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 10 kg object is dropped from a height. What is the force acting on it just before it hits the ground?
A.
10 N
B.
20 N
C.
30 N
D.
40 N
Show solution
Solution
The force acting on the object is its weight, F = mg = 10 kg * 10 m/s² = 100 N.
Correct Answer:
B
— 20 N
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Q. A 10 kg object is hanging at rest from a rope. What is the tension in the rope?
A.
0 N
B.
10 N
C.
100 N
D.
50 N
Show solution
Solution
The tension in the rope must balance the weight of the object. T = mg = 10 kg * 10 m/s² = 100 N.
Correct Answer:
C
— 100 N
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Q. A 10 kg object is hanging from a rope. What is the tension in the rope when the object is at rest?
A.
0 N
B.
10 N
C.
100 N
D.
50 N
Show solution
Solution
At rest, the tension in the rope equals the weight of the object: T = mg = 10 kg * 10 m/s² = 100 N.
Correct Answer:
C
— 100 N
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Q. A 10 kg object is lifted to a height of 10 m. How much work is done against gravity?
A.
0 J
B.
100 J
C.
200 J
D.
1000 J
Show solution
Solution
Work done against gravity is equal to the change in potential energy, W = mgh = 10 * 9.8 * 10 = 980 J.
Correct Answer:
D
— 1000 J
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Q. A 10 kg object is lifted to a height of 2 m. What is the work done against gravity?
A.
20 J
B.
40 J
C.
60 J
D.
80 J
Show solution
Solution
Work done = mass × g × height = 10 kg × 9.8 m/s² × 2 m = 196 J.
Correct Answer:
B
— 40 J
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Q. A 10 kg object is lifted to a height of 5 m. How much work is done against gravity?
A.
50 J
B.
100 J
C.
150 J
D.
200 J
Show solution
Solution
Work done = mgh = 10 * 9.8 * 5 = 490 J.
Correct Answer:
B
— 100 J
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Q. A 10 kg object is lifted to a height of 5 m. What is the potential energy gained by the object?
A.
50 J
B.
100 J
C.
150 J
D.
200 J
Show solution
Solution
Potential energy (PE) = mgh = 10 kg * 9.8 m/s² * 5 m = 490 J.
Correct Answer:
B
— 100 J
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Q. A 10 kg object is lifted to a height of 5 m. What is the work done against gravity?
A.
50 J
B.
100 J
C.
150 J
D.
200 J
Show solution
Solution
Work done against gravity = mgh = 10 * 9.8 * 5 = 490 J.
Correct Answer:
B
— 100 J
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Q. A 10 kg object is moving in a circular path of radius 5 m with a speed of 10 m/s. What is the centripetal force acting on the object?
A.
20 N
B.
10 N
C.
5 N
D.
50 N
Show solution
Solution
Centripetal force F = mv²/r = 10 kg * (10 m/s)² / 5 m = 200 N.
Correct Answer:
A
— 20 N
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Q. A 10 kg object is moving in a circular path of radius 5 m with a speed of 4 m/s. What is the centripetal force acting on the object?
A.
8 N
B.
16 N
C.
32 N
D.
40 N
Show solution
Solution
Centripetal force F = mv²/r = 10 kg * (4 m/s)² / 5 m = 32 N.
Correct Answer:
B
— 16 N
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Q. A 10 kg object is moving with a speed of 3 m/s. If it comes to a stop, how much work is done by the friction force?
A.
45 J
B.
90 J
C.
135 J
D.
180 J
Show solution
Solution
Work done = Change in Kinetic Energy = 0 - (0.5 × 10 kg × (3 m/s)²) = -45 J.
Correct Answer:
B
— 90 J
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Q. A 10 kg object is moving with a speed of 3 m/s. What is its kinetic energy?
A.
45 J
B.
30 J
C.
60 J
D.
90 J
Show solution
Solution
Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 10 kg × (3 m/s)² = 45 J.
Correct Answer:
A
— 45 J
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Q. A 10 kg object is moving with a speed of 5 m/s. What is its total mechanical energy?
A.
125 J
B.
250 J
C.
500 J
D.
1000 J
Show solution
Solution
Total mechanical energy = KE + PE. KE = 0.5 * 10 * (5)² = 125 J. Assuming PE = 0, total energy = 125 J.
Correct Answer:
B
— 250 J
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Q. A 10 kg object is moving with a velocity of 2 m/s. If it comes to rest, what is the change in momentum?
A.
20 kg·m/s
B.
10 kg·m/s
C.
5 kg·m/s
D.
0 kg·m/s
Show solution
Solution
Initial momentum = mv = 10 kg * 2 m/s = 20 kg·m/s. Final momentum = 0. Change in momentum = 20 kg·m/s - 0 = 20 kg·m/s.
Correct Answer:
A
— 20 kg·m/s
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Q. A 10 kg object is moving with a velocity of 3 m/s. What is its kinetic energy?
A.
15 J
B.
30 J
C.
45 J
D.
60 J
Show solution
Solution
Kinetic energy = 0.5 × m × v² = 0.5 × 10 kg × (3 m/s)² = 45 J.
Correct Answer:
C
— 45 J
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Q. A 10 kg object is moving with a velocity of 3 m/s. What is its momentum?
A.
10 kg·m/s
B.
30 kg·m/s
C.
3 kg·m/s
D.
0.3 kg·m/s
Show solution
Solution
Momentum p = mv = 10 kg * 3 m/s = 30 kg·m/s.
