Q. A force of 25 N is applied at an angle of 60 degrees to the horizontal while moving an object 3 m. What is the work done?
A.
37.5 J
B.
50 J
C.
75 J
D.
100 J
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Solution
Work done = Force × Distance × cos(θ) = 25 N × 3 m × cos(60°) = 37.5 J.
Correct Answer: A — 37.5 J
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Q. A force of 30 N is applied at an angle of 60 degrees to a lever arm of length 2 m. What is the torque about the pivot?
A.
15 Nm
B.
30 Nm
C.
60 Nm
D.
52 Nm
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Solution
Torque (τ) = F × r × sin(θ) = 30 N × 2 m × sin(60°) = 30 × 2 × (√3/2) = 30√3 Nm ≈ 51.96 Nm.
Correct Answer: D — 52 Nm
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Q. A force of 30 N is applied to a 3 kg object. What is the resulting acceleration? (2020)
A.
5 m/s²
B.
10 m/s²
C.
15 m/s²
D.
20 m/s²
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Solution
Using F = ma, a = F/m = 30 N / 3 kg = 10 m/s².
Correct Answer: B — 10 m/s²
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Q. A force of 30 N is applied to a 5 kg object. What is the object's acceleration?
A.
3 m/s²
B.
6 m/s²
C.
9 m/s²
D.
12 m/s²
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Solution
Using F = ma, a = F/m = 30 N / 5 kg = 6 m/s².
Correct Answer: B — 6 m/s²
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Q. A force of 30 N is applied to a 6 kg object. What is the object's acceleration? (2023)
A.
3 m/s²
B.
4 m/s²
C.
5 m/s²
D.
6 m/s²
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Solution
Using F = ma, we find a = F/m = 30 N / 6 kg = 5 m/s².
Correct Answer: B — 4 m/s²
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Q. A force of 40 N is applied at a distance of 0.5 m from the pivot. What is the torque?
A.
10 Nm
B.
15 Nm
C.
20 Nm
D.
25 Nm
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Solution
Torque = Force × Distance = 40 N × 0.5 m = 20 Nm.
Correct Answer: C — 20 Nm
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Q. A force of 40 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.
20 Nm
B.
40 Nm
C.
34.64 Nm
D.
69.28 Nm
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Solution
Torque (τ) = F × r × sin(θ) = 40 N × 2 m × sin(60°) = 40 × 2 × (√3/2) = 40√3 Nm ≈ 34.64 Nm.
Correct Answer: C — 34.64 Nm
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Q. A force of 50 N is applied at a distance of 0.5 m from the pivot at an angle of 60 degrees. What is the torque?
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 × 0.5 m × sin(60°) = 50 N × 0.5 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 30 degrees to the lever arm of length 2 m. What is the torque about the pivot?
A.
25 N·m
B.
50 N·m
C.
86.6 N·m
D.
100 N·m
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Solution
Torque (τ) = F × r × sin(θ) = 50 N × 2 m × sin(30°) = 50 N × 2 m × 0.5 = 50 N·m.
Correct Answer: C — 86.6 N·m
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Q. A force of 50 N is applied at an angle of 30 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.
86.6 Nm
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Solution
Torque (τ) = F × d × sin(θ) = 50 N × 1 m × sin(30°) = 50 N × 1 m × 0.5 = 25 Nm.
Correct Answer: B — 43.3 Nm
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Q. A force of 50 N is applied at an angle of 30° to the horizontal. What is the horizontal component of the force? (2020)
A.
25 N
B.
43.3 N
C.
50 N
D.
35 N
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Solution
Horizontal component = Fcos(θ) = 50 N × cos(30°) = 50 N × (√3/2) ≈ 43.3 N.
Correct Answer: B — 43.3 N
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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
Show solution
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
Show solution
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 force of 50 N is applied to a 25 kg object. What is the acceleration of the object? (2019)
A.
1 m/s²
B.
2 m/s²
C.
3 m/s²
D.
4 m/s²
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Solution
Using F = ma, a = F/m = 50 N / 25 kg = 2 m/s².
Correct Answer: B — 2 m/s²
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Q. A force of 50 N is applied to a 25 kg object. What is the object's acceleration? (2020)
A.
1 m/s²
B.
2 m/s²
C.
3 m/s²
D.
4 m/s²
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Solution
Using F = ma, a = F/m = 50 N / 25 kg = 2 m/s².
