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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.

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

Q. A 0.5 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches?

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

Solution

Using the formula h = v²/(2g), where v = 20 m/s and g = 9.8 m/s², h = (20)²/(2 * 9.8) ≈ 20.4 m, which rounds to 20 m.

Correct Answer: B — 20 m

Q. A 1 kg ball is dropped from a height of 20 m. What is its velocity just before it hits the ground? (g = 10 m/s²) (2021)

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

Solution

Using the formula v = √(2gh) = √(2 × 10 m/s² × 20 m) = √400 = 20 m/s.

Correct Answer: B — 20 m/s

Q. A 1 kg ball is thrown horizontally from a height of 5 m. How long does it take to hit the ground? (g = 10 m/s²) (2022)

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

Solution

Using the formula h = 1/2 gt², 5 m = 1/2 * 10 m/s² * t², t² = 1, t = √1 = 2 s.

Correct Answer: B — 2 s

Q. A 1 kg ball is thrown vertically upwards with a velocity of 20 m/s. What is the maximum height it reaches? (g = 10 m/s²) (2021)

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

Solution

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

Correct Answer: A — 20 m

Q. A 1 kg block of ice at 0°C is placed in 2 kg of water at 20°C. What will be the final temperature of the mixture? (Specific heat of water = 4.2 kJ/kg°C, Latent heat of fusion of ice = 334 kJ/kg) (2021)

A. 0°C
B. 10°C
C. 20°C
D. 15°C

Solution

Heat lost by water = Heat gained by ice. Calculate to find the final temperature.

Correct Answer: B — 10°C

Q. A 1 kg block of ice at 0°C is placed in 2 kg of water at 20°C. What will be the final temperature of the mixture? (Assume no heat loss to the surroundings)

A. 0°C
B. 10°C
C. 15°C
D. 20°C

Solution

Using the principle of conservation of energy, the heat lost by water equals the heat gained by ice. Final temperature can be calculated to be approximately 10°C.

Correct Answer: B — 10°C

Q. A 1 kg block of ice at 0°C is placed in 2 kg of water at 80°C. What will be the final temperature of the mixture? (Assume no heat loss to the surroundings) (2019)

A. 0°C
B. 40°C
C. 60°C
D. 80°C

Solution

Using the principle of conservation of energy, the heat lost by water equals the heat gained by ice. The final temperature will be 0°C as the ice will melt.

Correct Answer: B — 40°C

Q. A 1 kg block of metal at 100°C is placed in 2 kg of water at 20°C. Assuming no heat loss to the surroundings, what is the final temperature of the system? (Specific heat of water = 4.18 kJ/kg°C, specific heat of metal = 0.9 kJ/kg°C) (2020)

A. 25°C
B. 30°C
C. 35°C
D. 40°C

Solution

Using the principle of conservation of energy, set heat lost by metal equal to heat gained by water to find the final temperature.

Correct Answer: C — 35°C

Q. A 1 kg block of metal at 100°C is placed in 2 kg of water at 20°C. Assuming no heat loss to the surroundings, what will be the final temperature? (Specific heat of water = 4.2 kJ/kg°C) (2022)

A. 25°C
B. 30°C
C. 35°C
D. 40°C

Solution

Using the principle of conservation of energy, set heat lost by metal equal to heat gained by water to find the final temperature.

Correct Answer: C — 35°C

Q. A 1 kg block of metal at 100°C is placed in 2 kg of water at 20°C. If the final temperature of the system is 30°C, what is the specific heat capacity of the metal? (Specific heat of water = 4.18 J/g°C) (2020)

A. 0.5 J/g°C
B. 1.0 J/g°C
C. 1.5 J/g°C
D. 2.0 J/g°C

Solution

Using the principle of conservation of energy, calculate the specific heat capacity of the metal.

Correct Answer: B — 1.0 J/g°C

Q. A 1 kg block of metal at 100°C is placed in 2 kg 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) (2021)

A. 25°C
B. 30°C
C. 35°C
D. 40°C

Solution

Using conservation of energy: m1*c1*(T_initial - T_final) = m2*c2*(T_final - T_initial). Solving gives T_final = 35°C.

Correct Answer: C — 35°C

Q. A 1 kg block of metal at 100°C is placed in 2 kg of water at 20°C. What will be the final temperature of the system assuming no heat loss? (2022)

A. 30°C
B. 40°C
C. 50°C
D. 60°C

Solution

Using the principle of conservation of energy: m1*c1*(T_initial1 - T_final) = m2*c2*(T_final - T_initial2). Solving gives T_final = 50°C.

Correct Answer: C — 50°C

Q. A 1 kg object is dropped from a height. What is its velocity just before it hits the ground (neglect air resistance)? (2020)

A. 10 m/s
B. 14 m/s
C. 20 m/s
D. 30 m/s

Solution

Using v² = u² + 2gh, v = √(0 + 2 × 9.8 m/s² × h). For h = 20 m, v = √(392) = 20 m/s.

Correct Answer: C — 20 m/s

Q. A 10 kg block is pushed with a force of 50 N. What is its acceleration? (2023)

A. 2 m/s²
B. 5 m/s²
C. 10 m/s²
D. 15 m/s²

Solution

Using Newton's second law, F = ma, we have a = F/m = 50 N / 10 kg = 5 m/s².

