- UGC NET Electronic Science December 2021 Question Paper with Answer Key and Detailed Solutions
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1. (D)
2. A
3.C
4.A
5.B
Logical Breakdown of Silicon Oxidation
To understand the formula, we look at the Deal-Grove Model. This model describes how an oxide layer (SiO2) grows on a silicon wafer over time.
Step 1: Understanding the Variables
- t (Time): The actual duration the silicon stays in the oxidation furnace.
- τ (Tau): This accounts for the Initial Oxide Layer. Wafers usually have a tiny bit of oxide (native oxide) before we start. τ represents the "equivalent time" it would have taken to grow that pre-existing layer.
- B/A (Linear Rate Constant): This constant tells us how fast the silicon is reacting when the oxide is very thin.
Step 2: The General Equation
The full Deal-Grove equation for oxide thickness (xo) is:
(hG is the gas phase mass transfer coefficient)
Fundamental Logic
This question is about Mass Transport. It asks how fast oxygen gas moves from the furnace atmosphere to the surface of the silicon.
Step 1: Identify the "Push" (Driving Force)
Nature moves substances from high concentration to low concentration. The oxidant starts at CG (high) and moves toward CS (low). The "pushing force" is the difference: (CG - CS).
Step 2: Identify the "Ease of Movement"
The variable hG is the mass transfer coefficient. Think of it like "Conductivity." It tells us how easily the gas molecules can slip through the air to reach the surface.
Step 3: Combine into the Flux Law
In all transport physics, Flux = (Coefficient) × (Driving Force).
8) If Cs and Cl are the equilibrium concentration of impurities in the solid and liquid near the interface respectively, then for dilute solutions encountered in silicon growth, an equilibrium segregation coefficient k0 may be defined as:
Some impurities "prefer" to stay in the liquid.
Some "prefer" to move into the solid.
This "preference" is what we call segregation.
: The concentration of the impurity in the Solid phase.
Cl
: The concentration of the impurity in the Liquid phase.
9.A
9) The circuit shown in the following figure is initially under steady-state condition. The switch is moved from position 1 to position 2 at t = 0. The current after switching will be:
Solution
Before the switch moves, it is at position 1. In a DC steady-state condition, an inductor acts as a short circuit. The current flows only through the voltage source (20V) and resistor R1 (10Ω).
A fundamental rule of inductors is that the current cannot change instantaneously. Therefore, the current immediately after the switch moves (t = 0+) must be the same as it was before.
At position 2, the voltage source is removed. We now have a "source-free" RL circuit where the inductor discharges its stored energy through both resistors R1 and R2 in series.
- Total Resistance (Rtotal) = R1 + R2 = 10Ω + 10Ω = 20Ω
The standard formula for current decay in an RL circuit is:
Given the options provide an exponent of -5t, we can see that the ratio R/L must be 5. Substituting our initial current of 2A into the formula:
The current after switching is 2e-5t A. This corresponds to Option 1 [Option ID - 2865].
Solution
Note: This is a Linear Difference Equation (Recurrence Relation). It is solved similarly to a second-order linear differential equation by finding characteristic roots.
Initial Conditions: x(0) = 0, x(1) = 1
Assume a solution of the form x(n) = rn. Substituting this into the equation gives:
Factoring the quadratic equation:
r1 = -1, r2 = -2
For two distinct real roots, the general form is:
x(n) = A(-1)n + B(-2)n
Using x(0) = 0:
0 = A + B ⇒ A = -B
Using x(1) = 1:
1 = -A - 2B
Substitute A = -B into the second equation:
1 = B - 2B
1 = -B ⇒ B = -1
Since A = -B, A = 1
Substitute A and B back into the general equation:
x(n) = (-1)n - (-2)n
This matches Option 4.
Solution (State Equations)
In state-space analysis of circuits, we typically choose the voltage across the capacitor (Vc) and the current through the inductor (iL) as our state variables.
I = ic + iL
Since ic = C(dVc/dt):
I = C(dVc/dt) + iL
Rearranging for the derivative of the first state variable:
Vc = VL + VR
Since VL = L(diL/dt) and VR = iLR:
Vc = L(diL/dt) + iLR
Rearranging for the derivative of the second state variable:
The standard form is ẋ = Ax + Bu. Combining Eq. 1 and Eq. 2:
Comparing our derived matrices to the given choices, the values in the A matrix [0, -1/C; 1/L, -R/L] and the B matrix [1/C; 0] match perfectly with Option 1.
Solution
In Control Systems, the "Roots of a System" refer to the roots of the Characteristic Equation, which is the denominator of the Transfer Function. These are also known as Poles.
- Zeros: Roots of the Numerator (top).
- Poles (Roots): Roots of the Denominator (bottom).
To find the system roots (poles), we solve for s in the characteristic equation:
Using the zero-product property:
- s + 3 = 0 ⇒ s = -3
- s + 4 = 0 ⇒ s = -4
The value s = -2 (found in options 2 and 3) is a Zero of the system, not a root. Options featuring positive values (3 or 4) are mathematically incorrect based on the factors given.
The roots of the system are -3 and -4, which is Option 1.
Solution
To find the statement that is NOT correct, we evaluate each based on transistor theory:
| Statement | Engineering Fact | Status |
|---|---|---|
| 1. Emitter Follower output < input | Voltage gain is always slightly less than unity (approx. 0.98). | CORRECT |
| 2. CB has low input impedance | Input resistance is approximately re (very small). | CORRECT |
| 3. Unbypassed RE increases Rin | The unbypassed resistor adds βRE to the input impedance. | CORRECT |
| 4. Voltage Divider is less stable than Fixed Bias | Voltage Divider is the most stable bias; Fixed Bias is the least stable. | NOT CORRECT |
15. A
| AB \ CD | 00 | 01 | 11 | 10 |
|---|---|---|---|---|
| 00 | 0 | 0 | 1 | 0 |
| 01 | 1 | 1 | 1 | 1 |
| 11 | 0 | 1 | 1 | 1 |
| 10 | 0 | 1 | 0 | 0 |
Solution
To find the simplified Sum of Products (SOP) expression for the K-map, we follow these steps:
1. The Given K-Map
| AB \ CD | 00 | 01 | 11 | 10 |
|---|---|---|---|---|
| 00 | 0 | 0 | 1 | 0 |
| 01 | 1 | 1 | 1 | 1 |
| 11 | 0 | 1 | 1 | 1 |
| 10 | 0 | 1 | 0 | 0 |
2. Identify and Group the '1's
We group the '1's in powers of 2 (8, 4, 2, or 1) to find the simplest terms.
Cells: (01,01), (01,11), (11,01), (11,11).
Logic: Variable B is constant at 1, and Variable D is constant at 1. A and C change.
Term: BD
Cells: (01,00) and (01,10).
Logic: Row 01 means A=0, B=1. Columns 00 and 10 mean D=0. C changes.
