Engineering & Architecture Admissions MCQ & Objective Questions
Engineering & Architecture Admissions play a crucial role in shaping the future of aspiring students in India. With the increasing competition in entrance exams, mastering MCQs and objective questions is essential for effective exam preparation. Practicing these types of questions not only enhances concept clarity but also boosts confidence, helping students score better in their exams.
What You Will Practise Here
Key concepts in Engineering Mathematics
Fundamentals of Physics relevant to architecture and engineering
Important definitions and terminologies in engineering disciplines
Essential formulas for solving objective questions
Diagrams and illustrations for better understanding
Conceptual theories related to structural engineering
Analysis of previous years' important questions
Exam Relevance
The topics covered under Engineering & Architecture Admissions are highly relevant for various examinations such as CBSE, State Boards, NEET, and JEE. Students can expect to encounter MCQs that test their understanding of core concepts, application of formulas, and analytical skills. Common question patterns include multiple-choice questions that require selecting the correct answer from given options, as well as assertion-reason type questions that assess deeper comprehension.
Common Mistakes Students Make
Misinterpreting the question stem, leading to incorrect answers.
Overlooking units in numerical problems, which can change the outcome.
Confusing similar concepts or terms, especially in definitions.
Neglecting to review diagrams, which are often crucial for solving problems.
Rushing through practice questions without understanding the underlying concepts.
FAQs
Question: What are the best ways to prepare for Engineering & Architecture Admissions MCQs?Answer: Regular practice of objective questions, reviewing key concepts, and taking mock tests can significantly enhance your preparation.
Question: How can I improve my accuracy in solving MCQs?Answer: Focus on understanding the concepts thoroughly, practice regularly, and learn to eliminate incorrect options to improve accuracy.
Start your journey towards success by solving practice MCQs today! Test your understanding and strengthen your knowledge in Engineering & Architecture Admissions to excel in your exams.
Q. A 3 kg object is subjected to a net force of 12 N. What is its acceleration?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
Show solution
Solution
Using F = ma, a = F/m = 12 N / 3 kg = 4 m/s².
Correct Answer:
C
— 4 m/s²
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Q. A 4 kg object is at rest on a table. What is the force of static friction acting on it?
A.
0 N
B.
4 N
C.
40 N
D.
10 N
Show solution
Solution
If the object is at rest and no external force is applied, the static friction force is 0 N.
Correct Answer:
A
— 0 N
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Q. A 4 kg object is at rest on a table. What is the force of static friction if no external force is applied?
A.
0 N
B.
4 N
C.
40 N
D.
None of the above
Show solution
Solution
If no external force is applied, the force of static friction is 0 N.
Correct Answer:
A
— 0 N
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Q. A 4 kg object is dropped from a height. What is the force acting on it just before it hits the ground?
A.
0 N
B.
4 N
C.
40 N
D.
10 N
Show solution
Solution
The force acting on the object is its weight, F = mg = 4 kg * 10 m/s² = 40 N.
Correct Answer:
C
— 40 N
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Q. A 4 kg object is hanging from a rope. What is the tension in the rope if the object is at rest?
A.
0 N
B.
4 N
C.
40 N
D.
10 N
Show solution
Solution
The tension in the rope equals the weight of the object: T = mg = 4 kg * 10 m/s² = 40 N.
Correct Answer:
C
— 40 N
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Q. A 4 kg object is lifted to a height of 3 m. What is the change in gravitational potential energy?
A.
12 J
B.
24 J
C.
36 J
D.
48 J
Show solution
Solution
Change in potential energy = mgh = 4 kg × 9.8 m/s² × 3 m = 117.6 J.
Correct Answer:
B
— 24 J
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Q. A 4 kg object is lifted to a height of 3 m. What is the increase in gravitational potential energy? (g = 9.8 m/s²)
A.
117.6 J
B.
117 J
C.
120 J
D.
150 J
Show solution
Solution
Potential energy increase = mgh = 4 kg × 9.8 m/s² × 3 m = 117.6 J.
Correct Answer:
A
— 117.6 J
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Q. A 4 kg object is lifted to a height of 3 m. What is the increase in gravitational potential energy?
A.
12 J
B.
24 J
C.
36 J
D.
