Major Competitive Exams MCQ & Objective Questions
Major Competitive Exams play a crucial role in shaping the academic and professional futures of students in India. These exams not only assess knowledge but also test problem-solving skills and time management. Practicing MCQs and objective questions is essential for scoring better, as they help in familiarizing students with the exam format and identifying important questions that frequently appear in tests.
What You Will Practise Here
Key concepts and theories related to major subjects
Important formulas and their applications
Definitions of critical terms and terminologies
Diagrams and illustrations to enhance understanding
Practice questions that mirror actual exam patterns
Strategies for solving objective questions efficiently
Time management techniques for competitive exams
Exam Relevance
The topics covered under Major Competitive Exams are integral to various examinations such as CBSE, State Boards, NEET, and JEE. Students can expect to encounter a mix of conceptual and application-based questions that require a solid understanding of the subjects. Common question patterns include multiple-choice questions that test both knowledge and analytical skills, making it essential to be well-prepared with practice MCQs.
Common Mistakes Students Make
Rushing through questions without reading them carefully
Overlooking the negative marking scheme in MCQs
Confusing similar concepts or terms
Neglecting to review previous years’ question papers
Failing to manage time effectively during the exam
FAQs
Question: How can I improve my performance in Major Competitive Exams?Answer: Regular practice of MCQs and understanding key concepts will significantly enhance your performance.
Question: What types of questions should I focus on for these exams?Answer: Concentrate on important Major Competitive Exams questions that frequently appear in past papers and mock tests.
Question: Are there specific strategies for tackling objective questions?Answer: Yes, practicing under timed conditions and reviewing mistakes can help develop effective strategies.
Start your journey towards success by solving practice MCQs today! Test your understanding and build confidence for your upcoming exams. Remember, consistent practice is the key to mastering Major Competitive Exams!
Q. A 3 kg object is pushed with a force of 12 N over a distance of 4 m. What is the work done on the object?
A.
24 J
B.
36 J
C.
48 J
D.
60 J
Show solution
Solution
Work done (W) = Force (F) × Distance (d) = 12 N × 4 m = 48 J.
Correct Answer:
C
— 48 J
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Q. A 3 kg object is pushed with a force of 15 N over a distance of 4 m. If the object experiences a frictional force of 3 N, what is the net work done on the object?
A.
48 J
B.
60 J
C.
72 J
D.
84 J
Show solution
Solution
Net force = Applied force - Friction = 15 N - 3 N = 12 N. Work done = Net force × Distance = 12 N × 4 m = 48 J.
Correct Answer:
B
— 60 J
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Q. A 3 kg object is pushed with a force of 15 N over a distance of 4 m. What is the work done on the object?
A.
30 J
B.
45 J
C.
60 J
D.
75 J
Show solution
Solution
Work done = Force × Distance = 15 N × 4 m = 60 J.
Correct Answer:
C
— 60 J
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Q. A 3 kg object is subjected to a net force of 12 N. What is its acceleration?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
Show solution
Solution
Using F = ma, a = F/m = 12 N / 3 kg = 4 m/s².
Correct Answer:
C
— 4 m/s²
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Q. A 3 kg object is subjected to a net force of 12 N. What is the acceleration of the object? (2022)
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
Show solution
Solution
Using F = ma, a = F/m = 12 N / 3 kg = 4 m/s².
Correct Answer:
C
— 4 m/s²
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Q. A 3 kg object is thrown upwards with a velocity of 15 m/s. What is its maximum height? (2020)
A.
11.5 m
B.
17.5 m
C.
22.5 m
D.
10.5 m
Show solution
Solution
Using the formula v^2 = u^2 - 2gh, we find h = u^2 / (2g) = (15 m/s)^2 / (2 * 9.8 m/s²) = 11.5 m.
Correct Answer:
B
— 17.5 m
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Q. A 4 kg block is pulled with a force of 20 N. What is the acceleration of the block? (2020)
A.
5 m/s²
B.
4 m/s²
C.
3 m/s²
D.
2 m/s²
Show solution
Solution
Using F = ma, acceleration (a) = F/m = 20 N / 4 kg = 5 m/s².
