Engineering & Architecture Admissions
Q. A family has 3 children. What is the probability that at least one child is a girl given that at least one child is a boy?
A.
1/2
B.
2/3
C.
3/4
D.
1/4
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Solution
The only combinations with at least one boy are: BBB, BBG, BGB, GBB, BGG, GBG, GGB. Out of these, all combinations except BBB have at least one girl. Thus, P(At least one girl | At least one boy) = 6/7.
Correct Answer: B — 2/3
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Q. A family has 3 children. What is the probability that at least one of them is a girl given that at least one is a boy?
A.
1/2
B.
2/3
C.
3/4
D.
1
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Solution
The only scenario where there are no girls is if all are boys. The probability of at least one girl given at least one boy is 3/4.
Correct Answer: C — 3/4
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Q. A fiber optic cable uses total internal reflection to transmit light. What is the primary requirement for this to work effectively?
A.
The core must have a higher refractive index than the cladding
B.
The cladding must have a higher refractive index than the core
C.
The light must be monochromatic
D.
The cable must be straight
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Solution
For total internal reflection to occur in a fiber optic cable, the core must have a higher refractive index than the cladding.
Correct Answer: A — The core must have a higher refractive index than the cladding
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Q. A fiber optic cable uses total internal reflection. If the refractive index of the core is 1.6 and the cladding is 1.5, what is the critical angle?
A.
38.7°
B.
41.8°
C.
48.6°
D.
60.0°
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Solution
Critical angle θc = sin⁻¹(n2/n1) = sin⁻¹(1.5/1.6) ≈ 38.7°.
Correct Answer: A — 38.7°
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Q. A fiber optic cable uses total internal reflection. If the refractive index of the core is 1.5 and that of the cladding is 1.4, what is the critical angle?
A.
42.0°
B.
48.6°
C.
60.0°
D.
30.0°
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Solution
Critical angle θc = sin⁻¹(n2/n1) = sin⁻¹(1.4/1.5) ≈ 42.0°.
Correct Answer: A — 42.0°
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Q. A fiber optic cable uses total internal reflection. What is the minimum refractive index required for the core if the cladding has a refractive index of 1.45?
A.
1.50
B.
1.45
C.
1.60
D.
1.75
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Solution
For total internal reflection, the core must have a higher refractive index than the cladding, so it must be greater than 1.45.
Correct Answer: A — 1.50
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Q. A fiber optic cable uses total internal reflection. What is the role of the cladding?
A.
To increase the refractive index.
B.
To decrease the refractive index.
C.
To prevent light loss.
D.
To enhance light absorption.
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Solution
The cladding has a lower refractive index than the core, ensuring that light is kept within the core through total internal reflection.
Correct Answer: C — To prevent light loss.
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Q. A fiber optic cable uses total internal reflection. What is the role of the cladding in this context?
A.
To increase the speed of light.
B.
To provide structural support.
C.
To ensure light remains within the core.
D.
To change the wavelength of light.
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Solution
The cladding has a lower refractive index than the core, ensuring that light remains within the core by total internal reflection.
Correct Answer: C — To ensure light remains within the core.
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Q. A figure skater pulls in her arms while spinning. What happens to her angular momentum?
A.
Increases
B.
Decreases
C.
Remains constant
D.
Becomes zero
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Solution
Angular momentum is conserved; it remains constant, but her angular velocity increases.
Correct Answer: C — Remains constant
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Q. A figure skater pulls in her arms while spinning. What happens to her angular velocity?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
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Solution
By conservation of angular momentum, if the moment of inertia decreases, the angular velocity must increase.
Correct Answer: A — Increases
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Q. A figure skater spins with arms extended. When she pulls her arms in, what happens to her angular velocity?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
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Solution
By conservation of angular momentum, pulling arms in decreases moment of inertia, thus increasing angular velocity.
Correct Answer: A — Increases
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Q. A figure skater spins with arms extended. When she pulls her arms in, what happens to her angular momentum?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
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Solution
Angular momentum remains the same; however, her angular velocity increases due to a decrease in moment of inertia.
