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MHT-CET

Chemistry (MHT-CET) Mathematics (MHT-CET) Physics (MHT-CET)

Q. A 1000 kg car is moving at a speed of 15 m/s. What is its kinetic energy?

A. 112,500 J
B. 75,000 J
C. 50,000 J
D. 25,000 J

Solution

Kinetic Energy = 0.5 * m * v² = 0.5 * 1000 kg * (15 m/s)² = 112,500 J.

Correct Answer: A — 112,500 J

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Q. A 1000 kg car is moving at a speed of 20 m/s. What is its momentum? (2000)

A. 2000 kg m/s
B. 10000 kg m/s
C. 5000 kg m/s
D. 4000 kg m/s

Solution

Momentum (p) is calculated using the formula p = m * v. Here, p = 1000 kg * 20 m/s = 20000 kg m/s.

Correct Answer: B — 10000 kg m/s

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Q. A 1000 kg car is moving at a speed of 36 km/h. What is its kinetic energy? (1 km/h = 1/3.6 m/s) (2000)

A. 500 J
B. 1000 J
C. 2000 J
D. 5000 J

Solution

First convert speed: 36 km/h = 10 m/s. KE = 0.5 × m × v² = 0.5 × 1000 kg × (10 m/s)² = 0.5 × 1000 × 100 = 50000 J.

Correct Answer: C — 2000 J

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Q. A 1000 W electric motor runs for 2 hours. How much work does it do? (2023)

A. 720,000 J
B. 1,000,000 J
C. 2,000,000 J
D. 1,440,000 J

Solution

Work = Power × Time = 1000 W × (2 × 3600 s) = 7,200,000 J.

Correct Answer: D — 1,440,000 J

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Q. A 1000 W heater is used to heat water. How much heat is supplied in 10 minutes? (2020)

A. 600 kJ
B. 300 kJ
C. 100 kJ
D. 150 kJ

Solution

Heat (Q) = Power (P) * Time (t) = 1000 W * 600 s = 600 kJ.

Correct Answer: A — 600 kJ

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Q. A 100W bulb operates at 220V. What is the resistance of the bulb? (2021)

A. 484Ω
B. 220Ω
C. 110Ω
D. 100Ω

Solution

Using P = V^2/R, we rearrange to find R = V^2/P = (220V)^2 / 100W = 484Ω.

Correct Answer: A — 484Ω

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Q. A 12 kg object is pushed with a force of 36 N. What is its acceleration? (2023)

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

Solution

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

Correct Answer: C — 3 m/s²

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Q. A 12V battery is connected across a 4Ω resistor. What is the current flowing through the resistor? (2020)

A. 2A
B. 3A
C. 4A
D. 6A

Solution

Using Ohm's law, I = V/R = 12V / 4Ω = 3A.

Correct Answer: A — 2A

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Q. A 12V battery is connected to a resistor of 4Ω. What is the current flowing through the circuit? (2022)

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

Solution

Using Ohm's law, I = V/R = 12V / 4Ω = 3A.

Correct Answer: A — 2A

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Q. A 15 kg box is pulled with a force of 60 N at an angle of 30° to the horizontal. What is the horizontal component of the force? (2022)

A. 30 N
B. 60 N
C. 51.96 N
D. 15 N

Solution

Horizontal component = F * cos(θ) = 60 N * cos(30°) = 60 N * (√3/2) = 51.96 N.

Correct Answer: C — 51.96 N

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Q. A 15 kg object is dropped from a height. What is the force acting on it due to gravity? (2023)

A. 15 N
B. 30 N
C. 45 N
D. 150 N

Solution

Using F = mg, where g = 9.8 m/s², we have F = 15 kg * 9.8 m/s² = 147 N, which rounds to 150 N.

Correct Answer: B — 30 N

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Q. A 15 kg object is lifted to a height of 4 m. What is the work done against gravity? (g = 9.8 m/s²) (2021)

A. 588 J
B. 5880 J
C. 392 J
D. 294 J

Solution

Work = mgh = 15 kg * 9.8 m/s² * 4 m = 588 J.

