Physics
Q. A force of 10 N is applied to move a box 5 m across a floor. What is the work done by the force?
A.
10 J
B.
20 J
C.
50 J
D.
5 J
Show solution
Solution
Work done = Force × Distance = 10 N × 5 m = 50 J
Correct Answer: B — 20 J
Learn More →
Q. A force of 10 N is applied to move a box 5 m along a horizontal surface. What is the work done by the force? (2021)
A.
20 J
B.
50 J
C.
10 J
D.
5 J
Show solution
Solution
Work done = Force × Distance = 10 N × 5 m = 50 J
Correct Answer: B — 50 J
Learn More →
Q. A force of 10 N is applied to move an object 5 m in the direction of the force. What is the work done by the force?
A.
10 J
B.
20 J
C.
50 J
D.
5 J
Show solution
Solution
Work done (W) = Force (F) × Distance (d) × cos(θ). Here, θ = 0° (force and displacement are in the same direction). W = 10 N × 5 m × cos(0°) = 10 N × 5 m = 50 J.
Correct Answer: B — 20 J
Learn More →
Q. A force of 10 N is applied to move an object 5 m. What is the work done? (2023)
A.
50 J
B.
10 J
C.
5 J
D.
20 J
Show solution
Solution
Work done = Force × Distance = 10 N × 5 m = 50 J.
Correct Answer: A — 50 J
Learn More →
Q. A force of 15 N is applied to a 3 kg mass. What is the resulting acceleration? (2023)
A.
3 m/s²
B.
5 m/s²
C.
7 m/s²
D.
10 m/s²
Show solution
Solution
Using F = ma, acceleration a = F/m = 15 N / 3 kg = 5 m/s².
Correct Answer: B — 5 m/s²
Learn More →
Q. A force of 20 N is applied at an angle of 60° to the horizontal while moving an object 4 m. What is the work done?
A.
40 J
B.
80 J
C.
20 J
D.
60 J
Show solution
Solution
Work done (W) = F × d × cos(θ) = 20 N × 4 m × cos(60°) = 20 × 4 × 0.5 = 40 J.
Correct Answer: B — 80 J
Learn More →
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
Show solution
Solution
Using F = ma, we find a = F/m = 30 N / 6 kg = 5 m/s².
Correct Answer: B — 4 m/s²
Learn More →
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
Show solution
Solution
Work = Force * Distance = 50 N * 4 m = 200 J.
Correct Answer: C — 200 J
Learn More →
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
Show solution
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
Learn More →
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)
Show solution
Solution
Work done (W) = P * ΔV = P * (V2 - V1).
Correct Answer: A — P(V2 - V1)
Learn More →
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
Show solution
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
Learn More →
Q. A gas expands isothermally at 300 K from a volume of 1 m³ to 2 m³. If the pressure of the gas is 100 kPa, what is the work done by the gas? (2020)
A.
0 kJ
B.
10 kJ
C.
20 kJ
D.
30 kJ
Show solution
Solution
Work done (W) = P * ΔV = 100 kPa * (2 m³ - 1 m³) = 100 kPa * 1 m³ = 100 kJ = 10 kJ.
Correct Answer: B — 10 kJ
Learn More →
Q. A gas is compressed isothermally from a volume of 4 L to 1 L at a constant temperature of 300 K. If the initial pressure is 1 atm, what is the final pressure?
A.
4 atm
B.
3 atm
C.
2 atm
D.
1 atm
Show solution
Solution
Using Boyle's Law, P1V1 = P2V2, we find P2 = P1 * (V1/V2) = 1 atm * (4 L / 1 L) = 4 atm.
Correct Answer: A — 4 atm
Learn More →
Q. A light ray strikes a plane mirror at an angle of incidence of 30 degrees. What is the angle of reflection? (2021)
A.
30 degrees
B.
60 degrees
C.
90 degrees
D.
45 degrees
Show solution
Solution
According to the law of reflection, the angle of reflection is equal to the angle of incidence. Therefore, the angle of reflection is also 30 degrees.
Correct Answer: A — 30 degrees
Learn More →
Q. A machine does 1200 J of work in 60 seconds. What is its efficiency if the input energy is 2000 J? (2022)
A.
60%
B.
70%
C.
80%
D.
90%
Show solution
Solution
Efficiency = (Output Work / Input Energy) * 100 = (1200 J / 2000 J) * 100 = 60%.
Correct Answer: A — 60%
Learn More →
Q. A machine does 1500 J of work in 30 seconds. What is its power output?
A.
50 W
B.
30 W
C.
75 W
D.
100 W
Show solution
Solution
Power (P) = Work (W) / Time (t). P = 1500 J / 30 s = 50 W.
Correct Answer: A — 50 W
Learn More →
Q. A mass attached to a spring oscillates with a frequency of 2 Hz. What is the period of the oscillation?
A.
0.5 s
B.
1 s
C.
2 s
D.
4 s
Show solution
Solution
The period T is the reciprocal of frequency f. Thus, T = 1/f = 1/2 = 0.5 s.
