Engineering & Architecture Admissions - Exam Prep Guide

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Engineering & Architecture Admissions

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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.

JEE Main

Q. If the resistances in a Wheatstone bridge are R1 = 10Ω, R2 = 20Ω, R3 = 15Ω, what should R4 be for the bridge to be balanced?

A. 30Ω
B. 15Ω
C. 20Ω
D. 10Ω

Solution

Using the balance condition R1/R2 = R3/R4, we find R4 = (R2 * R3) / R1 = (20 * 15) / 10 = 30Ω.

Correct Answer: B — 15Ω

Q. If the resistances in a Wheatstone bridge are R1 = 20Ω, R2 = 30Ω, and R3 = 10Ω, what is the value of R4 for the bridge to be balanced?

A. 15Ω
B. 20Ω
C. 25Ω
D. 30Ω

Solution

Using the balance condition R1/R2 = R3/R4, we have 20/30 = 10/x, solving gives x = 20Ω.

Correct Answer: B — 20Ω

Q. If the resistances in a Wheatstone bridge are R1, R2, R3, and R4, what is the condition for balance?

A. R1/R2 = R3/R4
B. R1 + R2 = R3 + R4
C. R1 * R4 = R2 * R3
D. R1 - R2 = R3 - R4

Solution

The condition for balance in a Wheatstone bridge is R1/R2 = R3/R4.

Correct Answer: A — R1/R2 = R3/R4

Q. If the resistivity of a material is 1.5 x 10^-8 Ω·m, what is the resistance of a 3 m long wire with a cross-sectional area of 0.5 mm²?

A. 0.09 Ω
B. 0.18 Ω
C. 0.27 Ω
D. 0.36 Ω

Solution

Resistance R = ρ(L/A) = (1.5 x 10^-8)(3)/(0.5 x 10^-6) = 0.09 Ω.

Correct Answer: B — 0.18 Ω

Q. If the resistivity of a material is 2 x 10^-8 Ω·m and the wire has a length of 3 m and a cross-sectional area of 0.5 mm², what is the resistance?

A. 0.12 Ω
B. 0.15 Ω
C. 0.18 Ω
D. 0.20 Ω

Solution

Resistance R = ρ * (L / A) = 2 x 10^-8 * (3 / 0.5 x 10^-6) = 0.12 Ω.

Correct Answer: A — 0.12 Ω

Q. If the resistivity of a material is doubled, what happens to the resistance of a wire of constant length and cross-sectional area?

A. It doubles
B. It halves
C. It remains the same
D. It quadruples

Solution

Resistance R is directly proportional to resistivity ρ, so if ρ is doubled, R also doubles.

Correct Answer: A — It doubles

Q. If the resistivity of a material is doubled, what happens to the resistance of a wire of fixed length and cross-sectional area?

A. It doubles
B. It halves
C. It remains the same
D. It quadruples

Solution

Resistance R is directly proportional to resistivity; if resistivity doubles, resistance also doubles.

Correct Answer: A — It doubles

Q. If the resistivity of a material is halved, what happens to the resistance of a wire of fixed length and cross-sectional area?

A. Halved
B. Doubled
C. Remains the same
D. Quadrupled

Solution

Resistance is directly proportional to resistivity; halving resistivity halves the resistance.

Correct Answer: A — Halved

Q. If the resistivity of a material is halved, what happens to the resistance of a uniform wire of that material?

A. Halved
B. Doubled
C. Remains the same
D. Quadrupled

Solution

Resistance is directly proportional to resistivity; halving resistivity halves the resistance.

Correct Answer: A — Halved

Q. If the resistivity of a material is halved, what will happen to the resistance of a wire of fixed length and cross-sectional area?

A. Halved
B. Doubled
C. Remains the same
D. Quadrupled

Solution

Resistance is directly proportional to resistivity; halving resistivity halves the resistance.