Correct Answer: A — 588 J

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Q. A 15 kg object is moving at a speed of 4 m/s. What is its kinetic energy?

A. 120 J
B. 240 J
C. 480 J
D. 60 J

Solution

Kinetic Energy (KE) = 0.5 × m × v² = 0.5 × 15 kg × (4 m/s)² = 0.5 × 15 × 16 = 120 J.

Correct Answer: B — 240 J

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Q. A 15 kg object is moving with a velocity of 4 m/s. What is its momentum? (2019)

A. 60 kg m/s
B. 30 kg m/s
C. 45 kg m/s
D. 75 kg m/s

Solution

Momentum = mass × velocity = 15 kg × 4 m/s = 60 kg m/s.

Correct Answer: A — 60 kg m/s

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Q. A 1500 kg car accelerates from rest to a speed of 25 m/s. What is the work done on the car?

A. 468,750 J
B. 375,000 J
C. 250,000 J
D. 187,500 J

Solution

Work done = Change in Kinetic Energy = 0.5 * m * v² = 0.5 * 1500 kg * (25 m/s)² = 468,750 J.

Correct Answer: A — 468,750 J

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Q. A 1500 kg car is moving at a speed of 25 m/s. What is its kinetic energy? (2021)

A. 468,750 J
B. 1,875,000 J
C. 937,500 J
D. 750,000 J

Solution

Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 1500 kg × (25 m/s)² = 468,750 J

Correct Answer: B — 1,875,000 J

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Q. A 2 kg ball is thrown vertically upwards with a velocity of 20 m/s. What is the maximum height reached? (2020)

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

Solution

Using v² = u² + 2as, 0 = (20 m/s)² + 2(-9.8 m/s²)(h). Solving gives h = 20.4 m.

Correct Answer: A — 20 m

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Q. A 2 kg block is sliding down a frictionless incline of height 5 m. What is its speed at the bottom?

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

Solution

Using conservation of energy, potential energy at the top = kinetic energy at the bottom: mgh = 0.5mv². Solving gives v = √(2gh) = √(2 * 9.8 * 5) = 10 m/s.

Correct Answer: A — 10 m/s

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Q. A 2 kg block of ice at 0°C is converted to water at 0°C. How much heat is absorbed if the latent heat of fusion of ice is 334,000 J/kg?

A. 668,000 J
B. 334,000 J
C. 167,000 J
D. 500,000 J

Solution

Heat absorbed = mass * latent heat = 2 kg * 334,000 J/kg = 668,000 J.

Correct Answer: A — 668,000 J

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Q. A 2 kg block of metal at 100°C is placed in 1 kg of water at 20°C. What is the final temperature of the system? (Specific heat of water = 4.2 J/g°C, specific heat of metal = 0.9 J/g°C) (2021)

A. 30°C
B. 40°C
C. 50°C
D. 60°C

Solution

Using the heat transfer equation, the final temperature can be calculated to be 50°C.

Correct Answer: C — 50°C

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Q. A 2 kg object is heated from 20°C to 80°C. If the specific heat capacity is 0.5 kJ/kg°C, how much heat is absorbed? (2023)

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

Solution

Q = mcΔT = 2 kg * 0.5 kJ/kg°C * (80°C - 20°C) = 60 kJ.

Correct Answer: B — 40 kJ

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Q. A 2 kg object is lifted to a height of 5 m. What is the potential energy gained by the object? (g = 9.8 m/s²)

A. 98 J
B. 19.6 J
C. 39.2 J
D. 49 J

Solution

Potential Energy = mass × g × height = 2 kg × 9.8 m/s² × 5 m = 98 J

Correct Answer: A — 98 J

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Q. A 2 kg object is lifted to a height of 5 m. What is the potential energy gained by the object? (2022)

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

Solution

Potential Energy = mass × g × height = 2 kg × 10 m/s² × 5 m = 100 J

Correct Answer: B — 20 J

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Q. A 2 kg object is pushed with a force of 10 N over a distance of 5 m. What is the work done on the object?

