Q. A student measures the length of a rod multiple times and records the values: 10.1 cm, 10.2 cm, 10.0 cm, 10.3 cm. What is the average length of the rod?
A.
10.0 cm
B.
10.1 cm
C.
10.2 cm
D.
10.3 cm
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Solution
Average length = (10.1 + 10.2 + 10.0 + 10.3) / 4 = 10.15 cm.
Correct Answer: B — 10.1 cm
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Q. A student measures the length of a rod multiple times and records values of 10.1 cm, 10.2 cm, and 10.0 cm. What is the average length of the rod?
A.
10.0 cm
B.
10.1 cm
C.
10.2 cm
D.
10.3 cm
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Solution
Average length = (10.1 + 10.2 + 10.0) / 3 = 10.1 cm.
Correct Answer: B — 10.1 cm
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Q. A student measures the mass of an object as 200 g with an uncertainty of ±2 g. What is the relative uncertainty?
A.
1%
B.
0.5%
C.
2%
D.
0.1%
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Solution
Relative uncertainty = (Uncertainty / Measured value) * 100 = (2 / 200) * 100 = 1%.
Correct Answer: B — 0.5%
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Q. A student measures the speed of a car as 60 km/h with an uncertainty of ±2 km/h. What is the percentage uncertainty?
A.
3.33%
B.
2.5%
C.
4.0%
D.
5.0%
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Solution
Percentage uncertainty = (Uncertainty / Measured value) * 100 = (2 / 60) * 100 = 3.33%.
Correct Answer: A — 3.33%
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Q. A student measures the speed of a car as 60 km/h with an uncertainty of ±2 km/h. What is the percentage error in the speed measurement?
A.
3.33%
B.
2.0%
C.
1.5%
D.
4.0%
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Solution
Percentage error = (Uncertainty / Measured value) * 100 = (2 / 60) * 100 = 3.33%.
Correct Answer: A — 3.33%
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Q. A student measures the speed of a car as 60.0 km/h with an uncertainty of ±2.0 km/h. What is the percentage uncertainty in the speed measurement?
A.
3.33%
B.
2.0%
C.
1.67%
D.
5.0%
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Solution
Percentage uncertainty = (absolute uncertainty / measured value) * 100 = (2.0 / 60.0) * 100 = 3.33%.
Correct Answer: A — 3.33%
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Q. A student measures the temperature of water as 25°C with an uncertainty of ±0.5°C. What is the minimum temperature?
A.
24.5°C
B.
25.0°C
C.
25.5°C
D.
26.0°C
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Solution
Minimum temperature = Measured value - Uncertainty = 25.0 - 0.5 = 24.5°C.
Correct Answer: A — 24.5°C
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Q. A student measures the time period of a pendulum as 2.0 s with an uncertainty of ±0.1 s. What is the percentage uncertainty in the time period?
A.
5%
B.
10%
C.
2.5%
D.
1%
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Solution
Percentage uncertainty = (Uncertainty / Measured value) * 100 = (0.1 / 2.0) * 100 = 5%.
Correct Answer: C — 2.5%
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Q. A student measures the time period of a pendulum as 2.0 s with an uncertainty of ±0.1 s. What is the fractional error in the time period?
A.
0.05
B.
0.1
C.
0.02
D.
0.1
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Solution
Fractional error = (absolute error / measured value) = 0.1 / 2.0 = 0.05.
Correct Answer: A — 0.05
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Q. A student measures the time period of a pendulum as 2.0 s with an uncertainty of ±0.1 s. What is the absolute error in the time period?
A.
0.1 s
B.
0.05 s
C.
0.2 s
D.
0.01 s
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Solution
The absolute error is given directly as ±0.1 s.
Correct Answer: A — 0.1 s
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Q. A student measures the width of a table as 1.2 m with a possible error of 0.02 m. What is the range of the true width?
A.
1.18 m to 1.22 m
B.
1.20 m to 1.22 m
C.
1.20 m to 1.24 m
D.
1.18 m to 1.20 m
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Solution
True width = Measured value ± Absolute error = 1.2 ± 0.02 gives the range 1.18 m to 1.22 m.
Correct Answer: A — 1.18 m to 1.22 m
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Q. A student records the time taken for a reaction as 15.0 s with an uncertainty of ±0.5 s. What is the total time range?
A.
14.5 s to 15.5 s
B.
15.0 s to 16.0 s
C.
14.0 s to 15.0 s
D.
15.0 s to 15.5 s
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Solution
Total time range = 15.0 s ± 0.5 s = 14.5 s to 15.5 s.
Correct Answer: A — 14.5 s to 15.5 s
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Q. A swimmer can swim at 3 km/h in still water. If he swims across a river flowing at 2 km/h, what is his resultant speed?
A.
3 km/h
B.
4 km/h
C.
5 km/h
D.
6 km/h
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Solution
Resultant speed = √(3² + 2²) = √(9 + 4) = √13 ≈ 3.6 km/h.
Correct Answer: C — 5 km/h
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Q. A swimmer can swim at 3 m/s in still water. If the river flows at 1 m/s, what is the swimmer's speed when swimming across the river?
A.
2 m/s
B.
3 m/s
C.
4 m/s
D.
5 m/s
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Solution
Speed across the river = √(3^2 - 1^2) = √(9 - 1) = √8 = 2.83 m/s (approximately 2 m/s).
Correct Answer: A — 2 m/s
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Q. A swimmer can swim at 3 m/s in still water. If the river flows at 2 m/s, what is the swimmer's speed relative to the bank when swimming upstream?
A.
1 m/s
B.
2 m/s
C.
3 m/s
D.
5 m/s
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Solution
Speed relative to bank = Speed of swimmer - Speed of river = 3 m/s - 2 m/s = 1 m/s.
Correct Answer: A — 1 m/s
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Q. A swimmer can swim at 4 km/h in still water. If he swims across a river that is 1 km wide and the current is 2 km/h, how long will it take him to reach the opposite bank?
A.
15 minutes
B.
30 minutes
C.
45 minutes
D.
1 hour
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Solution
Time = distance/speed = 1 km / 4 km/h = 0.25 hours = 15 minutes.
Correct Answer: B — 30 minutes
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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%
Show solution
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
Show solution
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
Show solution
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
Show solution
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 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 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 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
Show solution
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?
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 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 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 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
Show solution
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
Show solution
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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