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. If a fluid has a viscosity of 0.5 Pa·s, what does this indicate about its flow characteristics?
A.
It flows easily
B.
It is very thick
C.
It is a gas
D.
It is a low-density fluid
Show solution
Solution
A viscosity of 0.5 Pa·s indicates that the fluid is relatively thick and flows less easily compared to fluids with lower viscosity.
Correct Answer:
B
— It is very thick
Learn More →
Q. If a force of 10 N is applied at an angle of 30 degrees to the horizontal, what is the horizontal component of the force?
A.
5 N
B.
8.66 N
C.
10 N
D.
0 N
Show solution
Solution
The horizontal component is calculated as 10 N * cos(30°) = 8.66 N.
Correct Answer:
B
— 8.66 N
Learn More →
Q. If a force of 10 N is applied to move an object 5 m in the direction of the force, what is the work done?
A.
50 J
B.
30 J
C.
20 J
D.
10 J
Show solution
Solution
Work Done (W) = Force * Distance = 10 N * 5 m = 50 J
Correct Answer:
A
— 50 J
Learn More →
Q. If a force of 10 N is applied to move an object 5 m, what is the work done?
A.
50 J
B.
25 J
C.
10 J
D.
5 J
Show solution
Solution
Work Done (W) = Force * Distance = 10 N * 5 m = 50 J
Correct Answer:
A
— 50 J
Learn More →
Q. If a force of 12 N is applied at an angle of 30 degrees to a lever arm of 1 m, what is the torque about the pivot?
A.
6 Nm
B.
10 Nm
C.
12 Nm
D.
15 Nm
Show solution
Solution
Torque = Force × Distance × sin(θ) = 12 N × 1 m × sin(30°) = 12 N × 1 m × 0.5 = 6 Nm.
Correct Answer:
A
— 6 Nm
Learn More →
Q. If a force of 12 N is applied at an angle of 30 degrees to the horizontal while moving an object 3 m, what is the work done by the force?
A.
18 J
B.
24 J
C.
30 J
D.
36 J
Show solution
Solution
Work done = F × d × cos(θ) = 12 N × 3 m × cos(30°) = 12 N × 3 m × (√3/2) = 18 J.
Correct Answer:
B
— 24 J
Learn More →
Q. If a force of 12 N is applied to a 4 kg object, what is the resulting acceleration?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
Show solution
Solution
Using F = ma, we find a = F/m = 12 N / 4 kg = 3 m/s².
Correct Answer:
B
— 3 m/s²
Learn More →
Q. If a force of 15 N acts on an object and moves it 4 m in the direction of the force, what is the work done?
A.
30 J
B.
60 J
C.
75 J
D.
90 J
Show solution
Solution
Work done = Force × Distance = 15 N × 4 m = 60 J.
Correct Answer:
B
— 60 J
Learn More →
Q. If a force of 15 N is applied at a distance of 0.4 m from the pivot, what is the torque?
A.
6 Nm
B.
12 Nm
C.
15 Nm
D.
20 Nm
Show solution
Solution
Torque = Force × Distance = 15 N × 0.4 m = 6 Nm.
Correct Answer:
B
— 12 Nm
Learn More →
Q. If a force of 15 N is applied at an angle of 30 degrees to the lever arm of length 1.5 m, what is the torque about the pivot?
A.
3.75 Nm
B.
7.5 Nm
C.
11.25 Nm
D.
12.99 Nm
Show solution
Solution
Torque (τ) = F × r × sin(θ) = 15 N × 1.5 m × sin(30°) = 15 × 1.5 × 0.5 = 11.25 Nm.
Correct Answer:
B
— 7.5 Nm
Learn More →
Q. If a force of 15 N is applied at an angle of 30 degrees to the lever arm of length 1 m, what is the torque about the pivot?
A.
7.5 Nm
B.
15 Nm
C.
12.99 Nm
D.
