Q. If 50 mL of a 3 M solution is diluted to 200 mL, what is the new molarity? (2020)
A.
0.75 M
B.
1.5 M
C.
2 M
D.
3 M
Show solution
Solution
Using M1V1 = M2V2, (3 M)(50 mL) = M2(200 mL) => M2 = (3 M × 50 mL) / 200 mL = 0.75 M.
Correct Answer:
A
— 0.75 M
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Q. If 500 J of heat is added to a system and 200 J of work is done by the system, what is the change in internal energy of the system?
A.
300 J
B.
500 J
C.
700 J
D.
200 J
Show solution
Solution
According to the first law of thermodynamics, ΔU = Q - W. Here, ΔU = 500 J - 200 J = 300 J.
Correct Answer:
A
— 300 J
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Q. If 500 J of heat is added to a system and it does 200 J of work, what is the change in internal energy? (2022)
A.
300 J
B.
200 J
C.
500 J
D.
700 J
Show solution
Solution
According to the first law of thermodynamics, ΔU = Q - W. Here, ΔU = 500 J - 200 J = 300 J.
Correct Answer:
A
— 300 J
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Q. If 5x = 25, what is the value of x? (2019)
Show solution
Solution
5x = 25, so x = 25/5 = 5.
Correct Answer:
A
— 5
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Q. If a 10 kg object is dropped from a height of 20 m, what is the velocity just before it hits the ground? (Assume g = 10 m/s²) (2019)
A.
10 m/s
B.
20 m/s
C.
14 m/s
D.
15 m/s
Show solution
Solution
Using v² = u² + 2gh, v = √(0 + 2 × 10 m/s² × 20 m) = √400 = 20 m/s.
Correct Answer:
B
— 20 m/s
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Q. If a 10 kg object is moving with a speed of 2 m/s, what is its kinetic energy?
A.
10 J
B.
20 J
C.
40 J
D.
80 J
Show solution
Solution
KE = 0.5 × m × v² = 0.5 × 10 kg × (2 m/s)² = 0.5 × 10 × 4 = 20 J.
Correct Answer:
B
— 20 J
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Q. If a 10 kg object is raised to a height of 5 m, what is the work done against gravity? (g = 9.8 m/s²)
A.
490 J
B.
980 J
C.
150 J
D.
250 J
Show solution
Solution
Work = m * g * h = 10 kg * 9.8 m/s² * 5 m = 490 J.
Correct Answer:
B
— 980 J
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Q. If a 10 N force is applied at an angle of 60 degrees to the horizontal while moving an object 3 m, what is the work done? (2020)
A.
15 J
B.
30 J
C.
25 J
D.
5 J
Show solution
Solution
Work done = Force × Distance × cos(θ) = 10 N × 3 m × cos(60°) = 10 N × 3 m × 0.5 = 15 J
Correct Answer:
A
— 15 J
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Q. If a 10 N force is applied at an angle of 60 degrees to the horizontal while moving an object 4 m, what is the work done? (2020)
A.
20 J
B.
40 J
C.
30 J
D.
50 J
Show solution
Solution
Work done = Force × Distance × cos(θ) = 10 N × 4 m × cos(60°) = 10 N × 4 m × 0.5 = 20 J
Correct Answer:
C
— 30 J
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Q. If a 1000 kg car is moving at a speed of 15 m/s, what is its kinetic energy?
A.
112,500 J
B.
225,000 J
C.
150,000 J
D.
75,000 J
Show solution
Solution
Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 1000 kg × (15 m/s)² = 112,500 J
Correct Answer:
B
— 225,000 J
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Q. If a 1000 kg car is moving with a speed of 15 m/s, what is its kinetic energy?
A.
112,500 J
B.
225,000 J
C.
75,000 J
D.
150,000 J
Show solution
Solution
Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 1000 kg × (15 m/s)² = 112,500 J
Correct Answer:
B
— 225,000 J
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Q. If a 1000 W motor runs for 2 hours, how much work does it do? (2000)
A.
7200 J
B.
2000 J
C.
3600000 J
D.
100000 J
Show solution
Solution
Work = Power × Time = 1000 W × (2 hours × 3600 s/hour) = 1000 W × 7200 s = 7200000 J
Correct Answer:
C
— 3600000 J
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Q. If a 100Ω resistor is connected in series with a 200Ω resistor, what is the total resistance? (2023)
A.
300Ω
B.
100Ω
C.
200Ω
D.
400Ω
Show solution
Solution
In series, R_total = R1 + R2 = 100Ω + 200Ω = 300Ω.
Correct Answer:
A
— 300Ω
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Q. If a 10Ω resistor is connected in parallel with a 5Ω resistor, what is the equivalent resistance? (2022)
A.
3.33Ω
B.
5Ω
C.
7.5Ω
D.
15Ω
Show solution
Solution
For parallel resistors, 1/R_eq = 1/R1 + 1/R2 = 1/10 + 1/5 = 1/10 + 2/10 = 3/10. Therefore, R_eq = 10/3 = 3.33Ω.
Correct Answer:
A
— 3.33Ω
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Q. If a 10Ω resistor is connected in parallel with a 5Ω resistor, what is the total resistance? (2022)
A.
3.33Ω
B.
5Ω
C.
7.5Ω
D.
15Ω
Show solution
Solution
For resistors in parallel, 1/R_eq = 1/R1 + 1/R2 = 1/10 + 1/5 = 1/10 + 2/10 = 3/10. Therefore, R_eq = 10/3 = 3.33Ω.
