Major Competitive Exams

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Major Competitive Exams

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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!

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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?

A. 2 m/s
B. 3 m/s
C. 4 m/s
D. 5 m/s

Solution

Work done = Force × Distance = 20 N × 3 m = 60 J. K.E = 0.5 × m × v², 60 J = 0.5 × 4 kg × v², v = 3.87 m/s.

Correct Answer: C — 4 m/s

Q. A 4 kg object is pushed with a force of 20 N. If the frictional force is 4 N, what is the acceleration of the object?

A. 4 m/s²
B. 5 m/s²
C. 6 m/s²
D. 7 m/s²

Solution

Net force = applied force - friction = 20 N - 4 N = 16 N. Acceleration a = F/m = 16 N / 4 kg = 4 m/s².

Correct Answer: B — 5 m/s²

Q. A 4 kg object is subjected to a net force of 12 N. What is the acceleration of the object?

A. 2 m/s²
B. 3 m/s²
C. 4 m/s²
D. 5 m/s²

Solution

Using F = ma, a = F/m = 12 N / 4 kg = 3 m/s².

Correct Answer: B — 3 m/s²

Q. A 4 kg object is subjected to a net force of 16 N. What is its acceleration?

A. 2 m/s²
B. 3 m/s²
C. 4 m/s²
D. 5 m/s²

Solution

Using F = ma, acceleration a = F/m = 16 N / 4 kg = 4 m/s².

Correct Answer: C — 4 m/s²

Q. A 5 kg block is placed on a surface with a coefficient of static friction of 0.6. What is the maximum static friction force before the block starts to move?

A. 30 N
B. 20 N
C. 15 N
D. 25 N

Solution

Maximum static friction force = μs * N = 0.6 * (5 kg * 9.8 m/s²) = 29.4 N, approximately 30 N.

Correct Answer: A — 30 N

Q. A 5 kg block is resting on a frictionless surface. A force of 10 N is applied to it. What is the acceleration of the block?

A. 1 m/s²
B. 2 m/s²
C. 3 m/s²
D. 4 m/s²

Solution

Using Newton's second law, F = ma, we have a = F/m = 10 N / 5 kg = 2 m/s².

Correct Answer: B — 2 m/s²

Q. A 5 kg block is resting on a frictionless surface. If a force of 10 N is applied to it, what will be its acceleration?

A. 1 m/s²
B. 2 m/s²
C. 5 m/s²
D. 10 m/s²

Solution

Using Newton's second law, F = ma, we have a = F/m = 10 N / 5 kg = 2 m/s².

Correct Answer: B — 2 m/s²

Q. A 5 kg block is resting on a horizontal surface. What is the normal force acting on the block?

A. 5 N
B. 10 N
C. 15 N
D. 20 N

Solution

The normal force equals the weight of the block, which is mass times gravity: 5 kg * 9.8 m/s² = 49 N.

Correct Answer: B — 10 N

Q. A 5 kg block of ice at 0°C is converted to water at 0°C. If the latent heat of fusion of ice is 334 kJ/kg, how much heat is absorbed?

A. 1670 kJ
B. 334 kJ
C. 167 kJ
D. 3340 kJ

Solution

Heat absorbed (Q) = m * Lf = 5 kg * 334 kJ/kg = 1670 kJ.

Correct Answer: B — 334 kJ

Q. A 5 kg block of ice at 0°C is placed in a calorimeter containing 2 kg of water at 80°C. What is the final temperature of the system when thermal equilibrium is reached? (Latent heat of fusion of ice = 334 kJ/kg)

A. 0°C
B. 20°C
C. 40°C
D. 60°C

Solution

The heat lost by water = heat gained by ice. Solving gives the final temperature as 0°C.

Correct Answer: B — 20°C

Q. A 5 kg block of metal is heated from 20°C to 100°C. If the specific heat capacity of the metal is 0.5 J/g°C, how much heat is required? (2023)

A. 2000 J
B. 2500 J
C. 3000 J
D. 3500 J

Solution

Heat (Q) = m * c * ΔT = 5000 g * 0.5 J/g°C * (100°C - 20°C) = 5000 * 0.5 * 80 = 200000 J = 2000 J.

Correct Answer: C — 3000 J

Q. A 5 kg object is at rest on a surface with a coefficient of static friction of 0.4. What is the maximum static friction force?

A. 10 N
B. 20 N
C. 30 N
D. 40 N

Solution

Maximum static friction force = μs * N = 0.4 * (5 kg * 9.8 m/s²) = 0.4 * 49 N = 19.6 N, approximately 20 N.

Correct Answer: B — 20 N

Q. A 5 kg object is at rest on a surface. If a horizontal force of 15 N is applied, what is the object's acceleration? (Assume friction is negligible)

A. 2 m/s²
B. 3 m/s²
C. 4 m/s²
D. 5 m/s²

Solution

Using F = ma, a = F/m = 15 N / 5 kg = 3 m/s².

Correct Answer: B — 3 m/s²

Q. A 5 kg object is dropped from a height of 10 m. What is its potential energy at the top? (g = 9.8 m/s²)

A. 490 J
B. 98 J
C. 50 J
D. 100 J

Solution

Potential Energy = mgh = 5 kg * 9.8 m/s² * 10 m = 490 J.