Correct Answer: A — Halved

Q. If the resistivity of a superconductor is zero, what can be said about its resistance?

A. Infinite
B. Zero
C. Depends on temperature
D. Undefined

Solution

A superconductor has zero resistivity, which means it has zero resistance.

Correct Answer: B — Zero

Q. If the resistivity of copper is 1.68 x 10^-8 Ω·m, what is the resistance of a copper wire of length 100 m and diameter 1 mm?

A. 0.168 Ω
B. 0.168 kΩ
C. 1.68 Ω
D. 1.68 kΩ

Solution

Resistance R = ρ * (L / A) = 1.68 x 10^-8 * (100 / (π * (0.5 x 10^-3)²)) = 0.168 Ω.

Correct Answer: A — 0.168 Ω

Q. If the RMS speed of a gas is 250 m/s, what is the temperature if the molar mass is 0.028 kg/mol?

A. 100 K
B. 200 K
C. 300 K
D. 400 K

Solution

Using v_rms = sqrt(3RT/M), we can rearrange to find T = (v_rms^2 * M) / (3R) = 300 K.

Correct Answer: C — 300 K

Q. If the RMS speed of a gas is 300 m/s and its molar mass is 28 g/mol, what is the temperature of the gas?

A. 300 K
B. 600 K
C. 900 K
D. 1200 K

Solution

Using the formula v_rms = sqrt((3RT)/M), we can rearrange to find T = (v_rms^2 * M)/(3R). Plugging in the values gives T = 600 K.

Correct Answer: B — 600 K

Q. If the RMS speed of a gas is 300 m/s at 300 K, what will be its RMS speed at 600 K?

A. 300 m/s
B. 600 m/s
C. 300√2 m/s
D. 600√2 m/s

Solution

The RMS speed is proportional to the square root of the temperature. Therefore, at 600 K, the RMS speed will be 300 * sqrt(2) m/s.

Correct Answer: C — 300√2 m/s

Q. If the RMS speed of a gas is 300 m/s at 400 K, what will be the RMS speed at 200 K?

A. 150 m/s
B. 300 m/s
C. 600 m/s
D. 100 m/s

Solution

The RMS speed is proportional to the square root of the temperature. Therefore, at 200 K, the RMS speed will be 300 * sqrt(200/400) = 150 m/s.

Correct Answer: A — 150 m/s

Q. If the RMS speed of a gas is 300 m/s at 400 K, what will be the RMS speed at 800 K?

A. 300 m/s
B. 600 m/s
C. 424 m/s
D. 848 m/s

Solution

RMS speed is proportional to the square root of temperature. v_rms at 800 K = 300 * sqrt(800/400) = 300 * sqrt(2) = 600 m/s.

Correct Answer: B — 600 m/s

Q. If the RMS speed of a gas is 300 m/s, what is the kinetic energy per molecule?

A. 0.5 * m * (300)^2
B. 0.5 * m * (150)^2
C. 0.5 * m * (600)^2
D. 0.5 * m * (100)^2

Solution

The kinetic energy per molecule is given by KE = 0.5 * m * v^2. Substituting v = 300 m/s gives the correct expression.

Correct Answer: A — 0.5 * m * (300)^2

Q. If the RMS speed of a gas is 300 m/s, what is the RMS speed of the same gas at double the temperature?

A. 300 m/s
B. 600 m/s
C. 300√2 m/s
D. 600√2 m/s

Solution

The RMS speed is proportional to the square root of the temperature. If the temperature is doubled, the RMS speed increases by a factor of sqrt(2). Therefore, the new RMS speed will be 300 * sqrt(2), which is approximately 600 m/s.

Correct Answer: B — 600 m/s

Q. If the RMS speed of a gas is 400 m/s and its molar mass is 16 g/mol, what is the temperature of the gas?

A. 200 K
B. 400 K
C. 800 K
D. 1600 K

Solution

Using the formula v_rms = sqrt((3RT)/M), we can rearrange to find T: T = (M * v_rms^2) / (3R). Substituting M = 0.016 kg/mol and v_rms = 400 m/s gives T = 400 K.

