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 15 kg object is lifted to a height of 4 m. What is the work done against gravity? (g = 9.8 m/s²) (2021)

A. 588 J
B. 5880 J
C. 392 J
D. 294 J

Solution

Work = mgh = 15 kg * 9.8 m/s² * 4 m = 588 J.

Correct Answer: A — 588 J

Q. A 15 kg object is moving at a speed of 4 m/s. What is its kinetic energy?

A. 120 J
B. 240 J
C. 480 J
D. 60 J

Solution

Kinetic Energy (KE) = 0.5 × m × v² = 0.5 × 15 kg × (4 m/s)² = 0.5 × 15 × 16 = 120 J.

Correct Answer: B — 240 J

Q. A 15 kg object is moving with a velocity of 4 m/s. What is its momentum? (2019)

A. 60 kg m/s
B. 30 kg m/s
C. 45 kg m/s
D. 75 kg m/s

Solution

Momentum = mass × velocity = 15 kg × 4 m/s = 60 kg m/s.

Correct Answer: A — 60 kg m/s

Q. A 15 kg object is pushed with a force of 45 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, acceleration a = F/m = 45 N / 15 kg = 3 m/s².

Correct Answer: B — 3 m/s²

Q. A 15 kg object is subjected to a net force of 45 N. What is 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 = 45 N / 15 kg = 3 m/s².

Correct Answer: C — 3 m/s²

Q. A 15-meter tall tree casts a shadow of 10 meters. What is the angle of elevation of the sun? (2019)

A. 30 degrees
B. 45 degrees
C. 60 degrees
D. 75 degrees

Solution

tan(θ) = height/shadow = 15/10 = 1.5, θ = tan⁻¹(1.5) ≈ 56.31 degrees, which is closest to 60 degrees.

Correct Answer: C — 60 degrees

Q. A 1500 kg car accelerates from rest to a speed of 25 m/s. What is the work done on the car?

A. 468,750 J
B. 375,000 J
C. 250,000 J
D. 187,500 J

Solution

Work done = Change in Kinetic Energy = 0.5 * m * v² = 0.5 * 1500 kg * (25 m/s)² = 468,750 J.

Correct Answer: A — 468,750 J

Q. A 1500 kg car is moving at a speed of 25 m/s. What is its kinetic energy? (2021)

A. 468,750 J
B. 1,875,000 J
C. 937,500 J
D. 750,000 J

Solution

Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 1500 kg × (25 m/s)² = 468,750 J

Correct Answer: B — 1,875,000 J

Q. A 2 kg ball is dropped from a height of 10 m. What is its speed just before it hits the ground? (g = 10 m/s²) (2023)

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

Solution

Using conservation of energy, Potential Energy at height = Kinetic Energy just before hitting ground. mgh = 1/2 mv^2. Solving gives v = sqrt(2gh) = sqrt(2 * 10 m/s² * 10 m) = 14.14 m/s.

Correct Answer: A — 10 m/s

Q. A 2 kg ball is dropped from a height of 10 m. What is its speed just before it hits the ground? (2022)

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

Solution

Using energy conservation, v = sqrt(2gh) = sqrt(2 * 9.8 m/s² * 10 m) = 14 m/s.

Correct Answer: A — 14 m/s

Q. A 2 kg ball is dropped from a height of 20 m. What is its speed just before it hits the ground? (g = 10 m/s²) (2023)

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

Solution

Using energy conservation, Potential Energy = Kinetic Energy at the ground: mgh = 0.5 mv^2; v = sqrt(2gh) = sqrt(2 * 10 m/s² * 20 m) = 20 m/s

Correct Answer: B — 20 m/s

Q. A 2 kg ball is thrown vertically upwards with a speed of 10 m/s. What is the maximum height it reaches?

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

Solution

Using energy conservation: mgh = 0.5mv², h = v²/(2g) = (10 m/s)²/(2 × 9.8 m/s²) = 5.1 m.

Correct Answer: A — 5 m

Q. A 2 kg ball is thrown vertically upwards with a speed of 15 m/s. What is the maximum height it reaches?

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

Solution

Using conservation of energy, KE at the bottom = PE at the top. 0.5mv^2 = mgh. Solving gives h = v^2/(2g) = (15^2)/(2*9.8) = 11.47 m.

Correct Answer: B — 10 m

Q. A 2 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)

A. 20.4 m
B. 30.4 m
C. 40.4 m
D. 50.4 m

Solution

Using conservation of energy, KE at the bottom = PE at the maximum height. 0.5mv² = mgh. Solving gives h = v²/(2g) = (20)²/(2 * 9.8) = 20.4 m.

Correct Answer: B — 30.4 m

Q. A 2 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches?

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

Solution

Using conservation of energy, KE at launch = PE at max height. 0.5mv² = mgh. Solving gives h = v²/(2g) = (20)²/(2*9.8) = 20.41 m.

Correct Answer: B — 20 m

Q. A 2 kg ball is thrown vertically upwards with a velocity of 20 m/s. What is the maximum height reached? (2020)

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

Solution

Using v² = u² + 2as, 0 = (20 m/s)² + 2(-9.8 m/s²)(h). Solving gives h = 20.4 m.

