Engineering & Architecture Admissions
Q. A force of 50 N is applied at an angle of 60 degrees to the lever arm of length 1 m. What is the torque about the pivot?
A.
25 Nm
B.
43.3 Nm
C.
50 Nm
D.
0 Nm
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Solution
Torque (τ) = F × r × sin(θ) = 50 N × 1 m × sin(60°) = 50 N × 1 m × (√3/2) ≈ 43.3 Nm.
Correct Answer: B — 43.3 Nm
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Q. A force of 50 N is applied at an angle of 60 degrees to the lever arm of length 2 m. What is the torque about the pivot?
A.
25 Nm
B.
50 Nm
C.
43.3 Nm
D.
100 Nm
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Solution
Torque (τ) = F × r × sin(θ) = 50 N × 2 m × sin(60°) = 50 × 2 × (√3/2) = 43.3 Nm.
Correct Answer: C — 43.3 Nm
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Q. A forced oscillator has a mass of 3 kg and is driven by a force of 12 N at a frequency of 2 Hz. What is the amplitude of the oscillation if the damping coefficient is 0.1 kg/s?
A.
0.1 m
B.
0.2 m
C.
0.3 m
D.
0.4 m
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Solution
Using F = mAω², we find A = F / (mω²) = 12 / (3*(2π*2)²) ≈ 0.2 m.
Correct Answer: B — 0.2 m
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Q. A gas at 300 K has an RMS speed of 400 m/s. What will be its RMS speed at 600 K?
A.
400 m/s
B.
400 sqrt(2) m/s
C.
800 m/s
D.
200 m/s
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Solution
The RMS speed is proportional to the square root of the temperature. Therefore, at 600 K, the RMS speed will be 400 * sqrt(600/300) = 400 * sqrt(2) m/s.
Correct Answer: B — 400 sqrt(2) m/s
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Q. A gas at 300 K has an RMS speed of 500 m/s. What will be its RMS speed at 600 K?
A.
500 m/s
B.
707 m/s
C.
1000 m/s
D.
250 m/s
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Solution
The RMS speed is proportional to the square root of the temperature. Therefore, v_rms at 600 K = 500 * sqrt(600/300) = 500 * sqrt(2) ≈ 707 m/s.
Correct Answer: B — 707 m/s
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Q. A gas expands isothermally at 300 K from a volume of 1 m³ to 2 m³. What is the work done by the gas?
A.
0 J
B.
300 J
C.
600 J
D.
150 J
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Solution
The work done in an isothermal expansion is W = nRT ln(Vf/Vi). Assuming 1 mole of gas, W = 1 * 8.31 * 300 * ln(2) ≈ 600 J.
Correct Answer: C — 600 J
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Q. A gas has an RMS speed of 500 m/s. If the molar mass of the gas is 0.02 kg/mol, what is the temperature of the gas?
A.
250 K
B.
500 K
C.
1000 K
D.
2000 K
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Solution
Using the formula v_rms = sqrt((3RT)/M), we can rearrange to find T = (v_rms^2 * M) / (3R). Substituting v_rms = 500 m/s and M = 0.02 kg/mol gives T = 500 K.
Correct Answer: B — 500 K
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Q. A height is measured as 180 cm with an error of 1 cm. What is the upper limit of the measurement?
A.
181 cm
B.
179 cm
C.
180 cm
D.
182 cm
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Solution
Upper limit = Measured value + Error = 180 + 1 = 181 cm
Correct Answer: A — 181 cm
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Q. A hollow cylinder with charge density λ is placed along the z-axis. What is the electric field at a point outside the cylinder?
A.
λ/(2πε₀r)
B.
λ/(4πε₀r²)
C.
Zero
D.
λ/(ε₀r)
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Solution
For a hollow cylinder, the electric field at a point outside is E = λ/(2πε₀r) using Gauss's law.