Term: ABD
Cells: (00,11) and (01,11).
Logic: Column 11 means C=1, D=1. Rows 00 and 01 mean A=0. B changes.
Term: ACD
Cells: (11,01) and (10,01).
Logic: Column 01 means C=0, D=1. Rows 11 and 10 mean A=1. B changes.
Term: ACD (Note: In the provided answer key, this is grouped as ABD using cells (11,01) and (11,11)).
Cells: (11,11) and (11,10).
Logic: Row 11 means A=1, B=1. Columns 11 and 10 mean C=1. D changes.
Term: ABC (Note: In the provided answer key, this cell is covered via ACD).
3. Final Combined Expression
Following the specific grouping logic shown in the exam's correct option (Option 1), we combine the terms:
19. Determine the output of the logic array given in the following figure. The Xs represent connected link
1. AB- + A-B
4. AB + A-B-
Answer: A
20. A
21. A
22) If the data segment register (DS) contains 4000H, what physical address will the instruction MOV, AL, [234BH] read?
1. 634BH
2.
3.
4. 4234BH
Answer: D
Solution
Solution
Here's how to calculate the physical address:
- DS (Data Segment Register): 4000H
- Offset: 234BH
The physical address is calculated using the formula:
Physical Address = (DS X 16) + Offset
- Convert DS to decimal and multiply by 16:
- 4000H in decimal is 16384.
- 16384 * 16 = 262144.
- Convert the result back to hexadecimal:
- 262144 in hexadecimal is 40000H.
- Add the offset:
- 40000H + 234BH = 4234BH
Therefore, the physical address is 4234BH.
So the answer is 4) 4234BH.
23) The three basic structure types used to write the algorithm for a program are sequence, selection and repetition. Which of the following is NOT an advantage of using only these structures when writing the algorithm for a program?
24) Which is lowest priority interrupt of 8086 microprocessor?
25) For parallel polarisation, for lossless dielectrics, the expression for Brewster angle for a wave traveling from medium 1 to the medium of refractive indices η1 and η2 respectively is
Solution
To find the expression for the Brewster angle (θB) for parallel polarization, follow these steps:
Brewster's angle occurs when the reflection coefficient for parallel polarization (p-polarization) is zero. For lossless dielectrics, this happens when the reflected ray and the refracted (transmitted) ray are perpendicular (90° apart).
Using Snell's Law for the interface between medium 1 and medium 2:
Substitute the condition from Step 1:
For non-magnetic dielectrics, the refractive index n is related to the relative permittivity ε by n = √ε. Therefore:
Since the options are in terms of sin-1, use the identity sin2θ = tan2θ / (1 + tan2θ):
Taking the square root and the inverse sine of both sides gives the standard expression:
26. Vector A̅ = ŷ.3 + ẑ.2 and B̅ = x̂.5 + ŷ.8 extend from the origin. Find A̅.B̅
1. 15
2.16
3.24
4.6
Answer: C
solution
Dot product A̅.B̅ = |A̅||B̅|cos(theta)
and x̂.x̂ = 1, x̂.ŷ = x̂.ẑ = 0 and so on
So, A̅.B̅ = 24
27) Water flowing in x direction has a rate of Jw = 3yz liters/minute/m2. The total flow or flux of water through the rectangular area with corners (0,0,0), (0,3,0), (0,0,2) and (0,3,2) m is
Flux is given as:
given :
correct option is 2.
28) The value of transmission line impedance of a coaxial line of internal diameter 'a' and external diameter 'b' is given by the expression:
29) In a multimode fiber (step index), number of modes passing at an operating wavelength of 1300 nm are 1000, the refractive index of the core is 1.50 and that of the cladding is 1.48. The value of core diameter is:
30. In optical fibers, the Rayleigh scattering is proportional to:
Why Rayleigh Scattering is Proportional to 1/λ⁴
An accelerating charge emits electromagnetic radiation. The radiated power from an oscillating dipole is proportional to the square of its acceleration:
P ∝ a²
If the oscillation frequency is f, then the acceleration scales as:
a ∝ f²
Therefore:
P ∝ f⁴
Convert Frequency to Wavelength
Since:
f = c / λ
substituting into the power relation gives:
P ∝ (1/λ)⁴
or
P ∝ 1/λ⁴
This is known as the Rayleigh scattering law.
Intuitive Way to Remember
Suppose the wavelength doubles:
λ → 2λ
Then the scattering becomes:
1/(2λ)⁴ = 1/(16λ⁴)
Therefore, doubling the wavelength reduces Rayleigh scattering by 16 times.
31. The absolute maximum operating frequency of a converter grade SCR whose turn on time and turn off time are 3 µs and 200 µs respectively is:
1. 9.4 Khz
2. 5.2 khz
3. 4.9 Khz
4. 300 KHz
Answer: C
Solution
1/{(3+200)*10^(-6)} = 4.9 khz
32. A single phase 220V, 1kW electric room radiator is connected across 220V supply through a triac. For a delay angle of 90°, the value of power dissipated by the radiator is:
1. 49.985 W
2. 2500 W
3. 398.9 W
4. 500 W
Answer: 4
Power Dissipated by a Radiator Controlled by a TRIAC
Given
- Supply Voltage, V = 220 V
- Rated Power of Radiator, Pfull = 1000 W
- Delay (Firing) Angle, α = 90°
- Load Type = Purely Resistive
Formula
For a single-phase AC voltage controller using a TRIAC feeding a resistive load:
P = Pfull × [1 − (α/π) + sin(2α)/(2π)]
Calculation
Convert the firing angle into radians:
α = 90° = π/2
Substitute into the power equation:
P = 1000 × [1 − (π/2)/π + sin(π)/(2π)]
Since:
- (π/2)/π = 1/2
- sin(π) = 0
Therefore:
P = 1000 × (1 − 1/2)
P = 1000 × 0.5
P = 500 W
Answer
Power dissipated by the radiator = 500 W
Quick Shortcut
For a resistive load with a firing angle of 90°:
sin(2α) = sin(180°) = 0
Hence:
P = Pfull × (1 − 1/2) = Pfull/2
Therefore, a 1 kW radiator dissipates:
500 W
33) If fm is modulating frequency and mf is modulation index, then by the Carson's rule, the bandwidth of an FM signal at the input of a conventional discriminator will be:
34) A 10kW carrier is simultaneously modulated by two modulating signals corresponding to a modulation index of 40% and 30%, respectively. The total radiated power will be:
35) If an amplitude modulated wave 10[1+0.6 cos 2π103t] cos 2π.106.t is to be detected by a linear diode detector, then the time constant will be:
36) If the in-phase and quadrature components in an M-ary PSK system are permitted to be independent, then this scheme becomes a:
Second-order instruments (like accelerometers, pressure transducers, or galvanometers) are described by their natural frequency (
Overdamped Systems (
):Underdamped Systems (
):Optimal Damping (
):
Why the other options are incorrect:
Option 2: Incorrect. The Time Constant (
)Option 3: Incorrect. The Mean Square Value (power) of a random signal is mathematically defined as the integral of the Power Spectral Density (PSD) over the frequency range. Therefore, it can be found from such a plot. Option 4: Generally Correct, but less specific. While compensation does indeed improve dynamic characteristics, in the context of specific technical exams, they often look for the specific physical mechanism (like adjusting the damping ratio) used to achieve that improvement. In this specific question paper, Option 1 is treated as the precise technical answer.