Correct Answer:
B
— 30 kg·m/s
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Q. A 10 kg object is moving with a velocity of 3 m/s. What is the total mechanical energy if it is at a height of 5 m?
A.
150 J
B.
180 J
C.
300 J
D.
450 J
Show solution
Solution
Total Mechanical Energy = Kinetic Energy + Potential Energy = 0.5 × 10 kg × (3 m/s)² + 10 kg × 9.8 m/s² × 5 m = 45 J + 490 J = 535 J.
Correct Answer:
C
— 300 J
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Q. A 10 kg object is moving with a velocity of 4 m/s. What is its momentum?
A.
40 kg m/s
B.
20 kg m/s
C.
10 kg m/s
D.
30 kg m/s
Show solution
Solution
Momentum (p) = m * v = 10 kg * 4 m/s = 40 kg m/s
Correct Answer:
A
— 40 kg m/s
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Q. A 10 kg object is moving with a velocity of 5 m/s. What is its momentum?
A.
10 kg·m/s
B.
25 kg·m/s
C.
50 kg·m/s
D.
75 kg·m/s
Show solution
Solution
Momentum p = mv = 10 kg * 5 m/s = 50 kg·m/s.
Correct Answer:
C
— 50 kg·m/s
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Q. A 10 kg object is pushed with a force of 30 N. If the frictional force is 10 N, what is the net force?
A.
10 N
B.
20 N
C.
30 N
D.
40 N
Show solution
Solution
Net force = applied force - friction = 30 N - 10 N = 20 N.
Correct Answer:
B
— 20 N
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Q. A 10 kg object is thrown upwards with a speed 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.
15.3 m
D.
10.0 m
Show solution
Solution
Using conservation of energy, KE at the bottom = PE at the top. 0.5mv² = mgh. Solving gives h = v²/(2g) = (15)²/(2*9.8) = 11.5 m.
Correct Answer:
A
— 11.5 m
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Q. A 10 kg object is thrown vertically upwards with a speed of 15 m/s. What is the maximum height it reaches? (g = 10 m/s²)
A.
11.25 m
B.
22.5 m
C.
15 m
D.
7.5 m
Show solution
Solution
Using energy conservation: KE_initial = PE_max; 1/2 * m * v^2 = m * g * h; h = v^2 / (2g) = (15^2) / (2 * 10) = 11.25 m
Correct Answer:
A
— 11.25 m
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Q. A 10 Ω resistor and a 20 Ω resistor are connected in series. What is the total resistance?
A.
10 Ω
B.
20 Ω
C.
30 Ω
D.
5 Ω
Show solution
Solution
In series, the total resistance R = R1 + R2 = 10 Ω + 20 Ω = 30 Ω.
Correct Answer:
C
— 30 Ω
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Q. A 10-ohm resistor and a 20-ohm resistor are connected in series. What is the total resistance?
A.
10 ohms
B.
20 ohms
C.
30 ohms
D.
5 ohms
Show solution
Solution
In series, the total resistance R = R1 + R2 = 10 + 20 = 30 ohms.
Correct Answer:
C
— 30 ohms
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Q. A 10-ohm resistor has a voltage of 20 volts across it. What is the current flowing through the resistor?
A.
1 A
B.
2 A
C.
3 A
D.
4 A
Show solution
Solution
Using Ohm's Law, I = V / R = 20 V / 10 Ω = 2 A.
Correct Answer:
B
— 2 A
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Q. A 10-ohm resistor has a voltage of 50 volts across it. What is the current flowing through the resistor?
A.
5 A
B.
10 A
C.
15 A
D.
20 A
Show solution
Solution
Using Ohm's Law, I = V / R = 50 V / 10 Ω = 5 A.
Correct Answer:
B
— 10 A
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Q. A 10-ohm resistor is connected to a 20-volt battery. What is the power dissipated by the resistor?
A.
40 W
B.
20 W
C.
10 W
D.
2 W
Show solution
Solution
Power (P) can be calculated using P = V^2 / R = 20^2 / 10 = 400 / 10 = 40 W.
Correct Answer:
A
— 40 W
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Q. A 1000 kg car accelerates from rest to a speed of 20 m/s in 10 seconds. What is the average power exerted by the engine?
A.
2000 W
B.
4000 W
C.
5000 W
D.
6000 W
Show solution
Solution
First, calculate the work done: W = (1/2)mv^2 = (1/2)(1000 kg)(20 m/s)^2 = 200000 J. Then, power is P = W/t = 200000 J / 10 s = 20000 W.
Correct Answer:
C
— 5000 W
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Q. A 1000 W heater operates for 1 hour. How much energy does it consume?
A.
3600 J
B.
1000 J
C.
3600000 J
D.
100000 J
Show solution
Solution
Energy consumed is E = P * t. Here, P = 1000 W and t = 1 hour = 3600 seconds. Thus, E = 1000 W * 3600 s = 3600000 J.
Correct Answer:
C
— 3600000 J
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Q. A 1000 W heater operates for 2 hours. How much energy does it consume?
A.
7200000 J
B.
2000000 J
C.
3600000 J
D.
1000000 J
Show solution
Solution
Energy consumed can be calculated using E = P * t. Here, P = 1000 W and t = 2 hours = 7200 seconds. Thus, E = 1000 W * 7200 s = 7200000 J.
Correct Answer:
C
— 3600000 J
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