Correct Answer: A — λ/(2πε₀r)
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Q. A hollow sphere has a charge +Q distributed uniformly on its surface. What is the electric field at a point inside the sphere?
A.
Q/(4πε₀r²)
B.
0
C.
Q/(4πε₀R²)
D.
Q/(4πε₀R)
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Solution
The electric field inside a hollow charged sphere is zero due to symmetry.
Correct Answer: B — 0
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Q. A hollow sphere has a charge Q distributed uniformly over its surface. What is the electric field inside the sphere?
A.
Q/(4πε₀R²)
B.
0
C.
Q/(4πε₀)
D.
Q/(4πε₀R)
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Solution
Inside a hollow charged sphere, the electric field is zero due to symmetry.
Correct Answer: B — 0
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Q. A hollow sphere has a charge Q uniformly distributed on its surface. What is the electric field inside the sphere?
A.
Q/4πε₀R²
B.
0
C.
Q/ε₀
D.
Q/4πε₀
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Solution
The electric field inside a hollow charged sphere is zero due to symmetry.
Correct Answer: B — 0
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Q. A hollow sphere rolls down a slope of height h. What fraction of its potential energy is converted into translational kinetic energy at the bottom?
A.
1/3
B.
1/2
C.
2/3
D.
1
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Solution
For a hollow sphere, I = (2/3)mr^2. Using energy conservation, the translational kinetic energy is 2/3 of the potential energy at the top.
Correct Answer: C — 2/3
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Q. A hollow sphere rolls down an incline. If it starts from rest, what fraction of its total energy is translational at the bottom?
A.
1/3
B.
2/3
C.
1/2
D.
1/4
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Solution
For a hollow sphere, the translational kinetic energy at the bottom is 2/3 of the total energy, hence the fraction is 2/3.
Correct Answer: B — 2/3
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Q. A hollow sphere rolls down an incline. If its mass is m and radius is R, what is its moment of inertia?
A.
(2/5)mR^2
B.
(1/2)mR^2
C.
(2/3)mR^2
D.
(3/5)mR^2
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Solution
The moment of inertia of a hollow sphere about its center is I = (2/3)mR^2.
Correct Answer: C — (2/3)mR^2
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Q. A hollow sphere with charge Q has a radius R. What is the electric field at a point inside the sphere?
A.
Q/(4πε₀R²)
B.
0
C.
Q/(4πε₀R)
D.
Q/(2πε₀R²)
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Solution
Inside a hollow charged sphere, the electric field is zero due to symmetry, as per Gauss's law.
Correct Answer: B — 0
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Q. A hollow spherical conductor carries a charge Q. What is the electric field inside the cavity?
A.
Q/(4πε₀r²)
B.
0
C.
Q/(4πε₀R²)
D.
Q/(4πε₀R)
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Solution
Inside the cavity of a hollow conductor, the electric field is zero due to the shielding effect of the conductor.
Correct Answer: B — 0
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Q. A jar contains 4 red, 5 green, and 6 blue marbles. If one marble is drawn at random, what is the probability that it is not green?
A.
1/3
B.
2/3
C.
1/2
D.
5/11
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Solution
Total marbles = 4 + 5 + 6 = 15. Non-green marbles = 4 + 6 = 10. Probability = 10/15 = 2/3.
Correct Answer: B — 2/3
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Q. A jar contains 4 red, 5 green, and 6 blue marbles. If one marble is drawn at random, what is the probability that it is not blue?
A.
1/3
B.
2/3
C.
1/2
D.
5/11
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Solution
Total marbles = 4 + 5 + 6 = 15. Non-blue marbles = 4 + 5 = 9. Probability = 9/15 = 3/5.
Correct Answer: B — 2/3
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Q. A jar contains 4 red, 5 green, and 6 blue marbles. If one marble is drawn at random, what is the probability that it is either red or green?
A.
4/15
B.
3/5
C.
9/15
D.