48 J
Show solution
Solution
Increase in potential energy = mgh = 4 kg × 9.8 m/s² × 3 m = 117.6 J.
Correct Answer:
B
— 24 J
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Q. A 4 kg object is moving in a circular path of radius 2 m with a speed of 6 m/s. What is the centripetal force acting on the object?
A.
9 N
B.
12 N
C.
18 N
D.
24 N
Show solution
Solution
Centripetal force F = mv²/r = 4 kg * (6 m/s)² / 2 m = 4 * 36 / 2 = 72 N / 2 = 36 N.
Correct Answer:
B
— 12 N
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Q. A 4 kg object is moving in a circular path of radius 2 m with a speed of 6 m/s. What is the centripetal force acting on it?
A.
9 N
B.
12 N
C.
18 N
D.
24 N
Show solution
Solution
Centripetal force F = mv²/r = 4 kg * (6 m/s)² / 2 m = 72 N / 2 = 36 N.
Correct Answer:
B
— 12 N
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Q. A 4 kg object is moving in a straight line with a velocity of 5 m/s. What is the kinetic energy of the object?
A.
50 J
B.
40 J
C.
20 J
D.
10 J
Show solution
Solution
Kinetic energy KE = 0.5 * m * v² = 0.5 * 4 kg * (5 m/s)² = 50 J.
Correct Answer:
A
— 50 J
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Q. A 4 kg object is moving with a speed of 5 m/s. If it comes to rest, what is the work done by friction?
A.
50 J
B.
75 J
C.
100 J
D.
125 J
Show solution
Solution
Work done = change in kinetic energy = 0 - 0.5 × 4 kg × (5 m/s)² = -50 J.
Correct Answer:
C
— 100 J
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Q. A 4 kg object is moving with a speed of 5 m/s. What is its kinetic energy?
A.
10 J
B.
20 J
C.
50 J
D.
100 J
Show solution
Solution
Kinetic energy = 0.5 × m × v² = 0.5 × 4 kg × (5 m/s)² = 50 J.
Correct Answer:
C
— 50 J
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Q. A 4 kg object is moving with a speed of 5 m/s. What is the total mechanical energy if it is at a height of 2 m?
A.
50 J
B.
60 J
C.
70 J
D.
80 J
Show solution
Solution
Total mechanical energy = Kinetic energy + Potential energy = 0.5 × 4 kg × (5 m/s)² + 4 kg × 9.8 m/s² × 2 m = 50 J + 78.4 J = 128.4 J.
Correct Answer:
C
— 70 J
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Q. A 4 kg object is moving with a velocity of 2 m/s. What is its kinetic energy?
A.
8 J
B.
4 J
C.
16 J
D.
2 J
Show solution
Solution
Kinetic Energy (KE) = 1/2 * m * v^2 = 1/2 * 4 * (2^2) = 8 J
Correct Answer:
A
— 8 J
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Q. A 4 kg object is moving with a velocity of 2 m/s. What is the change in momentum if the object comes to a stop?
A.
0 kg·m/s
B.
4 kg·m/s
C.
8 kg·m/s
D.
2 kg·m/s
Show solution
Solution
Change in momentum = final momentum - initial momentum = 0 - (4 kg * 2 m/s) = -8 kg·m/s.
Correct Answer:
C
— 8 kg·m/s
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Q. A 4 kg object is moving with a velocity of 3 m/s. What is its kinetic energy?
A.
18 J
B.
24 J
C.
36 J
D.
12 J
Show solution
Solution
Kinetic energy KE = 0.5 * m * v² = 0.5 * 4 kg * (3 m/s)² = 18 J.
Correct Answer:
B
— 24 J
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Q. A 4 kg object is moving with a velocity of 5 m/s. What is its momentum?
A.
10 kg·m/s
B.
20 kg·m/s
C.
15 kg·m/s
D.
25 kg·m/s
Show solution
Solution
Momentum p = mv = 4 kg * 5 m/s = 20 kg·m/s.
Correct Answer:
B
— 20 kg·m/s
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Q. A 4 kg object is pulled with a force of 20 N. If the frictional force is 4 N, what is the net force acting on the object?
A.
16 N
B.
20 N
C.
24 N
D.