■ \( Q_x = \frac{(C_1 - C_2)(Q_1 Q_2)}{(C_1 Q_1 - C_2 Q_2)} \) [Option ID = 2982]
Conceptual Solution for Question 38
Question 38 asks for the Quality Factor ($Q_x$) of a coil measured using the Series Substitution Method on a Q-Meter. Here is the fundamental breakdown:
1. The Fundamental Principle: Series Resonance
A Q-meter operates on the principle of series resonance. At the resonant frequency ($f$):
- Inductive reactance ($X_L$) exactly cancels out Capacitive reactance ($X_C$).
- The voltage across the capacitor is magnified by the Quality Factor: \( Q = \frac{X_L}{R} = \frac{1}{\omega C R} \).
2. Measurement Steps
The circuit is tuned to resonance using only the internal work coil.
- Resonance capacitance: \( C_1 \)
- Measured Quality Factor: \( Q_1 \)
- Circuit resistance is given by: \( R_1 = \frac{1}{\omega C_1 Q_1} \)
The unknown coil is added in series, and the capacitor is retuned to find a new resonance.
- New resonance capacitance: \( C_2 \)
- New Quality Factor: \( Q_2 \)
- Total resistance is now: \( R_{total} = R_1 + R_x = \frac{1}{\omega C_2 Q_2} \)
- The unknown resistance is: \( R_x = \frac{1}{\omega C_2 Q_2} - \frac{1}{\omega C_1 Q_1} \)
- The unknown reactance is: \( X_x = \frac{1}{\omega C_2} - \frac{1}{\omega C_1} \)
3. Deriving the Final Formula
The \( Q \) of the unknown coil is the ratio of its reactance to its resistance: \( Q_x = \frac{X_x}{R_x} \).
Substituting the values derived in Step 2:
By simplifying the common denominators and cancelling the \( \omega \) terms, we arrive at the final equation:
Summary of Logic: The formula represents the ratio of the "Reactance Shift" to the "Resistance Shift" caused by adding the unknown component into the series resonant circuit.
The question mentions:
Sampling rate: 20 kHz Time: 4 seconds
Summary Table
Solution
To determine the input torque (\(\tau_i\)) in a gear dynamometer, we analyze the relationship between power, efficiency, and torque equilibrium.
From the definition of efficiency:
Rearranging to solve for Output Torque (\(\tau_o\)):
In a gear dynamometer, the reaction torque (balancing torque) is created by the mass \(m\) at distance \(a\). This reaction torque must equal the difference between the output and input torques:
Substitute the expression for \(\tau_o\) from Step 1 into the equilibrium equation from Step 2:
Factor out the common term \(\tau_i\):
- A. Channel conductance is very low
- B. Channel conductance is very high
- C. Channel resistance is very low
- D. Channel resistance is very high
Choose the correct answer from the options given below:
- A. \( \frac{g_m}{2\pi} \)
- B. \( \frac{\mu_n V_d}{2\pi L^2} \)
- C. \( \frac{\mu_n L^2}{2\pi} \)
- D. \( \frac{\mu_n V_g}{2\pi L} \)
Choose the correct answer from the options given below:
- A. \( \frac{\partial n}{\partial t} = G_n - U_n + \frac{1}{q} \nabla \cdot J_n \)
- B. \( \frac{\partial p}{\partial t} = G_p - U_p + \frac{1}{q} \nabla \cdot J_p \)
- C. \( \frac{\partial p}{\partial t} = G_p - U_p - \frac{1}{q} \nabla \cdot J_p \)
- D. \( \frac{\partial n}{\partial t} = G_n - U_n + \frac{1}{q} \nabla J_n \)
Choose the correct answer from the options given below:
Holes have a positive charge (\(+q\)). Current density \(J_p\) flows in the same direction as holes. If holes leave a region (positive divergence, \(\nabla \cdot J_p > 0\)), the hole concentration decreases. Therefore, we use a negative sign:
(This matches Statement C)
Electrons have a negative charge (\(-q\)). Because of this negative charge, electron current \(J_n\) flows in the opposite direction to electron motion. If electrons leave a region, the math of the negative charge carrier flips the sign. Thus, for electrons, we use a positive sign:
- A. \( \sigma(x) E_x \)
- B. \( q N_D \mu E_x \)
- C. \( \frac{q}{2 \epsilon_s} N_D \mu \)
- D. \( \frac{N_D \mu}{2 \epsilon_s} \)
- A. \( \sqrt{\frac{V_G + V_{bi}}{V_P}} \)
- B. \( g_{max} \left( 1 - \sqrt{\frac{V_G + V_{bi}}{V_P}} \right) \)
- C. \( q N_D \mu \frac{2a W}{L} \left( 1 - \sqrt{\frac{V_G + V_{bi}}{V_P}} \right) \)
- D. \( \left( g_{max} - \sqrt{\frac{V_G + V_{bi}}{V_P}} \right) \)
Solution
1. Fundamental Physics of JFET
Transconductance (\(g_m\)) represents the relationship between the output drain current and the input gate voltage. In the saturation region, the depletion layer width is determined by the ratio of the applied reverse bias to the pinch-off voltage.
The maximum conductance (\(g_{max}\)) of the channel (when depletion width is zero) is defined by the physical dimensions and material properties:
2. Evaluating the Statements
The standard equation for JFET transconductance in the saturation region is derived from the square-root law of the depletion width:
-
Statement B: \( g_{max} \left( 1 - \sqrt{\frac{V_G + V_{bi}}{V_P}} \right) \)
✔ Correct: This is the standard theoretical expression for transconductance. -
Statement C: \( q N_D \mu \frac{2a W}{L} \left( 1 - \sqrt{\frac{V_G + V_{bi}}{V_P}} \right) \)
✔ Correct: This is identical to Statement B, but with the full physical definition of \(g_{max}\) substituted in. -
Statement A & D:
✘ Incorrect: Statement A is only a dimensionless ratio. Statement D is mathematically inconsistent because it tries to subtract a dimensionless ratio from a value with units of Conductance (Siemens).
Final Answer:
Since both statements B and C are mathematically and physically correct descriptions of the same formula:
Correct Option: 2 (B and C only) [Option ID = 3010]
- A. The system is linear.
- B. The system is non-linear.
- C. The system does not have homogeneity property.
- D. The system follows the additive property.
47) Which of the following statements regarding the Fourier series are correct?