Correct Answer: A — 20 m

Q. A 2 kg block is sliding down a frictionless incline of 30 degrees. What is the acceleration of the block?

A. 4.9 m/s²
B. 9.8 m/s²
C. 3.9 m/s²
D. 1.96 m/s²

Solution

The acceleration a = g sin(θ) = 9.8 m/s² * sin(30°) = 9.8 m/s² * 0.5 = 4.9 m/s².

Correct Answer: A — 4.9 m/s²

Q. A 2 kg block is sliding down a frictionless incline of height 5 m. What is its speed at the bottom?

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

Solution

Using conservation of energy, potential energy at the top = kinetic energy at the bottom: mgh = 0.5mv². Solving gives v = √(2gh) = √(2 * 9.8 * 5) = 10 m/s.

Correct Answer: A — 10 m/s

Q. A 2 kg block is sliding on a frictionless surface. If a force of 6 N is applied, what is its acceleration?

A. 2 m/s²
B. 3 m/s²
C. 4 m/s²
D. 6 m/s²

Solution

Using F = ma, acceleration a = F/m = 6 N / 2 kg = 3 m/s².

Correct Answer: B — 3 m/s²

Q. A 2 kg block of ice at 0°C is converted to water at 0°C. How much heat is absorbed if the latent heat of fusion of ice is 334,000 J/kg?

A. 668,000 J
B. 334,000 J
C. 167,000 J
D. 500,000 J

Solution

Heat absorbed = mass * latent heat = 2 kg * 334,000 J/kg = 668,000 J.

Correct Answer: A — 668,000 J

Q. A 2 kg block of metal at 100°C is placed in 1 kg of water at 20°C. What is the final temperature of the system? (Specific heat of water = 4.2 J/g°C, specific heat of metal = 0.9 J/g°C) (2021)

A. 30°C
B. 40°C
C. 50°C
D. 60°C

Solution

Using the heat transfer equation, the final temperature can be calculated to be 50°C.

Correct Answer: C — 50°C

Q. A 2 kg block slides down a frictionless incline of height 3 m. What is its speed at the bottom? (2023)

A. 6 m/s
B. 3 m/s
C. 4 m/s
D. 5 m/s

Solution

Using conservation of energy, Potential Energy at top = Kinetic Energy at bottom: mgh = 1/2 mv^2; v = sqrt(2gh) = sqrt(2 * 9.8 m/s² * 3 m) = 7.75 m/s (approx. 6 m/s)

Correct Answer: A — 6 m/s

Q. A 2 kg mass is dropped from a height. What is the force acting on it just before it hits the ground? (2022)

A. 2 N
B. 10 N
C. 20 N
D. 0 N

Solution

The force acting on it is its weight, F = mg = 2 kg × 10 m/s² = 20 N.

Correct Answer: B — 10 N

Q. A 2 kg object is dropped from a height of 10 m. What is its potential energy at the top?

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

Solution

Potential Energy = mass × g × height = 2 kg × 9.8 m/s² × 10 m = 196 J.

Correct Answer: B — 40 J

Q. A 2 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, PE = KE: mgh = 0.5mv². Solving gives v = √(2gh) = √(2 × 9.8 m/s² × 10 m) = 14 m/s.

Correct Answer: A — 14 m/s

Q. A 2 kg object is dropped from a height of 10 m. What is the speed of the object just before it hits the ground? (g = 9.8 m/s²)

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

Solution

Using energy conservation, mgh = 0.5mv²; v = sqrt(2gh) = sqrt(2 × 9.8 m/s² × 10 m) = 14 m/s.

Correct Answer: B — 14 m/s

Q. A 2 kg object is dropped from a height of 15 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)

A. 17.15 m/s
B. 12.25 m/s
C. 14.14 m/s
D. 10.0 m/s

Solution

Using conservation of energy, potential energy at height = kinetic energy just before hitting the ground. mgh = 0.5mv². Solving gives v = sqrt(2gh) = sqrt(2 * 9.8 * 15) = 17.15 m/s.

Correct Answer: A — 17.15 m/s

Q. A 2 kg object is dropped from a height of 20 m. What is its potential energy at the top? (g = 9.8 m/s²)

A. 39.2 J
B. 196 J
C. 78.4 J
D. 98.0 J

Solution

Potential Energy (PE) = m × g × h = 2 kg × 9.8 m/s² × 20 m = 392 J.

Correct Answer: B — 196 J

Q. A 2 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 = 0.5 mv². v = √(2gh) = √(2 × 9.8 m/s² × 20 m) = 19.8 m/s.

Correct Answer: B — 19.8 m/s

Q. A 2 kg object is dropped from a height of 20 m. What is its speed just before it hits the ground? (Assume g = 10 m/s²)

A. 10 m/s
B. 14 m/s
C. 20 m/s
D. 40 m/s

Solution

Using conservation of energy, Potential Energy at height = Kinetic Energy just before hitting ground. mgh = 0.5 mv². v = √(2gh) = √(2 × 10 m/s² × 20 m) = 20 m/s.

Correct Answer: C — 20 m/s

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