4 N
Show solution
Solution
Net force = applied force - frictional force = 20 N - 4 N = 16 N.
Correct Answer:
A
— 16 N
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Q. A 4 kg object is pushed with a force of 16 N. What is the acceleration of the object?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
Show solution
Solution
Using F = ma, we have a = F/m = 16 N / 4 kg = 4 m/s².
Correct Answer:
C
— 4 m/s²
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Q. A 4 kg object is pushed with a force of 20 N over a distance of 3 m. If the object starts from rest, what is its final speed? (Assume no friction)
A.
2 m/s
B.
3 m/s
C.
4 m/s
D.
5 m/s
Show solution
Solution
Work done = Force × Distance = 20 N × 3 m = 60 J. Kinetic energy = 0.5 × m × v². 60 J = 0.5 × 4 kg × v². v² = 30, v = √30 ≈ 5.48 m/s.
Correct Answer:
C
— 4 m/s
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Q. A 4 kg object is pushed with a force of 20 N over a distance of 3 m. If the object starts from rest, what is its final speed?
A.
2 m/s
B.
3 m/s
C.
4 m/s
D.
5 m/s
Show solution
Solution
Work done = Force × Distance = 20 N × 3 m = 60 J. K.E = 0.5 × m × v², 60 J = 0.5 × 4 kg × v², v = 3.87 m/s.
Correct Answer:
C
— 4 m/s
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Q. A 4 kg object is pushed with a force of 20 N. If the frictional force is 4 N, what is the acceleration of the object?
A.
4 m/s²
B.
5 m/s²
C.
6 m/s²
D.
7 m/s²
Show solution
Solution
Net force = applied force - friction = 20 N - 4 N = 16 N. Acceleration a = F/m = 16 N / 4 kg = 4 m/s².
Correct Answer:
B
— 5 m/s²
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Q. A 4 kg object is subjected to a net force of 12 N. What is the acceleration of the object?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
Show solution
Solution
Using F = ma, a = F/m = 12 N / 4 kg = 3 m/s².
Correct Answer:
B
— 3 m/s²
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Q. A 4 kg object is subjected to a net force of 16 N. What is its acceleration?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
Show solution
Solution
Using F = ma, acceleration a = F/m = 16 N / 4 kg = 4 m/s².
Correct Answer:
C
— 4 m/s²
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Q. A 5 kg block is placed on a surface with a coefficient of static friction of 0.6. What is the maximum static friction force before the block starts to move?
A.
30 N
B.
20 N
C.
15 N
D.
25 N
Show solution
Solution
Maximum static friction force = μs * N = 0.6 * (5 kg * 9.8 m/s²) = 29.4 N, approximately 30 N.
Correct Answer:
A
— 30 N
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Q. A 5 kg block is resting on a frictionless surface. A force of 10 N is applied to it. What is the acceleration of the block?
A.
1 m/s²
B.
2 m/s²
C.
3 m/s²
D.
4 m/s²
Show solution
Solution
Using Newton's second law, F = ma, we have a = F/m = 10 N / 5 kg = 2 m/s².
Correct Answer:
B
— 2 m/s²
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Q. A 5 kg block is resting on a frictionless surface. If a force of 10 N is applied to it, what will be its acceleration?
A.
1 m/s²
B.
2 m/s²
C.
5 m/s²
D.
10 m/s²
Show solution
Solution
Using Newton's second law, F = ma, we have a = F/m = 10 N / 5 kg = 2 m/s².
Correct Answer:
B
— 2 m/s²
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Q. A 5 kg block is resting on a horizontal surface. What is the normal force acting on the block?
A.
5 N
B.
10 N
C.
15 N
D.
20 N
Show solution
Solution
The normal force equals the weight of the block, which is mass times gravity: 5 kg * 9.8 m/s² = 49 N.
Correct Answer:
B
— 10 N
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Q. A 5 kg object is at rest on a surface with a coefficient of static friction of 0.4. What is the maximum static friction force?
A.
10 N
B.
20 N
C.
30 N
D.
40 N
Show solution
Solution
Maximum static friction force = μs * N = 0.4 * (5 kg * 9.8 m/s²) = 0.4 * 49 N = 19.6 N, approximately 20 N.
Correct Answer:
B
— 20 N
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