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. An object is in equilibrium. What can be said about the net force acting on it?
A.
It is zero
B.
It is equal to its weight
C.
It is equal to the applied force
D.
It is maximum
Show solution
Solution
For an object in equilibrium, the net force acting on it must be zero.
Correct Answer:
A
— It is zero
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Q. An object is lifted to a height of 10 m. If the mass of the object is 5 kg, what is the gravitational potential energy gained?
A.
50 J
B.
100 J
C.
150 J
D.
200 J
Show solution
Solution
Potential Energy = mass × g × height = 5 kg × 9.8 m/s² × 10 m = 490 J.
Correct Answer:
B
— 100 J
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Q. An object is lifted vertically 10 m against gravity. If the mass of the object is 5 kg, what is the work done against gravity?
A.
50 J
B.
100 J
C.
150 J
D.
200 J
Show solution
Solution
Work done = mass × g × height = 5 kg × 9.8 m/s² × 10 m = 490 J.
Correct Answer:
B
— 100 J
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Q. An object is lifted vertically 3 m against gravity. If the mass of the object is 4 kg, what is the work done against gravity? (g = 9.8 m/s²)
A.
117.6 J
B.
39.2 J
C.
29.4 J
D.
19.6 J
Show solution
Solution
Work done = mgh = 4 kg × 9.8 m/s² × 3 m = 117.6 J.
Correct Answer:
A
— 117.6 J
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Q. An object is lifted vertically to a height of 10 m. If the mass of the object is 5 kg, what is the work done against gravity?
A.
50 J
B.
100 J
C.
150 J
D.
200 J
Show solution
Solution
Work done = mass × g × height = 5 kg × 9.8 m/s² × 10 m = 490 J.
Correct Answer:
B
— 100 J
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Q. An object is moving in a circular path with a constant speed. What can be said about the work done by the centripetal force?
A.
Positive work
B.
Negative work
C.
Zero work
D.
Depends on speed
Show solution
Solution
Centripetal force does no work as it acts perpendicular to the direction of motion.
Correct Answer:
C
— Zero work
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Q. An object is moving in a circular path with a constant speed. What type of energy is primarily involved?
A.
Kinetic Energy
B.
Potential Energy
C.
Mechanical Energy
D.
Thermal Energy
Show solution
Solution
The object has Kinetic Energy due to its motion in the circular path.
Correct Answer:
A
— Kinetic Energy
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Q. An object is moving in a circular path with a radius of 10 m and a speed of 5 m/s. What is the net force acting on the object if its mass is 2 kg?
A.
1 N
B.
2 N
C.
5 N
D.
10 N
Show solution
Solution
Centripetal force (F_c) = mv²/r = 2 kg * (5 m/s)² / (10 m) = 5 N.
Correct Answer:
D
— 10 N
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Q. An object is moving in a circular path with a radius of 10 m at a speed of 5 m/s. What is the period of the motion?
A.
2π s
B.
4π s
C.
10 s
D.
20 s
Show solution
Solution
Period (T) = 2πr/v = 2π(10)/5 = 4π s.
Correct Answer:
A
— 2π s
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Q. An object is moving in a circular path with a radius of 3 m and an angular speed of 2 rad/s. What is the linear speed of the object?
A.
3 m/s
B.
6 m/s
C.
9 m/s
D.
12 m/s
Show solution
Solution
Linear speed (v) = rω = 3 m * 2 rad/s = 6 m/s.
Correct Answer:
B
— 6 m/s
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Q. An object is placed 10 cm in front of a convex mirror with a focal length of 5 cm. What is the nature of the image formed?
A.
Real and inverted
B.
Virtual and erect
C.
Real and erect
D.
Virtual and inverted
Show solution
Solution
Convex mirrors always produce virtual and erect images.
Correct Answer:
B
— Virtual and erect
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Q. An object is placed 25 cm from a convex lens of focal length 10 cm. Where is the image formed?
A.
10 cm
B.
15 cm
C.
20 cm
D.
30 cm
Show solution
Solution
Using the lens formula 1/f = 1/v - 1/u, we find v = 20 cm.
Correct Answer:
C
— 20 cm
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Q. An object is placed 30 cm from a concave lens of focal length 15 cm. What is the nature of the image formed?
A.
Real and inverted
B.
Virtual and erect
C.
Real and erect
D.
Virtual and inverted
Show solution
Solution
For a concave lens, the image formed is virtual and erect when the object is placed in front of it.
Correct Answer:
B
— Virtual and erect
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Q. An object is placed 40 cm from a convex lens with a focal length of 10 cm. Where is the image formed?
A.
At 10 cm
B.
At 20 cm
C.
At 30 cm
D.
At 40 cm
Show solution
Solution
Using the lens formula, the image is formed at 30 cm on the opposite side.
Correct Answer:
C
— At 30 cm
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Q. An object is placed 50 cm from a converging lens of focal length 25 cm. Where will the image be formed?
A.
16.67 cm
B.
33.33 cm
C.
25 cm
D.
20 cm
Show solution
Solution
Using the lens formula 1/f = 1/v - 1/u, we find v = 33.33 cm.
Correct Answer:
B
— 33.33 cm
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Q. An object is placed at a distance of 15 cm from a convex lens of focal length 10 cm. Where is the image formed?
A.
5 cm
B.
10 cm
C.
15 cm
D.
20 cm
Show solution
Solution
Using the lens formula, 1/f = 1/v - 1/u, we find v = 30 cm, meaning the image is formed 30 cm on the opposite side.
