Q. A tennis player serves the ball at a speed of 180 km/h. How many meters per second is this? (2022)
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Solution
To convert km/h to m/s, divide by 3.6. 180 km/h ÷ 3.6 = 50 m/s.
Correct Answer: B — 45
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Q. A tennis player serves the ball at a speed of 180 km/h. What is this speed in meters per second? (2023)
A.
50 m/s
B.
45 m/s
C.
40 m/s
D.
60 m/s
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Solution
To convert km/h to m/s, divide by 3.6. Speed in m/s = 180 km/h ÷ 3.6 = 50 m/s.
Correct Answer: B — 45 m/s
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Q. A tennis player wins 75% of the matches he plays. If he plays 40 matches, how many matches does he win? (2023)
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Solution
Matches won = 75% of 40 = 0.75 × 40 = 30 matches.
Correct Answer: A — 30
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Q. A tennis player won 60% of his matches. If he played 50 matches, how many matches did he win? (2023)
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Solution
Number of matches won = 60% of 50 = 0.6 * 50 = 30.
Correct Answer: A — 30
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Q. A tennis player won 80% of his matches. If he played 50 matches, how many matches did he lose? (2023)
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Solution
Matches won = 80% of 50 = 40. Matches lost = 50 - 40 = 10.
Correct Answer: A — 10
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Q. A thermometer has a least count of 1°C. If the reading is 25°C, what is the maximum possible error in the measurement?
A.
0.5°C
B.
1°C
C.
0.1°C
D.
2°C
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Solution
Maximum possible error = Least count = 1°C.
Correct Answer: B — 1°C
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Q. A thermometer reads 100.0 °C with an uncertainty of ±0.5 °C. What is the range of possible temperatures?
A.
99.5 °C to 100.5 °C
B.
99.0 °C to 100.0 °C
C.
100.0 °C to 101.0 °C
D.
100.5 °C to 101.5 °C
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Solution
Range = 100.0 ± 0.5 °C gives 99.5 °C to 100.5 °C.
Correct Answer: A — 99.5 °C to 100.5 °C
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Q. A thermometer reads 100.5 °C when the actual temperature is 100.0 °C. What is the percentage error in the measurement?
A.
0.5%
B.
1.0%
C.
0.1%
D.
0.2%
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Solution
Percentage error = (Absolute error / True value) * 100 = (0.5 / 100.0) * 100 = 0.5%.
Correct Answer: B — 1.0%
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Q. A thermometer reads 25.0 °C with an uncertainty of ±0.2 °C. What is the range of possible temperatures?
A.
24.8 °C to 25.2 °C
B.
24.5 °C to 25.5 °C
C.
25.0 °C to 25.4 °C
D.
24.0 °C to 26.0 °C
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Solution
Range = 25.0 ± 0.2 gives 24.8 °C to 25.2 °C.
Correct Answer: A — 24.8 °C to 25.2 °C
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Q. A thermometer reads 25.0 °C with an uncertainty of ±0.5 °C. If this temperature is used in a calculation, what is the uncertainty in the result if the temperature is multiplied by 2?
A.
1 °C
B.
0.5 °C
C.
0.25 °C
D.
0.1 °C
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Solution
When multiplying, the relative uncertainty doubles: 2 * 0.5 °C = 1 °C.
Correct Answer: A — 1 °C
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Q. A thermometer reads 25.0 °C with an uncertainty of ±0.5 °C. What is the range of possible true temperatures?
A.
24.5 °C to 25.5 °C
B.
25.0 °C to 26.0 °C
C.
24.0 °C to 25.0 °C
D.
25.0 °C to 25.5 °C
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Solution
Range = Measured value ± Uncertainty = 25.0 ± 0.5 = 24.5 °C to 25.5 °C.
Correct Answer: A — 24.5 °C to 25.5 °C
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Q. A thermometer reads 25.0 °C with an uncertainty of ±0.5 °C. What is the range of possible temperatures?
A.
24.5 °C to 25.5 °C
B.
25.0 °C to 26.0 °C
C.
24.0 °C to 25.0 °C
D.
25.0 °C to 25.5 °C
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Solution
Range = 25.0 ± 0.5 °C = 24.5 °C to 25.5 °C.
Correct Answer: A — 24.5 °C to 25.5 °C
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Q. A thermometer reads 25.0 °C with an uncertainty of ±0.5 °C. What is the range of possible true values?
A.
24.5 °C to 25.5 °C
B.
24.0 °C to 25.0 °C
C.
25.0 °C to 26.0 °C
D.
25.5 °C to 26.5 °C
Show solution
Solution
The range of possible true values is from 25.0 - 0.5 to 25.0 + 0.5, which is 24.5 °C to 25.5 °C.
Correct Answer: A — 24.5 °C to 25.5 °C
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Q. A thin film of oil (n = 1.5) is on water (n = 1.33). What is the condition for destructive interference for light of wavelength 600 nm in air? (2022)
A.
2t = (m + 1/2)λ
B.
2t = mλ
C.
2t = (m + 1)λ
D.
2t = (m - 1/2)λ
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Solution
For destructive interference in a thin film with a higher refractive index below, the condition is 2t = (m + 1/2)λ/n, where n is the refractive index of the film.
Correct Answer: A — 2t = (m + 1/2)λ
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Q. A thin lens has a focal length of 20 cm. What is the power of the lens?
A.
+2.5 D
B.
+5 D
C.
+10 D
D.
+15 D
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Solution
Power (P) is given by P = 1/f (in meters). Thus, P = 1/0.2 = +5 D.