A. For an even symmetry, only sine terms exist.
B. For an even symmetry, only cosine terms exist.
C. For an odd symmetry, only cosine terms exist.
D. For an odd symmetry, only sine terms exist.
Choose the correct answer
1. A and C only [Option ID = 3017]
2. B and D only [Option ID = 3018] 3.
3. C and D only [Option ID = 3019]
4. A and D only [Option ID = 3020
Answer: 2
- In order for a function to possess a Laplace transform, it must obey the condition ∫0⁻∞ |f(t)|e-σt dt > ∞, σ ∈ Re+
- In order for a function to possess a Laplace transform, it must obey the condition ∫0⁻∞ |f(t)|e-at dt < ∞, a ∈ Re+
- For a function to have a Fourier transform, it must obey the condition ∫-∞∞ |f(t)| dt > ∞
- For a function to have a Fourier transform, it must obey the condition ∫-∞∞ |f(t)|e-at dt < ∞
Choose the correct answer from the options given below:
- Schottky barriers are established by depositing a metal, such as Tungsten, on a p-type channel.
- The transfer characteristics of a depletion type MESFET are similar to those of a depletion type MOSFET.
- Maximum operating conditions are determined by the product of drain-to-source voltage and drain current.
- A complimentary MOSFET has negligibly small input impedance.
Choose the correct answer from the options given below:
- A multiplier can be used as a balanced demodulator or detector.
- Dynamic range compression is the process of reducing the peak-to-peak variation of a signal to a lower range.
- An antilogarithmic amplifier is useful for dynamic range compression.
- In a phase-locked loop circuit, the capture range is always larger than the lock range.
Choose the correct answer from the options given below:
- Y₁ = A ⊕ B ; Y₂ = AB̅ ; Y₃ = A̅B
- Y₁ = A ⊕ B ; Y₂ = AB ; Y₃ = A̅B̅
- Y₁ = A + B ; Y₂ = A ⊕ B ; Y₃ = AB
- Y₁ = A̅B̅ + AB ; Y₂ = AB̅ ; Y₃ = A̅B
Choose the correct answer from the options given below:
- When J = 0, K = 1, Qn+1 = 0
- When J = 1, K = 1, Qn+1 = 1
- When J = 1, K = 1, Qn+1 = Q-n
- When J = 1, K = 0, Qn+1 = 1
- When J = 1, K = 0, Qn+1 = 0
Choose the correct answer from the options given below:
Statement A: When J = 0, K = 1,
Evaluation: Correct. This is the Reset condition.
Statement B: When J = 1, K = 1, Evaluation: Incorrect. When both inputs are 1, the output toggles (
), it isn't fixed at 1.
Statement C: When J = 1, K = 1, Evaluation: Correct. This is the standard Toggle condition for a JK flip-flop.
Statement D: When J = 1, K = 0, Evaluation: Correct. This is the Set condition.
Statement E: When J = 1, K = 0, Evaluation: Incorrect.
sets the output to 1, not 0.
- STC
- CLC
- JNZ
- NOP
- CMC
Choose the correct answer from the options given below:
- Identify the symptoms and make a careful visual and tactical inspection.
- Check the power supply.
- Switch OFF and ON the system.
- Check the control signals such as RD, WR, ALE, RDY and RESET
Choose the correct answer from the options given below:
56. In 60‐degrees PWM inverter power supply,
A. Power devices are made 'ON' for 1/3 of the cycle.
B. Power devices are made 'OFF' for 1/3 of the cycle.
C. The phase voltage = 0.57735 V supply
D. Line voltage = V supply . E. Line voltage < Vsupply
1. B, C and E only [Option ID = 3053]
2. C and D only [Option ID = 3054]
3. B, C and D only [Option ID = 3055]
4. A, C and D only [Option ID = 3056
Answer: A
- Number of dipoles of unequal size.
- Number of dipoles equally spaced.
- Collinear dipoles.
- Dipoles in-phase.
- Dipoles are 90° out of phase.
Choose the correct answer from the options given below:
- ∇2V + με ∂2V / ∂t2 = -ρ/ε
- ∇2V - με ∂2V / ∂t2 = -ρ/ε
- ∇2A = -μJ + με ∂2A / ∂t2
- ∇2A - με ∂2A / ∂t2 = -μJ
Choose the correct answer from the options given below:
To solve this, we need to identify the standard inhomogeneous wave equations for electromagnetic potentials (V and A) in a time-varying field. These equations are derived from Maxwell's equations using the Lorenz gauge condition.
1. Theoretical Background
For time-varying fields, the scalar potential (V) and vector potential (A) satisfy wave equations of the form:
where □2 is the D'Alembertian operator, defined as:
2. Evaluating the Equations
For the Scalar Potential (V):
The standard equation is:
Comparing this with the given choices:
- A has a + sign (Incorrect).
- B has a − sign (Correct).
For the Vector Potential (A):
The standard equation is:
Comparing this with the given choices:
- C is written as
∇2A = -μJ + με (∂2A / ∂t2). While mathematically equivalent to the correct equation if rearranged, it is not the standard form usually paired with the scalar equation in textbooks. - D is written as
∇2A - με (∂2A / ∂t2) = -μJ. This is the standard textbook form of the inhomogeneous wave equation for the vector potential (Correct).
Conclusion
The correct equations representing the retarded time-varying fields are B and D.
59) Which of the following statements are correct?
A. DSB‐SC modulation is well suited for point to point communication involving one transmitter and one receiver.
B. VSB modulation is a linear modulation scheme.
C. SSB is a non‐linear modulation scheme.
D. FM is a linear modulation scheme.
Choose the correct answer from the options given below:
1. A and B only [Option ID = 3065]
2. A and C only [Option ID = 3066]
3. B and C only [Option ID = 3067]
4. B and D only [Option ID = 3068
Answer: 1
Solution:
Linear modulation schemes: AM, DSB, SSB, VSB
non-Linear modulation schemes: FM, PM
60. Which of the following statements are correct?
A. M‐ary modulation scheme is preferable where the bandwidth requirement is important.
B. M‐ary PSK system considers 'M' different phases in the range 'π/2'.
C. In M‐ary modulation scheme, only coherent detection is possible.
D. M‐ary QAM scheme uses 'm^2 ' carrier signals having the same frequency.
Choose the correct answer from the options given below:
1. A and B only [Option ID = 3069]
2. A and C only [Option ID = 3070]
3. B and C only [Option ID = 3071]
4. C and D only [Option ID = 3072
Answer: 2
61) Which of the following statements are correct?
A. If the intermediate frequency is too high, poor selectivity results even if sharp cutoff filters are used in the IF stage.
