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. A conical pendulum swings in a horizontal circle. If the angle of the string with the vertical is θ, what is the relationship between the tension in the string and the gravitational force acting on the pendulum bob?

A. T = mg
B. T = mg cos(θ)
C. T = mg sin(θ)
D. T = mg tan(θ)

Solution

Tension provides the vertical component to balance the weight: T cos(θ) = mg.

Correct Answer: B — T = mg cos(θ)

Q. A conical pendulum swings in a horizontal circle. If the angle of the string with the vertical is θ, what is the expression for the tension in the string?

A. T = mg
B. T = mg/cos(θ)
C. T = mg/sin(θ)
D. T = mg tan(θ)

Solution

The vertical component of tension balances the weight: T cos(θ) = mg, thus T = mg/cos(θ).

Correct Answer: B — T = mg/cos(θ)

Q. A conical pendulum swings with a constant speed. If the angle of the string with the vertical is θ, what is the expression for the tension in the string?

A. mg/cos(θ)
B. mg/sin(θ)
C. mg/tan(θ)
D. mg

Solution

Tension T = mg/cos(θ) to balance the vertical component of weight.

Correct Answer: A — mg/cos(θ)

Q. A convex lens has a focal length of 15 cm. What is the power of the lens?

A. +6.67 D
B. +7.5 D
C. +10 D
D. +15 D

Solution

Power (P) = 1/f (in meters). f = 0.15 m, so P = 1/0.15 = +6.67 D.

Correct Answer: B — +7.5 D

Q. A convex lens has a focal length of 20 cm. If an object is placed 60 cm from the lens, what is the distance of the image from the lens?

A. 15 cm
B. 30 cm
C. 45 cm
D. 60 cm

Solution

Using the lens formula 1/f = 1/v - 1/u, we find v = 30 cm.

Correct Answer: C — 45 cm

Q. A convex lens has a focal length of 20 cm. If an object is placed at a distance of 60 cm from the lens, what is the distance of the image from the lens?

A. 15 cm
B. 30 cm
C. 45 cm
D. 60 cm

Solution

Using the lens formula 1/f = 1/v - 1/u, we find v = 30 cm.

Correct Answer: C — 45 cm

Q. A convex lens has a focal length of 20 cm. If an object is placed at a distance of 40 cm from the lens, what is the distance of the image from the lens?

A. 20 cm
B. 40 cm
C. 60 cm
D. 80 cm

Solution

Using the lens formula 1/f = 1/v - 1/u, where f = 20 cm and u = -40 cm, we find v = 20 cm. The image is formed at 20 cm on the opposite side.

Correct Answer: A — 20 cm

Q. A convex lens has a focal length of 20 cm. If an object is placed at a distance of 30 cm from the lens, what is the distance of the image from the lens?

A. 60 cm
B. 15 cm
C. 30 cm
D. 10 cm

Solution

Using the lens formula 1/f = 1/v - 1/u, we find v = 60 cm.

Correct Answer: A — 60 cm

Q. A convex lens has a focal length of 20 cm. What is the power of the lens?

A. -5 D
B. 5 D
C. 20 D
D. 0.05 D

Solution

Power (P) = 1/f (in meters). f = 20 cm = 0.2 m, so P = 1/0.2 = 5 D.

Correct Answer: B — 5 D

Q. A copper wire has a resistivity of 1.68 x 10^-8 Ω·m. What is the resistance of a 100 m long wire with a cross-sectional area of 1 mm²?

A. 1.68 Ω
B. 0.168 Ω
C. 0.0168 Ω
D. 16.8 Ω

Solution

Resistance (R) = ρ * (L / A) = 1.68 x 10^-8 * (100 / 1 x 10^-6) = 1.68 Ω.

