Major Competitive Exams

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Q. If a 12V battery is connected across a 4 ohm resistor, what is the current flowing through the resistor?

A. 2 A
B. 3 A
C. 4 A
D. 6 A

Solution

Using Ohm's law, I = V/R = 12V / 4Ω = 3 A.

Correct Answer: A — 2 A

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Q. If a 12V battery is connected across a 4 ohm resistor, what is the power dissipated in the resistor?

A. 12 W
B. 24 W
C. 36 W
D. 48 W

Solution

Power (P) can be calculated using P = V^2 / R = 12V^2 / 4Ω = 144 / 4 = 36 W.

Correct Answer: B — 24 W

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Q. If a 12V battery is connected across a 4Ω resistor, what is the current flowing through the resistor?

A. 3A
B. 2A
C. 4A
D. 1A

Solution

Using Ohm's law, I = V/R = 12V / 4Ω = 3A.

Correct Answer: B — 2A

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Q. If a 12V battery is connected to a 4 ohm resistor, what is the power dissipated by the resistor?

A. 12 W
B. 24 W
C. 36 W
D. 48 W

Solution

Power (P) can be calculated using P = V^2 / R = 12V^2 / 4Ω = 36 W.

Correct Answer: B — 24 W

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Q. If a 1500 kg car is moving at a speed of 30 m/s, what is its kinetic energy?

A. 675,000 J
B. 450,000 J
C. 225,000 J
D. 900,000 J

Solution

Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 1500 kg × (30 m/s)² = 675,000 J

Correct Answer: B — 450,000 J

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Q. If a 15Ω resistor is connected in parallel with a 30Ω resistor, what is the equivalent resistance? (2021)

A. 10Ω
B. 7.5Ω
C. 5Ω
D. 12Ω

Solution

For resistors in parallel, 1/R_eq = 1/R1 + 1/R2 = 1/15 + 1/30 = 2/30 + 1/30 = 3/30. Therefore, R_eq = 30/3 = 10Ω.

Correct Answer: B — 7.5Ω

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Q. If a 15Ω resistor is connected in parallel with a 5Ω resistor, what is the equivalent resistance? (2023)

A. 3.75Ω
B. 4Ω
C. 5Ω
D. 10Ω

Solution

1/R_eq = 1/15 + 1/5 = 1/15 + 3/15 = 4/15; R_eq = 15/4 = 3.75Ω.

Correct Answer: A — 3.75Ω

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Q. If a 2 kg object is acted upon by a net force of 6 N, what is the object's acceleration?

A. 2 m/s²
B. 3 m/s²
C. 4 m/s²
D. 5 m/s²

Solution

Using F = ma, we find a = F/m = 6 N / 2 kg = 3 m/s².

Correct Answer: B — 3 m/s²

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Q. If a 2 kg object is dropped from a height of 5 m, what is its speed just before it hits the ground? (g = 10 m/s²)

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

Solution

Using energy conservation: PE_initial = KE_final; m * g * h = 1/2 * m * v^2; v = sqrt(2gh) = sqrt(2 * 10 * 5) = 10 m/s

Correct Answer: A — 10 m/s

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Q. If a 2 kg object is dropped from a height of 5 m, what is its velocity just before it hits the ground? (g = 10 m/s²)

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

Solution

Using energy conservation, v = sqrt(2gh) = sqrt(2 * 10 * 5) = 10 m/s

Correct Answer: A — 10 m/s

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Q. If a 2 kg object is dropped from a height, what is its velocity just before hitting the ground (g = 10 m/s²)? (2023)

A. 10 m/s
B. 20 m/s
C. 30 m/s
D. 40 m/s

Solution

Using v = √(2gh), for h = 10 m, v = √(2 * 10 m/s² * 10 m) = √200 = 20 m/s.

Correct Answer: B — 20 m/s

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Q. If a 2 kg object is dropped from a height, what is its velocity just before hitting the ground? (g = 9.8 m/s², h = 20 m) (2023)

A. 14 m/s
B. 19.6 m/s
C. 20 m/s
D. 28 m/s

Solution

Using v² = u² + 2gh, where u = 0, we find v = √(2 * 9.8 * 20) = 19.6 m/s.

Correct Answer: B — 19.6 m/s

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Q. If a 2 kg object is lifted to a height of 10 m, what is the potential energy gained by the object? (g = 9.8 m/s²)

A. 19.6 J
B. 98 J
C. 39.2 J
D. 78.4 J

Solution

Potential Energy (PE) = mgh = 2 kg × 9.8 m/s² × 10 m = 196 J.

Correct Answer: B — 98 J

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Q. If a 2 kg object is lifted to a height of 3 m, what is the potential energy gained by the object? (g = 9.8 m/s²)

A. 58.8 J
B. 19.6 J
C. 29.4 J
D. 39.2 J

Solution

Potential Energy (PE) = mgh = 2 kg × 9.8 m/s² × 3 m = 58.8 J.

Correct Answer: A — 58.8 J

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Q. If a 2 kg object is lifted to a height of 3 m, what is the potential energy gained by the object? (Take g = 9.8 m/s²) (2020)

A. 58.8 J
B. 19.6 J
C. 29.4 J
D. 39.2 J

Solution

Potential Energy (PE) = mgh = 2 kg × 9.8 m/s² × 3 m = 58.8 J.

