Engineering Entrance Exam Preparation Resources

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Engineering Entrance MCQ & Objective Questions

Preparing for Engineering Entrance exams is crucial for aspiring engineers in India. Mastering MCQs and objective questions not only enhances your understanding of key concepts but also boosts your confidence during exams. Regular practice with these questions helps identify important topics and improves your overall exam preparation.

What You Will Practise Here

Fundamental concepts of Physics and Mathematics
Key formulas and their applications in problem-solving
Important definitions and theorems relevant to engineering
Diagrams and graphical representations for better understanding
Conceptual questions that challenge your critical thinking
Previous years' question papers and their analysis
Time management strategies while solving MCQs

Exam Relevance

The Engineering Entrance syllabus is integral to various examinations like CBSE, State Boards, NEET, and JEE. Questions often focus on core subjects such as Physics, Chemistry, and Mathematics, with formats varying from direct MCQs to application-based problems. Understanding the common question patterns can significantly enhance your performance and help you tackle the exams with ease.

Common Mistakes Students Make

Overlooking the importance of units and dimensions in calculations
Misinterpreting questions due to lack of careful reading
Neglecting to review basic concepts before attempting advanced problems
Rushing through practice questions without thorough understanding

FAQs

Question: What are the best ways to prepare for Engineering Entrance MCQs?
Answer: Focus on understanding concepts, practice regularly with objective questions, and review previous years' papers.

Question: How can I improve my speed in solving MCQs?
Answer: Regular practice, time-bound mock tests, and familiarizing yourself with common question types can help improve your speed.

Start your journey towards success by solving Engineering Entrance MCQ questions today! Test your understanding and build a strong foundation for your exams.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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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