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Q. A 20 kg object is acted upon by a net force of 80 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

Acceleration = Net force / Mass = 80 N / 20 kg = 4 m/s².

Correct Answer: C — 4 m/s²

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Q. A 20 kg object is moving with a velocity of 5 m/s. What is its kinetic energy? (2021)

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

Solution

Kinetic energy KE = 0.5 * m * v² = 0.5 * 20 kg * (5 m/s)² = 0.5 * 20 * 25 = 250 J.

Correct Answer: C — 200 J

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Q. A 20 kg object is thrown vertically upward with a force of 100 N. What is the net force acting on the object? (2023)

A. 0 N
B. 20 N
C. 80 N
D. 100 N

Solution

Net force = Applied force - Weight = 100 N - (20 kg * 10 m/s²) = 100 N - 200 N = -100 N (downward).

Correct Answer: C — 80 N

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Q. A 25 kg object is pulled with a force of 100 N. If the frictional force is 25 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 = 100 N - 25 N = 75 N. Acceleration = Net force / Mass = 75 N / 25 kg = 3 m/s².

Correct Answer: C — 4 m/s²

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Q. A 3 kg object is dropped from a height of 12 m. What is its speed just before it hits the ground? (g = 10 m/s²) (2023)

A. 15 m/s
B. 10 m/s
C. 12 m/s
D. 20 m/s

Solution

Using conservation of energy, PE = KE: mgh = 1/2 mv²; v = sqrt(2gh) = sqrt(2 * 10 m/s² * 12 m) = 15.49 m/s

Correct Answer: A — 15 m/s

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Q. A 3 kg object is lifted to a height of 4 m. What is the gravitational potential energy gained? (2023)

A. 120 J
B. 60 J
C. 30 J
D. 90 J

Solution

Potential Energy (PE) = m * g * h = 3 kg * 9.8 m/s^2 * 4 m = 117.6 J (approximately 120 J)

Correct Answer: A — 120 J

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Q. A 3 kg object is moving with a velocity of 2 m/s. What is its kinetic energy? (2021)

A. 6 J
B. 12 J
C. 3 J
D. 9 J

Solution

Kinetic Energy (KE) = 1/2 * m * v^2 = 1/2 * 3 kg * (2 m/s)^2 = 6 J

Correct Answer: B — 12 J

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Q. A 3 kg object is moving with a velocity of 6 m/s. What is its momentum? (2021)

A. 18 kg m/s
B. 12 kg m/s
C. 9 kg m/s
D. 6 kg m/s

Solution

Momentum (p) = mv = 3 kg * 6 m/s = 18 kg m/s

Correct Answer: A — 18 kg m/s

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Q. A 3 kg object is subjected to a net force of 12 N. What is the acceleration of the object? (2022)

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

Solution

Using F = ma, a = F/m = 12 N / 3 kg = 4 m/s².

Correct Answer: C — 4 m/s²

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Q. A 5 kg object is moving with a constant velocity. What is the net force acting on it? (2022)

A. 0 N
B. 5 N
C. 10 N
D. 15 N

Solution

If the object is moving with constant velocity, the net force acting on it is 0 N according to Newton's first law.

Correct Answer: A — 0 N

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Q. A 5 kg object is moving with a velocity of 10 m/s. What is its momentum? (2021)

A. 50 kg m/s
B. 100 kg m/s
C. 150 kg m/s
D. 200 kg m/s

Solution

Momentum p = mv = 5 kg * 10 m/s = 50 kg m/s.

Correct Answer: B — 100 kg m/s

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Q. A 5 kg object is moving with a velocity of 4 m/s. What is its momentum? (2023)

A. 10 kg m/s
B. 15 kg m/s
C. 20 kg m/s
D. 25 kg m/s

Solution

Momentum = mass × velocity = 5 kg × 4 m/s = 20 kg m/s.

Correct Answer: C — 20 kg m/s

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Q. A 6 kg object is pushed with a force of 18 N. What is the resulting acceleration? (2023)

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

Solution

Using F = ma, a = F/m = 18 N / 6 kg = 3 m/s².

Correct Answer: B — 3 m/s²

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Q. A block of mass 5 kg is pulled along a horizontal surface with a force of 20 N. If the frictional force acting on the block is 5 N, what is the acceleration of the block? (2021)

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

Solution

Net force = Applied force - Frictional force = 20 N - 5 N = 15 N. 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 block of mass 5 kg is pulled on a horizontal surface with a force of 20 N. If the frictional force acting on the block is 5 N, what is the acceleration of the block? (2021)

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

Solution

Net force = Applied force - Frictional force = 20 N - 5 N = 15 N. Acceleration = Net force / Mass = 15 N / 5 kg = 3 m/s².

Correct Answer: B — 2 m/s²

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Q. A capacitor is charged to a potential difference of 10V. If the charge on the capacitor is 5μC, what is its capacitance? (2020)

A. 0.5μF
B. 1μF
C. 2μF
D. 5μF

Solution

Capacitance (C) is given by C = Q/V. Here, C = 5μC / 10V = 0.5μF.

