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Physics

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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

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

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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

Solution

Angular velocity ω = v/r.

Correct Answer: A — v/r

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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

Solution

Centripetal acceleration a_c = v²/r.

Correct Answer: A — v²/r

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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

Solution

Frequency f = number of oscillations / time = 20 / 40 s = 0.5 Hz.

Correct Answer: B — 1 Hz

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Q. A pendulum of length 1 m swings with a small amplitude. What is its approximate time period? (2023)

A. 1 s
B. 2 s
C. 0.5 s
D. 0.25 s

Solution

Time period (T) = 2π√(L/g) ≈ 2π√(1/9.8) ≈ 2 s.

Correct Answer: B — 2 s

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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

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

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Q. A pendulum swings with a maximum angle of 15 degrees. What is the approximate frequency of the pendulum if its length is 1 m? (2020)

A. 0.5 Hz
B. 1 Hz
C. 1.5 Hz
D. 2 Hz

Solution

Frequency (f) = 1 / (2π√(L/g)) = 1 / (2π√(1/9.8)) ≈ 1 Hz.

Correct Answer: B — 1 Hz

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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

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

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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

Solution

Work = mgh = 10 kg * 9.8 m/s² * 2 m = 196 J.

Correct Answer: A — 196 J

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Q. A projectile is launched at an angle of 30 degrees with an initial speed of 20 m/s. What is the maximum height reached? (g = 10 m/s²)

A. 5 m
B. 10 m
C. 15 m
D. 20 m

Solution

Maximum height h = (u^2 * sin²θ) / (2g) = (20² * (1/2)²) / (2 * 10) = 5 m.

Correct Answer: B — 10 m

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Q. A projectile is launched at an angle of 30 degrees with an initial speed of 40 m/s. What is the maximum height reached? (g = 10 m/s²)

A. 80 m
B. 60 m
C. 40 m
D. 20 m

Solution

Maximum height = (u^2 * sin²θ) / (2g) = (40^2 * (1/2)) / (2 * 10) = 80 m.

Correct Answer: B — 60 m

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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

Solution

Maximum height (H) = (u² * sin²θ) / (2g) = (30² * (1/2)) / (2 * 9.8) = 22.5 m.

Correct Answer: A — 22.5 m

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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 by the projectile? (2023)

A. 22.5 m
B. 30 m
C. 45 m
D. 15 m

Solution

Maximum height (H) = (u² * sin²θ) / (2g). Here, u = 30 m/s, θ = 45°, g = 9.81 m/s². H = (30² * (1/2)) / (2 * 9.81) = 22.5 m.

Correct Answer: A — 22.5 m

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Q. A ray of light passes from air into glass at an angle of 45 degrees. What is the angle of refraction if the refractive index of glass is 1.5? (2022)

A. 30 degrees
B. 45 degrees
C. 60 degrees
D. 90 degrees

Solution

Using Snell's law, n1 * sin(theta1) = n2 * sin(theta2). Here, n1 = 1 (air), theta1 = 45 degrees, n2 = 1.5 (glass). Thus, sin(theta2) = (1 * sin(45))/1.5 = 0.471, giving theta2 ≈ 30 degrees.

Correct Answer: A — 30 degrees

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Q. A refrigerator operates between 250 K and 300 K. What is its coefficient of performance (COP)? (2022)

A. 1.2
B. 2.0
C. 3.0
D. 4.0

Solution

COP = T_cold / (T_hot - T_cold) = 250 K / (300 K - 250 K) = 250 / 50 = 5.0.

Correct Answer: C — 3.0

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Q. A refrigerator removes heat from the inside at a rate of 200 J/s and expels it to the surroundings at a rate of 250 J/s. What is the coefficient of performance (COP) of the refrigerator? (2023)

A. 0.8
B. 1.25
C. 1.5
D. 2.0

Solution

COP = Q_in / W = 200 J/s / (250 J/s - 200 J/s) = 200 / 50 = 4.

Correct Answer: B — 1.25

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Q. A refrigerator removes heat from the inside at a rate of 200 J/s and expels it to the outside at a rate of 250 J/s. What is the coefficient of performance (COP) of the refrigerator? (2022)

A. 0.8
B. 1.25
C. 1.5
D. 2.0

Solution

COP = Q_c / W = 200 J/s / (250 J/s - 200 J/s) = 200 / 50 = 4.

Correct Answer: B — 1.25

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Q. A refrigerator removes heat from the interior at a rate of 200 W. If the coefficient of performance (COP) is 4, what is the power consumed by the refrigerator?