Correct Answer: C — Remains the same
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Q. A figure skater spins with arms extended. When she pulls her arms in, what happens to her rotational speed?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
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Solution
Pulling her arms in decreases her moment of inertia, causing her rotational speed to increase to conserve angular momentum.
Correct Answer: A — Increases
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Q. A fluid with a viscosity of 0.1 Pa·s flows through a pipe of radius 0.05 m. If the pressure difference across the pipe is 1000 Pa, what is the flow rate?
A.
0.01 m³/s
B.
0.02 m³/s
C.
0.03 m³/s
D.
0.04 m³/s
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Solution
Using Poiseuille's law, the flow rate Q = (π * r^4 * ΔP) / (8 * η * L). Assuming L = 1 m, Q = (π * (0.05)^4 * 1000) / (8 * 0.1 * 1) = 0.01 m³/s.
Correct Answer: A — 0.01 m³/s
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Q. A fluid with a viscosity of 0.1 Pa·s flows through a pipe of radius 0.05 m. What is the shear stress if the flow velocity is 1 m/s?
A.
0.1 Pa
B.
0.2 Pa
C.
0.4 Pa
D.
0.5 Pa
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Solution
Shear stress = viscosity × (velocity/radius) = 0.1 × (1/0.05) = 2 Pa.
Correct Answer: B — 0.2 Pa
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Q. A fluid with a viscosity of 0.1 Pa·s flows through a pipe of radius 0.05 m. What is the shear stress if the flow rate is 0.01 m³/s?
A.
0.4 Pa
B.
0.2 Pa
C.
0.1 Pa
D.
0.5 Pa
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Solution
Using the formula for shear stress, τ = (4 * η * Q) / (π * r^3), we find τ = 0.4 Pa.
Correct Answer: A — 0.4 Pa
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Q. A flywheel has a moment of inertia I and is rotating with an angular velocity ω. If a torque τ is applied for time t, what is the final angular velocity?
A.
ω + (τ/I)t
B.
ω - (τ/I)t
C.
ω + (I/τ)t
D.
ω - (I/τ)t
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Solution
Using the equation ω_f = ω + αt, where α = τ/I, we get ω_f = ω + (τ/I)t.
Correct Answer: A — ω + (τ/I)t
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Q. A flywheel has a moment of inertia I and is rotating with an angular velocity ω. If a torque τ is applied to it, what is the angular acceleration α?
A.
τ/I
B.
I/τ
C.
Iω/τ
D.
τω/I
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Solution
From Newton's second law for rotation, τ = Iα, thus α = τ/I.
Correct Answer: A — τ/I
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Q. A flywheel is rotating at 1000 rpm. If it is brought to rest in 10 seconds, what is the average angular deceleration?
A.
100 rad/s²
B.
10 rad/s²
C.
20 rad/s²
D.
50 rad/s²
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Solution
First convert rpm to rad/s: 1000 rpm = (1000 * 2π)/60 rad/s. Then use α = (ωf - ωi)/t = (0 - (1000 * 2π)/60) / 10.
Correct Answer: C — 20 rad/s²
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Q. A flywheel is rotating with an angular speed of 20 rad/s. If it comes to rest in 5 seconds, what is the angular deceleration?
A.
4 rad/s²
B.
5 rad/s²
C.
20 rad/s²
D.
0 rad/s²
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Solution
Angular deceleration α = (final angular speed - initial angular speed) / time = (0 - 20 rad/s) / 5 s = -4 rad/s².
Correct Answer: A — 4 rad/s²
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Q. A flywheel is rotating with an angular speed of 20 rad/s. If it experiences a torque of 5 Nm, what is the time taken to stop it?
A.
8 s
B.
4 s
C.
10 s
D.
5 s
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Solution
Using τ = Iα, we find α = τ/I. Assuming I = 1 kg·m², α = 5 rad/s². Time to stop = ω/α = 20 rad/s / 5 rad/s² = 4 s.
