MHT-CET Exam Prep Resources | Soulshift Education

MHT-CET

Chemistry Mathematics Physics

Q. A 3 kg object is dropped from a height of 5 m. What is the potential energy at the top? (g = 9.8 m/s²)

A. 14.7 J
B. 29.4 J
C. 39.2 J
D. 49.05 J

Solution

PE = mgh = 3 kg × 9.8 m/s² × 5 m = 147 J.

Correct Answer: C — 39.2 J

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Q. A 3 kg object is lifted to a height of 4 m. What is the work done against gravity? (g = 9.8 m/s²)

A. 117.6 J
B. 29.4 J
C. 39.2 J
D. 78.4 J

Solution

Work done (W) = mgh = 3 kg × 9.8 m/s² × 4 m = 117.6 J.

Correct Answer: A — 117.6 J

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Q. A 3 kg object is pushed with a force of 12 N over a distance of 4 m. What is the work done on the object?

A. 24 J
B. 36 J
C. 48 J
D. 60 J

Solution

Work done (W) = Force (F) × Distance (d) = 12 N × 4 m = 48 J.

Correct Answer: C — 48 J

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Q. A 4 kg block is pulled with a force of 20 N. What is the acceleration of the block? (2020)

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

Solution

Using F = ma, acceleration (a) = F/m = 20 N / 4 kg = 5 m/s².

Correct Answer: A — 5 m/s²

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Q. A 4 kg object is lifted to a height of 2 m. What is the work done against gravity? (2022)

A. 40 J
B. 80 J
C. 20 J
D. 60 J

Solution

Work done = mass × g × height = 4 kg × 10 m/s² × 2 m = 80 J

Correct Answer: B — 80 J

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Q. A 4 kg object is lifted to a height of 5 m. What is the work done against gravity? (g = 9.8 m/s²)

A. 19.6 J
B. 39.2 J
C. 49 J
D. 196 J

Solution

Work done against gravity = mgh = 4 kg × 9.8 m/s² × 5 m = 196 J.

Correct Answer: B — 39.2 J

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Q. A 4 kg object is moving with a velocity of 5 m/s. What is its kinetic energy? (2021)

A. 10 J
B. 20 J
C. 50 J
D. 100 J

Solution

Kinetic energy KE = 1/2 mv² = 1/2 * 4 kg * (5 m/s)² = 50 J.

Correct Answer: C — 50 J

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Q. A 5 kg block of ice at 0°C is converted to water at 0°C. If the latent heat of fusion of ice is 334 kJ/kg, how much heat is absorbed?

A. 1670 kJ
B. 334 kJ
C. 167 kJ
D. 3340 kJ

Solution

Heat absorbed (Q) = m * Lf = 5 kg * 334 kJ/kg = 1670 kJ.

Correct Answer: B — 334 kJ

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Q. A 5 kg block of ice at 0°C is placed in a calorimeter containing 2 kg of water at 80°C. What is the final temperature of the system when thermal equilibrium is reached? (Latent heat of fusion of ice = 334 kJ/kg)

A. 0°C
B. 20°C
C. 40°C
D. 60°C

Solution

The heat lost by water = heat gained by ice. Solving gives the final temperature as 0°C.

Correct Answer: B — 20°C

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Q. A 5 kg object is dropped from a height of 10 m. What is its potential energy at the top? (g = 9.8 m/s²)

A. 490 J
B. 98 J
C. 50 J
D. 100 J

Solution

Potential Energy = mgh = 5 kg * 9.8 m/s² * 10 m = 490 J.

Correct Answer: A — 490 J

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Q. A 5 kg object is dropped from a height of 10 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)

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

Solution

Using energy conservation, Potential Energy = Kinetic Energy at the ground: mgh = 0.5mv²; v = √(2gh) = √(2 × 9.8 m/s² × 10 m) = 14 m/s

Correct Answer: A — 14 m/s

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Q. A 5 kg object is lifted to a height of 2 m. How much work is done against gravity? (g = 9.8 m/s²) (2022)

A. 98 J
B. 196 J
C. 39 J
D. 49 J

Solution

Work done = mgh = 5 kg * 9.8 m/s² * 2 m = 98 J.

Correct Answer: A — 98 J

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Q. A 5 kg object is lifted to a height of 2 m. What is the work done against gravity?

A. 10 J
B. 20 J
C. 50 J
D. 100 J

Solution

Work done (W) against gravity is given by W = mgh. Here, m = 5 kg, g = 9.8 m/s², h = 2 m. W = 5 * 9.8 * 2 = 98 J.

Correct Answer: B — 20 J

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Q. A 50 kg crate is pushed with a force of 200 N. If the frictional force is 50 N, what is the acceleration? (2022)

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

Solution

Net force = applied force - friction = 200 N - 50 N = 150 N. Acceleration a = F/m = 150 N / 50 kg = 3 m/s².