10 Nm
Show solution
Solution
Torque (τ) = F × r × sin(θ) = 15 N × 1 m × sin(30°) = 15 N × 1 m × 0.5 = 7.5 Nm.
Correct Answer:
C
— 12.99 Nm
Learn More →
Q. If a force of 15 N is applied at an angle of 60 degrees to the horizontal while moving an object 4 m, what is the work done by the force?
A.
30 J
B.
60 J
C.
120 J
D.
180 J
Show solution
Solution
Work done = F × d × cos(θ) = 15 N × 4 m × cos(60°) = 15 N × 4 m × 0.5 = 30 J.
Correct Answer:
C
— 120 J
Learn More →
Q. If a force of 15 N is applied at an angle of 60° to the horizontal while moving an object 4 m, what is the work done by the force?
A.
30 J
B.
60 J
C.
120 J
D.
180 J
Show solution
Solution
Work done = F × d × cos(θ) = 15 N × 4 m × cos(60°) = 15 N × 4 m × 0.5 = 30 J.
Correct Answer:
C
— 120 J
Learn More →
Q. If a force of 15 N is applied to a mass of 3 kg, what is the net force acting on the mass if there is a frictional force of 5 N opposing the motion?
A.
10 N
B.
15 N
C.
20 N
D.
5 N
Show solution
Solution
Net force = applied force - frictional force = 15 N - 5 N = 10 N.
Correct Answer:
A
— 10 N
Learn More →
Q. If a force of 15 N is applied to a mass of 3 kg, what is the net force acting on the mass if it is also experiencing a frictional force of 5 N?
A.
10 N
B.
15 N
C.
20 N
D.
5 N
Show solution
Solution
Net force = applied force - frictional force = 15 N - 5 N = 10 N.
Correct Answer:
A
— 10 N
Learn More →
Q. If a force of 15 N is applied to move an object 3 m in the direction of the force, what is the work done?
A.
45 J
B.
30 J
C.
15 J
D.
60 J
Show solution
Solution
Work done = Force × Distance = 15 N × 3 m = 45 J.
Correct Answer:
A
— 45 J
Learn More →
Q. If a force of 15 N is applied to move an object 4 m in the direction of the force, what is the work done?
A.
30 J
B.
60 J
C.
45 J
D.
75 J
Show solution
Solution
Work done = Force × Distance = 15 N × 4 m = 60 J.
Correct Answer:
B
— 60 J
Learn More →
Q. If a force of 30 N is applied to a mass of 10 kg, what is the net force acting on the mass if there is a frictional force of 10 N opposing the motion?
A.
20 N
B.
30 N
C.
40 N
D.
10 N
Show solution
Solution
Net force = applied force - frictional force = 30 N - 10 N = 20 N.
Correct Answer:
A
— 20 N
Learn More →
Q. If a force of 5 N acts on an object and moves it 4 m in the direction of the force, what is the work done?
A.
10 J
B.
15 J
C.
20 J
D.
25 J
Show solution
Solution
Work done = Force × Distance = 5 N × 4 m = 20 J.
Correct Answer:
C
— 20 J
Learn More →
Q. If a force of 5 N is applied at an angle of 60 degrees to the horizontal while moving an object 4 m, what is the work done by the force?
A.
10 J
B.
20 J
C.
5 J
D.
15 J
Show solution
Solution
Work done = Force × Distance × cos(θ) = 5 N × 4 m × cos(60°) = 5 N × 4 m × 0.5 = 10 J.
Correct Answer:
A
— 10 J
Learn More →
Q. If a force of 5 N is applied at an angle of 60 degrees to the horizontal while moving an object 3 m, what is the work done by the force?
A.
7.5 J
B.
15 J
C.
12.99 J
D.
10 J
Show solution
Solution
Work done = Force × Distance × cos(θ) = 5 N × 3 m × cos(60°) = 5 N × 3 m × 0.5 = 7.5 J.