Correct Answer: A — 490 J

Q. A 5 kg object is dropped from a height of 10 m. What is its potential energy at the top? (Take g = 9.8 m/s²) (2020)

A. 49 J
B. 98 J
C. 39 J
D. 59 J

Solution

Potential Energy (PE) = mgh = 5 kg × 9.8 m/s² × 10 m = 490 J.

Correct Answer: B — 98 J

Q. A 5 kg object is dropped from a height of 10 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)

A. 14 m/s
B. 20 m/s
C. 10 m/s
D. 15 m/s

Solution

Using energy conservation, Potential Energy = Kinetic Energy at the ground: mgh = 0.5mv²; v = √(2gh) = √(2 × 9.8 m/s² × 10 m) = 14 m/s

Correct Answer: A — 14 m/s

Q. A 5 kg object is dropped from a height of 10 m. What is its speed just before it hits the ground? (Use g = 9.8 m/s²) (2021)

A. 14 m/s
B. 20 m/s
C. 10 m/s
D. 15 m/s

Solution

Using energy conservation, Potential Energy = Kinetic Energy. mgh = 1/2 mv². v = √(2gh) = √(2 × 9.8 m/s² × 10 m) = 14 m/s.

Correct Answer: A — 14 m/s

Q. A 5 kg object is dropped from a height of 10 m. What is the total mechanical energy just before it hits the ground?

A. 0 J
B. 50 J
C. 100 J
D. 200 J

Solution

Total mechanical energy is conserved. At height, PE = mgh = 5 * 9.8 * 10 = 490 J. Just before hitting the ground, total energy = 490 J.

Correct Answer: C — 100 J

Q. A 5 kg object is dropped from a height of 20 m. What is its velocity just before it hits the ground? (Take g = 10 m/s²)

A. 10 m/s
B. 20 m/s
C. 15 m/s
D. 25 m/s

Solution

Using the formula v² = u² + 2gh, where u = 0, we have v = √(0 + 2*10*20) = √400 = 20 m/s.

Correct Answer: B — 20 m/s

Q. A 5 kg object is dropped from a height of 20 m. What is the potential energy at the top? (g = 9.8 m/s²)

A. 980 J
B. 490 J
C. 196 J
D. 9800 J

Solution

Potential energy is given by PE = mgh. Here, PE = 5 * 9.8 * 20 = 980 J.

Correct Answer: A — 980 J

Q. A 5 kg object is dropped from a height of 20 m. What is the total mechanical energy just before it hits the ground?

A. 0 J
B. 100 J
C. 200 J
D. 500 J

Solution

Total mechanical energy is conserved. At height, PE = mgh = 5 * 9.8 * 20 = 980 J. Just before hitting the ground, total energy = 980 J.

Correct Answer: C — 200 J

Q. A 5 kg object is dropped from a height. What is the force acting on it just before it hits the ground?

A. 5 N
B. 10 N
C. 15 N
D. 20 N

Solution

The only force acting on it is its weight, W = mg = 5 kg * 10 m/s² = 50 N.

Correct Answer: B — 10 N

Q. A 5 kg object is dropped from a height. What is the force acting on it just before it hits the ground? (Assume g = 9.8 m/s²)

A. 5 N
B. 9.8 N
C. 49 N
D. 0 N

Solution

The force acting on the object is its weight, F = mg = 5 kg * 9.8 m/s² = 49 N.

Correct Answer: C — 49 N

Q. A 5 kg object is lifted to a height of 10 m. What is the gravitational potential energy gained?

A. 50 J
B. 100 J
C. 200 J
D. 500 J

Solution

Potential Energy (PE) = m * g * h = 5 kg * 10 m/s^2 * 10 m = 500 J

Correct Answer: B — 100 J

Q. A 5 kg object is lifted to a height of 10 m. What is the potential energy gained by the object?

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

Solution

Potential energy (PE) = mgh = 5 kg * 9.8 m/s² * 10 m = 490 J.

Correct Answer: B — 100 J

Q. A 5 kg object is lifted to a height of 10 m. What is the potential energy gained?

A. 50 J
B. 100 J
C. 200 J
D. 500 J

Solution

Potential Energy (PE) = m * g * h = 5 kg * 9.8 m/s^2 * 10 m = 490 J (approx. 500 J)

Correct Answer: B — 100 J

Q. A 5 kg object is lifted to a height of 10 m. What is the potential energy gained by the object? (Take g = 10 m/s²)

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

Solution

Potential energy PE = mgh = 5 kg * 10 m/s² * 10 m = 500 J.

Correct Answer: B — 100 J

Q. A 5 kg object is lifted to a height of 10 m. What is the work done against gravity?

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

Solution

Work done against gravity is W = mgh = 5 * 9.8 * 10 = 490 J.

Correct Answer: D — 200 J

Q. A 5 kg object is lifted to a height of 2 m. How much work is done against gravity? (g = 9.8 m/s²) (2022)

A. 98 J
B. 196 J
C. 39 J
D. 49 J

Solution

Work done = mgh = 5 kg * 9.8 m/s² * 2 m = 98 J.

Correct Answer: A — 98 J

Q. A 5 kg object is lifted to a height of 2 m. What is the work done against gravity?

A. 10 J
B. 20 J
C. 50 J
D. 100 J

Solution

Work done (W) against gravity is given by W = mgh. Here, m = 5 kg, g = 9.8 m/s², h = 2 m. W = 5 * 9.8 * 2 = 98 J.

Correct Answer: B — 20 J

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