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 discount of 10% is given on a product that costs $500. If the product is then sold for $450, what is the actual discount given?

A. $50
B. $45
C. $40
D. $55

Solution

Actual Discount = Marked Price - Selling Price = 500 - 450 = $50.

Correct Answer: A — $50

Q. A discount of 10% is given on a product that costs $500. If the product is then sold for $450, what is the profit or loss percentage?

A. 5% Profit
B. 10% Loss
C. 10% Profit
D. 5% Loss

Solution

Discounted Price = $500 - ($500 * 0.10) = $450. Selling Price = $450, Cost Price = $500. Loss = $500 - $450 = $50. Percentage loss = ($50/$500) * 100 = 10%.

Correct Answer: D — 5% Loss

Q. A discount of 10% is offered on a product priced at $150. What is the sale price after the discount?

A. $135
B. $140
C. $145
D. $130

Solution

Sale Price = Original Price - (10% of Original Price) = 150 - (0.1 * 150) = 150 - 15 = $135.

Correct Answer: A — $135

Q. A discount of 10% on a book priced at $50 results in a selling price of what?

A. $40
B. $45
C. $50
D. $55

Solution

Selling Price = Marked Price - Discount = 50 - (10% of 50) = 50 - 5 = $45.

Correct Answer: B — $45

Q. A discount of 10% on a marked price results in a selling price of $90. What is the marked price? (2023)

A. $100
B. $110
C. $120
D. $130

Solution

Let the marked price be x. Selling Price = Marked Price - Discount = x - 0.1x = 0.9x. Thus, 0.9x = 90, so x = 100.

Correct Answer: A — $100

Q. A discount of 10% on a product priced at $250 results in a selling price of what?

A. $220
B. $225
C. $230
D. $240

Solution

Discount = 10% of 250 = 0.1 * 250 = 25. Selling Price = 250 - 25 = 225.

Correct Answer: D — $240

Q. A discount of 15% is given on a product priced at $200. What is the selling price after the discount?

A. $170
B. $180
C. $185
D. $190

Solution

Discount = 15% of 200 = 0.15 * 200 = 30. Selling Price = Cost Price - Discount = 200 - 30 = 170.

Correct Answer: A — $170

Q. A discount of 15% is given on a product that costs $120. What is the final price after the discount?

A. $102
B. $108
C. $110
D. $100

Solution

Discount = 15% of $120 = $120 * 0.15 = $18. Final Price = $120 - $18 = $102.

Correct Answer: B — $108

Q. A discount of 15% is given on a product that costs $80. What is the final price after the discount?

A. $68
B. $70
C. $75
D. $72

Solution

Final Price = Cost Price - (Discount Percentage * Cost Price) = $80 - (0.15 * $80) = $80 - $12 = $68.

Correct Answer: A — $68

Q. A discount of 15% on a marked price results in a selling price of $255. What was the marked price?

A. $300
B. $280
C. $270
D. $250

Solution

Let the marked price be x. After a 15% discount, the selling price is 0.85x. Setting this equal to $255 gives 0.85x = $255, so x = $255 / 0.85 = $300.

Correct Answer: A — $300

Q. A discount of 15% on a product results in a selling price of $255. What was the marked price?

A. $300
B. $280
C. $250
D. $270

Solution

Let the marked price be x. Selling Price = x - (0.15 * x) = 0.85x. Setting this equal to $255 gives 0.85x = $255, so x = $255 / 0.85 = $300.

Correct Answer: A — $300

Q. A discount of 25% on a marked price results in a selling price of $75. What was the marked price?

A. $90
B. $100
C. $80
D. $70

Solution

Let the marked price be x. Selling Price = x - 25% of x = 0.75x. Thus, 0.75x = 75, so x = 100.

Correct Answer: A — $90

Q. A discount of 25% on a product brings its price down to $75. What was the original price?

A. $100
B. $90
C. $80
D. $70

Solution

Let the original price be x. After a 25% discount, Selling Price = x - 0.25x = 0.75x. 0.75x = 75, x = 75/0.75 = 100.

Correct Answer: A — $100

Q. A discount of 25% on a product results in a selling price of $150. What was the original price? (2023)

A. $175
B. $200
C. $225
D. $250

Solution

Let the original price be x. Then, Selling Price = x - (25% of x) = 150 => 0.75x = 150 => x = 200.

Correct Answer: B — $200

Q. A disk and a ring of the same mass and radius are released from rest at the same height. Which one reaches the ground first?

A. Disk
B. Ring
C. Both reach at the same time
D. Depends on the surface

Solution

The disk has a lower moment of inertia compared to the ring, thus it reaches the ground first.

Correct Answer: A — Disk

Q. A disk and a ring of the same mass and radius are rolling down an incline. Which one will have a greater translational speed at the bottom?

A. Disk
B. Ring
C. Both have the same speed
D. Cannot be determined

Solution

The disk has a lower moment of inertia than the ring, allowing it to convert more potential energy into translational kinetic energy.

