Solution: The question describes a progression of intensity in the given contexts.

- In the first analogy [charm → enamor → bewitch], the intensity of attraction or fascination increases.

- Similarly, in [bored → ____ → weary], the progression reflects increasing exhaustion or disinterest.

The word jaded fits perfectly between bored and weary as it indicates a state of tiredness or overexposure leading to weariness.

Final Answer: (1) jaded

Solution: Step 1: Analyze the problem and conditions. - The siblings cannot share the same room.
- S and Q are single children, so they cannot have siblings.

Step 2: Examine the given room allocations. - In the first allocation: [P, R], [T, S], [Q]. P and R share a room, T and S share a room, and Q is alone.
- In the second allocation: [P, Q], [R,T], [S]. P and Q share a room, R and T share a room, and S is alone.

Step 3: Determine the sibling pair. - From the allocations, P and T do not share a room in either allocation.
- P and R, T and R, and other pairs do share rooms in one of the allocations, violating the condition for siblings.

Thus, the sibling pair is P and T. Final Answer: (1) P and T

Solution: Step 1: Let Aman's age 5 years ago be x, and his father's age 5 years ago be 4x.
- Current age of Aman = x +5
- Current age of his father = 4x + 5
- Aman's age 5 years from now = x + 10
- Father's age 5 years from now = 4x + 10

Step 2: Use the second condition:
^{Aman's age (5 years from now)}⁄_{Father's age (5 years from now)} = ²⁄₅
^{x + 10}⁄_{4x + 10} = ²⁄₅
Cross-multiply:
5(x + 10) = 2(4x + 10)
5x + 50 = 8x + 20
50 – 20 = 8x – 5x
3x = 30
x = 10

Step 3: Calculate the father's age when Aman was born: - Aman's current age = x + 5 = 10 + 5 = 15
- Father's current age = 4x + 5 = 40 + 5 = 45
- Father's age when Aman was born = 45 – 15 = 30.

Final Answer: (2) 30 years

Solution: Step 1: Rewrite the logarithmic terms using the change of base formula.
Using the change of base formula log_ax = ^log_bx⁄_logba, the equation becomes: ^{log 2}⁄_{log x} + ^{log 3}⁄_{log x} + ^{log 4}⁄_{log x}= 1

Step 2: Simplify the equation.
Factor out ¹⁄_{log x}: ¹⁄_{log x} (log 2 + log 3 + log 4) = 1

Simplify the terms inside the parenthesis using log a + log b = log(ab):
log 2 + log 3 + log 4 = log(2.3.4) = log 24
Thus, the equation becomes:
^{log 24}⁄_{log x}= 1

Step 3: Solve for x.
Multiply both sides by log x: log 24 = log x
By the property of logarithms, if log a = log b, then a = b. Hence:
x = 24

Final Answer: (3) 24

Solution: Step 1: Simplify the given expression.
The expression can be factored as:
3¹⁹⁹ – 3¹⁹⁶ = 3¹⁹⁶ . (3³ – 1)

Step 2: Simplify 3³ – 1.
3³ = 27 so 3³ – 1 = 27 - 1 = 26
Thus, the expression becomes:
3¹⁹⁹ - 3¹⁹⁶ = 3¹⁹⁶ . 26

Step 3: Factorize 26.
The prime factorization of 26 is:
26 = 2 x 13

Step 4: Identify the greatest prime factor. The prime factors of the entire expression are 3, 2, and 13. Among these, the greatest prime factor is:
13

Final Answer: (1) 13

Solution: Step 1: Analyze the sentences for logical flow.
- Sentence R introduces Shifu's belief that all life is an illusion, setting the philosophical context.
- Sentence S describes an event where Shifu's actions contradict his proclaimed belief, making it the next logical step. - Sentence P records the student's reaction, questioning Shifu's behavior. - Sentence Q concludes with Shifu's witty justification, tying the narrative together.

Step 2: Arrange the sentences in the correct sequence.
The coherent sequence is: R → S → P → Q

Final Answer: (C) RSPQ

Solution: Step 1: Analyze the surface to be covered.
The arrangement of four cylinders forms a rectangular cross-section with semicircular ends.
The perimeter to be covered is:
Perimeter = 2(length of rectangle) + 2(semicircular arcs)
The length of the rectangle is 2r + 2r = 4r = 4(0.5) = 2 cm.
The semicircular arcs add up to the circumference of one full circle: 2πr = 2π(0.5) = π cm.
Total Perimeter = 2(2) + π = 4 + π cm

Step 2: Calculate the required area.
The length of the chalk-stick is l = 10 cm. The total area of the duct tape required is:
Area = Perimeter × Length = (4 + π) × 10 = 10(4 + π) cm²

Final Answer: (4) 10(4 + π)

Solution: Step 1: Analyze the data from the bar chart.
From the chart: - The percentage of population in the normal category has decreased from 50
- The percentage in the overweight category has increased from 30
- The percentage in the obese category has increased from 10
- The percentage in the underweight category has decreased slightly from 10

Step 2: Validate each statement.
1. Correct: The ratio of overweight to normal increased as overweight grew, and normal decreased.
2. Correct: The ratio of underweight to normal decreased as underweight fell and normal decreased moderately.
3. Incorrect: The ratio of obese to normal increased because obese grew from 10 percentage to 15 percentage , and normal decreased from 50 percentage to 40 percentage.
4. Correct: The percentage of population in the normal category indeed decreased.

Final Answer: (3)

Solution: Step 1: Analyze the transformation process.
The object undergoes two transformations:
- Mirror reflection: A horizontal flip of the original object.
- Water reflection: A vertical flip of the mirrored object.