Correct Answer: B — 1 s
Learn More →
Q. A mass attached to a spring oscillates with a period of 2 seconds. What is its frequency? (2014)
A.
0.5 Hz
B.
1 Hz
C.
2 Hz
D.
4 Hz
Show solution
Solution
Frequency f = 1/T = 1/2 s = 0.5 Hz.
Correct Answer: A — 0.5 Hz
Learn More →
Q. A mass m is attached to a string of length L and is swung in a vertical circle. What is the tension in the string at the top of the circle? (2023)
Show solution
Solution
At the top, T + mg = mv²/L. T = mv²/L - mg.
Correct Answer: A — mg
Learn More →
Q. A mass-spring system oscillates with a frequency of 1 Hz. What is the angular frequency? (2023)
A.
2π rad/s
B.
π rad/s
C.
1 rad/s
D.
4π rad/s
Show solution
Solution
Angular frequency ω is related to frequency f by ω = 2πf. Therefore, for f = 1 Hz, ω = 2π(1) = 2π rad/s.
Correct Answer: A — 2π rad/s
Learn More →
Q. A mass-spring system oscillates with a frequency of 5 Hz. What is the angular frequency? (2021)
A.
10π rad/s
B.
5π rad/s
C.
2π rad/s
D.
20π rad/s
Show solution
Solution
Angular frequency ω = 2πf = 2π × 5 Hz = 10π rad/s.
Correct Answer: A — 10π rad/s
Learn More →
Q. A mass-spring system oscillates with an amplitude of 0.1 m. What is the maximum speed of the mass if the angular frequency is 20 rad/s?
A.
1 m/s
B.
2 m/s
C.
3 m/s
D.
4 m/s
Show solution
Solution
The maximum speed v_max is given by v_max = Aω, where A is the amplitude and ω is the angular frequency. Thus, v_max = 0.1 * 20 = 2 m/s.
Correct Answer: B — 2 m/s
Learn More →
Q. A parallel plate capacitor has a capacitance of 5 µF. If the potential difference across it is increased from 10 V to 20 V, what is the change in stored energy?
A.
25 µJ
B.
50 µJ
C.
75 µJ
D.
100 µJ
Show solution
Solution
Energy U = 1/2 C V^2. Initial U = 1/2 * 5 × 10^-6 * 10^2 = 0.025 J; Final U = 1/2 * 5 × 10^-6 * 20^2 = 0.1 J. Change = 0.1 - 0.025 = 0.075 J = 75 µJ.
Correct Answer: B — 50 µJ
Learn More →
Q. A particle is moving in a circle of radius r with a constant speed v. What is the angular velocity? (2022)
A.
v/r
B.
r/v
C.
vr
D.
v²/r
Show solution
Solution
Angular velocity ω = v/r.
Correct Answer: A — v/r
Learn More →
Q. A particle moves in a circular path of radius r with a constant speed v. What is the centripetal acceleration? (2022)
A.
v²/r
B.
vr
C.
r/v
D.
v/r
Show solution
Solution
Centripetal acceleration a_c = v²/r.
Correct Answer: A — v²/r
Learn More →
Q. A pendulum completes 20 oscillations in 40 seconds. What is its frequency? (2022)
A.
0.5 Hz
B.
1 Hz
C.
2 Hz
D.
4 Hz
Show solution
Solution
Frequency f = number of oscillations / time = 20 / 40 s = 0.5 Hz.
Correct Answer: B — 1 Hz
Learn More →
Q. A pendulum swings back and forth. At which point is its potential energy maximum?
A.
At the lowest point
B.
At the highest point
C.
At the midpoint
D.
At all points
Show solution
Solution
The potential energy of a pendulum is maximum at the highest point of its swing, where its kinetic energy is minimum.
Correct Answer: B — At the highest point
Learn More →
Q. A pendulum swings with a maximum angle of 30 degrees. What is the approximate time period for small angles? (2022)
A.
1.0 s
B.
0.5 s
C.
2.0 s
D.
0.25 s
Show solution
Solution
For small angles, T ≈ 2π√(L/g). Assuming L = 1 m, T ≈ 2π√(1/9.8) ≈ 2.0 s.
Correct Answer: A — 1.0 s
Learn More →
Q. A person lifts a 10 kg box to a height of 2 m. What is the work done against gravity? (g = 9.8 m/s²)
A.
196 J
B.
98 J
C.
20 J
D.
10 J
Show solution
Solution
Work = mgh = 10 kg * 9.8 m/s² * 2 m = 196 J.
Correct Answer: A — 196 J
Learn More →
Q. A projectile is launched with an initial velocity of 30 m/s at an angle of 45 degrees. What is the maximum height reached?
A.
22.5 m
B.
30 m
C.
45 m
D.
15 m
Show solution
Solution
Maximum height (H) = (u² * sin²θ) / (2g) = (30² * (1/2)) / (2 * 9.8) = 22.5 m.
Correct Answer: A — 22.5 m
Learn More →
Showing 91 to 120 of 491 (17 Pages)