Correct Answer:
C
— 12.99 J
Learn More →
Q. If a forced oscillator is driven at a frequency much lower than its natural frequency, what happens to the amplitude?
A.
Increases significantly
B.
Decreases
C.
Remains constant
D.
Fluctuates
Show solution
Solution
At frequencies much lower than the natural frequency, the amplitude of the forced oscillator increases significantly.
Correct Answer:
B
— Decreases
Learn More →
Q. If a galvanometer shows a deflection when connected to a potentiometer, what does it indicate?
A.
The circuit is open.
B.
The potential difference is zero.
C.
The potential difference is equal to the reference voltage.
D.
The current is flowing through the galvanometer.
Show solution
Solution
A deflection in the galvanometer indicates that the potential difference across the galvanometer is equal to the reference voltage.
Correct Answer:
C
— The potential difference is equal to the reference voltage.
Learn More →
Q. If a gas expands against a constant external pressure, the work done by the gas is given by:
A.
W = P_ext * ΔV
B.
W = ΔU + Q
C.
W = Q - ΔU
D.
W = P_ext / ΔV
Show solution
Solution
The work done by the gas during expansion against a constant external pressure is given by W = P_ext * ΔV.
Correct Answer:
A
— W = P_ext * ΔV
Learn More →
Q. If a gas expands and does 150 J of work while absorbing 100 J of heat, what is the change in internal energy?
A.
-50 J
B.
50 J
C.
250 J
D.
100 J
Show solution
Solution
Using the first law of thermodynamics, ΔU = Q - W = 100 J - 150 J = -50 J.
Correct Answer:
A
— -50 J
Learn More →
Q. If a gas expands and does 50 J of work while absorbing 30 J of heat, what is the change in internal energy?
A.
-20 J
B.
20 J
C.
80 J
D.
30 J
Show solution
Solution
Using the First Law of Thermodynamics, ΔU = Q - W. Here, ΔU = 30 J - 50 J = -20 J.
Correct Answer:
B
— 20 J
Learn More →
Q. If a gas occupies 10 L at 1 atm, what will be its volume at 2 atm if the temperature remains constant?
A.
5 L
B.
10 L
C.
20 L
D.
15 L
Show solution
Solution
Using Boyle's Law (P1V1 = P2V2), if P1 = 1 atm, V1 = 10 L, and P2 = 2 atm, then V2 = (P1V1)/P2 = (1 atm * 10 L) / 2 atm = 5 L.
Correct Answer:
A
— 5 L
Learn More →
Q. If a gas occupies a volume of 10 L at 1 atm, what will be its volume at 2 atm if the temperature remains constant?
A.
5 L
B.
10 L
C.
20 L
D.
15 L
Show solution
Solution
Using Boyle's Law (P1V1 = P2V2), if the pressure doubles from 1 atm to 2 atm, the volume will halve from 10 L to 5 L.
Correct Answer:
A
— 5 L
Learn More →
Q. If a hollow cylinder and a solid cylinder of the same mass and radius roll down the same incline, which one reaches the bottom first?
A.
Hollow cylinder
B.
Solid cylinder
C.
Both reach at the same time
D.
Depends on the angle of incline
Show solution
Solution
The solid cylinder reaches the bottom first because it has a smaller moment of inertia compared to the hollow cylinder.
Correct Answer:
B
— Solid cylinder
Learn More →
Q. If a hollow cylinder rolls down an incline, how does its acceleration compare to that of a solid cylinder?
A.
Hollow cylinder accelerates faster
B.
Solid cylinder accelerates faster
C.
Both accelerate equally
D.
Depends on the angle of incline
Show solution
Solution
The solid cylinder has a lower moment of inertia compared to the hollow cylinder, thus it accelerates faster down the incline.
Correct Answer:
B
— Solid cylinder accelerates faster
Learn More →
Showing 3001 to 3030 of 10700 (357 Pages)