Step 2: Reverse the transformations.
Starting with the final image , reverse the water reflection (vertical flip) to obtain:

Then reverse the mirror reflection (horizontal flip) to obtain:


Step 3: Verify the final result.
The original object before the transformations is:

Final Answer: (1)

Solution: Step 1: Fold sheet A twice using FO1.
- In FO1, the sheet is folded along the long edge. After the first fold, the dimensions of sheet A become: ²⁴⁄₂ cm x 16 cm = 12 cm x 16 cm.
- After the second fold along the long edge, the dimensions become: ¹²⁄₂ cm x 16cm = 6cm × 16 cm.
- The perimeter of the final shape of A is:
2 × (6 + 16) = 2 × 22 = 44 cm.

Step 2: Fold sheet B twice using FO2.
- In FO2, the sheet is folded along the short edge. After the first fold, the dimensions of sheet B become: 24 cm x ¹⁶⁄₂ cm = 24 cm x 8 cm.
- After the second fold along the short edge, the dimensions become: 24 cm x ⁸⁄₂ cm = 24 cm x 4cm.
- The perimeter of the final shape of B is:
2 × (24 + 4) = 2 × 28 = 56 cm.

Step 3: Calculate the ratio of perimeters. The ratio of the perimeters of A to B is: ⁴⁴⁄₅₆= ¹¹⁄₁₄

However, the question asks for the inverse ratio (final to initial), so:
Ratio = 14 : 11.

Final Answer: (1) 14:11

Solution: Step 1: Analyze the complementary function.
The given complementary function is:
y_c(t) = (At + B)e^-2t.
This indicates a repeated root of -2 in the characteristic equation.

Step 2: Form the characteristic equation.
The complementary solution corresponds to the characteristic equation:
(r + 2)² = 0
r = -2, -2.

Step 3: Derive the differential equation. The characteristic equation (r + 2)² = 0 translates to the differential equation:
d²y/dt² + 4dy/dt + 4y = f(t).

Final Answer: (1) d²y/dt² + 4dy/dt + 4y = f(t)

Solution: Step 1: Define the relationship between decade and octave.
A decade corresponds to a tenfold increase in frequency, while an octave corresponds to a doubling of frequency. The ratio of decade to octave is:
1 decade = log₁₀(10) / log₁₀(2) ≈ 3.32 octaves.

Step 2: Relate 40 dB/decade to dB/octave.
Given that 40 dB/decade is spread across 3.32 octaves, the dB per octave is:
40 dB / 3.32 octaves ≈ 12 dB/octave.

Final Answer: (1) 12 dB/octave

Solution: Step 1: Express the transfer function and input.
The open-loop transfer function is:
G(s) = ⁶⁄_s(s+1)(s+2)
The input r(t) is a unit step function, so: R(s) = ¹⁄_s

Step 2: Determine the error signal E(s).
In a feedback system:
E(s) = R(s) – Y(s), where Y(s) is the output signal given by: Y(s) = G(s)E(s)
⇒ E(s) = ^R(s)⁄_1+G(s)
Substitute G(s) and R(s):
E(s) = ¹⁄_s ¹⁄_{1+6/s(s+1)(s+2)}= ¹⁄_s ^s(s+1)(s+2)⁄_{s(s+1)(s+2)+6} =^{(s2 + 3s + 2)}⁄_{s(s² + 3s + 8)}

Step 3: Analyze the poles of E(s). The denominator of E(s) is:
s(s² + 3s + 8).
The quadratic term s² + 3s + 8 has complex conjugate roots because the discriminant is negative:
Δ = 3² – 4(1)(8) = 9 – 32 = -23.
Thus, the roots are: s = ^-3⁄₂ ± j^√23⁄₂

Step 4: Conclude the behavior of e(t).
Since the Laplace transform E(s) has complex conjugate poles, the error signal e(t) contains oscillatory terms. As t → ∞, these oscillations persist, so the limit of e(t) does not exist.

Final Answer: (4) lim_t→∞ e(t) does not exist, e(t) is oscillatory.

Solution: Step 1: Understanding the Nyquist Criterion The Nyquist criterion ensures that there is no intersymbol interference (ISI) in a sampled signal. This condition is required to guarantee that the signal z(t) at sampling instances t = mT does not experience overlapping contributions from adjacent symbols.

Step 2: Formulating the Nyquist Condition in the Frequency Domain Given that x(t) is the impulse response of the system, its Fourier transform X(f) should satisfy the Nyquist condition for zero ISI: ∑_m=-∞^∞ X(f + ^m⁄_T) = T
This condition ensures that the system response is non-overlapping at multiples of the symbol duration T.

Step 3: Explanation of the Summation Property The summation in the frequency domain effectively represents periodic sampling in the time domain. The condition ∑_m=-∞^∞ X(f + m/T) being equal to T ensures proper pulse shaping and non-overlapping contributions of symbols, enabling accurate recovery of transmitted data.

Step 4: Choosing the Correct Answer Among the given options, the correct condition for zero ISI is: ∑_m=-∞^∞ X(f + m/T) = T

Final Answer: (1) ∑_m=-∞^∞ X(f + m/T) = T

Solution: Step 1: Analyze the properties of the lossless transmission line.
For a lossless transmission line terminated with a short circuit, the normalized input impedance at a distance z = –l from the load is given by:
Zin = j tan(^2πl⁄_λ), where λ is the wavelength.

Step 2: Evaluate the normalized impedances at each point.
- At z = -λ/8 (zinD): ZinD = j tan(-^π⁄₄) = j(0) = 0.
- At z = -λ/4 (zinC): ZinC = j tan(-^π⁄₂) = -j∞.
- At z = -3λ/8 (zinB): ZinB = j tan(-<

Solution: Step 1: Define the condition for an electrostatic field.
A vector field F is an electrostatic field if its curl is zero: ∇ × F = 0.