B. A high value of intermediate frequency increases tracking difficulties.
C. As the intermediate frequency is lowered, image frequency rejection becomes better.
D. A very low intermediate frequency can make the selectivity too sharp.
Choose the correct answer from the options given below:
1. A and B only [Option ID = 3073]
2. B and C only [Option ID = 3074]
3. C and D only [Option ID = 3075]
4. B and D only [Option ID = 3076
Answer: 4
62) A diac is
A. a two electrode, bidirectional avalanche diode.
B. a device which conducts below the breakover voltage.
C. a unidirectional resistive device.
D. designed to supply large gate current.
E. used to supply the voltage to the motors.
Choose the correct answer from the options given below:
1. A, B and C only [Option ID = 3077]
2. A, C and D only [Option ID = 3078]
3. A and D only [Option ID = 3079]
4. A, B and E only [Option ID = 3080]
Answer: 3
Solution
A. a two electrode, bidirectional avalanche diode.
A DIAC is indeed a two-terminal (electrode) semiconductor device. It is designed to conduct current in both directions (bidirectional) when the voltage across it exceeds a certain threshold voltage, known as the breakover voltage. The conduction occurs due to the avalanche breakdown mechanism. Thus, this statement accurately describes a DIAC.
B. a device which conducts below the breakover voltage.
This statement is incorrect. A DIAC is specifically designed to have very high resistance and conduct only a small leakage current when the voltage across it is below its breakover voltage. Significant conduction begins only when the voltage magnitude exceeds breakover voltage in either positive or negative direction.
C. a unidirectional resistive device.
This statement is incorrect. A DIAC is a bidirectional device, meaning it conducts in both directions above its breakover voltage. It is also not a purely resistive device; its behavior is non-linear and dependent on the voltage threshold (breakdown). Unidirectional devices conduct current primarily in one direction (like a standard diode).
D. designed to supply large gate current.
This statement refers to a common application of the DIAC. DIACs are frequently used as trigger devices for other power semiconductors, such as TRIACs and SCRs. When a DIAC breaks over, it rapidly conducts a pulse of current. This current pulse is often directed to the gate of a TRIAC or SCR to turn it on. In this sense, the DIAC is designed to provide the necessary current pulse (often referred to in this context as gate current for the triggered device) required to initiate conduction in the power control device. While not a continuous large current source, it provides the crucial trigger current pulse.
E. used to supply the voltage to the motors.
This statement is incorrect. DIACs are low-power trigger devices. They are not designed to handle the significant power required to supply voltage directly to motors. Power control for motors using semiconductor devices typically involves TRIACs or SCRs, which might be triggered by a DIAC, but the DIAC itself does not supply the main motor voltage.
63) Response time of photodiode has:
- A. dependance on transit time.
- B. dependance on diffusion time.
- C. RC time constant dependance.
- D. td = w / vd
- E. td = vd / w
Choose the correct answer from the options given below:
- A, B, C and D only
- B, C and E only
- A and B only
- A, C, and E only
64) Which of the following statements are correct?
A. For dynamic measurements using resistance strain gauges, temperature compensation is not necessary.
B. Eddy current type of transducer gives an output proportional to velocity.
C. A variable capacitance type transducer gives an output proportional to acceleration.
D. A piezoelectric transducer cannot be used to measure static variables.
Choose the correct answer from the options given below:
1. A and B only [Option ID = 3085]
2. B and D only [Option ID = 3086]
3. B and C only [Option ID = 3087]
4. A and D only [Option ID = 3088]
Answer: 4
65) Which of the following statements are correct?
A. Ultrasonic method for liquid level measurement is not preferred in industrial applications.
B. In a pH meter, the glass electrode is the reference electrode.
C. A chromatograph is used for analysing the composition of a gas.
D. The production of Korotkoff sounds in indirect blood pressure measurement is due to pressure pulses produced due to the difference in systolic and diastolic blood pressure.
Choose the correct answer from the options given below:
1. A and C only [Option ID = 3089]
2. B and C only [Option ID = 3090]
3. B and D only [Option ID = 3091]
4. C and D only [Option ID = 3092]
Answer: 4
66) Match List I with List II
| List I | List II |
|---|---|
| A. Ideal MIS diode (n-type) | I. L / vs |
| B. Extrinsic Debye length (hole) | II. μEy / (1 + μEy/vs) |
| C. Field dependent mobility | III. √(kTεs / pp0q2) |
| D. Transit time for velocity saturation | IV. φms = φm − (χ + Eg/2q − ψB) |
67) Match List I with List II
| List I | List II |
|---|---|
| A. Diffusion | I. PMMA |
| B. Oxidation | II. Proximity Printing |
| C. UV Resist | III. ∂C(x,t)/∂t = −∂J(x,t)/∂x |
| D. X-ray Lithography | IV. Deal and Grove's Model |
Choose the correct answer from the options given below:
68. D
69) Match List I with List II
| List I (JFET - Bias) | List II (Characteristic Equation) |
|---|---|
| A. Self-bias | I. ID = (VSS − VGS) / RS |
| B. Voltage-divider bias | II. ID = (VEE − VBE) / RE |
| C. Source bias | III. VGS = −IDRS |
| D. Current-source bias | IV. ID = (VG − VGS) / RS |
Choose the correct answer from the options given below:
- The Circuit: The gate is grounded through a resistor (VG = 0), and the source is connected to ground through a resistor RS.
- The Math: Since VG = 0 and the voltage at the source VS = ID ⋅ RS, the equation becomes:
VGS = 0 − ID ⋅ RS = −IDRS
- The Circuit: A resistive divider sets a constant voltage VG at the gate. The source is connected to ground through RS.
- The Math: VGS = VG − ID ⋅ RS. If you rearrange this to solve for ID:
ID ⋅ RS = VG − VGS
→ ID = (VG − VGS) / RS
- The Circuit: This typically uses two power supplies (+VDD and −VSS). The gate is grounded (VG = 0), and the source resistor RS is connected to the negative supply −VSS.
- The Math: The voltage drop across the source resistor is (VS − (−VSS)) = ID ⋅ RS, so VS = ID ⋅ RS − VSS.
VGS = VG − VS = 0 − (ID ⋅ RS − VSS) = VSS − ID ⋅ RS
Rearranging for ID:
ID = (VSS − VGS) / RS
- The Circuit: An active component, like a Bipolar Junction Transistor (BJT), is used to create a constant current source at the source lead of the JFET.
- The Math: You can tell a formula belongs to a BJT if it contains VBE (base-emitter voltage) and RE (emitter resistor). The current provided by the BJT is approximately:
ID ≈ IE = (VEE − VBE) / RE
70) Match List I with List II
| List I (Logic gate) | List II (Truth table) | |||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| A. Negative OR | I.
|
|||||||||||||||
| B. Exclusive NOR | II.
|
|||||||||||||||
| C. Exclusive OR | III.
|
|||||||||||||||
| D. Negative AND | IV.
|
Choose the correct answer from the options given below:
Logic Gate Truth Tables
1. Negative OR (NOR)
| A | B | F |
|---|---|---|
| 0 | 0 | 1 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 0 |
2. Exclusive NOR (XNOR)
| A | B | F |
|---|---|---|
| 0 | 0 | 1 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
3. Exclusive OR (XOR)
| A | B | F |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
4. Negative AND (NAND)
| A | B | F |
|---|---|---|
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
71) Match List I with List II
| List I (Assembly code) | List II (Content of register A after the code) |
|---|---|
A. MOV A, #37HCPL A |
I. 02H |
B. MOV A, #37HANL A, #8AH |
II. F7H |
C. MOV A, #37HORL A, #8AH |
III. FBH |
D. MOV A, #37HXRL A, #8AH |
IV. C8H |
Choose the correct answer from the options given below:
Solution
To solve this assembly code problem, we convert the hexadecimal values to 8-bit binary and apply bitwise logic operations.
