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 circle has a radius of 7 units. What is the length of a chord that is 4 units away from the center of the circle?

A. 8
B. 10
C. 12
D. 14

Solution

Using the formula: chord length = 2 * √(radius² - distance from center²) = 2 * √(7² - 4²) = 2 * √(49 - 16) = 2 * √33 ≈ 11.55.

Correct Answer: B — 10

Q. A circle has an area of 154 cm². What is the radius? (2019)

A. 7 cm
B. 14 cm
C. 21 cm
D. 28 cm

Solution

Area = πr². Therefore, r = √(Area/π) = √(154/3.14) ≈ 7 cm.

Correct Answer: B — 14 cm

Q. A circle has an area of 154 cm². What is the radius? (2019) 2019

A. 7 cm
B. 14 cm
C. 21 cm
D. 28 cm

Solution

Area = πr². Therefore, r = √(Area/π) = √(154/3.14) ≈ 7 cm.

Correct Answer: B — 14 cm

Q. A circle has an area of 78.5 cm². What is its radius? (Use π = 3.14) (2019)

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

Solution

Area = πr². Therefore, r² = Area / π = 78.5 cm² / 3.14 = 25. r = √25 = 5 cm.

Correct Answer: B — 7 cm

Q. A circle has an area of 78.5 cm². What is the radius? (Use π = 3.14) (2022)

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

Solution

Area = πr². Therefore, r² = Area / π = 78.5 cm² / 3.14 = 25. r = √25 = 5 cm.

Correct Answer: B — 7 cm

Q. A circle is defined by the equation x² + y² - 10x + 6y + 25 = 0. What is the radius of the circle?

A. 5
B. 6
C. 7
D. 4

Solution

Rearranging gives (x - 5)² + (y + 3)² = 9, so the radius is √9 = 3.

Correct Answer: A — 5

Q. A circle is inscribed in a square of side 10 cm. What is the area of the circle? (2019)

A. 78.5 cm²
B. 100 cm²
C. 50 cm²
D. 25 cm²

Solution

Radius = 10/2 = 5 cm; Area = πr² = π * 5² = 78.5 cm².

Correct Answer: A — 78.5 cm²

Q. A circle is inscribed in a square of side 10 cm. What is the area of the circle? (Use π = 3.14) (2020)

A. 78.5 cm²
B. 50 cm²
C. 100 cm²
D. 25 cm²

Solution

Radius of the circle = side/2 = 10 cm / 2 = 5 cm. Area = πr² = 3.14 * 5² = 78.5 cm².

Correct Answer: A — 78.5 cm²

Q. A circle is inscribed in a square of side 8 cm. What is the area of the circle? (2022)

A. 50.24 cm²
B. 64 cm²
C. 25.12 cm²
D. 32 cm²

Solution

Radius = 8/2 = 4 cm; Area = πr² = π * 4² = 50.24 cm².

Correct Answer: A — 50.24 cm²

Q. A circle is inscribed in a triangle with sides 7, 8, and 9. What is the radius of the inscribed circle?

A. 3
B. 4
C. 5
D. 6

Solution

Using the formula for the radius of the incircle r = A/s, where A is the area and s is the semi-perimeter.

Correct Answer: B — 4

Q. A circle is inscribed in a triangle with sides 7, 8, and 9. What is the radius of the circle?

A. 4
B. 5
C. 6
D. 3

Solution

Using the formula for the radius of the incircle r = A/s, where A is the area and s is the semi-perimeter.

Correct Answer: B — 5

Q. A circle is tangent to the x-axis at the point (4, 0). What is the equation of the circle if its radius is 3?

A. (x - 4)² + (y - 3)² = 9
B. (x - 4)² + (y + 3)² = 9
C. (x + 4)² + (y - 3)² = 9
D. (x + 4)² + (y + 3)² = 9

Solution

The center of the circle is (4, 3) since it is 3 units above the tangent point (4, 0).

Correct Answer: B — (x - 4)² + (y + 3)² = 9

Q. A circle passes through the points (1, 2), (3, 4), and (5, 6). What is the radius of the circle?

A. 2
B. 3
C. 4
D. 5

Solution

Using the distance formula, the radius can be calculated from the center found using the circumcircle method.

Correct Answer: B — 3

Q. A circuit consists of a 9V battery and two resistors of 3Ω and 6Ω in series. What is the current flowing through the circuit? (2021)

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

Solution

Total resistance R = 3Ω + 6Ω = 9Ω. Current I = V/R = 9V / 9Ω = 1A.

Correct Answer: B — 2A

Q. A circuit contains a 10Ω resistor and a 5Ω resistor in series with a 15V battery. What is the total current in the circuit?

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

Solution

Total resistance R = 10Ω + 5Ω = 15Ω. Current I = V/R = 15V/15Ω = 1A.

Correct Answer: B — 2A

Q. A circuit contains a 10Ω resistor and a 5Ω resistor in series. What is the total resistance?

A. 5Ω
B. 10Ω
C. 15Ω
D. 50Ω

Solution

In series, total resistance R = R1 + R2 = 10Ω + 5Ω = 15Ω.