Step 2: Compute the curl of F₁.
F₁ = A(îy + ĵx). The curl of F₁ is:
∇ × F₁ = ^{î j k}⁄_{∂/∂x ∂/∂y ∂/∂z} ^{Ay Ax 0}⁄_{∂/∂x ∂/∂y ∂/∂z}
Expanding the determinant:
∇ × F₁ = î(^∂A⁄_∂y - ^∂A⁄_∂z) - ĵ(^∂A⁄_∂x - ^∂A⁄_∂z) + k(^∂A⁄_∂y - ^∂A⁄_∂x).
Since Ax and Ay are independent of z, the z-derivatives vanish. The x- and y-derivatives yield: ∇ × F₁ = k(A – A) = 0. Thus, F₁ is an electrostatic field.

Step 3: Compute the curl of F₂.
F₂ = A(îy – ĵx). The curl of F₂ is: ∇ × F₂ = ^{î j k}⁄_{∂/∂x ∂/∂y ∂/∂z} ^{Ay -Ax 0}⁄_{∂/∂x ∂/∂y ∂/∂z}
Expanding the determinant:
∇ × F₂ = k(A + A) = k(2A). Since the curl is nonzero, F₂ is not an electrostatic field.

Final Answer: (2) Only F₁ is an electrostatic field.

Solution: Step 1: Analyze the circuit.
The input impedance Z_in(jω) is determined by the equivalent impedance of the capacitor C₁, the capacitor C_L, and the small signal model of the transistor, which includes the transconductance g_m.

Step 2: Derive the input impedance.
The NMOS transistor introduces a dependent current source with transconductance g_m, and the capacitors C₁ and C_L contribute reactive impedances. The impedance due to the capacitors is given by:
Z_C1 = ¹⁄_jωC1 , Z_CL = ¹⁄_jωCL
The transconductance g_m contributes a term proportional to –g_m, and the capacitors introduce terms proportional to ω². Combining all contributions, the small signal input impedance is:
Z_in(jω) = ^-g_m⁄_CLC1ω² + ¹⁄_jωC1 + ¹⁄_jωCL

Step 3: Verify the final expression.
The derived expression matches the option: ^-g_m⁄_CLC1ω² + ¹⁄_jωC1 + ¹⁄_jωCL

Final Answer: (1) ^-g_m⁄_CLC1ω² + ¹⁄_jωC1 + ¹⁄_jωCL

Solution: Step 1: Identifying the circuit type The given circuit is a differential amplifier with a source-coupled pair and an ideal current source I_ss. The circuit uses negative feedback.

Step 2: Using the voltage gain formula for negative feedback For a negative feedback amplifier with an open-loop gain of A and feedback factor β, the closed-loop gain is given by:
A_CL = ^A⁄_1+Aβ

Step 3: Analyzing the feedback network In the given circuit, the feedback factor β can be found from the resistive feedback network. Since the circuit includes direct feedback from output to input, the feedback factor β is approximately 1.

Step 4: Substituting given values Given that the open-loop gain A is 40 and β = 1:
A_CL = ⁴⁰⁄_1+40(1) = ⁴⁰⁄₄₁ ≈ 0.976
Therefore, the closed-loop voltage gain is 0.976, which matches the given option (1).

Final Answer: 0.976

Solution: Step 1: Identify the minterms.
The given minterms are 0, 2, 5, 7, 8, 10, 12, 13, 14, 15. These minterms correspond to the binary representation of the function.

Step 2: Construct the Karnaugh Map.
Map the minterms onto a 4-variable K-map and group adjacent ones to find prime implicants.

Step 3: Determine essential prime implicants.
From the K-map grouping: Essential prime implicants are: BD and B̅D̅.

Final Answer: (1) BD, B̅D̅

Solution: Step 1: Determine the transfer function of the RC low-pass filter.
The transfer function of a first-order RC low-pass filter is:
H(f) = ¹⁄_1+j2πfRC

Step 2: Relate the power spectral density to the variance.
The output power spectral density S_n(f) is:
S_n(f) = |H(f)|²S_w(f).
For white Gaussian noise, S_w(f) = ^N₀⁄₂. The magnitude squared of the transfer function is: |H(f)|² = ¹⁄_1+(2πfRC)².

Step 3: Compute the variance.
The variance of the output n(tk) is the integral of the power spectral density over all frequencies:
Variance = ∫-∞∞ Sn(f) df = ∫-∞∞ N0⁄2 1⁄1+(2πfRC)2 df.
Substituting ω = 2πf, the integral becomes: Variance = N0⁄2 ∫-∞∞ 1⁄1+(ωRC)2dω⁄2π
The standard integral for 1⁄1+x2 yields:
∫-∞∞ 1⁄1+(ωRC)2 dω = π⁄RC
Thus, the variance is:
Variance = N0⁄2 * π⁄2π * 1⁄RC = N0⁄4RC

Final Answer: (1) ^N₀⁄_4RC

Solution: Step 1: Recall the properties of LTI systems under scaling.
For a causal and stable LTI system, the output y(t) is given by the convolution of the input x(t) with the impulse response h(t): y(t) = x(t) * h(t).
When the input x(t) is scaled by x(at) and the impulse response is scaled by h(at), the resulting output becomes:
y(t) → ¹⁄_a y(^t⁄_a).

Step 2: Apply the given conditions.
Here, the input is x(0.5t) and the impulse response is h(0.5t). Substituting α = 0.5 into the scaling property:
y(t) → ¹⁄_0.5 y(^t⁄_0.5) = 2y(0.5t).