Correct Answer: B — +5 D
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Q. A thin rod of length L and mass M is rotated about an axis perpendicular to its length and passing through one end. What is its moment of inertia?
A.
1/3 ML^2
B.
1/12 ML^2
C.
1/2 ML^2
D.
ML^2
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Solution
The moment of inertia of a thin rod about an end is I = 1/3 ML^2.
Correct Answer: A — 1/3 ML^2
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Q. A thin rod of length L and mass M is rotated about an axis perpendicular to its length through one end. What is its moment of inertia?
A.
1/3 ML^2
B.
1/12 ML^2
C.
1/2 ML^2
D.
ML^2
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Solution
The moment of inertia of a thin rod about an end is I = 1/3 ML^2.
Correct Answer: A — 1/3 ML^2
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Q. A torque of 10 Nm is applied to a wheel with a moment of inertia of 2 kg·m². What is the angular acceleration of the wheel?
A.
5 rad/s²
B.
10 rad/s²
C.
2 rad/s²
D.
20 rad/s²
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Solution
Using τ = Iα, we have α = τ/I = 10 Nm / 2 kg·m² = 5 rad/s².
Correct Answer: A — 5 rad/s²
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Q. A torque of 10 Nm is applied to a wheel with a moment of inertia of 2 kg·m². What is the angular acceleration?
A.
5 rad/s²
B.
10 rad/s²
C.
2 rad/s²
D.
20 rad/s²
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Solution
Using τ = Iα, we have α = τ/I = 10 Nm / 2 kg·m² = 5 rad/s².
Correct Answer: A — 5 rad/s²
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Q. A torque of 10 Nm is applied to a wheel with a moment of inertia of 5 kg·m². What is the angular acceleration of the wheel?
A.
2 rad/s²
B.
5 rad/s²
C.
10 rad/s²
D.
20 rad/s²
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Solution
Using τ = Iα, we have α = τ/I = 10 Nm / 5 kg·m² = 2 rad/s².
Correct Answer: A — 2 rad/s²
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Q. A torque of 10 Nm is applied to a wheel. If the radius of the wheel is 0.2 m, what is the force applied tangentially?
A.
50 N
B.
20 N
C.
10 N
D.
5 N
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Solution
Torque (τ) = F × r; therefore, F = τ / r = 10 Nm / 0.2 m = 50 N.
Correct Answer: A — 50 N
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Q. A torque of 10 N·m is applied to a wheel with a moment of inertia of 2 kg·m². What is the angular acceleration of the wheel?
A.
5 rad/s²
B.
10 rad/s²
C.
2 rad/s²
D.
20 rad/s²
Show solution
Solution
Using τ = Iα, we have α = τ/I = 10 N·m / 2 kg·m² = 5 rad/s².
Correct Answer: A — 5 rad/s²
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Q. A torque of 12 Nm is applied to a lever arm of 0.4 m. What is the force applied?
A.
30 N
B.
25 N
C.
20 N
D.
15 N
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Solution
Force = Torque / Distance = 12 Nm / 0.4 m = 30 N.
Correct Answer: C — 20 N
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Q. A torque of 12 Nm is applied to a wheel of radius 0.4 m. What is the force applied at the edge of the wheel?
A.
30 N
B.
20 N
C.
15 N
D.
10 N
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Solution
Torque (τ) = r × F, thus F = τ / r = 12 Nm / 0.4 m = 30 N.
Correct Answer: B — 20 N
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Q. A torque of 12 Nm is applied to a wheel with a radius of 0.4 m. What is the force applied tangentially to the wheel?
A.
15 N
B.
30 N
C.
40 N
D.
50 N
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Solution
Using τ = F × r, we have F = τ / r = 12 Nm / 0.4 m = 30 N.
Correct Answer: B — 30 N
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Q. A torque of 12 Nm is produced by a force acting at a distance of 0.4 m from the pivot. What is the magnitude of the force?
A.
20 N
B.
30 N
C.
40 N
D.
50 N
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Solution
Force = Torque / Distance = 12 Nm / 0.4 m = 30 N.
Correct Answer: A — 20 N
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Q. A torque of 12 Nm is produced by a force acting at a distance of 4 m from the pivot. What is the magnitude of the force?
A.
2 N
B.
3 N
C.
4 N
D.
5 N
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Solution
Force = Torque / Distance = 12 Nm / 4 m = 3 N.
Correct Answer: B — 3 N
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Q. A torque of 15 N·m is applied to a wheel with a moment of inertia of 3 kg·m². What is the angular acceleration? (2023)
A.
3 rad/s²
B.
5 rad/s²
C.
10 rad/s²
D.
15 rad/s²
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Solution
Using τ = Iα, we have α = τ/I = 15/3 = 5 rad/s².
Correct Answer: B — 5 rad/s²
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Q. A torque of 15 N·m is applied to a wheel with a radius of 0.3 m. What is the force applied tangentially to the wheel?
A.
25 N
B.
50 N
C.
45 N
D.
30 N
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Solution
Torque (τ) = Force (F) × Radius (r) => F = τ / r = 15 N·m / 0.3 m = 50 N.
Correct Answer: B — 50 N
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Q. A torque of 25 Nm is applied to a wheel with a radius of 0.5 m. What is the force applied at the edge of the wheel?
A.
50 N
B.
25 N
C.
75 N
D.
100 N
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Solution
Force = Torque / Radius = 25 Nm / 0.5 m = 50 N.
Correct Answer: A — 50 N
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