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 sound wave travels in air at a speed of 340 m/s. If the frequency of the sound is 1700 Hz, what is the wavelength?
A.
0.2 m
B.
0.5 m
C.
2 m
D.
1 m
Show solution
Solution
Wavelength λ = v/f = 340 m/s / 1700 Hz = 0.2 m.
Correct Answer:
A
— 0.2 m
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Q. A sound wave travels in air at a speed of 340 m/s. If the frequency of the sound is 170 Hz, what is the wavelength?
A.
2 m
B.
1 m
C.
0.5 m
D.
0.2 m
Show solution
Solution
Wavelength λ = v/f = 340 m/s / 170 Hz = 2 m.
Correct Answer:
A
— 2 m
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Q. A sound wave travels through a medium with a speed of 340 m/s and has a frequency of 1700 Hz. What is its wavelength?
A.
0.2 m
B.
0.5 m
C.
2 m
D.
1 m
Show solution
Solution
Wavelength = Speed / Frequency = 340 m/s / 1700 Hz = 0.2 m.
Correct Answer:
B
— 0.5 m
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Q. A sound wave travels through air at a speed of 340 m/s. If the frequency of the sound is 1700 Hz, what is the wavelength?
A.
0.2 m
B.
0.5 m
C.
1 m
D.
2 m
Show solution
Solution
Using the formula v = fλ, we can rearrange to find λ = v/f. Thus, λ = 340 m/s / 1700 Hz = 0.2 m.
Correct Answer:
A
— 0.2 m
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Q. A sound wave travels through water at a speed of 1500 m/s. If the frequency of the sound wave is 300 Hz, what is the wavelength?
A.
2 m
B.
3 m
C.
4 m
D.
5 m
Show solution
Solution
Wavelength λ = v/f = 1500 m/s / 300 Hz = 5 m.
Correct Answer:
A
— 2 m
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Q. A speed is measured as 20 m/s with an uncertainty of ±0.5 m/s. If this speed is used to calculate kinetic energy, what is the percentage error in kinetic energy?
A.
5%
B.
2.5%
C.
1%
D.
10%
Show solution
Solution
K.E. = 0.5 * m * v², percentage error = 2 * (Δv/v) = 2 * (0.5/20) = 0.05 or 5%.
Correct Answer:
B
— 2.5%
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Q. A speed is measured as 20 m/s with an uncertainty of ±0.5 m/s. What is the absolute error in the speed measurement?
A.
0.5 m/s
B.
0.25 m/s
C.
1 m/s
D.
0.1 m/s
Show solution
Solution
The absolute error is given directly as ±0.5 m/s.
Correct Answer:
A
— 0.5 m/s
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Q. A speed is recorded as 60 km/h with an error of 2 km/h. What is the percentage error?
A.
3.33%
B.
2.5%
C.
1.67%
D.
4%
Show solution
Solution
Percentage error = (Absolute error / Measured value) * 100 = (2 / 60) * 100 = 3.33%
Correct Answer:
A
— 3.33%
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Q. A speed of 30 m/s is measured with an uncertainty of ±0.5 m/s. What is the total uncertainty if this speed is used to calculate kinetic energy?
A.
0.25 J
B.
0.5 J
C.
1 J
D.
2 J
Show solution
Solution
Kinetic energy = 0.5 * m * v²; uncertainty in KE = 2 * v * uncertainty in v = 2 * 30 * 0.5 = 30 J.
Correct Answer:
C
— 1 J
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Q. A sphere rolls down a ramp of height h. What is the total mechanical energy at the top?
A.
mgh
B.
1/2 mv^2
C.
mgh + 1/2 mv^2
D.
0
Show solution
Solution
The total mechanical energy at the top is purely potential energy, which is mgh.
Correct Answer:
A
— mgh
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Q. A sphere rolls down a ramp. If the height of the ramp is h, what is the speed of the sphere at the bottom assuming no energy loss?
A.
√(2gh)
B.
√(3gh)
C.
√(4gh)
D.
√(gh)
Show solution
Solution
Using conservation of energy, the potential energy at height h converts to kinetic energy at the bottom, giving speed √(2gh).
Correct Answer:
A
— √(2gh)
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Q. A sphere rolls on a flat surface with a speed v. What is the kinetic energy of the sphere?
A.
(1/2)mv^2
B.
(1/2)mv^2 + (1/5)mv^2
C.
(1/2)mv^2 + (2/5)mv^2
D.
(1/2)mv^2 + (3/5)mv^2
Show solution
Solution
The total kinetic energy of a rolling sphere is the sum of translational and rotational kinetic energy, which is (1/2)mv^2 + (2/5)mv^2.
Correct Answer:
C
— (1/2)mv^2 + (2/5)mv^2
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Q. A sphere rolls without slipping on a flat surface. If it has a radius R and rolls with a speed v, what is its angular speed?
A.
v/R
B.
2v/R
C.
v/2R
D.
v²/R
Show solution
Solution
The angular speed ω of a rolling sphere is given by ω = v/R.
Correct Answer:
A
— v/R
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Q. A spherical conductor has a charge Q. What is the electric potential inside the conductor?
A.
0
B.
Q/(4πε₀r)
C.
Q/(4πε₀R)
D.
Constant throughout
Show solution
Solution
The electric potential inside a charged spherical conductor is constant and equal to the potential on its surface, which is Q/(4πε₀R).
Correct Answer:
D
— Constant throughout
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Q. A spherical conductor has a radius R and carries a charge Q. What is the electric potential on its surface?
A.
kQ/R
B.
kQ/2R
C.
0
D.
kQ/R²
Show solution
Solution
The electric potential V on the surface of a charged spherical conductor is given by V = kQ/R.