• Register A (
37H) = 0011 0111• Operand (
0CAH) = 1100 1010
Flips every bit (NOT operation) of the original 37H.
| 37H | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 |
| Result | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 |
Binary 1100 1000 = C8H. Match: A → IV
Output is 1 only where both bits are 1.
| 37H | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 |
| 0CAH | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 0 |
| AND | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 |
Binary 0000 0010 = 02H. Match: B → I
Output is 1 if at least one bit is 1.
| 37H | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 |
| 0CAH | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 0 |
| OR | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
Binary 1111 1111 = FFH. Match: C → III
Output is 1 only if the bits are different.
| 37H | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 |
| 0CAH | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 0 |
| XOR | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 |
Binary 1111 1101 = FDH. Match: D → II
72) Match List I with List II
| List I (US New Military Bands for Microwave) | List II (Frequency range in GHz) |
|---|---|
| A. H band | I. 6.000-8.000 GHz |
| B. J band | II. 10.000-20.000 GHz |
| C. G band | III. 4.000-6.000 GHz |
| D. E band | IV. 2.000-3.000 GHz |
Choose the correct answer from the options given below:
73) Match List I with List II
| List I | List II |
|---|---|
| A. FDMA | I. Frequency Hopping |
| B. TDMA | II. Guard Band |
| C. CDMA | III. Onboard Switching |
| D. SDMA | IV. Guard Time |
Choose the correct answer from the options given below:
74) Match List I with List II
| List I (f(t)) | List II (F(s)) |
|---|---|
| A. e-at | i. s / (s2 + ω2) |
| B. te-at | ii. ω / (s2 + ω2) |
| C. sin ωt | iii. 1 / (s + a)2 |
| D. cos ωt | iv. 1 / (s + a) |
Choose the correct answer from the options given below:
75) Match List I with List II
| List I (Performance Characteristic) | List II (Definition) |
|---|---|
| A. Correction | I. Nearness of indicated value to the true value of the quantity being measured |
| B. Precision | II. Opposite of error |
| C. Accuracy | III. Closeness with which the same value of input quantity is measured under different conditions |
| D. Reproducibility | IV. Degree of refinement with which a measured value is stated |
Choose the correct answer from the options given below:
Answer: 4
76) Arrange the following in ascending order of their conductivity.
A. Silicon dioxide
B. Intrinsic Silicon
C. Extrinsic Silicon
D. Aluminium
Choose the correct answer from the options given below:
77) Arrange the following in ascending order of their bandgap (at ‐300K)
A. GaN
B. GaP
C. GaAs
D. Si
Choose the correct answer from the options given below
1. D, C, B, A [Option ID = 3137]
2. C, D, A, B [Option ID = 3138]
3. B, A, C, D [Option ID = 3139]
4. A, B, D, C [Option ID = 3140]
Answer: 1
Solution
Ge ----- .66 eV
Si ----- 1.1 eV
GaAs ----- 1.4 eV
GaP ----- 2.25 eV
GaN ----- 3.4 eV
78) Consider the following circuits and arrange them in ascending order of their complexity.
A. First order low pass filter
B. Sallen‐Key filter
C. Fourth order Butterworth low pass filter
D. Non‐inverting amplifier
Choose the correct answer from the options given below [Question ID = 786][Question Description = S2_qSNz_PG_ETS_Q78]
1. D, A, C, B [Option ID = 3141]
2. D, A, B, C [Option ID = 3142]
3. C, B, A, D [Option ID = 3143]
4. B, C, D, A [Option ID = 3144]
Correct Answer : D, A, C, B [Option ID = 3141
Answer: 2
Solution
Fourth order Butterworth low pass filter is the most complex. The correct order of complexity is as follows:Fourth order Butterworth low pass filter > First order low pass filter > Sallen‐Key filter > Non‐inverting amplifier
79) Arrange the following components of dual‐slot integrating A/D converter in order of their appearance while moving from input to output stage.
A. Comparator
B. Control
C. Integrator
D. Counter
Choose the correct answer from the options given below
1. C, A, D, B [Option ID = 3145]
2. C, A, B, D [Option ID = 3146]
3. C, B, A, D [Option ID = 3147]
4. C, D, A, B [Option ID = 3148]
Answer: 1
80) A number may be represented in various number systems. Arrange the following number systems in ascending order based on the number of bits needed to represent the same number.
A. Binary
B. Decimal
C. Hexadecimal
D. Octal
Choose the correct answer from the options given below
1. D, A, B, C [Option ID = 3149]
2. B, C, A, D [Option ID = 3150]
3. C, B, D, A [Option ID = 3151]
4. A, B, C, D [Option ID = 3152]
Answer: 3
81. Arrange the different operations to run instructions in a computer system, from the beginning to last stage
A. Instruction decode
B. Instruction fetch
C. Operand fetch
D. Execute
Choose the correct answer from the options given below
1. A, B, C, D [Option ID = 3153]
2. B, A, C, D [Option ID = 3154]
3. C, D, A, B [Option ID = 3155]
4. D, B, C, A [Option ID = 3156]
Answer: 2
82) Arrange the following antennas in ascending order of their radiation resistance.
- Short dipole (L = λ/10)(Iav = Io)
- Short dipole (L = λ/10)(Iav = Io/2)
- Linear λ/2 dipole (sinusoidal current distribution)
- Small Loop (square loop) single turn of(L = λ/10)
Choose the correct answer from the options given below
To understand the fundamentals behind this, we need to look at the formulas for Radiation Resistance (Rrad) of different antennas. Rrad represents the equivalent resistance that would dissipate the same amount of power as the antenna radiates.
1. Fundamental Formulas
Short Dipole (Length L ≪ λ):
The radiation resistance depends heavily on the current distribution along the length of the wire.
-
Case A: Uniform current (Iav = Io): This is a theoretical "Hertzian dipole."
Rrad = 80π2 (L/λ)2
-
Case B: Triangular current (Iav = Io/2): This is more realistic for a practical short dipole. Because the average current is halved, the power (which is proportional to I2) is divided by 4.
Rrad = 20π2 (L/λ)2
Small Square Loop (Side length L ≪ λ):
For a single-turn small loop with area A, the formula is:
If L is the side of the square, then area A = L2.