Correct Answer: A — 58.8 J

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Q. If a 2 kg object is lifted to a height of 3 m, what is the potential energy gained? (Use g = 9.8 m/s²) (2020)

A. 58.8 J
B. 19.6 J
C. 29.4 J
D. 39.2 J

Solution

Potential Energy = mass × g × height = 2 kg × 9.8 m/s² × 3 m = 58.8 J

Correct Answer: A — 58.8 J

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Q. If a 2 kg object is lifted to a height of 3 m, what is the potential energy gained by the object? (Use g = 9.8 m/s²)

A. 58.8 J
B. 19.6 J
C. 29.4 J
D. 39.2 J

Solution

Potential Energy (PE) = mgh = 2 kg × 9.8 m/s² × 3 m = 58.8 J.

Correct Answer: A — 58.8 J

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Q. If a 2 kg object is lifted to a height of 3 m, what is the work done against gravity? (Use g = 9.8 m/s²) (2023)

A. 58.8 J
B. 39.2 J
C. 19.6 J
D. 29.4 J

Solution

Work done = mgh = 2 kg × 9.8 m/s² × 3 m = 58.8 J.

Correct Answer: A — 58.8 J

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Q. If a 2 kg object is lifted to a height of 5 m, what is the work done against gravity? (g = 10 m/s²) (2019)

A. 20 J
B. 10 J
C. 50 J
D. 30 J

Solution

Work Done = mgh = 2 kg * 10 m/s² * 5 m = 100 J

Correct Answer: A — 20 J

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Q. If a 2 kg object is moving with a velocity of 15 m/s, what is its kinetic energy?

A. 225 J
B. 150 J
C. 300 J
D. 450 J

Solution

Kinetic Energy = 1/2 * m * v^2 = 1/2 * 2 kg * (15 m/s)^2 = 225 J.

Correct Answer: A — 225 J

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Q. If a 2 kg object is moving with a velocity of 3 m/s and a force of 6 N is applied in the opposite direction, what will be its final velocity after 2 seconds?

A. 0 m/s
B. 1 m/s
C. 2 m/s
D. 3 m/s

Solution

Net force = -6 N, acceleration = F/m = -6 N / 2 kg = -3 m/s². Final velocity = initial velocity + at = 3 m/s + (-3 m/s² * 2 s) = 0 m/s.

Correct Answer: B — 1 m/s

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Q. If a 2 kg object is moving with a velocity of 3 m/s, what is its kinetic energy?

A. 9 J
B. 6 J
C. 3 J
D. 12 J

Solution

Kinetic Energy = 0.5 * m * v² = 0.5 * 2 kg * (3 m/s)² = 9 J.

Correct Answer: A — 9 J

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Q. If a 2 kg object is pulled with a force of 10 N and experiences a frictional force of 4 N, what is its acceleration?

A. 3 m/s²
B. 5 m/s²
C. 2 m/s²
D. 1 m/s²

Solution

Net force = applied force - friction = 10 N - 4 N = 6 N. Acceleration a = F/m = 6 N / 2 kg = 3 m/s².

Correct Answer: A — 3 m/s²

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Q. If a 2 kg object is subjected to a net force of 10 N, what will be its acceleration? (2022)

A. 2 m/s²
B. 5 m/s²
C. 10 m/s²
D. 20 m/s²

Solution

Using F = ma, a = F/m = 10 N / 2 kg = 5 m/s².

Correct Answer: B — 5 m/s²

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Q. If a 2 kg object is subjected to a net force of 6 N, what is its acceleration?

A. 2 m/s²
B. 3 m/s²
C. 4 m/s²
D. 5 m/s²

Solution

Using F = ma, we have a = F/m = 6 N / 2 kg = 3 m/s².

Correct Answer: B — 3 m/s²

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Q. If a 3 kg object is in free fall, what is the force acting on it due to gravity?

A. 3 N
B. 9 N
C. 30 N
D. None of the above

Solution

Force due to gravity F = mg = 3 kg * 9.8 m/s² = 29.4 N, approximately 9 N for simplification.

Correct Answer: B — 9 N

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Q. If a 3 kg object is moving with a speed of 4 m/s and comes to a stop, what is the work done by friction?

A. -24 J
B. -48 J
C. 0 J
D. 24 J

Solution

Initial kinetic energy = 0.5 × m × v² = 0.5 × 3 kg × (4 m/s)² = 24 J. Work done by friction = -24 J.

Correct Answer: B — -48 J

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Q. If a 3 kg object is moving with a speed of 4 m/s and comes to a stop, what is the work done by the friction force?

A. -24 J
B. -48 J
C. -72 J
D. -96 J

Solution

Initial kinetic energy = 0.5 × 3 kg × (4 m/s)² = 24 J. Work done by friction = -24 J.

Correct Answer: B — -48 J

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Q. If a 3 kg object is moving with a speed of 4 m/s and comes to a stop, what is the work done by the friction?

A. -24 J
B. -48 J
C. -12 J
D. -36 J

Solution

Initial kinetic energy = 0.5 × 3 kg × (4 m/s)² = 24 J. Work done by friction = -24 J.

Correct Answer: B — -48 J

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Q. If a 3 kg object is moving with a speed of 4 m/s, what is its kinetic energy?

A. 12 J
B. 24 J
C. 36 J
D. 48 J

Solution

KE = 0.5 × m × v² = 0.5 × 3 kg × (4 m/s)² = 0.5 × 3 × 16 = 24 J.

Correct Answer: B — 24 J

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Major Competitive Exams

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