) is false but (R) is true

Answer: A

Solution

Assertion (A) is True: Power MOSFETs exhibit a positive temperature coefficient of resistance. As the temperature of a MOSFET increases, the mobility of the majority carriers decreases, which causes the "ON" resistance (
```
 $R_{D S (o n)}$ 
```
) to increase. This increase in resistance naturally limits the current flow, thereby preventing the uncontrolled heating cycle known as thermal runaway. This makes MOSFETs easy to connect in parallel.
Reason (R) is True: A negative temperature coefficient (found in devices like BJTs) means that as temperature increases, resistance decreases. This leads to higher current, more heat generation, and even lower resistance, creating a self-destructive loop called thermal runaway.
Connection: Reason (R) correctly explains why the positive coefficient mentioned in (A) is a beneficial property for avoiding the phenomenon described in (R). It provides the scientific context for why the behavior in Assertion (A) is desirable.

Conclusion:
Both (A) and (R) are true, and (R) is the correct explanation of (A).

91. D
92.A
93. C
94.B
95. B
96. C
97. D

98. B
99. C
100.B

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Linear (Triode) Region

Saturation Region

Solution:

Solution:

Solution

31. (D) Solution

Step 1: Understand the Parameters

Step 2: Determine the Number of Bits Required per Codeword

Step 3: Calculate Bits per Sample

Step 4: Match with the Given Options

Solution

Given

Ripple Current

ΔQ

Statement Analysis:

Why Statement (D) is Wrong:

Solution

Evaluation of Statements:

Analysis of the Statements:

Conclusion:

Analysis of the Steps:

Conclusion:

Analysis of the Statements:

Conclusion:

Analysis of the Statements:

Conclusion:

Evaluating the Statements:

Conclusion:

Evaluating the Statements:

Conclusion:

Breakdown of Switching Speeds:

Final Arrangement (Slowest to Fastest):

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31. (D)
Solution