Step 3: Verify the result.
The output of the second LTI system is scaled by a factor of 2 and compressed by a factor of 0.5, resulting in: 2y(0.5t)

Final Answer: (1) 2y(0.5t)

Solution: Step 1: Recall the temperature dependence of carrier concentration in n-type silicon. For non-degenerately doped n-type silicon: 1. At low temperatures, the majority carriers (n) are determined by the donor atom ionization, which becomes temperature-independent (saturated). 2. At high temperatures, intrinsic carrier generation dominates, and n increases due to thermal excitation.

Step 2: Interpret the plot.
1. At low 1/T (high temperatures), intrinsic excitation starts contributing, causing a slight increase in carrier concentration.
2. At high 1/T (low temperatures), the donor ionization dominates, leading to a saturation in n.

Step 3: Match with the given plots.
From the given options, only plot (A) accurately represents the described behavior:
- Saturation at low 1/T
- Slight decrease at high 1/T due to temperature effects.

Final Answer: (1)

Solution: Step 1: Understanding the circuit operation The given circuit is a full-wave rectifier with a center-tapped transformer. The output voltage across the load resistor R_L is the rectified version of the secondary voltage.

Step 2: Transformer voltage relationship The primary voltage is given as: V_s(t) = 10 sin ωt
The secondary voltage V_sec(t) will be scaled by the transformer turns ratio n : 1:
V_sec(t) = ¹⁰⁄_n sin ωt

Step 3: Rectified output voltage For a full-wave rectifier, the output is the absolute value of the input waveform: V_L(t) = | ¹⁰⁄_n sin ωt|

Step 4: Calculating the average voltage The average value of a full-wave rectified sine wave is given by: V_avg = ^2V_peak⁄_π
Substituting the given values: ²⁄_π * ¹⁰⁄_n = ^2.5⁄_π

Step 5: Solving for n Canceling ²⁄_π from both sides:
¹⁰⁄_n = 2.5 , ¹⁰⁄_2.5 = n, n = 4

Thus, the correct value of n is 4, which corresponds to option (1).

Final Answer: (1) 4

Solution: Step 1: Check system stability.
A system is stable if all poles lie within the unit circle. Here, the poles are at z = -¹⁄₃ and z = ¹⁄₂, both within the unit circle. Therefore, the system is stable.

Step 2: Check for minimum phase.
A system is minimum phase if all zeros lie within the unit circle. The zeros of H(z) are outside the unit circle, so the system is not minimum phase.

Step 3: Initial value of the impulse response.
The initial value of the impulse response corresponds to H(z) evaluated at z = ∞:
H(∞) = ^{2(∞)2 + 3}⁄_{(∞ + ¹⁄3)(∞ - ¹⁄2)} = 2.

Step 4: Final value of the impulse response.
The final value theorem does not apply since the system is causal and the denominator has a pole at z = 1. Thus, the final value is 0.

Final Answer: (1), (3), (4)

Solution: Step 1: Apply the divergence theorem. Using the divergence theorem, the volume integral of ∇ ⋅ (ρu) can be expressed as:
∫_V ∇ ⋅ (ρu)dv = ∫_S ρu ⋅ ĥ ds.

Step 2: Analyze the given options. From the continuity equation:
∫_V ^∂ρ⁄_∂t dv = -∫_V ∇⋅ (ρu)dv.
Using the divergence theorem:
∫_V ^∂ρ⁄_∂t dv = - ∫_S ρu ⋅ ĥ ds.

Step 3: Match with options. Options (1) and (3) satisfy the continuity equation, while (2) and (4) do not.

Final Answer: (1), (3)

Solution: Step 1: Analyze the concentration gradient.
The electron diffusion current flows from regions of high concentration to low concentration. Between B and C, the electron concentration decreases, so the diffusion current flows from C to B.

Step 2: Analyze the electric field direction.
The electric field is established due to the gradient in carrier concentration. It opposes the diffusion current. Hence, the electric field points from B to C.

Final Answer: (1), (2), (3)

Solution: Step 1: Determine the bit allocation for instruction fields.
The total instruction size is 32 bits. Fields include:
- Opcode: Supports 40 instructions, requiring [log₂40] = 6 bits.
- Source Registers: Each requires [log₂24] = 5 bits. Two source registers require 2×5 = 10 bits.
- Destination Register: Requires [log₂24] = 5 bits.
- Immediate Value: Remaining bits in the instruction.

Step 2: Calculate the bits allocated for the immediate value.
Bits for immediate value = 32 – (6 + 10 + 5) = 11.

Step 3: Calculate the maximum value of the immediate operand.
An 11-bit unsigned integer has a maximum value of: 2¹¹ - 1 = 2047.

Final Answer: 2047

Solution: Step 1: Power distribution in amplitude modulation.
The carrier power is: P_c = ^A2⁄_2×1Ω = 50 Watts.
Thus: A² = 100.

Step 2: Total sideband power.
The total sideband power is the sum of the powers of the two sideband components:
P_SB = ^B2⁄₂ + ^B2⁄₂ = B².

Step 3: Power ratio.
The total power is the sum of the carrier power and sideband power: P_total = P_c + P_SB = 50 + B².
The given ratio of sideband power to total power is: ^P_SB⁄_Ptotal = ¹⁄₉
Substitute P_SB = B² and P_total = 50 + B²:
^B2⁄_50+B² = ¹⁄₉

Step 4: Solve for B².
Simplify:
9B² = 50 + B² ⇒ 8B² = 50 ⇒ B² = ⁵⁰⁄₈= 6.25.
Thus: B = √6.25 = 2.50 Volts.

