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Q. A 0.5 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 the formula h = v²/(2g), where v = 20 m/s and g = 9.8 m/s², h = (20)²/(2 * 9.8) ≈ 20.4 m, which rounds to 20 m.

Correct Answer: B — 20 m

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Q. A 1 kg ball is dropped from a height of 20 m. What is its velocity just before it hits the ground? (g = 10 m/s²) (2021)

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

Solution

Using the formula v = √(2gh) = √(2 × 10 m/s² × 20 m) = √400 = 20 m/s.

Correct Answer: B — 20 m/s

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Q. A 1 kg ball is thrown horizontally from a height of 5 m. How long does it take to hit the ground? (g = 10 m/s²) (2022)

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

Solution

Using the formula h = 1/2 gt², 5 m = 1/2 * 10 m/s² * t², t² = 1, t = √1 = 2 s.

Correct Answer: B — 2 s

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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 upwards with a speed of 20 m/s. What is the maximum height it reaches? (g = 10 m/s²) (2021)

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

Solution

Maximum Height (h) = v^2 / (2g) = (20 m/s)^2 / (2 * 10 m/s²) = 20 m

Correct Answer: A — 20 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? (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?

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? (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 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? (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 velocity of 20 m/s. What is the maximum height it reaches? (g = 10 m/s²) (2021)

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

Solution

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

Correct Answer: A — 20 m

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Q. A 1 kg block of ice at 0°C is placed in 2 kg of water at 20°C. What will be the final temperature of the mixture? (Specific heat of water = 4.2 kJ/kg°C, Latent heat of fusion of ice = 334 kJ/kg) (2021)

A. 0°C
B. 10°C
C. 20°C
D. 15°C

Solution

Heat lost by water = Heat gained by ice. Calculate to find the final temperature.

Correct Answer: B — 10°C

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Q. A 1 kg block of ice at 0°C is placed in 2 kg of water at 20°C. What will be the final temperature of the mixture? (Assume no heat loss to the surroundings)

A. 0°C
B. 10°C
C. 15°C
D. 20°C

Solution

Using the principle of conservation of energy, the heat lost by water equals the heat gained by ice. Final temperature can be calculated to be approximately 10°C.

Correct Answer: B — 10°C

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Q. A 1 kg block of ice at 0°C is placed in 2 kg of water at 80°C. What will be the final temperature of the mixture? (Assume no heat loss to the surroundings) (2019)

A. 0°C
B. 40°C
C. 60°C
D. 80°C

Solution

Using the principle of conservation of energy, the heat lost by water equals the heat gained by ice. The final temperature will be 0°C as the ice will melt.

Correct Answer: B — 40°C

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Q. A 1 kg block of metal at 100°C is placed in 2 kg of water at 20°C. Assuming no heat loss to the surroundings, what is the final temperature of the system? (Specific heat of water = 4.18 kJ/kg°C, specific heat of metal = 0.9 kJ/kg°C) (2020)

A. 25°C
B. 30°C
C. 35°C
D. 40°C

Solution

Using the principle of conservation of energy, set heat lost by metal equal to heat gained by water to find the final temperature.

Correct Answer: C — 35°C

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Q. A 1 kg block of metal at 100°C is placed in 2 kg of water at 20°C. Assuming no heat loss to the surroundings, what will be the final temperature? (Specific heat of water = 4.2 kJ/kg°C) (2022)

A. 25°C
B. 30°C
C. 35°C
D. 40°C

Solution

Using the principle of conservation of energy, set heat lost by metal equal to heat gained by water to find the final temperature.

Correct Answer: C — 35°C

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Q. A 1 kg block of metal at 100°C is placed in 2 kg of water at 20°C. If the final temperature of the system is 30°C, what is the specific heat capacity of the metal? (Specific heat of water = 4.18 J/g°C) (2020)

A. 0.5 J/g°C
B. 1.0 J/g°C
C. 1.5 J/g°C
D. 2.0 J/g°C

Solution

Using the principle of conservation of energy, calculate the specific heat capacity of the metal.

Correct Answer: B — 1.0 J/g°C

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Q. A 1 kg block of metal at 100°C is placed in 2 kg of water at 20°C. If the final temperature is 30°C, what is the specific heat capacity of the metal? (2023)

A. 0.5 J/kg°C
B. 1 J/kg°C
C. 2 J/kg°C
D. 4 J/kg°C

Solution

Using conservation of energy: m1c1(T1 - Tf) = m2c2(Tf - T2), we find c1 = 2 J/kg°C.

Correct Answer: C — 2 J/kg°C

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Q. A 1 kg block of metal at 100°C is placed in 2 kg of water at 20°C. What is the final temperature of the system? (Specific heat of water = 4.18 J/g°C, specific heat of metal = 0.9 J/g°C) (2021)

A. 25°C
B. 30°C
C. 35°C
D. 40°C

Solution

Using conservation of energy: m1*c1*(T_initial - T_final) = m2*c2*(T_final - T_initial). Solving gives T_final = 35°C.

Correct Answer: C — 35°C

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Q. A 1 kg block of metal at 100°C is placed in 2 kg of water at 20°C. What will be the final temperature of the system assuming no heat loss? (2022)

A. 30°C
B. 40°C
C. 50°C
D. 60°C

Solution

Using the principle of conservation of energy: m1*c1*(T_initial1 - T_final) = m2*c2*(T_final - T_initial2). Solving gives T_final = 50°C.

Correct Answer: C — 50°C

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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 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 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 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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Major Competitive Exams

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