Correct Answer:
A
— 5 m/s²
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Q. A 4 kg object is at rest on a table. What is the force of static friction acting on it?
A.
0 N
B.
4 N
C.
40 N
D.
10 N
Show solution
Solution
If the object is at rest and no external force is applied, the static friction force is 0 N.
Correct Answer:
A
— 0 N
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Q. A 4 kg object is at rest on a table. What is the force of static friction if no external force is applied?
A.
0 N
B.
4 N
C.
40 N
D.
None of the above
Show solution
Solution
If no external force is applied, the force of static friction is 0 N.
Correct Answer:
A
— 0 N
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Q. A 4 kg object is dropped from a height. What is the force acting on it just before it hits the ground?
A.
0 N
B.
4 N
C.
40 N
D.
10 N
Show solution
Solution
The force acting on the object is its weight, F = mg = 4 kg * 10 m/s² = 40 N.
Correct Answer:
C
— 40 N
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Q. A 4 kg object is hanging from a rope. What is the tension in the rope if the object is at rest?
A.
0 N
B.
4 N
C.
40 N
D.
10 N
Show solution
Solution
The tension in the rope equals the weight of the object: T = mg = 4 kg * 10 m/s² = 40 N.
Correct Answer:
C
— 40 N
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Q. A 4 kg object is lifted to a height of 2 m. What is the work done against gravity? (2022)
A.
40 J
B.
80 J
C.
20 J
D.
60 J
Show solution
Solution
Work done = mass × g × height = 4 kg × 10 m/s² × 2 m = 80 J
Correct Answer:
B
— 80 J
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Q. A 4 kg object is lifted to a height of 3 m. What is the change in gravitational potential energy?
A.
12 J
B.
24 J
C.
36 J
D.
48 J
Show solution
Solution
Change in potential energy = mgh = 4 kg × 9.8 m/s² × 3 m = 117.6 J.
Correct Answer:
B
— 24 J
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Q. A 4 kg object is lifted to a height of 3 m. What is the increase in gravitational potential energy?
A.
12 J
B.
24 J
C.
36 J
D.
48 J
Show solution
Solution
Increase in potential energy = mgh = 4 kg × 9.8 m/s² × 3 m = 117.6 J.
Correct Answer:
B
— 24 J
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Q. A 4 kg object is lifted to a height of 3 m. What is the increase in gravitational potential energy? (g = 9.8 m/s²)
A.
117.6 J
B.
117 J
C.
120 J
D.
150 J
Show solution
Solution
Potential energy increase = mgh = 4 kg × 9.8 m/s² × 3 m = 117.6 J.
Correct Answer:
A
— 117.6 J
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Q. A 4 kg object is lifted to a height of 5 m. What is the work done against gravity? (g = 9.8 m/s²)
A.
19.6 J
B.
39.2 J
C.
49 J
D.
196 J
Show solution
Solution
Work done against gravity = mgh = 4 kg × 9.8 m/s² × 5 m = 196 J.
Correct Answer:
B
— 39.2 J
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Q. A 4 kg object is moving in a circular path of radius 2 m with a speed of 6 m/s. What is the centripetal force acting on the object?
A.
9 N
B.
12 N
C.
18 N
D.
24 N
Show solution
Solution
Centripetal force F = mv²/r = 4 kg * (6 m/s)² / 2 m = 4 * 36 / 2 = 72 N / 2 = 36 N.
Correct Answer:
B
— 12 N
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Q. A 4 kg object is moving in a circular path of radius 2 m with a speed of 6 m/s. What is the centripetal force acting on it?
A.
9 N
B.
12 N
C.
18 N
D.
24 N
Show solution
Solution
Centripetal force F = mv²/r = 4 kg * (6 m/s)² / 2 m = 72 N / 2 = 36 N.
Correct Answer:
B
— 12 N
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Q. A 4 kg object is moving in a straight line with a velocity of 5 m/s. What is the kinetic energy of the object?
A.
50 J
B.
40 J
C.
20 J
D.
10 J
Show solution
Solution
Kinetic energy KE = 0.5 * m * v² = 0.5 * 4 kg * (5 m/s)² = 50 J.