Half-Wave (λ/2) Dipole:
This is a resonant antenna with a sinusoidal current distribution. Its radiation resistance is a standard derived value:
2. Comparative Calculations (using L = λ/10)
Let's plug the values into the formulas to find the relative magnitudes:
Rrad ≈ 20 × (3.14)2 × (1/10)2 ≈ 20 × 9.87 × 0.01 ≈ 1.97 Ω
Rrad = 31,200 × [(λ/10)2 / λ2]2 = 31,200 × (1/100)2 = 31,200 / 10,000 = 3.12 Ω
Rrad ≈ 80 × (3.14)2 × (1/10)2 ≈ 80 × 9.87 × 0.01 ≈ 7.9 Ω
Rrad ≈ 73.13 Ω
3. Conclusion (Ascending Order)
Comparing the calculated values:
This gives us the sequence: B, D, A, C.
83) Arrange the following modulation systems in decreasing order of the bandwidth requirements.
A. Single Side Band
B. Double Side Band
C. Vestigial Side Band
D. Frequency Modulation
Choose the correct answer from the options given below
1. A, C, D, B [Option ID = 3161]
2. C, A, B, D [Option ID = 3162]
3. D, B, A, C [Option ID = 3163]
4. D, B, C, A [Option ID = 3164]
Answer: 4
Solution
AM & DSB --> 2*fm
FM & PM ---> 2fm(β+1)
SSB ---> fm
ssB < VSB <DSB
84) Arrange the following system types in ascending order of their intersection with 0 dB axis.
A. Type N system
B. Type 3 system
C. Type 2 system
D. Type 4 system
Choose the correct answer from the options given below
1. A, D, B, C [Option ID = 3165]
2. C, B, D, A [Option ID = 3166]
3. C, B, A, D [Option ID = 3167] 4.
D, C, B, A [Option ID = 3168]
Answer: A
Control Systems: Bode Plots and System Types
This question involves understanding how the System Type affects the magnitude plot on a Bode diagram, specifically where it crosses the 0 dB axis (Gain Crossover Frequency).
1. What defines "System Type"?
The "Type" of a system is defined by the number of integrators (poles at the origin) in the open-loop transfer function G(s)H(s). This determines the initial slope of the magnitude plot.
| System Type | Initial Slope on Bode Plot |
|---|---|
| Type 2 (C) | -40 dB/decade |
| Type 3 (B) | -60 dB/decade |
| Type 4 (D) | -80 dB/decade |
| Type N (A) | -20N dB/decade (Steepest) |
2. Intersection Logic
- Type N (A): Having the highest number of integrators (assuming N > 4), it has the steepest drop and intersects 0 dB at the lowest frequency.
- Type 4 (D): Intersects next.
- Type 3 (B): Intersects after Type 4.
- Type 2 (C): Having the shallowest slope among these, it takes the longest to reach 0 dB, intersecting at the highest frequency.
Imagine four cars parked at the top of a cliff. Type N is a vertical drop (it hits the ground immediately). Type 2 is a gentle walking path (it takes a long distance to reach the bottom). When arranging by "intersection point" from first to last, the steepest one always comes first.
3. Ascending Order of Intersection
Ordering from the smallest frequency (first intersection) to the largest frequency (last intersection):
A (Type N) → D (Type 4) → B (Type 3) → C (Type 2)
85) Arrange the following manometric liquid in ascending order of their specific gravity at 20 º C.
A. Water
B. Carbon tetrachloride
C. Transformer Oil
D. Mercury
E. Dibutylphthalate
Choose the correct answer from the options given below
1. A, C, B, E, D [Option ID = 3169]
2. A, D, B, C, E [Option ID = 3170]
3. C, A, E, B, D [Option ID = 3171]
4. B, C, A, E, D [Option ID = 3172]
Answer: 3
Solution
| Name | Specific gravity (20°C) |
|---|---|
| Water | 1 |
| Carbon tetrachloride (CCl4) | 1.594 |
| Transformer Oil | 0.856 |
| Mercury | 13.55 |
| Dibutyl Phthalate | 1.04 |
86) Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R
Assertion A: When MOSFET is in conductive state, the saturation current flow after the channel is pinchedoff.
Reason R: The substrate bias affects the threshold voltage of a MOSFET.
In light of the above statements, choose the most appropriate answer from the options given below
1. Both A and R are correct and R is the correct explanation of A [Option ID = 3173]
2. Both A and R are correct but R is NOT the correct explanation of A [Option ID = 3174]
3. A is correct but R is not correct [Option ID = 3175]
4. A is not correct but R is correct [Option ID = 3176]
Answer: 2
Explanation:
Analysis of Assertion A:This statement is Correct. In a MOSFET, when the drain-to-source voltage (VDS) reaches the overdrive voltage (VGS - Vth), the inversion layer at the drain end disappears; this is called "pinch-off." Despite the channel being pinched off, current continues to flow because electrons are swept across the depletion region by the high electric field. This is the definition of the Saturation Region.
Analysis of Reason R:This statement is also Correct. This phenomenon is known as the Body Effect. When a voltage is applied to the substrate (body) relative to the source, it changes the amount of charge in the depletion layer, which directly modifies the threshold voltage (Vth) required to create a conducting channel.
Relationship:While both statements are scientifically true, the fact that substrate bias affects the threshold voltage (Reason R) does not explain why current flows after pinch-off (Assertion A). The pinch-off current flow is a result of carrier transport physics in the saturation region, whereas Reason R describes the control of the channel's turn-on point.
87) Given below are two statements
Statement I: The scaling theorem relates scale changes in frequency domain to the consequent changes in scale in the time domain.
Statement II: The scaling theorem is not applicable to an impulse signal.
In light of the above statements, choose the correct answer from the options given below
1. Both Statement I and Statement II are true [Option ID = 3177]
2. Both Statement I and Statement II are false [Option ID = 3178]
3. Statement I is true but Statement II is false [Option ID = 3179]
4. Statement I is false but Statement II is true [Option ID = 3180]
Answer: 1
Solution
The Scaling Theorem (or Time-Scaling Property) states that if a signal is compressed in time, its frequency spectrum expands, and vice-versa. This is known as the reciprocal relationship between time and frequency.
Then x(at) ↔ (1/|a|) * X(ω/a)
88) Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R
Assertion A: A reflex klystron is a single cavity high power generator of efficiency 90%.
Reason R: The loop gain is unity with a phase shift of multiple of 2π.
In light of the above statements, choose the most appropriate answer from the options given below
1. Both A and R are correct and R is the correct explanation of A [Option ID = 3181]
2. Both A and R are correct but R is NOT the correct explanation of A [Option ID = 3182]
3. A is correct but R is not correct [Option ID = 3183]
4. A is not correct but R is correct [Option ID = 3184]
Answer: 4
Solution
Statement 2: This is the Barkhausen criterion for sustained oscillations:
and
where
89) Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R
Assertion A: Major advantage of DRAM is that the capacitor cell never loses its charge and hence there is no need to refresh periodically.
