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 1 kg ball is thrown upwards with a force of 10 N. What is the net force acting on the ball at the peak of its motion?

A. 0 N
B. 10 N
C. 5 N
D. 20 N

Solution

At the peak, the only force acting on the ball is its weight (10 N down), so the net force is 0 N.

Correct Answer: A — 0 N

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Q. A 1 kg ball is thrown vertically upwards with a speed of 10 m/s. What is the maximum height it reaches? (g = 10 m/s²)

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

Solution

Using energy conservation, initial kinetic energy = mgh. 0.5 × 1 kg × (10 m/s)² = 1 kg × 10 m/s² × h. h = 5 m.

Correct Answer: B — 10 m

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Q. A 1 kg ball is thrown vertically upwards with a speed of 10 m/s. What is the maximum height it reaches?

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

Solution

Using energy conservation, mgh = 0.5 mv², h = v²/(2g) = (10 m/s)²/(2 × 9.8 m/s²) = 5.1 m.

Correct Answer: A — 5 m

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Q. A 1 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)

A. 20.4 m
B. 30.4 m
C. 40.8 m
D. 50.0 m

Solution

Using energy conservation, initial kinetic energy = mgh; 0.5 × 1 kg × (20 m/s)² = 9.8 m × h; h = 20.4 m.

Correct Answer: A — 20.4 m

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Q. A 1 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches?

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

Solution

Using energy conservation: KE_initial = PE_max; 0.5 × m × v² = mgh; h = v²/(2g) = (20 m/s)²/(2 × 9.8 m/s²) = 20.4 m.

Correct Answer: B — 20 m

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Q. A 1 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches? (Assume g = 10 m/s²)

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

Solution

Using the formula h = v²/(2g), h = (20 m/s)² / (2 * 10 m/s²) = 400 / 20 = 20 m.

Correct Answer: B — 30 m

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Q. A 1 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches? (g = 10 m/s²)

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

Solution

Using energy conservation: Initial K.E. = Potential Energy at max height. 0.5mv² = mgh. h = v²/(2g) = (20 m/s)² / (2 × 10 m/s²) = 20 m.

Correct Answer: B — 30 m

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Q. A 1 kg mass is attached to a spring and compressed by 0.2 m. If the spring constant is 100 N/m, what is the potential energy stored in the spring?

A. 2 J
B. 4 J
C. 6 J
D. 8 J

Solution

Potential energy in a spring is given by PE = 0.5kx². Thus, PE = 0.5 * 100 * (0.2)² = 2 J.

Correct Answer: B — 4 J

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Q. A 1 kg mass is attached to a spring with a spring constant of 200 N/m. What is the force exerted by the spring when it is compressed by 0.1 m?

A. 2 N
B. 5 N
C. 10 N
D. 20 N

Solution

Using Hooke's Law, F = kx = 200 N/m * 0.1 m = 20 N.

Correct Answer: C — 10 N

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Q. A 1 kg mass is attached to a spring with a spring constant of 200 N/m. What is the maximum force exerted by the spring when it is compressed by 0.1 m?

A. 2 N
B. 5 N
C. 20 N
D. 10 N

Solution

Using Hooke's law, F = kx = 200 N/m * 0.1 m = 20 N.

Correct Answer: C — 20 N

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Q. A 1 kg mass is attached to a spring with a spring constant of 200 N/m. What is the maximum extension of the spring when the mass is released from rest?

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

Solution

Using Hooke's law, F = kx, where F = mg = 1 kg * 9.8 m/s² = 9.8 N. Thus, x = F/k = 9.8 N / 200 N/m = 0.049 m.

Correct Answer: A — 0.5 m

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Q. A 1 kg mass is dropped from a height of 1 m. What is its speed just before it hits the ground?

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

Solution

Using conservation of energy, v = √(2gh) = √(2 * 9.8 * 1) = 4.43 m/s.

Correct Answer: B — 2 m/s

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Q. A 1 kg mass is dropped from a height of 1 m. What is the speed just before it hits the ground?

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

Solution

Using conservation of energy, v = sqrt(2gh) = sqrt(2 * 9.8 * 1) = 4.43 m/s.

Correct Answer: B — 2 m/s

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Q. A 1 kg mass is dropped from a height of 10 m. What is its speed just before it hits the ground?

A. 5 m/s
B. 10 m/s
C. 14 m/s
D. 20 m/s

Solution

Using conservation of energy, v = √(2gh) = √(2 * 9.8 * 10) = 14 m/s.

Correct Answer: C — 14 m/s

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Q. A 1 kg mass is dropped from a height of 10 m. What is the speed just before it hits the ground?

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

Solution

Using conservation of energy, potential energy at height = kinetic energy just before hitting the ground. mgh = 0.5mv^2. Solving gives v = sqrt(2gh) = sqrt(2*9.8*10) = 14 m/s.