Correct Answer: B — 3 m/s²

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Q. A 50 kg crate is pushed with a force of 200 N. If the frictional force is 50 N, what is the acceleration of the crate? (2022)

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

Solution

Net force = 200 N - 50 N = 150 N. Acceleration a = F/m = 150 N / 50 kg = 3 m/s².

Correct Answer: B — 3 m/s²

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Q. A 50 kg crate is pushed with a force of 200 N. What is the acceleration of the crate? (Assume no friction) (2021)

A. 2 m/s²
B. 4 m/s²
C. 5 m/s²
D. 10 m/s²

Solution

Using F = ma, acceleration a = F/m = 200 N / 50 kg = 4 m/s².

Correct Answer: B — 4 m/s²

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Q. A 50 kg object is dropped from a height of 20 m. What is the speed just before it hits the ground? (g = 10 m/s²) (2020)

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

Solution

Using v² = u² + 2gh, v² = 0 + 2 * 10 m/s² * 20 m = 400, v = √400 = 20 m/s.

Correct Answer: B — 20 m/s

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Q. A 50 kg object is moving with a velocity of 4 m/s. What is its kinetic energy?

A. 100 J
B. 200 J
C. 400 J
D. 800 J

Solution

Kinetic Energy (KE) = 0.5 * m * v² = 0.5 * 50 kg * (4 m/s)² = 400 J.

Correct Answer: C — 400 J

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Q. A 50 kg object is moving with a velocity of 5 m/s. What is its momentum? (2023)

A. 100 kg·m/s
B. 150 kg·m/s
C. 200 kg·m/s
D. 250 kg·m/s

Solution

Momentum p = mv = 50 kg * 5 m/s = 250 kg·m/s.

Correct Answer: A — 100 kg·m/s

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Q. A 50 kg object is pulled with a force of 200 N. What is its acceleration? (2023)

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

Solution

Using F = ma, we find a = F/m = 200 N / 50 kg = 4 m/s².

Correct Answer: D — 5 m/s²

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Q. A 50 kg object is pushed with a force of 200 N. What is the frictional force if the object moves with constant velocity? (2023)

A. 0 N
B. 50 N
C. 200 N
D. 100 N

Solution

If the object moves with constant velocity, the frictional force equals the applied force, which is 200 N.

Correct Answer: C — 200 N

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Q. A 9V battery is connected to a circuit with a total resistance of 3Ω. What is the total current flowing in the circuit? (2023)

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

Solution

Using Ohm's law, I = V/R = 9V / 3Ω = 3A.

Correct Answer: B — 2A

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Q. A ball is thrown horizontally from a height of 20 m. How long does it take to hit the ground? (2019)

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

Solution

Using the equation of motion for free fall: h = (1/2)gt². 20 m = (1/2)(9.8)t². Solving gives t ≈ 2 s.

Correct Answer: B — 2 s

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Q. A ball is thrown upwards with a speed of 20 m/s. How long will it take to reach the maximum height? (Take g = 10 m/s²) (2023)

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

Solution

Time to reach maximum height = initial velocity / g = 20 / 10 = 2 s.

Correct Answer: B — 2 s

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Q. A ball is thrown upwards with a speed of 25 m/s. How long will it take to reach the maximum height? (Assume g = 10 m/s²)

A. 2.5 s
B. 3 s
C. 4 s
D. 5 s

Solution

Time to reach maximum height = initial velocity / g = 25 / 10 = 2.5 s.

Correct Answer: A — 2.5 s

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Q. A ball is thrown upwards with a velocity of 20 m/s. How high will it go before coming to rest? (2023)

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

Solution

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

Correct Answer: C — 30 m

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Q. A ball is thrown upwards with an initial velocity of 20 m/s. How long will it take to reach the maximum height? (g = 10 m/s²)

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

Solution

Using the formula: t = v/g = 20/10 = 2 s.

Correct Answer: B — 2 s

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Q. A ball is thrown vertically upwards with a speed of 15 m/s. How high will it rise before coming to a stop? (2023)

A. 11.25 m
B. 15 m
C. 22.5 m
D. 30 m

Solution

Using the formula: height = (initial velocity^2) / (2 * g), where g = 9.8 m/s². Height = (15^2) / (2 * 9.8) = 11.25 m.

Correct Answer: A — 11.25 m

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Q. A ball is thrown vertically upwards with a speed of 20 m/s. How high will it go? (g = 10 m/s²) (2023)

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

Solution

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

Correct Answer: B — 20 m

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Q. A beam of light passes through a narrow slit and produces a diffraction pattern. What is the angle of the first minimum? (2019)

A. λ/a
B. λ/2a
C. 2λ/a
D. 3λ/a

Solution

The angle of the first minimum in single-slit diffraction is given by sin(θ) = λ/a, where a is the width of the slit.

Correct Answer: A — λ/a

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MHT-CET

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