Engineering Entrance Exam Preparation Resources

My Learning Cart

Engineering Entrance

Download Q&A

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 cyclist is moving at a speed of 15 m/s. If the total mass of the cyclist and the bicycle is 75 kg, what is the kinetic energy?

A. 843.75 J
B. 562.5 J
C. 1687.5 J
D. 1125 J

Solution

Kinetic Energy (KE) = 0.5 × m × v² = 0.5 × 75 kg × (15 m/s)² = 0.5 × 75 × 225 = 8437.5 J.

Correct Answer: A — 843.75 J

Q. A cyclist is moving up a hill and gains 3000 J of potential energy. If the mass of the cyclist and the bicycle is 75 kg, what is the height of the hill? (g = 9.8 m/s²)

A. 4.08 m
B. 3.06 m
C. 2.04 m
D. 1.5 m

Solution

Height (h) = PE / (mg) = 3000 J / (75 kg * 9.8 m/s²) = 4.08 m.

Correct Answer: A — 4.08 m

Q. A cyclist of mass 70 kg accelerates at 2 m/s². What is the force exerted by the cyclist on the ground? (2021)

A. 70 N
B. 140 N
C. 210 N
D. 280 N

Solution

Using F = ma, F = 70 kg * 2 m/s² = 140 N.

Correct Answer: B — 140 N

Q. A cyclist travels at a speed of 10 m/s for 5 seconds. What distance does he cover? (2023)

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

Solution

Distance = speed * time = 10 m/s * 5 s = 50 m.

Correct Answer: B — 50 m

Q. A cyclist travels at a speed of 15 m/s for 2 minutes. How far does he travel?

A. 1800 m
B. 1200 m
C. 900 m
D. 600 m

Solution

Distance = speed * time. Time in seconds = 2 * 60 = 120 s. Distance = 15 m/s * 120 s = 1800 m.

Correct Answer: A — 1800 m

Q. A cyclist travels at a speed of 15 m/s for 2 minutes. How far does the cyclist travel?

A. 1800 m
B. 1200 m
C. 900 m
D. 600 m

Solution

Distance = speed * time. Time in seconds = 2 minutes = 120 seconds. Distance = 15 m/s * 120 s = 1800 m.

Correct Answer: A — 1800 m

Q. A cyclist travels at a speed of 15 m/s for 20 seconds. What distance does he cover?

A. 200 m
B. 300 m
C. 400 m
D. 500 m

Solution

Distance = speed * time = 15 m/s * 20 s = 300 m.

Correct Answer: C — 400 m

Q. A cyclist travels at a speed of 15 m/s for 20 seconds. What distance does the cyclist cover?

A. 150 m
B. 300 m
C. 400 m
D. 600 m

Solution

Distance = speed * time = 15 m/s * 20 s = 300 m.

Correct Answer: B — 300 m

Q. A cylinder rolls down an incline without slipping. If its mass is M and radius is R, what is the acceleration of its center of mass? (2020)

A. g/2
B. g/3
C. g/4
D. g/5

Solution

For rolling without slipping, a_cm = (2/3)g sin(θ).

Correct Answer: B — g/3

Q. A cylindrical can is to be made with a fixed volume of 1000 cm³. What dimensions minimize the surface area? (2022) 2022

A. 10, 10
B. 5, 20
C. 8, 15
D. 6, 18

Solution

Using the formula for surface area and volume, the optimal dimensions are found to be radius = 5 and height = 20.

Correct Answer: C — 8, 15

Q. A cylindrical can is to be made with a fixed volume of 1000 cm³. What dimensions minimize the surface area? (2022)

A. 10 cm height, 10 cm radius
B. 5 cm height, 15.87 cm radius
C. 8 cm height, 12.5 cm radius
D. 12 cm height, 8.33 cm radius

Solution

Using the formula for surface area and volume, the optimal dimensions are found to be 5 cm height and approximately 15.87 cm radius.

Correct Answer: B — 5 cm height, 15.87 cm radius

Q. A cylindrical can is to be made with a volume of 1000 cm³. What dimensions minimize the surface area? (2021)

A. 10, 10
B. 5, 20
C. 8, 15
D. 6, 18

Solution

Using the formula for volume and surface area, the optimal dimensions are found to be radius = 5 cm and height = 20 cm.

Correct Answer: B — 5, 20

Q. A dipole consists of two equal and opposite charges separated by a distance 'd'. What is the dipole moment? (2023)

A. qd
B. q/d
C. q^2d
D. qd^2

Solution

The dipole moment (p) is defined as p = qd, where q is the charge and d is the separation distance.

Correct Answer: A — qd

Q. A dipole is placed in a uniform electric field. What is the torque experienced by the dipole? (2023)

A. pE
B. pE sin(θ)
C. pE cos(θ)
D. Zero

Solution

The torque (τ) experienced by a dipole in a uniform electric field is given by τ = pE sin(θ), where p is the dipole moment and θ is the angle between p and E.

Correct Answer: B — pE sin(θ)

Q. A disc of radius R and mass M is rotating about its axis with an angular velocity ω. What is its rotational kinetic energy? (2020)

A. (1/2)Iω²
B. (1/2)Mω²
C. Iω
D. Mω²

Solution

Rotational kinetic energy K.E. = (1/2)Iω². For a disc, I = (1/2)MR², thus K.E. = (1/4)MR²ω².