A. 25 J
B. 50 J
C. 75 J
D. 100 J

Solution

Work done = Force * Distance = 10 N * 5 m = 50 J.

Correct Answer: B — 50 J

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Q. A 2 kg object is subjected to a net force of 8 N. What is its acceleration? (2020)

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

Solution

Using F = ma, a = F/m = 8 N / 2 kg = 4 m/s².

Correct Answer: C — 4 m/s²

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Q. A 20 kg cart is pulled with a force of 80 N. What is the acceleration of the cart? (2023)

A. 2 m/s²
B. 4 m/s²
C. 6 m/s²
D. 8 m/s²

Solution

Using F = ma, we find a = F/m = 80 N / 20 kg = 4 m/s².

Correct Answer: B — 4 m/s²

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Q. A 200 g piece of metal at 100°C is placed in 300 g 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) (2022)

A. 22.5°C
B. 30°C
C. 35°C
D. 40°C

Solution

Using the principle of conservation of energy: m1c1(T_initial - T_final) = m2c2(T_final - T_initial). Solving gives T_final = 35°C.

Correct Answer: C — 35°C

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Q. A 25 kg object is acted upon by a force of 100 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 = 100 N / 25 kg = 4 m/s².

Correct Answer: D — 5 m/s²

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Q. A 3 kg object is dropped from a height of 10 m. What is its potential energy at the top? (g = 9.8 m/s²)

A. 294 J
B. 30 J
C. 39.2 J
D. 19.6 J

Solution

Potential Energy (PE) = m × g × h = 3 kg × 9.8 m/s² × 10 m = 294 J.

Correct Answer: A — 294 J

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Q. A 3 kg object is dropped from a height of 10 m. What is the potential energy at the top? (2022)

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

Solution

Potential Energy = mass × g × height = 3 kg × 10 m/s² × 10 m = 300 J

Correct Answer: C — 90 J

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Showing 31 to 60 of 2530 (85 Pages)

MHT-CET MCQ & Objective Questions

The MHT-CET exam is a crucial stepping stone for students aspiring to pursue engineering and pharmacy courses in Maharashtra. Mastering the MHT-CET MCQ format is essential, as it not only tests your knowledge but also enhances your exam preparation strategy. Practicing objective questions helps in identifying important concepts and improves your chances of scoring better in this competitive exam.

What You Will Practise Here

Fundamental concepts in Physics, Chemistry, and Mathematics
Key formulas and their applications in problem-solving
Important definitions and terminologies relevant to MHT-CET
Diagrams and illustrations for better conceptual understanding
Practice questions that mirror the exam pattern
Analysis of previous years' MHT-CET questions
Techniques for tackling tricky MCQs effectively

Exam Relevance

The MHT-CET exam is aligned with the syllabus of CBSE, State Boards, and is also relevant for students preparing for NEET and JEE. Many concepts from the MHT-CET syllabus appear in these competitive exams, often in the form of application-based questions or conceptual MCQs. Understanding the common question patterns can significantly enhance your preparation and performance.

Common Mistakes Students Make

Misinterpreting questions due to lack of clarity in reading
Neglecting to review fundamental concepts before attempting MCQs
Overlooking units and dimensions in Physics and Chemistry problems
Rushing through practice questions without thorough understanding
Failing to manage time effectively during the exam

FAQs

Question: What are the best resources for MHT-CET MCQ questions?
Answer: Utilizing online platforms like SoulShift, which offer a variety of practice questions and mock tests, can be very beneficial.

Question: How can I improve my speed in solving MHT-CET objective questions?
Answer: Regular practice and timed mock tests can help enhance your speed and accuracy in solving MCQs.

Start your journey towards success by solving MHT-CET practice MCQs today! Test your understanding and build your confidence for the exam ahead.

MHT-CET

MHT-CET MCQ & Objective Questions

What You Will Practise Here

Exam Relevance

Common Mistakes Students Make

FAQs

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