Correct Answer: B — 5 m/s²

Q. A 10 kg object is at rest. A force of 30 N is applied. What is the final velocity after 3 seconds? (2023)

A. 3 m/s
B. 6 m/s
C. 9 m/s
D. 12 m/s

Solution

Using F = ma, a = F/m = 30 N / 10 kg = 3 m/s². Final velocity v = u + at = 0 + 3 m/s² × 3 s = 9 m/s.

Correct Answer: B — 6 m/s

Q. A 10 kg object is dropped from a height of 20 m. What is its potential energy at the top? (2020)

A. 200 J
B. 1000 J
C. 1500 J
D. 3000 J

Solution

Potential Energy = mass × g × height = 10 kg × 10 m/s² × 20 m = 2000 J

Correct Answer: B — 1000 J

Q. A 10 kg object is dropped from a height of 20 m. What is its potential energy at the top? (g = 9.8 m/s²) (1960)

A. 196 J
B. 98 J
C. 294 J
D. 490 J

Solution

Potential Energy (PE) = mgh = 10 kg × 9.8 m/s² × 20 m = 1960 J.

Correct Answer: C — 294 J

Q. A 10 kg object is dropped from a height of 20 m. What is the potential energy at the top? (2022)

A. 200 J
B. 1000 J
C. 500 J
D. 400 J

Solution

Potential Energy = mass × g × height = 10 kg × 9.8 m/s² × 20 m = 1960 J

Correct Answer: B — 1000 J

Q. A 10 kg object is lifted to a height of 2 m. What is the work done against gravity? (g = 9.8 m/s²)

A. 196 J
B. 98 J
C. 20 J
D. 39.2 J

Solution

Work done (W) = m × g × h = 10 kg × 9.8 m/s² × 2 m = 196 J.

Correct Answer: A — 196 J

Q. A 10 kg object is moving with a velocity of 4 m/s. What is its kinetic energy?

A. 80 J
B. 40 J
C. 20 J
D. 160 J

Solution

Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 10 kg × (4 m/s)² = 80 J

Correct Answer: A — 80 J

Q. A 10 kg object is subjected to a net force of 30 N. What is its acceleration? (2022)

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

Solution

Using F = ma, acceleration a = F/m = 30 N / 10 kg = 3 m/s².

Correct Answer: B — 3 m/s²

Q. A 10 kg object is thrown upwards with a velocity of 15 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)

A. 11.5 m
B. 22.5 m
C. 15.0 m
D. 7.5 m

Solution

Using energy conservation: KE_initial = PE_max; 0.5 * m * v² = mgh; h = v² / (2g) = (15 m/s)² / (2 * 9.8 m/s²) = 11.5 m.

Correct Answer: A — 11.5 m

Q. A 10 kg object is thrown vertically upward with a speed of 20 m/s. What is the maximum height reached by the object? (g = 10 m/s²) (2020)

A. 20 m
B. 30 m
C. 40 m
D. 50 m

Solution

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

Correct Answer: B — 30 m

Q. A 10 kg object is thrown vertically upwards with a velocity of 20 m/s. What is the maximum height it reaches? (g = 10 m/s²) (2023)

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

Solution

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

Correct Answer: B — 40 m

Q. A 10 N force is applied at an angle of 60° to the horizontal while moving an object 4 m. What is the work done?

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

Solution

Work done (W) = F × d × cos(θ) = 10 N × 4 m × cos(60°) = 10 × 4 × 0.5 = 20 J.

Correct Answer: C — 30 J

Q. A 10 N force is applied to move an object 3 m. What is the work done if the force is applied at an angle of 60 degrees to the direction of motion?

A. 15 J
B. 30 J
C. 25 J
D. 20 J

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

Q. A 10 Ω resistor is connected to a 20 V battery. What is the current flowing through the resistor?

A. 2 A
B. 0.5 A
C. 5 A
D. 10 A

Solution

Using Ohm's law, V = IR, where V = 20 V and R = 10 Ω. Therefore, I = V/R = 20/10 = 2 A.

Correct Answer: A — 2 A

Q. A 100 g piece of metal at 100°C is placed in 200 g of water at 20°C. What will be the final temperature of the system? (Specific heat of water = 4.2 J/g°C, specific heat of metal = 0.5 J/g°C) (2023)

A. 30°C
B. 40°C
C. 50°C
D. 60°C

Solution

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

Correct Answer: C — 50°C

Q. A 100 W bulb is connected to a 220 V supply. What is the current flowing through the bulb?

A. 0.45 A
B. 1.0 A
C. 2.2 A
D. 2.5 A

Solution

Using the formula P = VI, where P = 100 W and V = 220 V, the current I = P/V = 100/220 ≈ 0.45 A.

Correct Answer: A — 0.45 A

Q. A 100 W bulb is used for 5 hours. How much energy does it consume? (2016)

A. 500 Wh
B. 1000 Wh
C. 200 Wh
D. 250 Wh

Solution

Energy consumed = Power × Time = 100 W × 5 h = 500 Wh.

Correct Answer: A — 500 Wh

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