Correct Answer:
D
— 20 cm
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Q. An object is placed at a distance of 30 cm from a convex lens of focal length 15 cm. What is the distance of the image from the lens?
A.
10 cm
B.
15 cm
C.
20 cm
D.
30 cm
Show solution
Solution
Using the lens formula, 1/f = 1/v - 1/u. Here, f = 15 cm, u = -30 cm. Solving gives v = 10 cm.
Correct Answer:
C
— 20 cm
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Q. An object is placed at a distance of 40 cm from a convex lens of focal length 20 cm. Where will the image be formed?
A.
10 cm
B.
20 cm
C.
30 cm
D.
40 cm
Show solution
Solution
Using the lens formula 1/f = 1/v - 1/u, we find v = 10 cm.
Correct Answer:
A
— 10 cm
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Q. An object is placed at the focus of a concave lens. What type of image is formed?
A.
Real and inverted
B.
Virtual and erect
C.
Real and erect
D.
Virtual and inverted
Show solution
Solution
A concave lens always forms a virtual and erect image regardless of the object position.
Correct Answer:
B
— Virtual and erect
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Q. An object is placed at the focus of a concave mirror. What type of image is formed?
A.
Real and inverted
B.
Virtual and upright
C.
No image
D.
Real and upright
Show solution
Solution
When the object is at the focus of a concave mirror, the rays are parallel after reflection, resulting in no image being formed.
Correct Answer:
C
— No image
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Q. An object is projected at an angle of 60 degrees with an initial speed of 30 m/s. What is the horizontal component of its velocity?
A.
15 m/s
B.
25 m/s
C.
30 m/s
D.
20 m/s
Show solution
Solution
Horizontal component (v_x) = u * cos(θ) = 30 * (1/2) = 15 m/s.
Correct Answer:
B
— 25 m/s
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Q. An object is projected at an angle of 60 degrees with an initial speed of 30 m/s. What is the vertical component of its velocity?
A.
15 m/s
B.
25 m/s
C.
30 m/s
D.
20 m/s
Show solution
Solution
Vertical component (v_y) = u * sin(θ) = 30 * (√3/2) = 25.98 m/s.
Correct Answer:
B
— 25 m/s
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Q. An object is projected at an angle of 60 degrees with an initial velocity of 30 m/s. What is the horizontal component of its velocity?
A.
15 m/s
B.
25 m/s
C.
30 m/s
D.
20 m/s
Show solution
Solution
Horizontal component Vx = u * cos(θ) = 30 * (1/2) = 15 m/s.
Correct Answer:
B
— 25 m/s
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Q. An object is projected at an angle of 60 degrees with an initial velocity of 30 m/s. What is the time of flight?
A.
3 s
B.
5 s
C.
6 s
D.
10 s
Show solution
Solution
Time of flight (T) = (2u * sin(θ)) / g = (2 * 30 * √3/2) / 9.8 = 5.18 s.
Correct Answer:
C
— 6 s
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Q. An object is thrown at an angle of 30 degrees with a speed of 40 m/s. What is the time of flight until it returns to the same vertical level?
A.
4 s
B.
5 s
C.
6 s
D.
8 s
Show solution
Solution
Time of flight (T) = (2 * u * sin(θ)) / g = (2 * 40 * 0.5) / 9.8 ≈ 4.08 s.
Correct Answer:
C
— 6 s
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Q. An object is thrown horizontally from the top of a 45 m high cliff. How far from the base of the cliff will it land if the initial speed is 10 m/s?
A.
10 m
B.
20 m
C.
30 m
D.
40 m
Show solution
Solution
Time of flight (t) = √(2h/g) = √(2*45/9.8) = 3.03 s. Horizontal distance = speed * time = 10 * 3.03 = 30.3 m.
Correct Answer:
B
— 20 m
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Q. An object is thrown horizontally from the top of a 45 m high cliff. How far from the base of the cliff will it land if it is thrown with a speed of 10 m/s?
A.
20 m
B.
30 m
C.
40 m
D.
50 m
Show solution
Solution
Time of flight (t) = √(2h/g) = √(2*45/9.8) = 3.03 s. Horizontal distance = speed * time = 10 * 3.03 = 30.3 m.
Correct Answer:
B
— 30 m
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Q. An object is thrown horizontally from the top of a cliff 80 m high. How long does it take to hit the ground?
A.
2 s
B.
4 s
C.
5 s
D.
8 s
Show solution
Solution
Using h = (1/2)gt^2, we have 80 = (1/2)(9.8)t^2, solving gives t ≈ 4 s.
Correct Answer:
B
— 4 s
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Q. An object is thrown horizontally from the top of a cliff of height 80 m. How long does it take to hit the ground?
A.
2 s
B.
4 s
C.
5 s
D.
8 s
Show solution
Solution
Using h = (1/2)gt^2, we have 80 = (1/2)(9.8)t^2, solving gives t ≈ 4 s.
Correct Answer:
B
— 4 s
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Q. An object is thrown horizontally from the top of a tower 80 m high. How long will it take to hit the ground?
A.
4 s
B.
5 s
C.
3 s
D.
2 s
Show solution
Solution
Using the equation h = (1/2)gt², where h = 80 m and g = 9.8 m/s², we have 80 = (1/2)*9.8*t². Solving gives t² = 16.33, so t ≈ 4 s.
Correct Answer:
B
— 5 s
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