Correct Answer: C — 15Ω

Q. A circuit contains a 12 V battery and two resistors of 4 ohms and 8 ohms in series. What is the current flowing through the circuit?

A. 0.5 A
B. 1 A
C. 1.5 A
D. 2 A

Solution

Total resistance R = 4 + 8 = 12 ohms. Current I = V/R = 12 V / 12 ohms = 1 A.

Correct Answer: B — 1 A

Q. A circuit contains a 12V battery and two resistors of 4 ohms and 8 ohms in series. What is the total current in the circuit?

A. 1 A
B. 0.5 A
C. 2 A
D. 3 A

Solution

Total resistance R = R1 + R2 = 4 + 8 = 12 ohms. Current I = V/R = 12V / 12 ohms = 1 A.

Correct Answer: B — 0.5 A

Q. A circuit contains a 12V battery and two resistors of 4Ω and 6Ω in parallel. What is the total current supplied by the battery? (2019)

A. 1.2 A
B. 2 A
C. 3 A
D. 4 A

Solution

First, find the equivalent resistance R_eq = 1/(1/4 + 1/6) = 2.4Ω. Then, I = V/R_eq = 12V/2.4Ω = 5 A.

Correct Answer: C — 3 A

Q. A circuit contains a 9V battery and two resistors in series: 3Ω and 6Ω. What is the voltage across the 6Ω resistor?

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

Solution

The total resistance R_total = 3Ω + 6Ω = 9Ω. The current I = V/R_total = 9V / 9Ω = 1A. Voltage across 6Ω, V = I * R = 1A * 6Ω = 6V.

Correct Answer: A — 6V

Q. A circuit contains a 9V battery and two resistors of 3 ohms and 6 ohms in series. What is the voltage drop across the 6 ohm resistor?

A. 3 V
B. 6 V
C. 9 V
D. 4.5 V

Solution

Total resistance R_total = 3 + 6 = 9 ohms. Current I = V/R_total = 9V / 9Ω = 1 A. Voltage drop across 6 ohm resistor = I * R = 1 A * 6Ω = 6 V.

Correct Answer: B — 6 V

Q. A circuit contains a 9V battery and two resistors of 3 ohms and 6 ohms in series. What is the voltage across the 6 ohm resistor?

A. 6V
B. 3V
C. 9V
D. 4.5V

Solution

Total resistance R_total = 3 + 6 = 9 ohms. Current I = V/R_total = 9V / 9Ω = 1 A. Voltage across 6 ohm resistor = I * R = 1 A * 6Ω = 6V.

Correct Answer: A — 6V

Q. A circuit contains a 9V battery and two resistors of 3Ω and 6Ω in series. What is the voltage across the 6Ω resistor?

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

Solution

The total resistance R_total = 3Ω + 6Ω = 9Ω. The current I = V/R_total = 9V / 9Ω = 1A. Voltage across 6Ω, V = I * R = 1A * 6Ω = 6V.

Correct Answer: A — 6V

Q. A circuit has a 15V battery and a 5Ω resistor. What is the voltage drop across the resistor? (2021)

A. 3V
B. 5V
C. 15V
D. 10V

Solution

The voltage drop across the resistor is equal to the battery voltage, which is 15V.

Correct Answer: C — 15V

Q. A circuit has a 15V battery and a total resistance of 5Ω. What is the total current flowing through the circuit? (2023)

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

Solution

Using Ohm's law, I = V/R = 15V/5Ω = 3A.

Correct Answer: C — 3A

Q. A circuit has a 24V battery and a total resistance of 12Ω. What is the current flowing through the circuit? (2020)

A. 2A
B. 4A
C. 6A
D. 8A

Solution

Using Ohm's law, I = V/R = 24V / 12Ω = 2A.

Correct Answer: B — 4A

Q. A circuit has a 24V battery and a total resistance of 6Ω. What is the current flowing through the circuit? (2020)

A. 4A
B. 6A
C. 8A
D. 2A

Solution

Using Ohm's law, I = V/R = 24V / 6Ω = 4A.

Correct Answer: A — 4A

Q. A circuit has a 5Ω resistor and a 10Ω resistor in parallel. What is the total current if the voltage across them is 20V? (2021)

A. 2A
B. 4A
C. 6A
D. 8A

Solution

First, find the equivalent resistance: 1/R_eq = 1/5 + 1/10 = 3/10, so R_eq = 10/3Ω. Then, I = V/R_eq = 20V / (10/3)Ω = 6A.

Correct Answer: D — 8A

Q. A circuit has a 9V battery and a 3Ω resistor. What is the voltage drop across the resistor? (2022)

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

Solution

The voltage drop across the resistor is equal to the battery voltage in a simple circuit: 9V.

Correct Answer: C — 9V

Q. A circuit has a 9V battery and two resistors of 3Ω and 6Ω in series. What is the current flowing through the circuit? (2022)

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

Solution

Total resistance R = 3Ω + 6Ω = 9Ω. Current I = V/R = 9V / 9Ω = 1A.

Correct Answer: B — 2A

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