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. If the quadratic equation x^2 - 4x + k = 0 has equal roots, what is the value of k?

A. 0
B. 4
C. 8
D. 16

Solution

For equal roots, the discriminant must be zero: (-4)^2 - 4*1*k = 0, leading to k = 4.

Correct Answer: B — 4

Q. If the quadratic equation x^2 - 8x + 15 = 0 is solved, what are the roots? (2022)

A. 3 and 5
B. 2 and 6
C. 1 and 7
D. 4 and 4

Solution

Factoring gives (x - 3)(x - 5) = 0, hence the roots are 3 and 5.

Correct Answer: A — 3 and 5

Q. If the quadratic equation x^2 - kx + 9 = 0 has equal roots, what is the value of k?

A. 6
B. 9
C. 3
D. 0

Solution

For equal roots, the discriminant must be zero: k^2 - 36 = 0, hence k = 6.

Correct Answer: A — 6

Q. If the quadratic equation x² + 5x + k = 0 has roots -2 and -3, find k. (2020)

A. 6
B. 5
C. 7
D. 8

Solution

The product of the roots gives k = (-2)(-3) = 6.

Correct Answer: A — 6

Q. If the quadratic equation x² - 4x + k = 0 has one real root, what must be the value of k?

A. 4
B. 0
C. -4
D. 8

Solution

For the equation to have one real root, the discriminant must be zero. Thus, k must equal 4.

Correct Answer: A — 4

Q. If the quadratic equation x² - 5x + 6 = 0 is factored, which of the following pairs of numbers represents the roots?

A. 2 and 3
B. 1 and 6
C. 0 and 6
D. 3 and 2

Solution

Factoring the equation gives (x - 2)(x - 3) = 0, thus the roots are 2 and 3.

Correct Answer: A — 2 and 3

Q. If the radius of a charged sphere is halved while keeping the charge constant, what happens to the electric field at the surface?

A. It remains the same
B. It doubles
C. It halves
D. It quadruples

Solution

The electric field at the surface of a sphere is given by E = Q/(4πε₀R²). If R is halved, E increases by a factor of 4.

Correct Answer: B — It doubles

Q. If the radius of a circle is 4 cm, what is the length of a chord that is 3 cm from the center? (2014)

A. 5 cm
B. 6 cm
C. 7 cm
D. 8 cm

Solution

Using Pythagoras theorem: chord length = 2√(r² - d²) = 2√(4² - 3²) = 2√(16 - 9) = 2√7 ≈ 5.29 cm.

Correct Answer: A — 5 cm

Q. If the radius of a circle is 7 cm, what is its circumference? (2023)

A. 14π cm
B. 21π cm
C. 28π cm
D. 35π cm

Solution

Circumference = 2πr = 2π × 7 cm = 14π cm.

Correct Answer: B — 21π cm

Q. If the radius of a circle is 7 cm, what is its circumference? (Use π = 22/7) (2014)

A. 44 cm
B. 22 cm
C. 14 cm
D. 28 cm

Solution

Circumference = 2πr = 2 × (22/7) × 7 = 44 cm.

Correct Answer: A — 44 cm

Q. If the radius of a circle is 7 cm, what is its circumference? (Use π ≈ 3.14)

A. 43.96 cm
B. 21.98 cm
C. 31.4 cm
D. 14 cm

Solution

Circumference = 2πr = 2 × 3.14 × 7 cm = 43.96 cm.

Correct Answer: A — 43.96 cm

Q. If the radius of a circle is 7 cm, what is the length of a tangent drawn from a point 10 cm away from the center?

A. 5
B. 7
C. 10
D. 12

Solution

Using the tangent length formula: length = √(distance from center² - radius²) = √(10² - 7²) = √(100 - 49) = √51, which is approximately 7.14.

Correct Answer: A — 5

Q. If the radius of a circle is 7 units, what is its circumference?

A. 14π units
B. 21π units
C. 28π units
D. 7π units

Solution

Circumference = 2 * π * radius = 2 * π * 7 = 14π units.

Correct Answer: A — 14π units

Q. If the radius of a circle is 7 units, what is the length of a tangent drawn from a point 10 units away from the center?

A. 5
B. 7
C. 10
D. 12

Solution

Using the tangent length formula: length = √(distance from center² - radius²) = √(10² - 7²) = √(100 - 49) = √51 ≈ 7.14.

Correct Answer: A — 5

Q. If the radius of a circle is decreased by 2 cm, how does the area change? (Original radius is 10 cm) (2021)

A. Decreases by 12.56 cm²
B. Decreases by 25.12 cm²
C. Decreases by 31.4 cm²
D. Decreases by 50.24 cm²

Solution

Original area = π(10)² = 314 cm²; New area = π(8)² = 201.06 cm². Change = 314 - 201.06 = 112.94 cm².