Correct Answer:
A
— 50 J
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Q. A 4 kg object is moving with a speed of 5 m/s. If it comes to rest, what is the work done by friction?
A.
50 J
B.
75 J
C.
100 J
D.
125 J
Show solution
Solution
Work done = change in kinetic energy = 0 - 0.5 × 4 kg × (5 m/s)² = -50 J.
Correct Answer:
C
— 100 J
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Q. A 4 kg object is moving with a speed of 5 m/s. What is its kinetic energy?
A.
10 J
B.
20 J
C.
50 J
D.
100 J
Show solution
Solution
Kinetic energy = 0.5 × m × v² = 0.5 × 4 kg × (5 m/s)² = 50 J.
Correct Answer:
C
— 50 J
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Q. A 4 kg object is moving with a speed of 5 m/s. What is the total mechanical energy if it is at a height of 2 m?
A.
50 J
B.
60 J
C.
70 J
D.
80 J
Show solution
Solution
Total mechanical energy = Kinetic energy + Potential energy = 0.5 × 4 kg × (5 m/s)² + 4 kg × 9.8 m/s² × 2 m = 50 J + 78.4 J = 128.4 J.
Correct Answer:
C
— 70 J
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Q. A 4 kg object is moving with a velocity of 2 m/s. What is its kinetic energy?
A.
8 J
B.
4 J
C.
16 J
D.
2 J
Show solution
Solution
Kinetic Energy (KE) = 1/2 * m * v^2 = 1/2 * 4 * (2^2) = 8 J
Correct Answer:
A
— 8 J
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Q. A 4 kg object is moving with a velocity of 2 m/s. What is the change in momentum if the object comes to a stop?
A.
0 kg·m/s
B.
4 kg·m/s
C.
8 kg·m/s
D.
2 kg·m/s
Show solution
Solution
Change in momentum = final momentum - initial momentum = 0 - (4 kg * 2 m/s) = -8 kg·m/s.
Correct Answer:
C
— 8 kg·m/s
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Q. A 4 kg object is moving with a velocity of 3 m/s. What is its kinetic energy?
A.
18 J
B.
24 J
C.
36 J
D.
12 J
Show solution
Solution
Kinetic energy KE = 0.5 * m * v² = 0.5 * 4 kg * (3 m/s)² = 18 J.
Correct Answer:
B
— 24 J
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Q. A 4 kg object is moving with a velocity of 5 m/s. What is its kinetic energy? (2021)
A.
10 J
B.
20 J
C.
50 J
D.
100 J
Show solution
Solution
Kinetic energy KE = 1/2 mv² = 1/2 * 4 kg * (5 m/s)² = 50 J.
Correct Answer:
C
— 50 J
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Q. A 4 kg object is moving with a velocity of 5 m/s. What is its momentum?
A.
10 kg·m/s
B.
20 kg·m/s
C.
15 kg·m/s
D.
25 kg·m/s
Show solution
Solution
Momentum p = mv = 4 kg * 5 m/s = 20 kg·m/s.
Correct Answer:
B
— 20 kg·m/s
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Q. A 4 kg object is pulled with a force of 20 N. If the frictional force is 4 N, what is the net force acting on the object?
A.
16 N
B.
20 N
C.
24 N
D.
4 N
Show solution
Solution
Net force = applied force - frictional force = 20 N - 4 N = 16 N.
Correct Answer:
A
— 16 N
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Q. A 4 kg object is pushed with a force of 16 N. What is the acceleration of the object?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
Show solution
Solution
Using F = ma, we have a = F/m = 16 N / 4 kg = 4 m/s².
Correct Answer:
C
— 4 m/s²
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Q. A 4 kg object is pushed with a force of 20 N over a distance of 3 m. If the object starts from rest, what is its final speed? (Assume no friction)
A.
2 m/s
B.
3 m/s
C.
4 m/s
D.
5 m/s
Show solution
Solution
Work done = Force × Distance = 20 N × 3 m = 60 J. Kinetic energy = 0.5 × m × v². 60 J = 0.5 × 4 kg × v². v² = 30, v = √30 ≈ 5.48 m/s.
Correct Answer:
C
— 4 m/s
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