Correct Answer: A — 0.5μF

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Q. A capacitor is charged to a potential difference of 12V. If it is disconnected from the battery and the plates are moved apart, what happens to the potential difference? (2021)

A. Increases
B. Decreases
C. Remains the same
D. Becomes zero

Solution

When the plates of a disconnected capacitor are moved apart, the capacitance decreases, leading to an increase in potential difference.

Correct Answer: A — Increases

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Q. A capacitor is charged to a potential difference of 12V. If the charge on the capacitor is 24μC, what is its capacitance? (2019)

A. 2μF
B. 4μF
C. 6μF
D. 8μF

Solution

Capacitance C is given by C = Q/V. Here, C = 24μC / 12V = 2μF.

Correct Answer: B — 4μF

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Q. A capacitor is charged to a potential difference of V. If the charge on the capacitor is Q, what is the capacitance? (2023)

A. Q/V
B. V/Q
C. QV
D. V^2/Q

Solution

The capacitance C is defined as C = Q/V.

Correct Answer: A — Q/V

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Q. A capacitor is charged to a potential of 12 V. If the capacitance is 4 µF, what is the charge stored in the capacitor? (2019)

A. 12 µC
B. 48 µC
C. 24 µC
D. 36 µC

Solution

Charge (Q) = Capacitance (C) × Voltage (V) = 4 µF × 12 V = 48 µC.

Correct Answer: B — 48 µC

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Q. A capacitor is charged to a potential of 12 V. If the charge on the capacitor is 6 µC, what is its capacitance? (2022)

A. 0.5 µF
B. 1 µF
C. 2 µF
D. 3 µF

Solution

Capacitance (C) = Charge (Q) / Voltage (V) = 6 µC / 12 V = 0.5 µF.

Correct Answer: B — 1 µF

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Q. A capacitor is charged to a potential of 12V and then disconnected from the battery. If the distance between the plates is doubled, what is the new potential difference? (2022)

A. 6V
B. 12V
C. 24V
D. 0V

Solution

When the distance is doubled, the potential difference across the capacitor also doubles, resulting in 24V.

Correct Answer: C — 24V

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Q. A capacitor is charged to a potential of 12V and then disconnected from the battery. If the plate area is doubled, what will be the new potential difference? (2022)

A. 6V
B. 12V
C. 24V
D. It cannot be determined

Solution

Once disconnected, the charge remains constant. Doubling the area increases capacitance but does not change the potential difference since the charge is fixed.

Correct Answer: B — 12V

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Q. A capacitor is charged to a voltage of 12V and then disconnected from the battery. If the distance between the plates is doubled, what happens to the voltage across the capacitor? (2023)

A. It remains the same
B. It doubles
C. It halves
D. It becomes zero

Solution

When the distance between the plates is doubled, the capacitance decreases, which causes the voltage to double since Q remains constant.

Correct Answer: B — It doubles

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Q. A capacitor of capacitance C is charged to a voltage V. If the charge is then removed, what is the potential difference across the capacitor? (2023)

A. 0
B. V
C. C
D. CV

Solution

If the charge is removed, the potential difference across the capacitor becomes 0 volts.

Correct Answer: A — 0

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Q. A capacitor of capacitance C is charged to a voltage V. What is the charge stored in the capacitor? (2023)

A. C/V
B. CV
C. V/C
D. C^2V

Solution

The charge Q stored in a capacitor is given by Q = CV.

Correct Answer: B — CV

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Q. A car accelerates from rest to a speed of 20 m/s. If its mass is 1000 kg, what is the work done on the car? (2022)

A. 200,000 J
B. 100,000 J
C. 50,000 J
D. 400,000 J

Solution

Work Done = Change in Kinetic Energy = 1/2 * m * (v^2 - u^2) = 1/2 * 1000 kg * (20 m/s)^2 = 200,000 J

Correct Answer: A — 200,000 J

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Q. A car moving at 15 m/s comes to a stop in 5 seconds. What is its acceleration? (2019)

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

Solution

Using the formula: acceleration = (final velocity - initial velocity) / time. Acceleration = (0 - 15) / 5 = -3 m/s².

Correct Answer: A — -2 m/s²

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Q. A charge of +1 μC is placed in an electric field of strength 1000 N/C. What is the force experienced by the charge? (2023)

A. 0.001 N
B. 0.01 N
C. 1 N
D. 10 N

Solution

Force F = qE = (1 × 10^-6 C)(1000 N/C) = 0.001 N.

Correct Answer: B — 0.01 N

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Q. A charge of +10 μC is placed in an electric field of 2000 N/C. What is the force experienced by the charge? (2023)

A. 20 N
B. 200 N
C. 2 N
D. 0.2 N

Solution

Force F = qE = (10 × 10^-6 C) * (2000 N/C) = 0.02 N = 20 N.

Correct Answer: A — 20 N

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Physics

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