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 lens has a focal length of 40 cm. If an object is placed 80 cm from the lens, what is the image distance?
A.
40 cm
B.
60 cm
C.
80 cm
D.
100 cm
Show solution
Solution
Using the lens formula, we find the image distance to be 40 cm.
Correct Answer:
A
— 40 cm
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Q. A lens has a focal length of 50 cm. If an object is placed at 100 cm, 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
Since the object distance is greater than the focal length, a real and inverted image is formed.
Correct Answer:
A
— Real and inverted
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Q. A lens has a power of +2 diopters. What is its focal length?
A.
0.5 m
B.
1 m
C.
2 m
D.
3 m
Show solution
Solution
Power (P) is given by P = 1/f (in meters). Thus, f = 1/P = 1/2 = 0.5 m.
Correct Answer:
B
— 1 m
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Q. A lens has a power of +2.0 D. What is its focal length?
A.
50 cm
B.
25 cm
C.
20 cm
D.
10 cm
Show solution
Solution
Power (P) = 1/f (in meters), so f = 1/2.0 = 0.5 m = 50 cm.
Correct Answer:
B
— 25 cm
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Q. A lens has a power of +2.5 D. What is its focal length?
A.
40 cm
B.
25 cm
C.
50 cm
D.
20 cm
Show solution
Solution
Power (P) = 1/f (in meters). Therefore, f = 1/2.5 = 0.4 m = 40 cm.
Correct Answer:
B
— 25 cm
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Q. A lens has a power of +2.5 D. What is the focal length of the lens in meters?
A.
0.4 m
B.
0.5 m
C.
0.6 m
D.
0.7 m
Show solution
Solution
Power (P) = 1/f, so f = 1/P = 1/2.5 = 0.4 m.
Correct Answer:
B
— 0.5 m
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Q. A lens has a power of +5 diopters. What is its focal length?
A.
20 cm
B.
25 cm
C.
30 cm
D.
15 cm
Show solution
Solution
Power (P) = 1/f (in meters). Therefore, f = 1/P = 1/5 = 0.2 m = 20 cm.
Correct Answer:
A
— 20 cm
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Q. A lens has a power of -4 D. What is the type of lens?
A.
Convex
B.
Concave
C.
Bifocal
D.
Plano-convex
Show solution
Solution
A negative power indicates a concave lens.
Correct Answer:
B
— Concave
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Q. A lens has a power of -4 D. What type of lens is it?
A.
Convex lens
B.
Concave lens
C.
Bifocal lens
D.
Plano-convex lens
Show solution
Solution
A negative power indicates that the lens is a concave lens.
Correct Answer:
B
— Concave lens
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Q. A lens produces a magnification of 3. If the object height is 2 cm, what is the image height?
A.
4 cm
B.
6 cm
C.
3 cm
D.
2 cm
Show solution
Solution
Magnification (m) = h'/h, thus h' = m * h = 3 * 2 = 6 cm.
Correct Answer:
A
— 4 cm
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Q. A lens produces a virtual image at a distance of 12 cm when the object is placed at 8 cm. What type of lens is it?
A.
Convex
B.
Concave
C.
Biconvex
D.
Biconcave
Show solution
Solution
A virtual image is formed by a concave lens when the object is placed in front of it.
Correct Answer:
B
— Concave
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Q. A light bulb uses 60 W of power. How much energy does it consume in 1 hour?
A.
3600 J
B.
216000 J
C.
60000 J
D.
180000 J
Show solution
Solution
Energy consumed can be calculated using E = P * t. Here, P = 60 W and t = 1 hour = 3600 seconds. Thus, E = 60 W * 3600 s = 216000 J.
Correct Answer:
B
— 216000 J
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Q. A light bulb uses 60 W of power. How much energy does it consume in 2 hours?
A.
120 J
B.
7200 J
C.
432000 J
D.
360 J
Show solution
Solution
Energy consumed can be calculated using E = P * t. Here, P = 60 W and t = 2 hours = 7200 seconds. Thus, E = 60 W * 7200 s = 432000 J.
Correct Answer:
C
— 432000 J
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Q. A light bulb uses 60 W of power. How much energy does it consume in 5 hours?
A.
18000 J
B.
108000 J
C.
300000 J
D.
360000 J
Show solution
Solution
Energy consumed can be calculated using E = P * t. Here, E = 60 W * (5 * 3600 s) = 108000 J.
Correct Answer:
B
— 108000 J
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Q. A light bulb uses 60 Watts of power. How much energy does it consume in 2 hours?
A.
120 Joules
B.
7200 Joules
C.
432000 Joules
D.
360000 Joules
Show solution
Solution
Energy consumed = Power × Time = 60 W × (2 × 3600 s) = 432000 J.
Correct Answer:
C
— 432000 Joules
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Q. A light ray enters a glass prism with a refractive index of 1.5. If the angle of incidence is 30 degrees, what is the angle of refraction?
A.
15 degrees
B.
20 degrees
C.
25 degrees
D.
30 degrees
Show solution
Solution
Using Snell's law, n1 * sin(i) = n2 * sin(r). Here, n1 = 1 (air), n2 = 1.5, i = 30 degrees. Solving gives r = 19.2 degrees.
Correct Answer:
B
— 20 degrees
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Q. A light ray in glass (n=1.5) strikes the boundary with air at an angle of 30°. Will it undergo total internal reflection?
A.
Yes
B.
No
C.
Depends on the angle
D.
Not enough information
Show solution
Solution
Since the angle of incidence is less than the critical angle (θc = sin⁻¹(1/1.5) ≈ 41.8°), it will not undergo total internal reflection.
Correct Answer:
B
— No
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Q. A light ray passes from air into glass with a refractive index of 1.5. If the angle of incidence is 30 degrees, what is the angle of refraction?