Correct Answer: B — 4 s
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Q. A flywheel is rotating with an angular velocity of 10 rad/s. If it is subjected to a torque of 5 Nm, what is the angular acceleration?
A.
0.5 rad/s²
B.
2 rad/s²
C.
0.2 rad/s²
D.
1 rad/s²
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Solution
Using τ = Iα, where τ is torque, I is moment of inertia, and α is angular acceleration. Assuming I = 25 kg·m², α = τ/I = 5/25 = 0.2 rad/s².
Correct Answer: B — 2 rad/s²
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Q. A flywheel is rotating with an angular velocity of 15 rad/s. If it comes to rest in 3 seconds, what is the angular deceleration?
A.
5 rad/s²
B.
10 rad/s²
C.
15 rad/s²
D.
20 rad/s²
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Solution
Angular deceleration = (final angular velocity - initial angular velocity) / time = (0 - 15) / 3 = -5 rad/s², so the magnitude is 5 rad/s².
Correct Answer: B — 10 rad/s²
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Q. A flywheel is rotating with an angular velocity of 15 rad/s. If it experiences a torque of 3 N·m, what is the angular acceleration?
A.
0.2 rad/s²
B.
0.5 rad/s²
C.
1 rad/s²
D.
5 rad/s²
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Solution
Angular acceleration α = Torque/I. Assuming I = 6 kg·m², α = 3 N·m / 6 kg·m² = 0.5 rad/s².
Correct Answer: B — 0.5 rad/s²
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Q. A flywheel is rotating with an angular velocity of 20 rad/s. If it comes to rest in 5 seconds, what is the angular deceleration?
A.
4 rad/s²
B.
5 rad/s²
C.
20 rad/s²
D.
0 rad/s²
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Solution
Angular deceleration α = (ω_f - ω_i) / t = (0 - 20 rad/s) / 5 s = -4 rad/s², so the magnitude is 4 rad/s².
Correct Answer: B — 5 rad/s²
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Q. A flywheel is rotating with an angular velocity of 20 rad/s. If it experiences a constant torque that reduces its angular velocity to 10 rad/s in 5 seconds, what is the magnitude of the torque if the moment of inertia is 4 kg·m²?
A.
8 N·m
B.
4 N·m
C.
2 N·m
D.
10 N·m
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Solution
The angular deceleration α = (ω_final - ω_initial) / time = (10 - 20) / 5 = -2 rad/s². Torque τ = Iα = 4 kg·m² * (-2 rad/s²) = -8 N·m, so the magnitude is 8 N·m.
Correct Answer: B — 4 N·m
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Q. A force is measured as 100 N with an uncertainty of ±2 N. What is the maximum possible value of the force?
A.
102 N
B.
98 N
C.
100 N
D.
104 N
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Solution
Maximum possible value = measured value + uncertainty = 100 + 2 = 102 N.
Correct Answer: A — 102 N
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Q. A force is measured as 50 N with an uncertainty of ±1 N. What is the percentage uncertainty in the force measurement?
A.
2%
B.
1%
C.
0.5%
D.
0.1%
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Solution
Percentage uncertainty = (absolute uncertainty / measured value) * 100 = (1 / 50) * 100 = 2%.
Correct Answer: B — 1%
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Q. A force is measured as 50 N with an uncertainty of ±2 N. What is the percentage uncertainty in the force measurement?
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Solution
Percentage uncertainty = (absolute uncertainty / measured value) * 100 = (2 / 50) * 100 = 4%.
Correct Answer: A — 4%
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Q. A force is measured as 50 N with an uncertainty of ±2 N. What is the relative uncertainty in this force measurement?
A.
0.04
B.
0.02
C.
0.01
D.
0.05
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Solution
Relative uncertainty = (absolute uncertainty / measured value) = 2 / 50 = 0.04 or 4%.
Correct Answer: B — 0.02
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