Correct Answer:
A
— 3.33Ω
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Q. If a 1500 kg car is moving at a speed of 30 m/s, what is its kinetic energy?
A.
675,000 J
B.
450,000 J
C.
225,000 J
D.
900,000 J
Show solution
Solution
Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 1500 kg × (30 m/s)² = 675,000 J
Correct Answer:
B
— 450,000 J
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Q. If a 15Ω resistor is connected in parallel with a 30Ω resistor, what is the equivalent resistance? (2021)
A.
10Ω
B.
7.5Ω
C.
5Ω
D.
12Ω
Show solution
Solution
For resistors in parallel, 1/R_eq = 1/R1 + 1/R2 = 1/15 + 1/30 = 2/30 + 1/30 = 3/30. Therefore, R_eq = 30/3 = 10Ω.
Correct Answer:
B
— 7.5Ω
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Q. If a 15Ω resistor is connected in parallel with a 5Ω resistor, what is the equivalent resistance? (2023)
A.
3.75Ω
B.
4Ω
C.
5Ω
D.
10Ω
Show solution
Solution
1/R_eq = 1/15 + 1/5 = 1/15 + 3/15 = 4/15; R_eq = 15/4 = 3.75Ω.
Correct Answer:
A
— 3.75Ω
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Q. If a 2 kg object is dropped from a height, what is its velocity just before hitting the ground? (g = 9.8 m/s², h = 20 m) (2023)
A.
14 m/s
B.
19.6 m/s
C.
20 m/s
D.
28 m/s
Show solution
Solution
Using v² = u² + 2gh, where u = 0, we find v = √(2 * 9.8 * 20) = 19.6 m/s.
Correct Answer:
B
— 19.6 m/s
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Q. If a 2 kg object is dropped from a height, what is its velocity just before hitting the ground (g = 10 m/s²)? (2023)
A.
10 m/s
B.
20 m/s
C.
30 m/s
D.
40 m/s
Show solution
Solution
Using v = √(2gh), for h = 10 m, v = √(2 * 10 m/s² * 10 m) = √200 = 20 m/s.
Correct Answer:
B
— 20 m/s
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Q. If a 2 kg object is lifted to a height of 10 m, what is the potential energy gained by the object? (g = 9.8 m/s²)
A.
19.6 J
B.
98 J
C.
39.2 J
D.
78.4 J
Show solution
Solution
Potential Energy (PE) = mgh = 2 kg × 9.8 m/s² × 10 m = 196 J.
Correct Answer:
B
— 98 J
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Q. If a 2 kg object is lifted to a height of 3 m, what is the potential energy gained by the object? (g = 9.8 m/s²)
A.
58.8 J
B.
19.6 J
C.
29.4 J
D.
39.2 J
Show solution
Solution
Potential Energy (PE) = mgh = 2 kg × 9.8 m/s² × 3 m = 58.8 J.
Correct Answer:
A
— 58.8 J
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Q. If a 2 kg object is moving with a velocity of 15 m/s, what is its kinetic energy?
A.
225 J
B.
150 J
C.
300 J
D.
450 J
Show solution
Solution
Kinetic Energy = 1/2 * m * v^2 = 1/2 * 2 kg * (15 m/s)^2 = 225 J.
Correct Answer:
A
— 225 J
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Q. If a 2 kg object is moving with a velocity of 3 m/s, what is its kinetic energy?
A.
9 J
B.
6 J
C.
3 J
D.
12 J
Show solution
Solution
Kinetic Energy = 0.5 * m * v² = 0.5 * 2 kg * (3 m/s)² = 9 J.
Correct Answer:
A
— 9 J
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Q. If a 3 kg object is moving with a speed of 4 m/s, what is its kinetic energy?
A.
12 J
B.
24 J
C.
36 J
D.
48 J
Show solution
Solution
KE = 0.5 × m × v² = 0.5 × 3 kg × (4 m/s)² = 0.5 × 3 × 16 = 24 J.
Correct Answer:
B
— 24 J
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Q. If a 3 kg object is subjected to a net force of 12 N, what is its acceleration? (2023)
A.
4 m/s²
B.
3 m/s²
C.
2 m/s²
D.
1 m/s²
Show solution
Solution
Acceleration a = F/m = 12 N / 3 kg = 4 m/s².
Correct Answer:
A
— 4 m/s²
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Q. If a 4 kg object is lifted to a height of 2 m, what is the work done against gravity? (2021)
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. If a 4 kg object is moving with a speed of 5 m/s, what is its kinetic energy?
A.
50 J
B.
40 J
C.
100 J
D.
80 J
Show solution
Solution
Kinetic Energy (KE) = 0.5 × m × v² = 0.5 × 4 kg × (5 m/s)² = 0.5 × 4 × 25 = 50 J.
Correct Answer:
D
— 80 J
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Q. If a 5 kg object is dropped from a height of 10 m, what is its potential energy just before it hits the ground? (g = 9.8 m/s²)
A.
490 J
B.
980 J
C.
245 J
D.
2450 J
Show solution
Solution
Potential Energy (PE) = m × g × h = 5 kg × 9.8 m/s² × 10 m = 490 J.
Correct Answer:
B
— 980 J
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Q. If a 5 kg object is dropped from a height of 10 m, what is its potential energy just before it hits the ground? (2021)
A.
50 J
B.
100 J
C.
150 J
D.
200 J
Show solution
Solution
PE = m × g × h = 5 kg × 9.8 m/s² × 10 m = 490 J.
Correct Answer:
B
— 100 J
Learn More →
Showing 811 to 840 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.
