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 1 kg mass is lifted vertically 10 m. How much work is done against gravity? (g = 9.8 m/s²)

A. 9.8 J
B. 19.6 J
C. 29.4 J
D. 39.2 J

Solution

Work done = mgh = 1 kg × 9.8 m/s² × 10 m = 98 J.

Correct Answer: B — 19.6 J

Q. A 1 kg object is dropped from a height. What is its velocity just before it hits the ground (neglect air resistance)? (2020)

A. 10 m/s
B. 14 m/s
C. 20 m/s
D. 30 m/s

Solution

Using v² = u² + 2gh, v = √(0 + 2 × 9.8 m/s² × h). For h = 20 m, v = √(392) = 20 m/s.

Correct Answer: C — 20 m/s

Q. A 1 kg object is moving with a velocity of 4 m/s. What is its kinetic energy? (2021)

A. 8 J
B. 4 J
C. 16 J
D. 2 J

Solution

Kinetic Energy (KE) = 1/2 * m * v^2 = 1/2 * 1 kg * (4 m/s)^2 = 8 J.

Correct Answer: A — 8 J

Q. A 1 kg object is pushed with a force of 10 N over a distance of 3 m. What is the work done?

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

Solution

Work done = Force × Distance = 10 N × 3 m = 30 J.

Correct Answer: C — 30 J

Q. A 1 kg object is pushed with a force of 10 N. If the frictional force is 4 N, what is the net force acting on the object?

A. 6 N
B. 10 N
C. 4 N
D. 0 N

Solution

Net force = applied force - frictional force = 10 N - 4 N = 6 N.

Correct Answer: A — 6 N

Q. A 10 kg block is at rest on a horizontal surface. If a horizontal force of 30 N is applied, what will be the acceleration of the block?

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

Solution

Using F = ma, acceleration a = F/m = 30 N / 10 kg = 3 m/s².

Correct Answer: C — 3 m/s²

Q. A 10 kg block is at rest on a horizontal surface. If a horizontal force of 30 N is applied, what will be its acceleration?

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

Solution

Using F = ma, acceleration a = F/m = 30 N / 10 kg = 3 m/s².

Correct Answer: C — 3 m/s²

Q. A 10 kg block is at rest on a horizontal surface. If a horizontal force of 30 N is applied, what will be its acceleration? (Assume no friction)

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

Solution

Using F = ma, a = F/m = 30 N / 10 kg = 3 m/s².

Correct Answer: C — 3 m/s²

Q. A 10 kg block is pushed with a force of 50 N. What is its acceleration? (2023)

A. 2 m/s²
B. 5 m/s²
C. 10 m/s²
D. 15 m/s²

Solution

Using Newton's second law, F = ma, we have a = F/m = 50 N / 10 kg = 5 m/s².

Correct Answer: B — 5 m/s²

Q. A 10 kg block slides down a frictionless incline of height 5 m. What is its speed at the bottom? (g = 10 m/s²) (2020)

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

Solution

Using conservation of energy, PE at top = KE at bottom: mgh = 1/2 mv²; v = sqrt(2gh) = sqrt(2 * 10 m/s² * 5 m) = 10 m/s

Correct Answer: A — 10 m/s

Q. A 10 kg block slides down a frictionless incline of height 5 m. What is its speed at the bottom? (2020)

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

Solution

Using conservation of energy, PE at top = KE at bottom. mgh = 1/2 mv^2. 10 kg * 9.8 m/s^2 * 5 m = 1/2 * 10 kg * v^2. Solving gives v = 10 m/s.

Correct Answer: B — 10 m/s

Q. A 10 kg box is at rest on a frictionless surface. A force of 30 N is applied. What is the final velocity after 3 seconds? (2023)

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

Solution

Acceleration = F/m = 30 N / 10 kg = 3 m/s². Final velocity = initial velocity + (acceleration × time) = 0 + (3 m/s² × 3 s) = 9 m/s.

Correct Answer: B — 10 m/s

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

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

Solution

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

Correct Answer: B — 100 J

Q. A 10 kg object falls freely from a height of 20 m. What is its potential energy at the top?

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

Solution

Potential energy is given by PE = mgh = 10 * 9.8 * 20 = 1960 J.

Correct Answer: B — 200 J

Q. A 10 kg object falls freely from a height of 20 m. What is its speed just before hitting the ground? (g = 9.8 m/s²)

A. 10 m/s
B. 14 m/s
C. 20 m/s
D. 28 m/s

Solution

Using conservation of energy, v = √(2gh) = √(2 * 9.8 * 20) = 28 m/s.

Correct Answer: D — 28 m/s

Q. A 10 kg object falls from a height of 20 m. What is its potential energy at the top?

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

Solution

Potential energy is given by PE = mgh = 10 * 9.8 * 20 = 1960 J.

