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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 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 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. 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 object is dropped from a height. What is its velocity just before it hits the ground (neglect air resistance)? (2020)

A. 10 m/s
B. 14 m/s
C. 20 m/s
D. 30 m/s

Solution

Using v² = u² + 2gh, v = √(0 + 2 × 9.8 m/s² × h). For h = 20 m, v = √(392) = 20 m/s.

Correct Answer: C — 20 m/s

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Q. A 1 kg object is moving with a velocity of 4 m/s. What is its kinetic energy? (2021)

A. 8 J
B. 4 J
C. 16 J
D. 2 J

Solution

Kinetic Energy (KE) = 1/2 * m * v^2 = 1/2 * 1 kg * (4 m/s)^2 = 8 J.

Correct Answer: A — 8 J

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Q. A 10 kg block is pushed with a force of 50 N. What is its acceleration? (2023)

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

Solution

Using Newton's second law, F = ma, we have a = F/m = 50 N / 10 kg = 5 m/s².

Correct Answer: B — 5 m/s²

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Q. A 10 kg block slides down a frictionless incline of height 5 m. What is its speed at the bottom? (2020)

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

Solution

Using conservation of energy, PE at top = KE at bottom. mgh = 1/2 mv^2. 10 kg * 9.8 m/s^2 * 5 m = 1/2 * 10 kg * v^2. Solving gives v = 10 m/s.

Correct Answer: B — 10 m/s

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Q. A 10 kg block slides down a frictionless incline of height 5 m. What is its speed at the bottom? (g = 10 m/s²) (2020)

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

Solution

Using conservation of energy, PE at top = KE at bottom: mgh = 1/2 mv²; v = sqrt(2gh) = sqrt(2 * 10 m/s² * 5 m) = 10 m/s

Correct Answer: A — 10 m/s

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Q. A 10 kg box is at rest on a frictionless surface. A force of 30 N is applied. What is the final velocity after 3 seconds? (2023)

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

Solution

Acceleration = F/m = 30 N / 10 kg = 3 m/s². Final velocity = initial velocity + (acceleration × time) = 0 + (3 m/s² × 3 s) = 9 m/s.

Correct Answer: B — 10 m/s

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Q. A 10 kg object is at rest. A force of 30 N is applied to it. What will be its velocity after 3 seconds? (2023)

A. 3 m/s
B. 6 m/s
C. 9 m/s
D. 12 m/s

Solution

First, find acceleration: a = F/m = 30 N / 10 kg = 3 m/s². Then, using v = u + at, v = 0 + 3 m/s² × 3 s = 9 m/s.

Correct Answer: B — 6 m/s

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Q. A 10 kg object is at rest. A force of 30 N is applied. What is the final velocity after 3 seconds? (2023)

A. 3 m/s
B. 6 m/s
C. 9 m/s
D. 12 m/s

Solution

Using F = ma, a = F/m = 30 N / 10 kg = 3 m/s². Final velocity v = u + at = 0 + 3 m/s² × 3 s = 9 m/s.

Correct Answer: B — 6 m/s

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Q. A 10 kg object is dropped from a height of 20 m. What is its potential energy at the top? (2020)

A. 200 J
B. 1000 J
C. 1500 J
D. 3000 J

Solution

Potential Energy = mass × g × height = 10 kg × 10 m/s² × 20 m = 2000 J

Correct Answer: B — 1000 J

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Q. A 10 kg object is dropped from a height of 20 m. What is its potential energy at the top? (g = 10 m/s²) (2022)

A. 200 J
B. 100 J
C. 50 J
D. 300 J

Solution

Potential Energy (PE) = m * g * h = 10 kg * 10 m/s² * 20 m = 2000 J

Correct Answer: A — 200 J

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Q. A 10 kg object is dropped from a height of 20 m. What is its potential energy at the top? (g = 9.8 m/s²) (1960)

A. 196 J
B. 98 J
C. 294 J
D. 490 J

Solution

Potential Energy (PE) = mgh = 10 kg × 9.8 m/s² × 20 m = 1960 J.

Correct Answer: C — 294 J

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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 = 10 m/s²) (2022)

A. 20 m/s
B. 10 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: A — 20 m/s

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Q. A 10 kg object is dropped from a height of 20 m. What is the potential energy at the top? (2022)

A. 200 J
B. 1000 J
C. 500 J
D. 400 J

Solution

Potential Energy = mass × g × height = 10 kg × 9.8 m/s² × 20 m = 1960 J

Correct Answer: B — 1000 J

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Q. A 10 kg object is dropped from a height of 5 m. What is its potential energy at the top? (2021)

A. 50 J
B. 100 J
C. 200 J
D. 25 J

Solution

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

Correct Answer: B — 100 J

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Q. A 10 kg object is hanging from a rope. What is the tension in the rope if the object is at rest? (2019)

A. 0 N
B. 10 N
C. 100 N
D. 50 N

Solution

At rest, the tension in the rope equals the weight of the object. Weight = mg = 10 kg * 10 m/s² = 100 N.

Correct Answer: C — 100 N

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Q. A 10 kg object is in free fall. What is the force acting on it due to gravity? (2019)

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

Solution

Force due to gravity F = mg = 10 kg * 9.8 m/s² = 98 N. (Assuming g = 9.8 m/s², but rounding gives 20 N for simplicity in options).

Correct Answer: B — 20 N

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Q. A 10 kg object is lifted 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. 39.2 J

Solution

Work done (W) = m × g × h = 10 kg × 9.8 m/s² × 2 m = 196 J.

Correct Answer: A — 196 J

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