Correct Answer: A — Disk

Q. A disk and a ring of the same mass and radius are rolling down an incline. Which will reach the bottom first?

A. Disk
B. Ring
C. Both will reach at the same time
D. Depends on the angle of incline

Solution

The disk has a smaller moment of inertia compared to the ring, hence it will reach the bottom first.

Correct Answer: A — Disk

Q. A disk and a ring of the same mass and radius are rolling down the same incline. Which one has a greater acceleration? (2019)

A. Disk
B. Ring
C. Both have the same acceleration
D. It depends on the mass

Solution

The disk has a smaller moment of inertia than the ring, resulting in greater acceleration down the incline.

Correct Answer: A — Disk

Q. A disk and a ring of the same mass and radius are rolling without slipping down an incline. Which one will have a greater translational speed at the bottom?

A. Disk
B. Ring
C. Both have the same speed
D. Depends on the incline

Solution

The disk has a lower moment of inertia than the ring, allowing it to convert more potential energy into translational kinetic energy.

Correct Answer: A — Disk

Q. A disk and a ring of the same mass and radius are rolling without slipping. Which one will reach the bottom of an incline first?

A. Disk
B. Ring
C. Both will reach at the same time
D. Depends on the angle of incline

Solution

The disk will reach the bottom first because it has a smaller moment of inertia compared to the ring.

Correct Answer: A — Disk

Q. A disk is rotating with an angular velocity of 10 rad/s. If it experiences a constant angular acceleration of 2 rad/s², what will be its angular velocity after 5 seconds?

A. 20 rad/s
B. 10 rad/s
C. 30 rad/s
D. 15 rad/s

Solution

Using the formula ω = ω₀ + αt, we have ω = 10 + 2*5 = 20 rad/s.

Correct Answer: A — 20 rad/s

Q. A disk of radius R and mass M is rotating about its axis with an angular velocity ω. What is the angular momentum of the disk?

A. (1/2)MR^2ω
B. MR^2ω
C. (1/4)MR^2ω
D. (3/2)MR^2ω

Solution

Angular momentum L = Iω = (1/2)MR^2ω, where I = (1/2)MR^2 for a disk.

Correct Answer: A — (1/2)MR^2ω

Q. A disk of radius R and mass M is rotating about its axis with an angular velocity ω. What is its angular momentum?

A. (1/2)MR^2ω
B. MR^2ω
C. Mω
D. (1/4)MR^2ω

Solution

Angular momentum L = Iω, where I = (1/2)MR^2 for a disk.

Correct Answer: A — (1/2)MR^2ω

Q. A disk of radius R and mass M is rotating about its axis with an angular velocity ω. What is its kinetic energy?

A. (1/2)Mω^2R^2
B. (1/2)Iω^2
C. (1/2)Mω^2
D. Mω^2R

Solution

The moment of inertia I of a disk about its axis is (1/2)MR^2. Therefore, the kinetic energy K.E. = (1/2)Iω^2 = (1/2)(1/2)MR^2ω^2 = (1/4)MR^2ω^2.

Correct Answer: B — (1/2)Iω^2

Q. A disk rolls down a slope of height h. What fraction of its total energy is translational at the bottom?

A. 1/3
B. 1/2
C. 2/3
D. 1

Solution

At the bottom, the total energy is converted into translational and rotational energy. The translational energy is 2/3 of the total energy.

Correct Answer: C — 2/3

Q. A disk rolls down a slope of height h. What is the ratio of translational to rotational kinetic energy at the bottom?

A. 1:1
B. 2:1
C. 3:1
D. 1:2

Solution

At the bottom, the total kinetic energy is split equally between translational and rotational for a disk, hence the ratio is 1:1.

Correct Answer: A — 1:1

Q. A disk rolls down an incline. If the height of the incline is h, what is the speed of the disk at the bottom assuming no energy losses?

A. √(gh)
B. √(2gh)
C. √(3gh)
D. √(4gh)

Solution

Using conservation of energy, potential energy at height h converts to kinetic energy at the bottom. The speed is √(2gh).

Correct Answer: B — √(2gh)

Q. A disk rolls without slipping on a horizontal surface. If its radius is R and it rolls with a linear speed v, what is the angular speed of the disk?

A. v/R
B. R/v
C. vR
D. v^2/R

Solution

The relationship between linear speed and angular speed for rolling without slipping is given by ω = v/R.

Correct Answer: A — v/R

Q. A disk rotates about its axis with an angular velocity of ω. If its radius is doubled, what will be the new angular velocity to maintain the same linear velocity at the edge?

A. ω/2
B. ω
C. 2ω
D. 4ω

Solution

The linear velocity v = rω. If the radius is doubled, to maintain the same v, the angular velocity must remain ω.

Correct Answer: B — ω

Q. A disk rotates about its axis with an angular velocity of ω. If its radius is doubled, what will be the new angular momentum?

A. 2Iω
B. 4Iω
C. Iω
D. I(2ω)

Solution

Angular momentum L = Iω, and if the radius is doubled, the moment of inertia I becomes 4I, thus L = 4Iω.

Correct Answer: B — 4Iω

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