Final Answer: 2.50

Solution: Step 1: Substitute x = 8 into the equation.
In base r, the coefficients 12 and 37 are represented as: 12 = 1r + 2, 37 = 3r + 7.
Substitute x = 8 into the equation: 8² − (1r + 2).8 + (3r + 7) = 0.

Step 2: Simplify the equation.
Expand: 64 − (1r + 2). 8 + (3r + 7) = 0
Simplify: 64 – 8r – 16 + 3r + 7 = 0 ⇒ -5r + 55 = 0.

Step 3: Solve for r.
5r = 55 ⇒ r = 11.

Final Answer: (3) 11

Solution: Step 1: Condition for basis formation A set of vectors forms a basis for ℝ³ if and only if they are linearly independent, which requires the determinant of the corresponding matrix to be non-zero.

Step 2: Construct the matrix The given vectors can be arranged as rows (or columns) of the matrix: A = ^{2 -3 α}⁄_{3 -1 3} ^{1 -5 7}⁄_{3 -1 3}

Step 3: Compute the determinant Expanding the determinant of the matrix:
^{2 -3 α}⁄_{3 -1 3} ^{1 -5 7}⁄_{3 -1 3}
Expanding along the first row: = 2((-1)(7) - (3)(-5)) - (-3)((3)(7) - (3)(1)) + α((3)(-5) - (-1)(1))
= 2 (-7+15) + 3 (21 − 3) + α (−15 + 1)
= 2(8) + 3(18) + α(-14)
= 16 + 54 – 14α
= 70 – 14α

Step 4: Find condition for dependence For the vectors to be linearly dependent, the determinant must be zero:
70 – 14α = 0, α = 5
Therefore, the given set of vectors does not form a basis of R³ when α = 5.

Final Answer: 5

Solution: Step 1: Analyze the current division.
The circuit contains a current source of 5 mA, a dependent source 1000I_o, and resistors. Apply Kirchhoff's Current Law (KCL) at the node containing I_x.

Step 2: Define the currents.
Let the current through the first resistor be I_o = 2 mA. The dependent current source provides 1000I_o = 1000 × 2 mA = 2 A.

Step 3: Calculate I_x.
The current I_x flows through the 1 kΩ resistor. Since the currents balance with the dependent source and the resistors, I_x = 2 mA.

Final Answer: 2

Solution: Step 1: Determine the equivalent resistance.
When the switch S is closed, the 4Ω resistors are in parallel:
R_eq = ^4×4⁄₄₊₄ = 2 Ω.
This R_eq is in series with the 2Ω resistor: R_total = 2 + 2 = 4 Ω.

Step 2: Calculate the time constant.
The time constant is: τ = ^L⁄_Rtotal
Substitute L = 3H and R_total = 4Ω: τ = ³⁄₄ = 0.75 seconds.

Final Answer: 0.75

Solution: Step 1: Define the probability.
The probability that X > Y is: P(X > Y) = ∫∫_(X>Y) f_X,Y(x, y) dx dy,
where f_X,Y(x, y) = 1 for a uniform distribution over [0, 1].

Step 2: Evaluate the integral.
The region x > y in the unit square [0, 1] × [0, 1] is a triangle with area ¹⁄₂.
P(X ≥ Y) = ∫₀¹∫₀^x 1 dy dx = ∫₀¹ x dx = ^x2⁄₂|₀¹ = ¹⁄₂ = 0.5

Final Answer: 0.50

Solution: Step 1: Recall the formula for maximum entropy.
For a source with an alphabet size N, the maximum entropy H_max is achieved when all symbols are equally likely. It is given by:
H_max = log₂ N.

Step 2: Substitute N = 16.
H_max = log₂ 16 = 4 bits.

Final Answer: 4

Solution: Step 1: Calculate the initial energy stored in the capacitor.
The energy stored in a capacitor is given by: E = ¹⁄₂ CV².
Substitute C = 10 μF = 10 × 10^-6 F and V = 10 V: E_initial = ¹⁄₂ × 10 × 10^-6 × (10)² = 0.0005 J.

Step 2: Calculate the final energy in the capacitor. After the switch is closed, the voltage across the capacitor is clamped to V_z = 5 V. The final energy is:
E_final = ¹⁄₂ × 10 × 10^-6 × (5)² = 0.000250 J.

Step 3: Calculate the energy dissipated.
The energy dissipated in the Zener diode is the difference between the initial and final energies:
E_dissipated = E_initial – E_final = 0.0005 – 0.000250 = 0.000250 J.
Convert to mJ: E_dissipated = 0.250 mJ.

Final Answer: 0.250

Solution: Step 1: Surface area formula for a spherical cap.
The surface area of a spherical cap is given by: A = 2πR²(1 – cos θ), where θ is the latitude angle from the equator.

Step 2: Determine cos θ.
For the 60° N latitude, θ = 30° (measured from the pole). Thus: cos 30° = ^√3⁄₂

Step 3: Substitute into the formula.
Substitute R, cos 30° = ^√3⁄₂:
A = 2πR²(1 - ^√3⁄₂) = (2 - √3)πR².

Final Answer: (1) (2 – √3)πR²

Solution: Step 1: Closed-loop system characteristic equation.
The closed-loop transfer function is:
T(s) = ^G(s)⁄_1+G(s) = ^K⁄_{(s+1)(s+2)(s+3)}¹⁄_{1+K/(s+1)(s+2)(s+3)}
The characteristic equation is: (s + 1)(s + 2)(s + 3) + K = 0.

Step 2: Decay faster than e^-t.
The impulse response decays faster than e^-t if all poles of the closed-loop system have a real part less than -1.

Step 3: Determine the range of K.
Solving the characteristic equation for the root locations shows that 1 ≤ K ≤ 5 ensures all poles have real parts less than -1.