1/3
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Solution
Total marbles = 4 + 5 + 6 = 15. Probability of red or green = (4 + 5)/15 = 9/15 = 3/5.
Correct Answer: B — 3/5
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Q. A jar contains 4 red, 5 green, and 6 blue marbles. What is the probability of drawing a green marble?
A.
1/3
B.
5/15
C.
5/15
D.
1/5
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Solution
Total marbles = 4 + 5 + 6 = 15. Probability of green = 5/15 = 1/3.
Correct Answer: A — 1/3
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Q. A jar contains 4 red, 5 green, and 6 blue marbles. What is the probability of picking a green marble?
A.
5/15
B.
1/3
C.
1/5
D.
1/2
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Solution
The total number of marbles is 4 + 5 + 6 = 15. The probability of picking a green marble is 5/15 = 1/3.
Correct Answer: B — 1/3
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Q. A jar contains 4 red, 5 green, and 6 blue marbles. What is the probability of randomly selecting a green marble?
A.
1/3
B.
5/15
C.
5/15
D.
1/5
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Solution
The total number of marbles is 4 + 5 + 6 = 15. The probability of selecting a green marble is 5/15 = 1/3.
Correct Answer: A — 1/3
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Q. A jar contains 5 red, 3 green, and 2 blue marbles. If one marble is drawn at random, what is the probability that it is either red or green?
A.
1/2
B.
2/5
C.
4/5
D.
3/5
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Solution
Total marbles = 5 + 3 + 2 = 10. Favorable outcomes (red or green) = 5 + 3 = 8. Probability = 8/10 = 4/5.
Correct Answer: C — 4/5
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Q. A jar contains 5 red, 3 green, and 2 blue marbles. What is the probability of drawing a green marble?
A.
1/5
B.
1/4
C.
3/10
D.
1/2
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Solution
Total marbles = 5 + 3 + 2 = 10. Probability of green = 3/10.
Correct Answer: C — 3/10
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Q. A jar contains 5 red, 3 green, and 2 yellow marbles. If one marble is drawn at random, what is the probability that it is either red or green?
A.
1/2
B.
4/5
C.
2/5
D.
3/5
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Solution
The total number of marbles is 5 + 3 + 2 = 10. The number of favorable outcomes (red or green) is 5 + 3 = 8. Thus, the probability is 8/10 = 4/5.
Correct Answer: B — 4/5
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Q. A kite is flying at a height of 100 m. If the angle of elevation from a point on the ground to the kite is 30 degrees, how far is the point from the base of the kite?
A.
100 m
B.
200 m
C.
300 m
D.
400 m
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Solution
Using tan(30°) = height/distance, we have 1/√3 = 100/distance. Therefore, distance = 100√3 ≈ 173.2 m.
Correct Answer: B — 200 m
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Q. A kite is flying at a height of 100 meters. If the angle of depression from the kite to a point on the ground is 30 degrees, how far is the point from the point directly below the kite?
A.
50 m
B.
60 m
C.
70 m
D.
80 m
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Solution
Using tan(30°) = 100/distance, we have 1/√3 = 100/distance. Therefore, distance = 100√3 ≈ 173.21 m.
Correct Answer: A — 50 m
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Q. A kite is flying at a height of 30 m. If the angle of elevation from a point on the ground to the kite is 60 degrees, how far is the point from the base of the kite?
A.
15√3 m
B.
30 m
C.
10√3 m
D.
20 m
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Solution
Using tan(60°) = height/distance, we have distance = height/tan(60°) = 30/√3 = 15√3 m.
Correct Answer: A — 15√3 m
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Q. A kite is flying at a height of 30 meters. If the angle of elevation from a point on the ground to the kite is 45 degrees, how far is the point from the base of the kite?
A.
15 m
B.
30 m
C.
45 m
D.
60 m
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Solution
Using tan(45°) = height/distance, we have 1 = 30/distance. Therefore, distance = 30 m.
Correct Answer: B — 30 m
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