Physics Syllabus (JEE Main)

Current Electricity Electrostatics Gravitation Kinematics Kinetic Theory of Gases Laws of Motion Magnetism & EMI Modern Physics Optics Oscillations & Waves Properties of Matter Rotational Motion Thermodynamics Units & Measurement Work, Energy & Power

Q. A 5 kg object is at rest on a surface. If a horizontal force of 15 N is applied, what is the object's acceleration? (Assume friction is negligible)

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

Solution

Using F = ma, a = F/m = 15 N / 5 kg = 3 m/s².

Correct Answer: B — 3 m/s²

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Q. A 5 kg object is dropped from a height of 10 m. What is the total mechanical energy just before it hits the ground?

A. 0 J
B. 50 J
C. 100 J
D. 200 J

Solution

Total mechanical energy is conserved. At height, PE = mgh = 5 * 9.8 * 10 = 490 J. Just before hitting the ground, total energy = 490 J.

Correct Answer: C — 100 J

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Q. A 5 kg object is dropped from a height of 20 m. What is the potential energy at the top? (g = 9.8 m/s²)

A. 980 J
B. 490 J
C. 196 J
D. 9800 J

Solution

Potential energy is given by PE = mgh. Here, PE = 5 * 9.8 * 20 = 980 J.

Correct Answer: A — 980 J

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Q. A 5 kg object is dropped from a height of 20 m. What is the total mechanical energy just before it hits the ground?

A. 0 J
B. 100 J
C. 200 J
D. 500 J

Solution

Total mechanical energy is conserved. At height, PE = mgh = 5 * 9.8 * 20 = 980 J. Just before hitting the ground, total energy = 980 J.

Correct Answer: C — 200 J

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Q. A 5 kg object is dropped from a height. What is the force acting on it just before it hits the ground?

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

Solution

The only force acting on it is its weight, W = mg = 5 kg * 10 m/s² = 50 N.

Correct Answer: B — 10 N

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Q. A 5 kg object is dropped from a height. What is the force acting on it just before it hits the ground? (Assume g = 9.8 m/s²)

A. 5 N
B. 9.8 N
C. 49 N
D. 0 N

Solution

The force acting on the object is its weight, F = mg = 5 kg * 9.8 m/s² = 49 N.

Correct Answer: C — 49 N

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Q. A 5 kg object is lifted to a height of 10 m. What is the gravitational potential energy gained?

A. 50 J
B. 100 J
C. 200 J
D. 500 J

Solution

Potential Energy (PE) = m * g * h = 5 kg * 10 m/s^2 * 10 m = 500 J

Correct Answer: B — 100 J

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Q. A 5 kg object is lifted to a height of 10 m. What is the potential energy gained by the object?

A. 50 J
B. 100 J
C. 150 J
D. 200 J

Solution

Potential energy (PE) = mgh = 5 kg * 9.8 m/s² * 10 m = 490 J.

Correct Answer: B — 100 J

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Q. A 5 kg object is lifted to a height of 10 m. What is the potential energy gained?

A. 50 J
B. 100 J
C. 200 J
D. 500 J

Solution

Potential Energy (PE) = m * g * h = 5 kg * 9.8 m/s^2 * 10 m = 490 J (approx. 500 J)

Correct Answer: B — 100 J

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Q. A 5 kg object is lifted to a height of 10 m. What is the work done against gravity?

A. 50 J
B. 100 J
C. 150 J
D. 200 J

Solution

Work done against gravity is W = mgh = 5 * 9.8 * 10 = 490 J.

Correct Answer: D — 200 J

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Q. A 5 kg object is lifted to a height of 3 m. What is the work done against gravity?

A. 15 J
B. 30 J
C. 45 J
D. 60 J

Solution

Work done against gravity is W = mgh = 5 * 9.8 * 3 = 147 J.

Correct Answer: B — 30 J

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Q. A 5 kg object is moving with a velocity of 10 m/s. What is its kinetic energy?

A. 25 J
B. 50 J
C. 100 J
D. 250 J

Solution

Kinetic energy (KE) = 0.5 * mass * velocity² = 0.5 * 5 * (10)² = 250 J.

Correct Answer: C — 100 J

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Q. A 5 kg object is moving with a velocity of 3 m/s. What is its kinetic energy?

A. 22.5 J
B. 30 J
C. 45 J
D. 60 J

Solution

Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 5 kg × (3 m/s)² = 22.5 J.

Correct Answer: A — 22.5 J

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Q. A 5 kg object is pulled with a force of 25 N. What is the acceleration of the object?

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

Solution

Using F = ma, acceleration a = F/m = 25 N / 5 kg = 5 m/s².

Correct Answer: C — 4 m/s²

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Q. A 5 kg object is pushed with a force of 15 N. If the frictional force is 5 N, what is the acceleration of the object?

A. 1 m/s²
B. 2 m/s²
C. 3 m/s²
D. 4 m/s²

Solution

Net force = applied force - friction = 15 N - 5 N = 10 N. Acceleration = F/m = 10 N / 5 kg = 2 m/s².