Correct Answer: D — 400 J

Q. A 10 kg object falls from a height of 20 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)

A. 10 m/s
B. 14 m/s
C. 20 m/s
D. 28 m/s

Solution

Using conservation of energy, v = √(2gh) = √(2 * 9.8 * 20) = 19.8 m/s.

Correct Answer: B — 14 m/s

Q. A 10 kg object falls from a height of 20 m. What is its speed just before it hits the ground?

A. 10 m/s
B. 14 m/s
C. 20 m/s
D. 28 m/s

Solution

Using conservation of energy, v = √(2gh) = √(2 * 9.8 * 20) = 20 m/s.

Correct Answer: C — 20 m/s

Q. A 10 kg object is at rest on a horizontal surface. If a force of 30 N is applied, what is the object's acceleration assuming a frictional force of 10 N?

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

Solution

Net force = 30 N - 10 N = 20 N. Using F = ma, a = F/m = 20 N / 10 kg = 2 m/s².

Correct Answer: B — 3 m/s²

Q. A 10 kg object is at rest on a surface. If a force of 30 N is applied, what is the object's acceleration assuming the friction is negligible?

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

Solution

Using F = ma, acceleration a = F/m = 30 N / 10 kg = 3 m/s².

Correct Answer: C — 3 m/s²

Q. A 10 kg object is at rest. A force of 30 N is applied to it. What will be its velocity after 3 seconds? (2023)

A. 3 m/s
B. 6 m/s
C. 9 m/s
D. 12 m/s

Solution

First, find acceleration: a = F/m = 30 N / 10 kg = 3 m/s². Then, using v = u + at, v = 0 + 3 m/s² × 3 s = 9 m/s.

Correct Answer: B — 6 m/s

Q. A 10 kg object is at rest. A force of 30 N is applied. What is the final velocity after 3 seconds? (2023)

A. 3 m/s
B. 6 m/s
C. 9 m/s
D. 12 m/s

Solution

Using F = ma, a = F/m = 30 N / 10 kg = 3 m/s². Final velocity v = u + at = 0 + 3 m/s² × 3 s = 9 m/s.

Correct Answer: B — 6 m/s

Q. A 10 kg object is dropped from a height of 20 m. What is its potential energy at the top? (g = 10 m/s²) (2022)

A. 200 J
B. 100 J
C. 50 J
D. 300 J

Solution

Potential Energy (PE) = m * g * h = 10 kg * 10 m/s² * 20 m = 2000 J

Correct Answer: A — 200 J

Q. A 10 kg object is dropped from a height of 20 m. What is its potential energy at the top? (g = 9.8 m/s²) (1960)

A. 196 J
B. 98 J
C. 294 J
D. 490 J

Solution

Potential Energy (PE) = mgh = 10 kg × 9.8 m/s² × 20 m = 1960 J.

Correct Answer: C — 294 J

Q. A 10 kg object is dropped from a height of 20 m. What is its potential energy at the top? (2020)

A. 200 J
B. 1000 J
C. 1500 J
D. 3000 J

Solution

Potential Energy = mass × g × height = 10 kg × 10 m/s² × 20 m = 2000 J

Correct Answer: B — 1000 J

Q. A 10 kg object is dropped from a height of 20 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)

A. 14 m/s
B. 19.8 m/s
C. 20 m/s
D. 28 m/s

Solution

Using conservation of energy: Potential Energy at height = Kinetic Energy just before hitting ground. mgh = 0.5mv². v = √(2gh) = √(2 × 9.8 m/s² × 20 m) = 19.8 m/s.

Correct Answer: B — 19.8 m/s

Q. A 10 kg object is dropped from a height of 20 m. What is its speed just before it hits the ground? (g = 10 m/s²) (2022)

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

Solution

Using the formula v = √(2gh) = √(2 * 10 m/s² * 20 m) = √400 = 20 m/s.

Correct Answer: A — 20 m/s

Q. A 10 kg object is dropped from a height of 20 m. What is the potential energy at the top? (2022)

A. 200 J
B. 1000 J
C. 500 J
D. 400 J

Solution

Potential Energy = mass × g × height = 10 kg × 9.8 m/s² × 20 m = 1960 J

Correct Answer: B — 1000 J

Q. A 10 kg object is dropped from a height of 5 m. What is its potential energy at the top? (2021)

A. 50 J
B. 100 J
C. 200 J
D. 25 J

Solution

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

Correct Answer: B — 100 J

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

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

Solution

Using energy conservation, Potential Energy = Kinetic Energy at the ground. mgh = (1/2)mv². v = sqrt(2gh) = sqrt(2 × 9.8 m/s² × 5 m) ≈ 10 m/s.

Correct Answer: B — 15 m/s

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