Correct Answer: B — 10 m/s

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Q. A 1 kg mass is lifted to a height of 5 m. How much work is done against gravity?

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

Solution

Work done against gravity = mgh = 1 * 9.8 * 5 = 49 J.

Correct Answer: B — 10 J

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Q. A 1 kg mass is lifted vertically 10 m. How much work is done against gravity? (g = 9.8 m/s²)

A. 9.8 J
B. 19.6 J
C. 29.4 J
D. 39.2 J

Solution

Work done = mgh = 1 kg × 9.8 m/s² × 10 m = 98 J.

Correct Answer: B — 19.6 J

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Q. A 1 kg object is pushed with a force of 10 N over a distance of 3 m. What is the work done?

A. 10 J
B. 20 J
C. 30 J
D. 40 J

Solution

Work done = Force × Distance = 10 N × 3 m = 30 J.

Correct Answer: C — 30 J

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Q. A 1 kg object is pushed with a force of 10 N. If the frictional force is 4 N, what is the net force acting on the object?

A. 6 N
B. 10 N
C. 4 N
D. 0 N

Solution

Net force = applied force - frictional force = 10 N - 4 N = 6 N.

Correct Answer: A — 6 N

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Q. A 10 kg block is at rest on a horizontal surface. If a horizontal force of 30 N is applied, what will be its acceleration?

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

Solution

Using F = ma, acceleration a = F/m = 30 N / 10 kg = 3 m/s².

Correct Answer: C — 3 m/s²

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Q. A 10 kg block is at rest on a horizontal surface. If a horizontal force of 30 N is applied, what will be the acceleration of the block?

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

Solution

Using F = ma, acceleration a = F/m = 30 N / 10 kg = 3 m/s².

Correct Answer: C — 3 m/s²

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Q. A 10 kg block is at rest on a horizontal surface. If a horizontal force of 30 N is applied, what will be its acceleration? (Assume no friction)

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

Solution

Using F = ma, a = F/m = 30 N / 10 kg = 3 m/s².

Correct Answer: C — 3 m/s²

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Q. A 10 kg object falls freely from a height of 20 m. What is its potential energy at the top?

A. 100 J
B. 200 J
C. 300 J
D. 400 J

Solution

Potential energy is given by PE = mgh = 10 * 9.8 * 20 = 1960 J.

Correct Answer: B — 200 J

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Q. A 10 kg object falls freely from a height of 20 m. What is its speed just before hitting the ground? (g = 9.8 m/s²)

A. 10 m/s
B. 14 m/s
C. 20 m/s
D. 28 m/s

Solution

Using conservation of energy, v = √(2gh) = √(2 * 9.8 * 20) = 28 m/s.

Correct Answer: D — 28 m/s

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Q. A 10 kg object falls from a height of 20 m. What is its potential energy at the top?

A. 100 J
B. 200 J
C. 300 J
D. 400 J

Solution

Potential energy is given by PE = mgh = 10 * 9.8 * 20 = 1960 J.

Correct Answer: D — 400 J

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Q. A 10 kg object falls from a height of 20 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)

A. 10 m/s
B. 14 m/s
C. 20 m/s
D. 28 m/s

Solution

Using conservation of energy, v = √(2gh) = √(2 * 9.8 * 20) = 19.8 m/s.

Correct Answer: B — 14 m/s

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Q. A 10 kg object falls from a height of 20 m. What is its speed just before it hits the ground?

A. 10 m/s
B. 14 m/s
C. 20 m/s
D. 28 m/s

Solution

Using conservation of energy, v = √(2gh) = √(2 * 9.8 * 20) = 20 m/s.

Correct Answer: C — 20 m/s

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Q. A 10 kg object is at rest on a horizontal surface. If a force of 30 N is applied, what is the object's acceleration assuming a frictional force of 10 N?

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

Solution

Net force = 30 N - 10 N = 20 N. Using F = ma, a = F/m = 20 N / 10 kg = 2 m/s².

Correct Answer: B — 3 m/s²

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Q. A 10 kg object is at rest on a surface. If a force of 30 N is applied, what is the object's acceleration assuming the friction is negligible?

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

Solution

Using F = ma, acceleration a = F/m = 30 N / 10 kg = 3 m/s².

Correct Answer: C — 3 m/s²

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Q. A 10 kg object is dropped from a height of 20 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)

A. 14 m/s
B. 19.8 m/s
C. 20 m/s
D. 28 m/s

Solution

Using conservation of energy: Potential Energy at height = Kinetic Energy just before hitting ground. mgh = 0.5mv². v = √(2gh) = √(2 × 9.8 m/s² × 20 m) = 19.8 m/s.

Correct Answer: B — 19.8 m/s

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Physics Syllabus (JEE Main)

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