A.
18.4 degrees
B.
20 degrees
C.
22 degrees
D.
25 degrees
Show solution
Solution
Using Snell's law, n1 * sin(theta1) = n2 * sin(theta2). Here, n1 = 1 (air), n2 = 1.5 (glass), and theta1 = 30 degrees. Thus, sin(theta2) = (1 * sin(30))/1.5 = 0.333, giving theta2 ≈ 19.1 degrees.
Correct Answer:
A
— 18.4 degrees
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Q. A light ray passes from diamond (n=2.42) to air. What is the critical angle?
A.
24.4°
B.
30.0°
C.
36.9°
D.
42.0°
Show solution
Solution
Critical angle θc = sin⁻¹(1.00/2.42) ≈ 24.4°.
Correct Answer:
A
— 24.4°
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Q. A light ray passes through the center of curvature of a concave mirror. What will be the angle of reflection?
A.
0 degrees
B.
30 degrees
C.
45 degrees
D.
90 degrees
Show solution
Solution
When a light ray passes through the center of curvature, it strikes the mirror perpendicularly, resulting in an angle of reflection of 0 degrees.
Correct Answer:
A
— 0 degrees
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Q. A light ray strikes a glass slab at an angle of incidence of 45 degrees. If the refractive index of glass is 1.5, what is the angle of refraction?
A.
30 degrees
B.
45 degrees
C.
60 degrees
D.
90 degrees
Show solution
Solution
Using Snell's law, n1 * sin(i) = n2 * sin(r). Here, sin(r) = (1 * sin(45)) / 1.5 = 0.471, giving r ≈ 28 degrees.
Correct Answer:
A
— 30 degrees
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Q. A light ray strikes a glass surface at an angle of 60 degrees. If the refractive index of glass is 1.5, what is the angle of refraction?
A.
30 degrees
B.
40 degrees
C.
60 degrees
D.
90 degrees
Show solution
Solution
Using Snell's law, n1 * sin(i) = n2 * sin(r). Here, n1 = 1 (air), n2 = 1.5 (glass), i = 60 degrees. Solving gives r = 40 degrees.
Correct Answer:
B
— 40 degrees
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Q. A light ray strikes a glass surface at an angle of incidence of 30 degrees. If the refractive index of glass is 1.5, what is the angle of refraction?
A.
20 degrees
B.
30 degrees
C.
40 degrees
D.
50 degrees
Show solution
Solution
Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n1 = 1 (air), θ1 = 30 degrees, n2 = 1.5. Thus, sin(θ2) = (1 * sin(30))/1.5 = 1/3, giving θ2 ≈ 20 degrees.
Correct Answer:
A
— 20 degrees
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Q. A light ray strikes a plane mirror at an angle of 45 degrees. What is the angle of reflection?
A.
0 degrees
B.
45 degrees
C.
90 degrees
D.
30 degrees
Show solution
Solution
According to the law of reflection, the angle of reflection equals the angle of incidence, so it is 45 degrees.
Correct Answer:
B
— 45 degrees
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Q. A light ray traveling in a medium with a refractive index of 1.6 strikes a boundary with air at an angle of 50°. What will be the outcome?
A.
Total internal reflection occurs.
B.
Light is refracted into the air.
C.
Light is absorbed.
D.
Light is scattered.
Show solution
Solution
The critical angle for this scenario is approximately 38.7°. Since 50° is greater than the critical angle, total internal reflection occurs.
Correct Answer:
A
— Total internal reflection occurs.
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Q. A light ray traveling in a medium with n=1.5 strikes the boundary with n=1.0 at 50°. What occurs?
A.
Total internal reflection
B.
Partial reflection and refraction
C.
Complete refraction
D.
None of the above
Show solution
Solution
Since 50° is greater than the critical angle (θc ≈ 41.8°), total internal reflection occurs.
Correct Answer:
A
— Total internal reflection
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Q. A light ray traveling in a medium with n=1.6 strikes the boundary with air at an angle of 50°. What will happen?
A.
Total internal reflection
B.
Partial reflection and refraction
C.
Complete absorption
D.
No reflection
Show solution
Solution
Calculate critical angle: θc = sin^(-1)(1/n) = sin^(-1)(1/1.6) ≈ 38.7°. Since 50° > 38.7°, total internal reflection occurs.
Correct Answer:
A
— Total internal reflection
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Q. A light ray traveling in a medium with n=2.0 strikes a boundary with air at an angle of incidence of 45°. What will be the angle of refraction in air?
A.
22.5°
B.
45°
C.
60°
D.
90°
Show solution
Solution
Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n1 = 2.0, θ1 = 45°, and n2 = 1.0 (air). Thus, 2.0 * sin(45°) = 1.0 * sin(θ2) leads to sin(θ2) = √2, which gives θ2 = 90°.
Correct Answer:
D
— 90°
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Q. A light ray traveling in diamond (n=2.42) strikes the diamond-air interface. What is the critical angle?
A.
24.4°
B.
36.9°
C.
42.5°
D.
49.5°
Show solution
Solution
Using sin(θc) = n2/n1, where n1 = 2.42 (diamond) and n2 = 1.0 (air), we find sin(θc) = 1.0/2.42, leading to θc ≈ 24.4°.
Correct Answer:
A
— 24.4°
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Q. A light ray traveling in diamond (n=2.42) strikes the diamond-air interface. What is the critical angle for total internal reflection?
A.
24.4°
B.
30.0°
C.
36.0°
D.
42.0°
Show solution
Solution
Using Snell's law, sin(θc) = n2/n1 = 1.00/2.42. Thus, θc ≈ 24.4°.
Correct Answer:
A
— 24.4°
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