Correct Answer: B — 400 K

Q. If the RMS speed of a gas is 400 m/s at 300 K, what will be the RMS speed at 600 K?

A. 400 m/s
B. 800 m/s
C. 400√2 m/s
D. 800√2 m/s

Solution

The RMS speed increases with the square root of the temperature. Therefore, at 600 K, the RMS speed will be 400 * sqrt(2), which is approximately 800 m/s.

Correct Answer: B — 800 m/s

Q. If the RMS speed of a gas is 400 m/s, what is the kinetic energy per molecule at 300 K?

A. 0.5 mJ
B. 0.4 mJ
C. 0.2 mJ
D. 0.1 mJ

Solution

Kinetic energy per molecule = (1/2)mv^2. Using v = 400 m/s and m = M/N_A, we find KE = 0.5 mJ.

Correct Answer: A — 0.5 mJ

Q. If the RMS speed of a gas is 400 m/s, what is the speed of the gas molecules in terms of average speed?

A. 400 m/s
B. 300 m/s
C. 500 m/s
D. 600 m/s

Solution

The average speed is related to RMS speed by the relation v_avg = (2/3) * v_rms. Thus, v_avg = (2/3) * 400 = 267 m/s.

Correct Answer: B — 300 m/s

Q. If the RMS speed of a gas is 400 m/s, what is the speed of the molecules in terms of average speed?

A. 400 m/s
B. 300 m/s
C. 500 m/s
D. 600 m/s

Solution

The average speed of gas molecules is approximately 0.8 times the RMS speed, so average speed = 0.8 * 400 m/s = 320 m/s.

Correct Answer: B — 300 m/s

Q. If the RMS speed of a gas is 500 m/s, what is the speed of the gas molecules at 1/2 of the RMS speed?

A. 250 m/s
B. 500 m/s
C. 1000 m/s
D. 125 m/s

Solution

The speed at 1/2 of the RMS speed is simply 500 m/s / 2 = 250 m/s.

Correct Answer: A — 250 m/s

Q. If the RMS speed of a gas is 500 m/s, what is the speed of the gas molecules in terms of average speed?

A. 500 m/s
B. 250 m/s
C. 400 m/s
D. 600 m/s

Solution

The average speed of gas molecules is related to the RMS speed by the relation v_avg = (v_rms * sqrt(8/3)). Therefore, the average speed is approximately 400 m/s.

Correct Answer: C — 400 m/s

Q. If the RMS speed of a gas is 500 m/s, what is the speed of the molecules in the gas?

A. 500 m/s
B. 250 m/s
C. 1000 m/s
D. It varies

Solution

The RMS speed is an average measure; individual molecular speeds will vary around this value.

Correct Answer: D — It varies

Q. If the RMS speed of a gas is 500 m/s, what is the temperature if the molar mass is 0.028 kg/mol?

A. 200 K
B. 300 K
C. 400 K
D. 500 K

Solution

Using v_rms = sqrt(3RT/M), we rearrange to find T = (v_rms^2 * M) / (3R). Plugging in values gives T approximately 300 K.

Correct Answer: B — 300 K

Q. If the roots of the equation ax^2 + bx + c = 0 are 3 and -2, what is the value of a if b = 5 and c = -6?

A. 1
B. 2
C. 3
D. 4

Solution

Using Vieta's formulas, a = 1 since the product of the roots (3 * -2) = -6 and sum (3 + -2) = 1.

Correct Answer: A — 1

Q. If the roots of the equation ax^2 + bx + c = 0 are 3 and -2, what is the value of b if a = 1 and c = -6?

A. -1
B. 1
C. 5
D. -5

Solution

Using the sum of roots (-b/a = 3 + (-2) = 1), we find b = -1.

Correct Answer: A — -1

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