Correct Answer: A — 1.68 Ω

Q. A cube encloses a charge Q at its center. What is the electric flux through one face of the cube?

A. Q/ε₀
B. Q/6ε₀
C. Q/3ε₀
D. Zero

Solution

The total flux through the cube is Q/ε₀, and since there are 6 faces, the flux through one face is Q/6ε₀.

Correct Answer: B — Q/6ε₀

Q. A cube of side length a has a charge Q at one of its corners. What is the electric flux through one face of the cube?

A. Q/(6ε₀)
B. Q/(12ε₀)
C. Q/(8ε₀)
D. Q/(4ε₀)

Solution

The total flux through the cube is Q/ε₀, and since the charge is at one corner, the flux through one face is Q/(12ε₀).

Correct Answer: B — Q/(12ε₀)

Q. A cube of side length a has a charge Q at one of its corners. What is the total electric flux through the cube?

A. Q/ε₀
B. Q/(6ε₀)
C. Q/(12ε₀)
D. 0

Solution

The charge at one corner contributes 1/8 of its flux to the cube. Thus, total flux = (1/8)Q/ε₀ = Q/(12ε₀).

Correct Answer: C — Q/(12ε₀)

Q. A current-carrying conductor experiences a force in a magnetic field. This phenomenon is known as?

A. Electromagnetic induction
B. Lorentz force
C. Faraday's law
D. Ampere's law

Solution

The force experienced by a current-carrying conductor in a magnetic field is known as the Lorentz force.

Correct Answer: B — Lorentz force

Q. A current-carrying conductor experiences a force in a magnetic field. What is the direction of this force given by?

A. Right-hand rule
B. Left-hand rule
C. Ampere's law
D. Faraday's law

Solution

The direction of the force on a current-carrying conductor in a magnetic field is given by the right-hand rule.

Correct Answer: A — Right-hand rule

Q. A current-carrying wire is placed in a magnetic field. What is the direction of the force acting on the wire?

A. Parallel to the wire
B. Perpendicular to the wire and field
C. Opposite to the field
D. In the direction of the field

Solution

The force on a current-carrying wire in a magnetic field is given by the right-hand rule and is perpendicular to both the wire and the magnetic field.

Correct Answer: B — Perpendicular to the wire and field

Q. A cyclist accelerates from rest at a rate of 1 m/s². How far will he travel in 10 seconds?

A. 50 m
B. 100 m
C. 150 m
D. 200 m

Solution

Using the formula: distance = initial velocity * time + 0.5 * acceleration * time². Distance = 0 + 0.5 * 1 * 10² = 50 m.

Correct Answer: B — 100 m

Q. A cyclist accelerates from rest at a rate of 1 m/s². How far will he travel in 8 seconds?

A. 32 m
B. 40 m
C. 48 m
D. 64 m

Solution

Using the formula: distance = initial velocity * time + 0.5 * acceleration * time². Distance = 0 + 0.5 * 1 * 8² = 32 m.

Correct Answer: C — 48 m

Q. A cyclist accelerates from rest at a rate of 2 m/s². How far will he travel in 10 seconds?

A. 100 m
B. 50 m
C. 200 m
D. 150 m

Solution

Using the formula: distance = ut + 0.5at². Here, u = 0, a = 2 m/s², t = 10 s. Distance = 0 + 0.5 * 2 * 10² = 100 m.

Correct Answer: A — 100 m

Q. A cyclist accelerates from rest to a speed of 10 m/s in 5 seconds. What is the average power output if the cyclist has a mass of 70 kg?

A. 140 W
B. 280 W
C. 560 W
D. 700 W

Solution

Kinetic Energy = 0.5 × m × v² = 0.5 × 70 kg × (10 m/s)² = 3500 J. Power = Work / Time = 3500 J / 5 s = 700 W.

Correct Answer: C — 560 W

Q. A cyclist accelerates from rest to a speed of 15 m/s. If the mass of the cyclist and the bicycle is 75 kg, what is the work done by the cyclist?