Correct Answer: A — (1/2)Iω²

Q. A farmer wants to fence a rectangular area of 200 m^2. What dimensions will minimize the fencing required? (2021)

A. 10, 20
B. 14, 14.28
C. 15, 13.33
D. 20, 10

Solution

For a fixed area, the minimum perimeter occurs when the rectangle is a square. Thus, dimensions are approximately 14 m by 14.28 m.

Correct Answer: B — 14, 14.28

Q. A farmer wants to fence a rectangular field with 100 m of fencing. What dimensions will maximize the area? (2022)

A. 25 m by 25 m
B. 30 m by 20 m
C. 40 m by 10 m
D. 50 m by 0 m

Solution

For maximum area, the rectangle should be a square. Each side = 100/4 = 25 m.

Correct Answer: A — 25 m by 25 m

Q. A flywheel has a moment of inertia I and is rotating with an angular velocity ω. If a torque τ is applied, what is the angular acceleration? (2021)

A. τ/I
B. I/τ
C. ω/τ
D. Iω

Solution

From Newton's second law for rotation, τ = Iα, thus α = τ/I.

Correct Answer: A — τ/I

Q. A force of 10 N is applied to an object, causing it to accelerate at 2 m/s². What is the mass of the object? (2020)

A. 5 kg
B. 10 kg
C. 15 kg
D. 20 kg

Solution

Using F = ma, mass m = F/a = 10 N / 2 m/s² = 5 kg.

Correct Answer: A — 5 kg

Q. A force of 10 N is applied to move a box 5 m across a floor. What is the work done on the box?

A. 10 J
B. 20 J
C. 50 J
D. 5 J

Solution

Work done = Force × Distance = 10 N × 5 m = 50 J

Correct Answer: B — 20 J

Q. A force of 10 N is applied to move a box 5 m across a floor. What is the work done by the force?

A. 10 J
B. 20 J
C. 50 J
D. 5 J

Solution

Work done = Force × Distance = 10 N × 5 m = 50 J

Correct Answer: B — 20 J

Q. A force of 10 N is applied to move a box 5 m along a horizontal surface. What is the work done by the force? (2021)

A. 20 J
B. 50 J
C. 10 J
D. 5 J

Solution

Work done = Force × Distance = 10 N × 5 m = 50 J

Correct Answer: B — 50 J

Q. A force of 10 N is applied to move an object 5 m in the direction of the force. What is the work done by the force?

A. 10 J
B. 20 J
C. 50 J
D. 5 J

Solution

Work done (W) = Force (F) × Distance (d) × cos(θ). Here, θ = 0° (force and displacement are in the same direction). W = 10 N × 5 m × cos(0°) = 10 N × 5 m = 50 J.

Correct Answer: B — 20 J

Q. A force of 10 N is applied to move an object 5 m. What is the work done? (2023)

A. 50 J
B. 10 J
C. 5 J
D. 20 J

Solution

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

Correct Answer: A — 50 J

Q. A force of 15 N is applied to a 3 kg mass. What is the resulting acceleration? (2023)

A. 3 m/s²
B. 5 m/s²
C. 7 m/s²
D. 10 m/s²

Solution

Using F = ma, acceleration a = F/m = 15 N / 3 kg = 5 m/s².

Correct Answer: B — 5 m/s²

Q. A force of 20 N is applied at an angle of 60° to the horizontal while moving an object 4 m. What is the work done?

A. 40 J
B. 80 J
C. 20 J
D. 60 J

Solution

Work done (W) = F × d × cos(θ) = 20 N × 4 m × cos(60°) = 20 × 4 × 0.5 = 40 J.

Correct Answer: B — 80 J

Q. A force of 30 N is applied to a 6 kg object. What is the object's acceleration? (2023)

A. 3 m/s²
B. 4 m/s²
C. 5 m/s²
D. 6 m/s²

Solution

Using F = ma, we find a = F/m = 30 N / 6 kg = 5 m/s².

Correct Answer: B — 4 m/s²

Q. A force of 50 N is applied to move an object 4 m. How much work is done?

A. 100 J
B. 150 J
C. 200 J
D. 250 J

Solution

Work = Force * Distance = 50 N * 4 m = 200 J.

Correct Answer: C — 200 J

Q. A gas expands from 2 L to 5 L at a constant pressure of 1 atm. How much work is done by the gas? (2023)

A. 3 L·atm
B. 5 L·atm
C. 2 L·atm
D. 1 L·atm

Solution

Work done by the gas during expansion at constant pressure is W = PΔV. Here, ΔV = 5 L - 2 L = 3 L, so W = 1 atm * 3 L = 3 L·atm.

Correct Answer: A — 3 L·atm

Q. A gas expands from a volume of 2 m³ to 5 m³ at a constant pressure of 100 kPa. What is the work done by the gas? (2022)

A. 300 kJ
B. 500 kJ
C. 700 kJ
D. 3000 kJ

Solution

Work done (W) = P * ΔV = 100 kPa * (5 m³ - 2 m³) = 100 kPa * 3 m³ = 300 kJ.

Correct Answer: A — 300 kJ

|<
<
3
4
5
6
7
8
9
10
11
>
>|

Showing 181 to 210 of 2530 (85 Pages)

School

College

Degree

Competitve

Skills

Soulshift Feedback ×

On a scale of 0–10, how likely are you to recommend The Soulshift Academy?

Not likely Very likely