Correct Answer: B — Decreases by 25.12 cm²

Q. If the radius of a circle is decreased by 2 cm, how does the area change? (Use π = 3.14) (2020)

A. Decreases by 12.56 cm²
B. Decreases by 25.12 cm²
C. Increases by 12.56 cm²
D. Remains the same

Solution

Area change = π[(r-2)² - r²] = π[-4r + 4] = 3.14 * (-4r + 4).

Correct Answer: B — Decreases by 25.12 cm²

Q. If the radius of a circle is doubled, how does the area change? (2021)

A. It doubles
B. It triples
C. It quadruples
D. It remains the same

Solution

Area = πr²; if r is doubled, area = π(2r)² = 4πr², so it quadruples.

Correct Answer: C — It quadruples

Q. If the radius of a circle is doubled, what happens to its area? (2020)

A. It remains the same
B. It doubles
C. It triples
D. It quadruples

Solution

The area of a circle is given by A = πr². If the radius is doubled (r becomes 2r), the new area is A' = π(2r)² = 4πr², which is four times the original area.

Correct Answer: D — It quadruples

Q. If the radius of a circle is halved, by what factor does the circumference decrease? (2020)

A. 1/2
B. 1/4
C. 1/3
D. 1/6

Solution

Circumference = 2πr; If r is halved, new circumference = πr; Factor = 1/2.

Correct Answer: A — 1/2

Q. If the radius of a circle is halved, how does the circumference change? (2021)

A. Halved
B. Remains the same
C. Doubled
D. Tripled

Solution

Circumference = 2πr. If radius is halved, new circumference = 2π(r/2) = πr, which is halved.

Correct Answer: A — Halved

Q. If the radius of a circle is halved, how does the circumference change? (2022) 2022

A. Halved
B. Remains the same
C. Doubled
D. Quadrupled

Solution

Circumference = 2πr. If radius is halved, new circumference = 2π(r/2) = πr, which is halved.

Correct Answer: A — Halved

Q. If the radius of a circle is increased by 50%, by what percentage does the area of the circle increase?

A. 25%
B. 50%
C. 75%
D. 100%

Solution

The area of a circle is given by A = πr². If the radius is increased by 50%, the new radius is 1.5r. The new area is A' = π(1.5r)² = 2.25πr². The increase in area is (2.25 - 1) = 1.25 times the original area, which is a 125% increase.

Correct Answer: C — 75%

Q. If the radius of a circle is increased by 50%, what happens to the area of the circle?

A. It increases by 50%.
B. It increases by 100%.
C. It increases by 125%.
D. It increases by 150%.

Solution

If the radius increases by 50%, the new radius is 1.5r, and the area becomes (1.5r)² = 2.25r², which is a 125% increase.

Correct Answer: C — It increases by 125%.

Q. If the radius of a circle is tripled, by what factor does the area increase? (2021)

A. 3
B. 6
C. 9
D. 12

Solution

Area increases by a factor of (3r)²/r² = 9.

Correct Answer: C — 9

Q. If the radius of a circle is tripled, how does the area change? (2019)

A. Increases by 3 times
B. Increases by 6 times
C. Increases by 9 times
D. Remains the same

Solution

Area = πr². If radius is tripled, new area = π(3r)² = 9πr², which is 9 times the original area.

Correct Answer: C — Increases by 9 times

Q. If the radius of a circular loop carrying current is doubled, how does the magnetic field at the center change?

A. It doubles
B. It halves
C. It remains the same
D. It quadruples

Solution

The magnetic field at the center of a circular loop is inversely proportional to the radius; thus, doubling the radius halves the magnetic field.

Correct Answer: B — It halves

Q. If the radius of a circular loop carrying current is doubled, what happens to the magnetic field at the center of the loop?

A. It doubles
B. It halves
C. It remains the same
D. It quadruples

Solution

The magnetic field at the center of a circular loop is given by B = (μ₀I)/(2r). If the radius is doubled, the magnetic field strength is halved.

Correct Answer: B — It halves

Q. If the radius of a circular loop carrying current is halved, how does the magnetic field at the center change?

A. Remains the same
B. Doubles
C. Halves
D. Quadruples

Solution

The magnetic field at the center is inversely proportional to the radius, so it quadruples.

Correct Answer: D — Quadruples

Q. If the radius of a disc is doubled while keeping its mass constant, how does its moment of inertia change?

A. It remains the same
B. It doubles
C. It quadruples
D. It halves

Solution

The moment of inertia of a disc is I = 1/2 MR^2. If R is doubled, I becomes 1/2 M(2R)^2 = 2MR^2, which is quadrupled.

Correct Answer: C — It quadruples

Q. If the radius of a disk is doubled while keeping its mass constant, how does its moment of inertia change?

A. Increases by a factor of 2
B. Increases by a factor of 4
C. Remains the same
D. Decreases by a factor of 4

Solution

The moment of inertia of a disk is I = 1/2 MR^2. If R is doubled, I becomes 1/2 M(2R)^2 = 2MR^2, which is 4 times the original.

Correct Answer: B — Increases by a factor of 4

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