A. 800 J
B. 900 J
C. 1000 J
D. 1200 J

Solution

Work done = Change in Kinetic Energy = 0.5 × mass × (final velocity² - initial velocity²) = 0.5 × 75 kg × (15 m/s)² = 843.75 J.

Correct Answer: C — 1000 J

Q. A cyclist accelerates from rest to a speed of 15 m/s. If the mass of the cyclist and the bicycle is 75 kg, what is the kinetic energy at that speed?

A. 500 J
B. 750 J
C. 1000 J
D. 1250 J

Solution

Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 75 kg × (15 m/s)² = 8437.5 J.

Correct Answer: C — 1000 J

Q. A cyclist does 300 J of work to climb a hill. If the height of the hill is 5 m, what is the effective weight of the cyclist?

A. 30 kg
B. 60 kg
C. 90 kg
D. 120 kg

Solution

Weight = Work / Height = 300 J / 5 m = 60 N; mass = Weight / g = 60 N / 9.8 m/s² ≈ 6.12 kg.

Correct Answer: B — 60 kg

Q. A cyclist exerts a force of 100 N to maintain a speed of 5 m/s. What is the power output of the cyclist?

A. 200 W
B. 500 W
C. 1000 W
D. 250 W

Solution

Power can be calculated using P = F * v. Here, F = 100 N and v = 5 m/s. Thus, P = 100 N * 5 m/s = 500 W.

Correct Answer: B — 500 W

Q. A cyclist is moving around a circular track of radius 100 m. If he completes one lap in 40 seconds, what is his average speed?

A. 5 m/s
B. 10 m/s
C. 15 m/s
D. 20 m/s

Solution

Circumference = 2πr = 2π(100 m). Average speed = distance/time = (2π(100 m))/40 s ≈ 15.7 m/s.

Correct Answer: B — 10 m/s

Q. A cyclist is moving at 15 m/s and a pedestrian at 5 m/s in the same direction. How fast does the cyclist appear to be moving to the pedestrian?

A. 10 m/s
B. 15 m/s
C. 20 m/s
D. 5 m/s

Solution

Relative speed = Speed of cyclist - Speed of pedestrian = 15 m/s - 5 m/s = 10 m/s.

Correct Answer: A — 10 m/s

Q. A cyclist is moving at 15 m/s and a pedestrian is walking at 5 m/s in the same direction. What is the speed of the cyclist relative to the pedestrian?

A. 10 m/s
B. 15 m/s
C. 5 m/s
D. 20 m/s

Solution

Relative speed = Speed of cyclist - Speed of pedestrian = 15 m/s - 5 m/s = 10 m/s.

Correct Answer: A — 10 m/s

Q. A cyclist is moving at 15 m/s and a pedestrian is walking at 5 m/s in the same direction. What is the relative speed of the pedestrian with respect to the cyclist?

A. 10 m/s
B. 5 m/s
C. 20 m/s
D. 15 m/s

Solution

Relative speed = Speed of pedestrian - Speed of cyclist = 5 m/s - 15 m/s = -10 m/s (10 m/s behind).

Correct Answer: A — 10 m/s

Q. A cyclist is moving at 15 m/s and passes a stationary observer. If the observer starts moving at 5 m/s in the same direction, what is the speed of the cyclist relative to the observer?

A. 10 m/s
B. 15 m/s
C. 20 m/s
D. 5 m/s

Solution

Relative speed = Speed of cyclist - Speed of observer = 15 m/s - 5 m/s = 10 m/s.

Correct Answer: A — 10 m/s

Q. A cyclist is moving at 15 m/s towards the east while a car is moving at 25 m/s towards the west. What is the relative speed of the cyclist with respect to the car?

A. 10 m/s
B. 15 m/s
C. 40 m/s
D. 25 m/s

Solution

Relative speed = Speed of cyclist + Speed of car = 15 m/s + 25 m/s = 40 m/s.

Correct Answer: C — 40 m/s

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