Reason R: When DRAM is being refreshed, the data cannot be accessed.
In 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 [Option ID = 3185]
2. Both A and R are true but R is NOT the correct explanation of A [Option ID = 3186]
3. A is true but R is false [Option ID = 3187]
4. A is false but R is true [Option ID = 3188]
Answer: 4
Detailed Explanation:
DRAM (Dynamic RAM) stores data in capacitors. Capacitors naturally leak charge over time. To prevent data loss, the memory controller must perform a Refresh Cycle typically every 64ms or less. This is actually a major disadvantage compared to SRAM.
During a refresh operation, the specific memory bank or row is locked. Because the refresh process requires reading and rewriting the data to the capacitor, the data bus and internal logic are busy, making the memory temporarily unavailable for CPU access.
90) Given below are two statements
Statement I: By bringing together the computational capability of microelectronics with the perception and control capabilities of microminiaturized sensors and actuators, MEMS technologies are enabling smart systems on a chip to be mass produced.
Statement II: While considering IoT deployment, data integrity and authenticity are no more security concerns at the sensor node level.
In light of the above statements, choose the correct answer from the options given below
1. Both Statement I and Statement II are true [Option ID = 3189]
2. Both Statement I and Statement II are false [Option ID = 3190]
3. Statement I is true but Statement II is false [Option ID = 3191]
4. Statement I is false but Statement II is true [Option ID = 3192]
Answer: 3
91. The drain current density including both drift and diffusion component is given by:
Answer :- 4
Solution
Drift Current: Drift current is caused by electric fields. The direction of the drift current is always in the direction of the electric field. The drift current density is given by
Diffusion Current: Diffusion current is caused by variation in the carrier concentration and it's direction is from a higher concentration region to a lower concentration region.
The drain current density including both drift and diffusion component is given by
92. The threshold voltage under idealised conditions is:
Correct Answer :- 3
- 2ψB: The surface potential needed to reach "strong inversion."
- The Square Root Term: The voltage drop across the oxide layer required to balance the fixed depletion region charge.
93. The channel resistance of an elemental section dy is
Correct Answer :- 2
Resistance \( R = \frac{L}{\sigma A} \)
\( L = dy \), \( A = Z \cdot t \)
\( \sigma = q n \mu_n \)
\( |Q_n| = q n t \)
The Fundamental: The channel resistance must depend on the inversion charge density \(|Q_n(y)|\). Without this term, the gate would have no control over the transistor's resistance.
94. In n-channel MOSFET, the charge induced in the inversion layer per unit area at a distance y from the source is:
Correct Answer :- 2
Fundamental Solution: Inversion Charge
- VG − ψs: The actual voltage drop across the oxide layer.
- −[VG − ψs] Ci: The total semiconductor charge required to balance the gate.
- − QB(y): Subtracting the depletion charge leaves only the mobile inversion charge.
95. The one dimensional Poisson equation for the surface space charge region at the drain is:
The Fundamental: Poisson's Equation relates electrostatic potential (Ψ) to net charge density (ρ) as: ∇2Ψ = −ρ / ε
Brief Logic: In the semiconductor, the net charge density ρ is the sum of positive charges minus negative charges:
ρ = q(ND+ + p − NA− − n)
Substituting this into the 1D Poisson equation gives the negative-sign expression in Option 1.
96. Which of the following statements is NOT correct for a Chebyshev Filter?
1. Chebyshev filter tends to exhibit ringing effect with transient signals. [Option ID = 3213]
2. Time delay and phase characteristics of Chebyshev filter are comparitively better than that of Butterworth filter. [Option ID = 3214]
3. Chebyshev filters have a sharper slope than Butterworth filters. [Option ID = 3215]
4. Chebyshev filters are capable of achieving more attenuation in stopband. [Option ID = 3216]
Correct Answer : B
Solution
Chebyshev filters achieve a sharper cutoff by allowing ripple (usually in the passband for Type I).This sharper response comes at the cost of poorer phase linearity and more variation in group delay than a Butterworth filter.
Therefore, Butterworth filters generally have better phase and time-delay characteristics.
97. Which of the following statements is NOT correct about Butterworth filter?
1. Butterworth is the most popular alignment type. [Option ID = 3217]
2. Butterworth is characterised by its moderate amplitude and phase response. [Option ID = 3218]
3. It exhibits the slowest roll‐off of any monoatomic (single‐slope) filter. [Option ID = 3219]
4. This is the only filter whose 3dB down frequency equals its critical frequency (f =f ). [Option ID = 3220]
Correct Answer : 3
Solution
Butterworth filter has a slower roll-off compared to the Chebyshev or Elliptic filters, it does not have the slowest roll-off of all monotonic filters. The Bessel filter actually has a much slower roll-off than the Butterworth filter for the same order (n). The Butterworth is "middle-of-the-road" in terms of roll-off speed.
98. A filter is needed to remove induced 60 Hz hum from a transducer signal. The rejection bandwidth is 2 Hz. The correct combination of R damping and C values for this filter are:
1. 30 Ω, 32.5 milli Farads [Option ID = 3221]
2. 68 Ω, 30.5 milli Farads [Option ID = 3222]
3. 82 Ω, 3.25 milli Farads [Option ID = 3223]
4. 89 Ω, 2.65 milli Farads [Option ID = 3224]
Correct Answer : 4
Solution
To remove the 60 Hz frequency, we use a notch filter with a center frequency (f0) of 60 Hz and a rejection bandwidth (BW) of 2 Hz.
The Q factor determines the sharpness of the notch:
Using the damping relationship for this specific filter topology:
Using the frequency component relationship:
Conclusion:
The correct combination is 89 Ω and 2.65 milli Farads.
99. The characteristic equation for the output voltage of an All-pass filter is given by:
Solution
Based on the question, the correct characteristic equation is Option 1:
Comparison: How others look
To identify filters quickly, look at the numerator (the top part of the fraction):
100. Which of the following statements are correct for active filters?
- A. Multiple-feedback are preferred for Q ≤ 10.
- B. For Q > 10, state-variable filter is used.
- C. The upper and lower cut-off frequencies have arithmetic symmetry about centre frequency fc for smaller Q values.
- D. As Q is increased, the filter becomes less selective and both upper and lower cut-off frequencies approximate geometric symmetry.
Choose the correct answer from the options given below:
- Statement A: Correct. Multiple-feedback (MFB) circuits are simple but get unstable if the "Q" (sharpness) is too high. They are best for Q ≤ 10.
- Statement B: Correct. State-variable filters use more Op-Amps to stay stable at high Q values (Q > 10).
- Statement C: Incorrect. Arithmetic symmetry is an approximation used for High Q, not low Q.
- Statement D: Incorrect. Higher Q means MORE selectivity, not less.