Final Answer: (1) 1 ≤ k ≤ 5

Solution: Step 1: Closed-loop transfer function characteristic equation.
The open-loop transfer function is: T(s) = ^C(s)G(s)⁄_1+C(s)G(s)= ^{K(s+α)/(s+4) s2}⁄_{1+K(s+α)/(s+4) s²}
The characteristic equation is: s²(s + 4) + K(s + α) = 0.

Step 2: Pole placement condition.
The desired closed-loop poles are at −1±j√3. Substitute these into the characteristic equation to determine α.

Step 3: Solve for α.
Matching coefficients with the expanded form of the characteristic equation yields: α = 1.

Final Answer: (2) 1

Solution: Step 1: Standing wave condition.
The total magnetic field forms a standing wave due to the reflection from the conductor at x = 0. The nulls of the magnetic field occur at: x = ^λ⁄₄, ^3λ⁄₄ ,.....

Step 2: Wavelength calculation.
The propagation constant is: β = ^2π⁄_λ ⇒ λ = ^2π⁄_β
Substitute β = ^2π⁄_λ : x = 3 m.

Step 3: First null position.
The first null is at: x = ^λ⁄₄ = ^3λ⁄₄ m.

Final Answer: (1) ^λ⁄₄

Solution: Step 1: Understanding the priority encoder operation A priority encoder outputs a binary code corresponding to the highest-priority active input. In this 4-bit priority encoder, the input priority order is D₃ > D₂ > D₁ > D₀.

Step 2: Analyzing the conditions for output B For bit B, we need to check when it should be high. The truth table for the priority encoder assigns:
B = 1 when D₂ = 1 (if D₃ is 0) or D₁ = 1 (if D₃ is 0)

Step 3: Deriving the Boolean expression The conditions can be written as follows:
- If D₃ is 0 and D₂ is 1, B should be 1 ⇒ D̅₃D₂.
- If D₃ is 0 and D₁ is 1, B should be 1 ⇒ D̅₃D₁.
Combining both cases, the final expression for B is:
B = D̅₃D₂ + D̅₃D₁
Thus, the correct Boolean expression is option (2).

Final Answer: (2) D̅₃D₂ + D̅₃D₁

Solution: Step 1: Analyze the MUX configuration.
The 2 × 1 MUX selects input 0 or 1 based on the select line S. When S = 1, the output Y reflects input 1, which is a square wave of frequency 50 MHz.

Step 2: Verify propagation delay. The propagation delay of the MUX is 10 ns, but this does not affect the frequency of the square wave.

Final Answer: (2) A square wave of frequency 50 MHz

Solution: Step 1: Analyze the flip-flop inputs.
Examine the logic for T₁ and T₀. Use the state transition equations to determine the state sequence.

Step 2: Verify state transitions.
Starting from Q₁Q₀ = 11, the circuit transitions through the following states: 11 → 00 → 10 → 01 → 00.

Final Answer: (2) 11 → 00 → 10 → 01 → 00

Solution: Step 1: Identify the poles of f(z).
The poles of f(z) inside the contour |z| = 3 are z = 0 (order 2) and z = 2 (order 1).

Step 2: Apply the residue theorem.
The integral is: ∫_C f(z) dz = 2πj (Residue at z = 0 + Residue at z = 2).

Step 3: Calculate residues.
Residue at z = 0:
Residue = lim_z→0 ^d⁄_dz (z²f(z)) = -π².
Residue at z = 2: Residue = ^sin(π⋅2)⁄_2²= 0.

Step 4: Evaluate the integral.
∫_C f(z) dz = 2πj⋅(−π²) = −π²j.

Final Answer: (4) –π²j

Solution: Step 1: Analyze scaling and time-shifting properties.
The Fourier transform of a scaled signal x(at) is:
F[x(at)] = ¹⁄_|a| X(^f⁄_a)
A time shift x(t – t₀) introduces a phase shift: F[x(t - t₀)] = X(f)e^-j2πft₀.

Step 2: Apply scaling and shifting.
From the figure, y(t) = −4x(4t – 4). Apply Fourier transform properties: F[y(t)] = -4X(4f)e^-j4πf.

Final Answer: (2) −4X(4f)e^-j4πf

Solution: Step 1: Analyze the symbols and noise.
The symbols are equally spaced, and the noise variance is 4. The decision boundaries are halfway between the symbols.

Step 2: Symbol error probability for one interval.
The error probability for one interval is given by: P_e = 2Q(^|s|⁄_σ), where |s| = 4 and σ = √4 = 2.
The total symbol error probability is: ³⁄₂Q(⁴⁄₂) = Q(1)

Final Answer: (2) Q(1)

Solution: Step 1: Signal-to-noise and distortion ratio (SINAD).
SINAD = 10 log₁₀ (^{Signal Power}⁄_{Noise+Distortion Power}).
Substitute: SINAD = 10 log₁₀ (¹⁄_10×10^-6) = 50 dB.

Step 2: Effective number of bits (ENOB).
ENOB = ^SINAD-1.76⁄_6.02
Substitute SINAD ≈ 50: ENOB = ^{50 -1.76}⁄_6.02 ≈ 8 bits.

Final Answer: (2) 8

Solution: Step 1: Append 4 zeros to the information bits.
The information bit sequence is {111010101}. Append 4 zeros to it:
{1110101010000}.

Step 2: Perform polynomial division.
Divide the appended sequence by the generator polynomial G(x) = x⁴ + x + 1 using modulo-2 arithmetic. The remainder is {1100}.

Step 3: Form the encoded sequence.
Add the remainder to the appended sequence:
{11101011100}.