Correct Answer: B — 2 m/s²

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Q. A 5 kg object is subjected to a net force of 15 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, a = F/m = 15 N / 5 kg = 3 m/s².

Correct Answer: B — 3 m/s²

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Q. A 5 kg object is subjected to a net force of 15 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, a = F/m = 15 N / 5 kg = 3 m/s².

Correct Answer: B — 3 m/s²

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Q. A 5 kg object is subjected to a net force of 25 N. What is its acceleration?

A. 2 m/s²
B. 5 m/s²
C. 10 m/s²
D. 15 m/s²

Solution

Using F = ma, a = F/m = 25 N / 5 kg = 5 m/s².

Correct Answer: C — 10 m/s²

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Q. A 5 kg object is thrown vertically upward with a speed of 20 m/s. What is the maximum height reached by the object? (g = 10 m/s²)

A. 20 m
B. 30 m
C. 40 m
D. 50 m

Solution

Using the formula h = v²/(2g) = (20 m/s)² / (2 * 10 m/s²) = 20 m.

Correct Answer: B — 30 m

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Q. A 5 ohm resistor and a 10 ohm resistor are connected in series. What is the total resistance?

A. 15 ohms
B. 5 ohms
C. 10 ohms
D. 2 ohms

Solution

In series, the total resistance is R_total = R1 + R2 = 5 + 10 = 15 ohms.

Correct Answer: A — 15 ohms

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Q. A 50 kg crate is at rest on a flat surface. What is the normal force acting on the crate?

A. 0 N
B. 50 N
C. 500 N
D. 1000 N

Solution

The normal force equals the weight of the crate, which is N = mg = 50 kg * 10 m/s² = 500 N.

Correct Answer: C — 500 N

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Q. A 50 kg object is at rest on a surface. What is the normal force acting on it?

A. 0 N
B. 50 N
C. 100 N
D. 500 N

Solution

The normal force equals the weight of the object, which is F = mg = 50 kg * 9.8 m/s² = 490 N.

Correct Answer: C — 100 N

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Q. A 50 kg object is in free fall. What is the net force acting on it?

A. 50 N
B. 100 N
C. 200 N
D. 500 N

Solution

The net force is equal to the weight of the object, F = mg = 50 kg * 9.8 m/s² = 490 N (approximately 500 N).

Correct Answer: B — 100 N

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Q. A 50 kg object is moving with a constant velocity. What can be said about the net force acting on it?

A. It is zero
B. It is equal to its weight
C. It is equal to the applied force
D. It is maximum

Solution

If the object is moving with constant velocity, the net force acting on it is zero.

Correct Answer: A — It is zero

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Q. A 50 kg object is moving with a velocity of 10 m/s. What is its momentum?

A. 500 kg m/s
B. 1000 kg m/s
C. 1500 kg m/s
D. 2000 kg m/s

Solution

Momentum = mass × velocity = 50 kg × 10 m/s = 500 kg m/s.

Correct Answer: B — 1000 kg m/s

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Q. A 50 kg object is pulled with a force of 200 N. What is the acceleration of the object?

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 = 200 N / 50 kg = 4 m/s².

Correct Answer: D — 5 m/s²

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Q. A 50 kg object is pushed with a force of 200 N over a distance of 10 m. What is the work done? (2000)

A. 2000 J
B. 1000 J
C. 500 J
D. 3000 J

Solution

Work Done = Force * Distance = 200 N * 10 m = 2000 J

Correct Answer: A — 2000 J

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Q. A 50 kg person jumps off a diving board with a speed of 5 m/s. What is the momentum of the person just before hitting the water?

A. 100 kg·m/s
B. 200 kg·m/s
C. 250 kg·m/s
D. 300 kg·m/s

Solution

Momentum p = mv = 50 kg * 5 m/s = 250 kg·m/s.

Correct Answer: A — 100 kg·m/s

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Q. A 500 g block of ice at -10°C is heated until it melts completely and the water is at 0°C. How much heat is required? (Specific heat of ice = 2.1 J/g°C, Latent heat of fusion = 334 J/g) (2000)

A. 1050 J
B. 1340 J
C. 1500 J
D. 2000 J

Solution

Heat to raise ice to 0°C = 500g * 2.1 J/g°C * 10°C = 10500 J. Heat to melt = 500g * 334 J/g = 167000 J. Total = 1050 J + 167000 J = 1340 J.

Correct Answer: B — 1340 J

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Q. A 6 kg object is at rest on a horizontal surface. If a horizontal force of 12 N is applied, what is the acceleration assuming no friction?

A. 1 m/s²
B. 2 m/s²
C. 3 m/s²
D. 4 m/s²

Solution

Using F = ma, acceleration a = F/m = 12 N / 6 kg = 2 m/s².

Correct Answer: B — 2 m/s²

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Physics Syllabus (JEE Main)

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