Correct Answer: B — 2 m/s²
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Q. A force of 50 N is applied to move an object 4 m. How much work is done?
A.
100 J
B.
150 J
C.
200 J
D.
250 J
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Solution
Work = Force * Distance = 50 N * 4 m = 200 J.
Correct Answer: C — 200 J
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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 foundation awards $5000 to be distributed among 4 projects in the ratio 1:2:3:4. How much does the third project receive? (2023)
A.
$500
B.
$1000
C.
$1500
D.
$2000
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Solution
Total parts = 1 + 2 + 3 + 4 = 10. Third project's share = (3/10) * 5000 = $1500.
Correct Answer: C — $1500
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Q. A fruit seller has 120 apples. He sells 25% of them. How many apples does he have left?
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Solution
If the seller sells 25% of 120 apples, he sells 0.25 * 120 = 30 apples. Therefore, the number of apples left is 120 - 30 = 90.
Correct Answer: A — 90
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Q. A fruit seller has apples and oranges in the ratio of 5:3. If he has 40 apples, how many oranges does he have?
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Solution
If the ratio of apples to oranges is 5:3, then for every 5 apples, there are 3 oranges. If there are 40 apples, we can set up the proportion: 5/3 = 40/x. Cross-multiplying gives us 5x = 120, so x = 24. Therefore, there are 24 oranges.
Correct Answer: B — 30
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Q. A gardener has 36 red roses and 48 yellow roses. He wants to plant them in rows with the same number of each type of rose in each row. What is the maximum number of rows he can plant? (2023)
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Solution
The HCF of 36 and 48 is 12, which is the maximum number of rows he can plant.
Correct Answer: B — 12
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Q. A gardener has two types of plants, one type has a height of 3 feet and the other 5 feet. What is the minimum height at which both types can be tied together? (2023)
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Solution
The minimum height is the LCM of 3 and 5, which is 15 feet.
Correct Answer: C — 60
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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 from 2 L to 5 L at a constant pressure of 1 atm. How much work is done by the gas? (2023)
A.
3 L·atm
B.
5 L·atm
C.
2 L·atm
D.
1 L·atm
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Solution
Work done by the gas during expansion at constant pressure is W = PΔV. Here, ΔV = 5 L - 2 L = 3 L, so W = 1 atm * 3 L = 3 L·atm.
Correct Answer: A — 3 L·atm
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Q. A gas expands from a volume of 2 m³ to 5 m³ at a constant pressure of 100 kPa. What is the work done by the gas? (2022)
A.
300 kJ
B.
500 kJ
C.
700 kJ
D.
3000 kJ
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Solution
Work done (W) = P * ΔV = 100 kPa * (5 m³ - 2 m³) = 100 kPa * 3 m³ = 300 kJ.
Correct Answer: A — 300 kJ
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Q. A gas expands from volume V1 to V2 at constant pressure P. What is the work done by the gas? (2022)
A.
P(V2 - V1)
B.
P(V1 + V2)
C.
P(V1 * V2)
D.
P(V1 - V2)
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Solution
Work done (W) = P * ΔV = P * (V2 - V1).
Correct Answer: A — P(V2 - V1)
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Q. A gas expands from volume V1 to V2 at constant temperature. If the initial pressure is P1, what is the final pressure P2? (2020)
A.
P1(V1/V2)
B.
P1(V2/V1)
C.
P1
D.
P1/2
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Solution
According to Boyle's Law, P1V1 = P2V2, thus P2 = P1(V1/V2).
Correct Answer: A — P1(V1/V2)
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Q. A gas expands isothermally at 300 K and absorbs 600 J of heat. What is the work done by the gas? (2023)
A.
600 J
B.
300 J
C.
900 J
D.
0 J
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Solution
In an isothermal process, the work done by the gas is equal to the heat absorbed. Therefore, work done = 600 J.
Correct Answer: A — 600 J
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Q. A gas expands isothermally at 300 K from a volume of 1 m³ to 2 m³. If the pressure at the initial state is 100 kPa, what is the work done by the gas?
A.
0 kJ
B.
100 kJ
C.
150 kJ
D.
200 kJ
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Solution
Work done (W) = nRT ln(Vf/Vi). Here, nRT = PVi = 100 kPa * 1 m³ = 100 kJ. W = 100 kJ * ln(2) ≈ 69.3 kJ.
Correct Answer: B — 100 kJ
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