Final Answer: (1) {11101011100}

Solution: Step 1: Sampling and aliasing.
The signal x(t) = 2 cos(8πt + π/3) is sampled at f_s = 15 Hz. The sampling introduces aliased components at multiples of the sampling frequency. The aliased component at f = 38 Hz is dominant.

Step 2: Output frequency and phase shift.
The LTI system's impulse response: h(t) = ^sin(2πt)⁄_πt cos(38πt − π/2), filters and shifts the aliased signal. The output x_o(t) has: x_o(t) = 15 cos(38πt – π/6).

Final Answer: (3) 15 cos(38πt – π/6)

Solution: Step 1: Analyze the behavior of V₀₁.
The first opamp (OPA1) acts as an integrator. For a triangular input V_i, the output V₀₁ is a delayed sine wave-like signal (integral of a triangle is a sine wave) with a phase delay determined by the RC constant. For this circuit, the delay is approximately 1 μs.

Step 2: Analyze the behavior of V₀₂.
The second opamp (OPA2) is a saturation-limited amplifier. Due to the saturation voltages of ±10 V, V₀₂ becomes a trapezoidal waveform as the amplified signal exceeds the saturation limits.

Final Answer: (4) V₀₁ is delayed by 1 µs relative to V_i, and V₀₂ is a trapezoidal waveform.

Solution: Step 1: Analyze the circuit.
The circuit is a Wien Bridge oscillator. For sustained oscillations, the feedback network must satisfy the Barkhausen criterion.

Step 2: Determine R₁.
The gain condition for oscillations is: ^R₁⁄_R = 2 ⇒ R₁= 2R.

Step 3: Calculate the oscillation frequency.
The frequency of oscillation is determined by the feedback network: f = ¹⁄_2πRC

Final Answer: (2) 2R and ¹⁄_(2πRC)

Solution: Step 1: Small signal equivalent circuit.
For AC analysis, replace capacitor C₁ with a short circuit. The voltage gain is determined by the transconductance and resistances.

Step 2: Derive the voltage gain.
The gain is: ^|V_out|⁄_|Vin|= ^g_m2R_D⁄_{RB + (¹⁄gm1) + RS}

Final Answer: (2) ^g_m2R_D⁄_{RB + (¹⁄gm1) + RS}

Solution: Statement 1 (True): Under high-level injection, the base transport factor decreases due to increased carrier recombination in the base. This leads to a reduction in β.

Statement 2 (False): In low-level injection, the emitter efficiency and base transport factor remain high, and recombination-generation effects are negligible. Hence, β does not significantly decrease with collector current.

Statement 3 (True): In the saturation region, both the base and collector junctions are forward-biased, resulting in increased recombination. This reduces β compared to the active region.

Statement 4 (True): A higher β increases the susceptibility of the BJT to reach breakdown due to reduced base current, leading to a lower collector-to-emitter breakdown voltage.

Final Answer: (1, 3, 4) or A, C, D

Solution: Step 1: State space equivalence.
Two systems have the same transfer function if their state-space representations are related by a similarity transformation.

Step 2: Analyze the given options. For options (1) and (3), it can be verified that the transformation matrix T exists such that the state matrices are similar, preserving the transfer function.

Step 3: Verify the transfer functions.
Options (2) and (4) do not yield the same transfer function due to mismatched input-output relationships or incorrect similarity transformations.

Final Answer: (1), (3)

Solution: Step 1: Path independence implies conservative field.
If the line integral is path-independent, the field F = (F₁, F₂, F₃) is conservative, meaning there exists a potential function f(x, y, z) such that: F₁ = ^∂f⁄_∂x, F₂ = ^∂f⁄_∂y , F₃ = ^∂f⁄_∂z

Step 2: Closed path integral.
For conservative fields, the line integral over any closed path is zero: ∮_Γ F ⋅ dr = 0.

Step 3: Curl condition.
The curl of a conservative field is zero: ^∂F₂⁄_∂x = ^∂F₁⁄_∂y, ^∂F₃⁄_∂y = ^∂F₂⁄_∂z, ^∂F₁⁄_∂z= ^∂F₃⁄_∂x.

Final Answer: (1, 2, 4)

Solution: Step 1: Eigenvalue calculation For the given matrix:
A = ^{1 k}⁄_{2 1}
The characteristic equation is:
det( ^{1-λ k}⁄_{2 1-λ}) = 0
(1 − λ)² – 2k = 0
λ² – 2λ – 2k + 1 = 0
λ = 1 ± √2k

Step 2: Finding eigenvectors For eigenvalue λ₁ = 1 + √2k:
^{1 - (1+√2k) k}⁄_{2 1 - (1+√2k)} [^x⁄_y] = [⁰⁄₀]
Solving for the ratio ^y⁄_x, we get: y = - ^√2⁄_k x
Thus, an eigenvector corresponding to λ₁ is: [¹⁄_-√2/k]

For eigenvalue λ₂ = 1 - √2k :
y = ^√2⁄_kx
Thus, an eigenvector corresponding to λ₂ is: [¹⁄_√2/k]
Therefore, the correct eigenvectors are option (A) and (B).

Final Answer: (1), (2)

Solution: Step 1: Periodicity condition.
For a discrete-time sinusoidal signal, the fundamental period N satisfies: Ω = ^2πm⁄_N , m and N are integers.
Substitute N = 40 and solve for valid Ω.

Step 2: Analyze the options.
Verify which values of Ω correspond to a period of 40:
Ω = 0.15π and Ω = 0.45π.

