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
Show solution
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
Show solution
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
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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
Show solution
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
Show solution
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
Show solution
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
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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
Show solution
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
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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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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)
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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 for the first minimum? (2022)
A.
λ/a
B.
2λ/a
C.
λ/2a
D.
a/λ
Show solution
Solution
The angle for the first minimum in single-slit diffraction is given by a sin θ = λ, hence θ = λ/a.
Correct Answer:
A
— λ/a
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Showing 61 to 90 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.