A. 50 W
B. 100 W
C. 200 W
D. 800 W

Solution

COP = Qc/W, thus W = Qc/COP = 200 W / 4 = 50 W.

Correct Answer: B — 100 W

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Q. A resistor of 10 Ω and a capacitor of 100 μF are connected in series to an AC source of frequency 50 Hz. What is the capacitive reactance? (2020)

A. 3.18 Ω
B. 31.8 Ω
C. 0.318 Ω
D. 318 Ω

Solution

Capacitive reactance Xc = 1/(2πfC) = 1/(2π(50)(100 × 10^-6)) ≈ 3.18 Ω.

Correct Answer: A — 3.18 Ω

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Q. A resistor of 10Ω is connected across a 20V battery. What is the power consumed by the resistor? (2021)

A. 40W
B. 20W
C. 10W
D. 5W

Solution

Power (P) is given by P = V^2/R = 20V^2 / 10Ω = 400W / 10 = 40W.

Correct Answer: A — 40W

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Q. A resistor of 10Ω is connected to a 20V battery. What is the power consumed by the resistor? (2021)

A. 40W
B. 20W
C. 10W
D. 2W

Solution

Power (P) is given by P = V^2/R = (20V)^2 / 10Ω = 400/10 = 40W.

Correct Answer: A — 40W

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Q. A resistor of 10Ω is connected to a 20V power supply. What is the power consumed by the resistor? (2021)

A. 40W
B. 20W
C. 10W
D. 2W

Solution

Power (P) is given by P = V^2/R = (20V)^2 / 10Ω = 400/10 = 40W.

Correct Answer: A — 40W

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Q. A rotating object has a moment of inertia I and an angular velocity ω. What is its rotational kinetic energy? (2020)

A. (1/2)Iω
B. (1/2)Iω²
C. Iω²
D. I/2ω²

Solution

Rotational kinetic energy K.E. = (1/2)Iω².

Correct Answer: B — (1/2)Iω²

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Q. A rotating object has an angular momentum L. If its moment of inertia is halved and angular velocity is doubled, what is the new angular momentum? (2022)

A. L
B. 2L
C. 3L
D. 4L

Solution

L = Iω, if I is halved and ω is doubled, L' = (1/2)(2ω) = L.

Correct Answer: B — 2L

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Q. A runner completes a 400 m lap in 50 seconds. What is his average velocity?

A. 6 m/s
B. 7 m/s
C. 8 m/s
D. 9 m/s

Solution

Average velocity = total distance / total time = 400 m / 50 s = 8 m/s.

Correct Answer: A — 6 m/s

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Q. A runner completes a 400 m lap in 50 seconds. What is the runner's average velocity?

A. 8 m/s
B. 6 m/s
C. 4 m/s
D. 2 m/s

Solution

Average velocity = total displacement / total time. Since the runner returns to the starting point, displacement = 0. Average velocity = 0 m/s.

Correct Answer: A — 8 m/s

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Q. A satellite is in a circular orbit around the Earth. If its orbital radius is tripled, how does the gravitational force acting on it change?

A. It triples
B. It halves
C. It becomes one-ninth
D. It remains the same

Solution

Gravitational force is inversely proportional to the square of the radius. If radius is tripled, force becomes 1/3² = 1/9.

Correct Answer: C — It becomes one-ninth

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Q. A simple pendulum completes 20 oscillations in 40 seconds. What is its time period? (2021)

A. 1 s
B. 2 s
C. 0.5 s
D. 0.25 s

Solution

Time period (T) = Total time / Number of oscillations = 40 s / 20 = 2 s.

Correct Answer: B — 2 s

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Q. A simple pendulum has a length of 1 m. What is its time period? (2023)

A. 1.0 s
B. 2.0 s
C. 0.5 s
D. 0.25 s

Solution

The time period T of a simple pendulum is given by T = 2π√(L/g). For L = 1 m and g = 9.8 m/s², T = 2π√(1/9.8) ≈ 2.0 s.

Correct Answer: A — 1.0 s

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Q. A simple pendulum oscillates with a period of 2 seconds. What is the length of the pendulum? (2021)

A. 0.5 m
B. 1 m
C. 2 m
D. 4 m

Solution

The period T of a simple pendulum is given by T = 2π√(L/g). Rearranging gives L = (T^2 * g) / (4π^2). Using g = 9.8 m/s² and T = 2 s, we find L = (2^2 * 9.8) / (4π^2) ≈ 1 m.

Correct Answer: B — 1 m

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Physics

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