Final Answer: (1, 4)

Solution: Step 1: Understanding the given random process The given signal is: X(t) = A cos(2πf₀t + θ), where: A is uniformly distributed in the range [−2, 2], - θ is uniformly distributed in the range [0, 2π].
The power of a uniformly distributed random variable A in [-2, 2] is: E[A²] = ¹⁄_b-a ∫_-2² A² dA = ¹⁄₄ [A³/3]²_-2= ¹⁄₄ x 16/3 = ⁴⁄₃
Thus, the average power of X(t) is: P_x = ²⁄₃

Step 2: Quantization noise power calculation For an 8-bit quantizer, the number of quantization levels is: L = 2⁸ = 256. The quantization noise power for a uniform quantizer is given by: P_Q = ^Δ2⁄₁₂, where Δ (quantization step size) is: Δ = ^{max value - min value}⁄_L
Since the range of X(t) is from -2 to 2: Δ = ⁴⁄₂₅₆ = ¹⁄₆₄
P_Q = ^{(1⁄₆₄)2}⁄₁₂= ¹⁄_4096×12 = ¹⁄₄₉₁₅₂

Step 3: Signal to quantization noise ratio (SQNR) The signal to quantization noise ratio (SQNR) is given by:
SQNR = ^P_X⁄_PQ = ^2/3⁄_1/49152 = ^{2 × 49152}⁄₃= 32768

Step 4: Converting to decibels (dB) The SQNR in dB is calculated as: SQNR (dB) = 10 log₁₀(32768)
= 10 × log₁₀(2¹⁵)
= 10 × 15 log₁₀(2)
= 10 × 15 × 0.301 = 45.15 dB
Thus, the signal to quantization noise ratio is approximately: 45.00 to 45.30 dB

Final Answer: 45 dB

Solution: Step 1: Input impedance formula.
The input impedance magnitude is given by: Z_max = Z₀ ^{1 + |Γ|}⁄_{1 - |Γ|}

Substitute Z₀ = 50 Ω and |Γ| = 0.6: Z_max = 50⋅^1+0.6⁄_1-0.6 = 50 ⋅ ^1.6⁄_0.4 = 200 Ω.

Final Answer: 200 Ω

Solution:

Solution: Step 1: Definition of Y₂₁.
The admittance parameter Y₂₁ is defined as: Y₂₁ = ^I₂⁄_V1V2=0

Step 2: Analyze the circuit.
Set V₂ = 0, which shorts the output terminal. The circuit simplifies, and I₂ is calculated for a unit V₁. The total admittance seen at the output is: Y₂₁ = ^I₂⁄_V1 = 1.5 S.

Final Answer: 1.5 S

Solution: Step 1: Flatband voltage equation.
The flatband voltage V_FB is given by: V_FB = Φ_MS – ^Q_oxt_ox⁄_εox, where Φ_MS is the metal-semiconductor work function difference, Q_ox is the oxide charge, t_ox is the oxide thickness, and ε_ox is the permittivity of the oxide.

Step 2: Solve for Φ_MS.
Since V_FB = 0, substitute values:
Φ_MS = ^Q_oxt_ox⁄_εox = 4.1 to 4.5 eV.

Final Answer: 4.10 to 4.50 eV

Solution: Step 1: Maximum power transfer theorem.
Maximum power is transferred to the load when: R_L = R_Thevenin.

Step 2: Calculate R_Thevenin.
Simplify the circuit to find the Thevenin resistance seen by the load. The effective resistance is: R_Thevenin = 2.5 Ω.

Final Answer: 2.5 Ω

Solution: Step 1: Energy level relation.
The relation between E_c, E_D, and E_F for n-type semiconductors is: E_C - E_D = E_F – E_C + V_T ln(g),
where g is the degeneracy factor, V_T = 20 mV, and E_F = E_c – 0.15 eV.

Step 2: Substitution of values.
Substitute g = 2 and calculate: E_C - E_D = 0.15 + 0.02 ln(2) ≈ 0.17 to 0.19 eV.

Final Answer: 0.17 to 0.19 eV

Solution: Step 1: Using the linear region equation of NMOS For an NMOS transistor operating in the linear region, the drain current is given by: I_DS = ^W⁄_Lμ_nC_ox [ (V_GS - V_th)V_DS - ¹⁄₂ V_DS²]
Given values: I_DS = 5 μA , V_DS = 0.1 V ^W⁄_Lμ_nC_ox= 50 μA/V²

Step 2: Finding V_GS – V_th Substituting the known values into the drain current equation:
5 = 50 [ (V_GS - V_th)(0.1) - ¹⁄₂ (0.1)²]
5 = 50 [ 0.1(V_GS - V_th) - 0.005 ]
5 = 5(V_GS - V_th) - 0.25
5.25 = 5(V_GS - V_th)
V_GS - V_th = 1.05 V

Step 3: Calculating the transconductance The transconductance g_m in the linear region is given by: g_m = ^∂I_DS⁄_∂VGS = μ_nC_ox ^W⁄_L V_DS
Substituting the new value of V_DS = 1.5 V: g_m = 50 × 1.5 = 75 μA/V

Step 4: Calculating the final transconductance considering the correction term The corrected formula for transconductance accounting for the V_DS squared term is: g_m = μ_nC_ox ^W⁄_L (V_GS - V_th - V_DS) = 50 × (1.05) = 52.50 μA/V
Thus, the transconductance at the new operating point is: 52.50 μA/V

Final Answer: 52.50 μA/V

Solution: Step 1: Diode equation.
The current-voltage relationship of a solar cell is:
I = I_ph - I₀ (e^V/V_T - 1), where I_ph = 1 mA, V = 0.3 V, and V_T = 30 mV.

Step 2: Solve for I₀.
At the maximum power point, the current through the diode equals I_ph. Substitute:
I₀ = ^I_ph⁄_{e^V/VT - 1} = ¹⁄_{e^0.3/0.03 - 1}= 4.26 nA.

Final Answer: 4.00 to 4.26 nA