Correct Answer:
A
— kQ/R
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Q. A spherical Gaussian surface of radius R encloses a charge Q. What is the electric field at a distance 2R from the center?
A.
Q/4πε₀R²
B.
Q/4πε₀(2R)²
C.
0
D.
Q/ε₀(2R)²
Show solution
Solution
The electric field outside a spherical charge distribution is given by E = Q/4πε₀r². At 2R, it becomes Q/4πε₀(2R)².
Correct Answer:
B
— Q/4πε₀(2R)²
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Q. A spherical shell of radius R carries a total charge Q. What is the electric field at a point outside the shell?
A.
0
B.
Q/(4πε₀R²)
C.
Q/(4πε₀R)
D.
Q/(4πε₀R³)
Show solution
Solution
For a spherical shell, the electric field outside the shell behaves as if all the charge were concentrated at the center, so E = Q/(4πε₀R²).
Correct Answer:
B
— Q/(4πε₀R²)
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Q. A spherical shell of radius R carries a uniform charge Q. What is the electric field inside the shell?
A.
Q/(4πε₀R²)
B.
0
C.
Q/(4πε₀R)
D.
Q/(4πε₀)
Show solution
Solution
By Gauss's law, the electric field inside a uniformly charged spherical shell is zero.
Correct Answer:
B
— 0
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Q. A spherical shell of radius R carries a uniform surface charge density σ. What is the electric field inside the shell?
A.
0
B.
σ/ε₀
C.
σ/2ε₀
D.
σ/4ε₀
Show solution
Solution
By Gauss's law, the electric field inside a uniformly charged spherical shell is zero.
Correct Answer:
A
— 0
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Q. A spring obeys Hooke's law. If the spring constant is doubled, what happens to the elongation for the same applied force?
A.
Elongation doubles
B.
Elongation halves
C.
Elongation remains the same
D.
Elongation quadruples
Show solution
Solution
According to Hooke's law, elongation is inversely proportional to the spring constant; thus, if the spring constant is doubled, the elongation halves.
Correct Answer:
B
— Elongation halves
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Q. A spring stretches 5 cm when a load of 10 N is applied. What is the spring constant?
A.
200 N/m
B.
100 N/m
C.
50 N/m
D.
25 N/m
Show solution
Solution
Using Hooke's law, k = F/x = 10 N / 0.05 m = 200 N/m.
Correct Answer:
B
— 100 N/m
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Q. A spring with a spring constant of 200 N/m is compressed by 0.1 m. What is the potential energy stored in the spring?
A.
1 J
B.
2 J
C.
3 J
D.
4 J
Show solution
Solution
Potential Energy = 0.5 × k × x² = 0.5 × 200 N/m × (0.1 m)² = 1 J.
Correct Answer:
A
— 1 J
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Q. A spring with a spring constant of 200 N/m is compressed by 0.1 m. What is the work done in compressing the spring?
A.
1 J
B.
2 J
C.
3 J
D.
4 J
Show solution
Solution
Work done = 0.5 × k × x^2 = 0.5 × 200 N/m × (0.1 m)^2 = 1 J.
Correct Answer:
A
— 1 J
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Q. A spring with a spring constant of 200 N/m is compressed by 0.5 m. What is the potential energy stored in the spring?
A.
25 J
B.
50 J
C.
100 J
D.
200 J
Show solution
Solution
Potential Energy in spring = 0.5 × k × x² = 0.5 × 200 N/m × (0.5 m)² = 25 J.
Correct Answer:
B
— 50 J
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Q. A spring with a spring constant of 200 N/m is compressed by 0.5 m. What is the work done in compressing the spring?
A.
25 J
B.
50 J
C.
100 J
D.
200 J
Show solution
Solution
Work done = 0.5 × k × x² = 0.5 × 200 N/m × (0.5 m)² = 25 J.
Correct Answer:
B
— 50 J
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Q. A stone is dropped from a height of 20 m. How long does it take to reach the ground?
A.
2 s
B.
1 s
C.
3 s
D.
4 s
Show solution
Solution
Using h = (1/2)gt², where h = 20 m and g = 9.8 m/s², we have 20 = (1/2)*9.8*t². Solving gives t² = 4.08, so t ≈ 2 s.
Correct Answer:
A
— 2 s
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Q. A stone is dropped from a height of 20 m. How long will it take to reach the ground?
A.
2 s
B.
1 s
C.
3 s
D.
4 s
Show solution
Solution
Using h = (1/2)gt², where h = 20 m and g = 9.8 m/s², we have 20 = (1/2)*9.8*t². Solving gives t² = 4.08, so t ≈ 2 s.
Correct Answer:
A
— 2 s
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Q. A stone is dropped from a height of 45 m. How far will it have fallen after 2 seconds?
A.
10 m
B.
20 m
C.
30 m
D.
40 m
Show solution
Solution
Distance fallen (s) = (1/2)gt² = (1/2)(9.8)(2²) = 19.6 m.
Correct Answer:
C
— 30 m
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Q. A stone is dropped from a height of 45 m. How far will it have fallen after 3 seconds?
A.
10.5 m
B.
20 m
C.
30 m
D.
40.5 m
Show solution
Solution
Distance fallen (s) = (1/2)gt² = (1/2)(9.8)(3²) = 44.1 m.
Correct Answer:
D
— 40.5 m
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Q. A stone is dropped from a height of 45 m. How far will it travel horizontally if it is thrown horizontally with a speed of 10 m/s?
A.
20 m
B.
30 m
C.
40 m
D.
50 m
Show solution
Solution
Time to fall (t) = √(2h/g) = √(2*45/9.8) ≈ 3 s; horizontal distance = v*t = 10*3 = 30 m.
Correct Answer:
B
— 30 m
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