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Biology Chemistry Physics

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 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 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 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 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 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 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 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 moving with a velocity of 4 m/s. What is its total mechanical energy if it is at a height of 5 m? (2020)

A. 120 J
B. 140 J
C. 100 J
D. 80 J

Solution

Total Mechanical Energy = Kinetic Energy + Potential Energy = (1/2 * 10 kg * (4 m/s)²) + (10 kg * 9.8 m/s² * 5 m) = 80 J + 490 J = 570 J

Correct Answer: B — 140 J

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Q. A 10 kg object is subjected to a net force of 30 N. What is the object's acceleration? (2019)

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-ohm resistor is connected to a 20-volt battery. What is the power consumed by the resistor? (2021)

A. 20 W
B. 40 W
C. 10 W
D. 2 W

Solution

Power (P) can be calculated using the formula P = V^2 / R. Here, P = 20^2 / 10 = 400 / 10 = 40 W.

Correct Answer: B — 40 W

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Q. A 12 kg object is at rest. A force of 36 N is applied to it. What is the acceleration of the object? (2021)

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

Solution

Using F = ma, a = F/m = 36 N / 12 kg = 3 m/s².

Correct Answer: C — 3 m/s²

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Q. A 12 kg object is at rest. A force of 36 N is applied to it. What will be its velocity after 3 seconds? (2020)

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

Solution

Acceleration = Force / Mass = 36 N / 12 kg = 3 m/s². Velocity = Initial velocity + (Acceleration × Time) = 0 + (3 m/s² × 3 s) = 9 m/s.

Correct Answer: C — 3 m/s

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Q. A 12 kg object is moving with a velocity of 3 m/s. What is its momentum? (2019)

A. 36 kg·m/s
B. 12 kg·m/s
C. 9 kg·m/s
D. 24 kg·m/s

Solution

Momentum p = mv = 12 kg * 3 m/s = 36 kg·m/s.

Correct Answer: A — 36 kg·m/s

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Q. A 12 kg object is pulled with a force of 48 N. If the frictional force is 12 N, what is the acceleration of the object? (2023)

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

Solution

Net force = 48 N - 12 N = 36 N. Acceleration a = F/m = 36 N / 12 kg = 3 m/s².

Correct Answer: B — 3 m/s²

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Q. A 15 kg box is pushed with a force of 60 N. If the frictional force is 15 N, what is the acceleration of the box? (2023)

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

Solution

Net force = 60 N - 15 N = 45 N. Acceleration a = F/m = 45 N / 15 kg = 3 m/s².

Correct Answer: B — 3 m/s²

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Q. A 2 kg ball is dropped from a height of 10 m. What is its speed just before it hits the ground? (g = 10 m/s²) (2023)

A. 10 m/s
B. 14 m/s
C. 20 m/s
D. 15 m/s

Solution

Using conservation of energy, Potential Energy at height = Kinetic Energy just before hitting ground. mgh = 1/2 mv^2. Solving gives v = sqrt(2gh) = sqrt(2 * 10 m/s² * 10 m) = 14.14 m/s.

Correct Answer: A — 10 m/s

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Q. A 2 kg ball is dropped from a height of 10 m. What is its speed just before it hits the ground? (2022)

A. 14 m/s
B. 10 m/s
C. 20 m/s
D. 15 m/s

Solution

Using energy conservation, v = sqrt(2gh) = sqrt(2 * 9.8 m/s² * 10 m) = 14 m/s.

Correct Answer: A — 14 m/s

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Q. A 2 kg ball is dropped from a height of 20 m. What is its speed just before it hits the ground? (g = 10 m/s²) (2023)

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

Solution

Using energy conservation, Potential Energy = Kinetic Energy at the ground: mgh = 0.5 mv^2; v = sqrt(2gh) = sqrt(2 * 10 m/s² * 20 m) = 20 m/s

Correct Answer: B — 20 m/s

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Q. A 2 kg block slides down a frictionless incline of height 3 m. What is its speed at the bottom? (2023)

A. 6 m/s
B. 3 m/s
C. 4 m/s
D. 5 m/s

Solution

Using conservation of energy, Potential Energy at top = Kinetic Energy at bottom: mgh = 1/2 mv^2; v = sqrt(2gh) = sqrt(2 * 9.8 m/s² * 3 m) = 7.75 m/s (approx. 6 m/s)

Correct Answer: A — 6 m/s

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Q. A 2 kg mass is dropped from a height. What is the force acting on it just before it hits the ground? (2022)

A. 2 N
B. 10 N
C. 20 N
D. 0 N

Solution

The force acting on it is its weight, F = mg = 2 kg × 10 m/s² = 20 N.

Correct Answer: B — 10 N

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Q. A 2 kg object is dropped from a height of 20 m. What is its speed just before it hits the ground? (g = 10 m/s²) (2023)

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 2 kg object is dropped from a height. What is the force acting on it just before it hits the ground? (2022)

A. 2 N
B. 9.8 N
C. 19.6 N
D. 29.4 N

Solution

The force acting on the object is its weight, F = mg = 2 kg × 9.8 m/s² = 19.6 N.

Correct Answer: C — 19.6 N

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

A. 58.8 J
B. 19.6 J
C. 29.4 J
D. 39.2 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. A 2 kg object is lifted to a height of 3 m. What is the work done against gravity? (2020)

A. 60 J
B. 30 J
C. 20 J
D. 10 J

Solution

Work Done (W) = m * g * h = 2 kg * 9.8 m/s^2 * 3 m = 58.8 J (approx. 60 J)

Correct Answer: B — 30 J

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Q. A 2 kg object is moving with a velocity of 10 m/s. What is its momentum? (2021)

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

Solution

Momentum = Mass × Velocity = 2 kg × 10 m/s = 20 kg·m/s.

Correct Answer: D — 20 kg·m/s

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Q. A 2 kg object is moving with a velocity of 4 m/s. What is its kinetic energy? (2023)

A. 16 J
B. 8 J
C. 4 J
D. 32 J

Solution

Kinetic Energy (KE) = 1/2 * m * v^2 = 1/2 * 2 kg * (4 m/s)^2 = 16 J.

Correct Answer: A — 16 J

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