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!
Q. A thermometer reads 25.0 °C with an uncertainty of ±0.5 °C. What is the range of possible temperatures?
A.
24.5 °C to 25.5 °C
B.
25.0 °C to 26.0 °C
C.
24.0 °C to 25.0 °C
D.
25.0 °C to 25.5 °C
Show solution
Solution
Range = 25.0 ± 0.5 °C = 24.5 °C to 25.5 °C.
Correct Answer:
A
— 24.5 °C to 25.5 °C
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Q. A thermometer reads 25.0 °C with an uncertainty of ±0.5 °C. What is the range of possible true values?
A.
24.5 °C to 25.5 °C
B.
24.0 °C to 25.0 °C
C.
25.0 °C to 26.0 °C
D.
25.5 °C to 26.5 °C
Show solution
Solution
The range of possible true values is from 25.0 - 0.5 to 25.0 + 0.5, which is 24.5 °C to 25.5 °C.
Correct Answer:
A
— 24.5 °C to 25.5 °C
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Q. A thin film of oil (n = 1.5) is on water (n = 1.33). What is the condition for destructive interference for light of wavelength 600 nm in air? (2022)
A.
2t = (m + 1/2)λ
B.
2t = mλ
C.
2t = (m + 1)λ
D.
2t = (m - 1/2)λ
Show solution
Solution
For destructive interference in a thin film with a higher refractive index below, the condition is 2t = (m + 1/2)λ/n, where n is the refractive index of the film.
Correct Answer:
A
— 2t = (m + 1/2)λ
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Q. A thin lens has a focal length of 20 cm. What is the power of the lens?
A.
+2.5 D
B.
+5 D
C.
+10 D
D.
+15 D
Show solution
Solution
Power (P) is given by P = 1/f (in meters). Thus, P = 1/0.2 = +5 D.
Correct Answer:
B
— +5 D
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Q. A thin rod of length L and mass M is rotated about an axis perpendicular to its length and passing through one end. What is its moment of inertia?
A.
1/3 ML^2
B.
1/12 ML^2
C.
1/2 ML^2
D.
ML^2
Show solution
Solution
The moment of inertia of a thin rod about an end is I = 1/3 ML^2.
Correct Answer:
A
— 1/3 ML^2
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Q. A thin rod of length L and mass M is rotated about an axis perpendicular to its length through one end. What is its moment of inertia?
A.
1/3 ML^2
B.
1/12 ML^2
C.
1/2 ML^2
D.
ML^2
Show solution
Solution
The moment of inertia of a thin rod about an end is I = 1/3 ML^2.
Correct Answer:
A
— 1/3 ML^2
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Q. A ticket price is increased by 15%. If the original price is $120, what is the new price?
A.
$138
B.
$140
C.
$135
D.
$130
Show solution
Solution
New Price = Original Price + (15% of Original Price) = 120 + (0.15 * 120) = 120 + 18 = $138.
Correct Answer:
A
— $138
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Q. A torque of 10 Nm is applied to a wheel with a moment of inertia of 2 kg·m². What is the angular acceleration of the wheel?
A.
5 rad/s²
B.
10 rad/s²
C.
2 rad/s²
D.
20 rad/s²
Show solution
Solution
Using τ = Iα, we have α = τ/I = 10 Nm / 2 kg·m² = 5 rad/s².
Correct Answer:
A
— 5 rad/s²
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Q. A torque of 10 Nm is applied to a wheel with a moment of inertia of 2 kg·m². What is the angular acceleration?
A.
5 rad/s²
B.
10 rad/s²
C.
2 rad/s²
D.
20 rad/s²
Show solution
Solution
Using τ = Iα, we have α = τ/I = 10 Nm / 2 kg·m² = 5 rad/s².
Correct Answer:
A
— 5 rad/s²
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Q. A torque of 10 Nm is applied to a wheel with a moment of inertia of 5 kg·m². What is the angular acceleration of the wheel?
A.
2 rad/s²
B.
5 rad/s²
C.
10 rad/s²
D.
20 rad/s²
Show solution
Solution
Using τ = Iα, we have α = τ/I = 10 Nm / 5 kg·m² = 2 rad/s².
Correct Answer:
A
— 2 rad/s²
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Q. A torque of 10 Nm is applied to a wheel. If the radius of the wheel is 0.2 m, what is the force applied tangentially?
A.
50 N
B.
20 N
C.
10 N
D.
5 N
Show solution
Solution
Torque (τ) = F × r; therefore, F = τ / r = 10 Nm / 0.2 m = 50 N.
Correct Answer:
A
— 50 N
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Q. A torque of 10 N·m is applied to a wheel with a moment of inertia of 2 kg·m². What is the angular acceleration of the wheel?
A.
5 rad/s²
B.
10 rad/s²
C.
2 rad/s²
D.
20 rad/s²
Show solution
Solution
Using τ = Iα, we have α = τ/I = 10 N·m / 2 kg·m² = 5 rad/s².
Correct Answer:
A
— 5 rad/s²
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Q. A torque of 12 Nm is applied to a lever arm of 0.4 m. What is the force applied?
A.
30 N
B.
25 N
C.
20 N
D.
15 N
Show solution
Solution
Force = Torque / Distance = 12 Nm / 0.4 m = 30 N.
Correct Answer:
C
— 20 N
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Q. A torque of 12 Nm is applied to a wheel of radius 0.4 m. What is the force applied at the edge of the wheel?
A.
30 N
B.
20 N
C.
15 N
D.
10 N
Show solution
Solution
Torque (τ) = r × F, thus F = τ / r = 12 Nm / 0.4 m = 30 N.
Correct Answer:
B
— 20 N
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Q. A torque of 12 Nm is applied to a wheel with a radius of 0.4 m. What is the force applied tangentially to the wheel?
A.
15 N
B.
30 N
C.
40 N
D.
50 N
Show solution
Solution
Using τ = F × r, we have F = τ / r = 12 Nm / 0.4 m = 30 N.
Correct Answer:
B
— 30 N
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Q. A torque of 12 Nm is produced by a force acting at a distance of 0.4 m from the pivot. What is the magnitude of the force?
A.
20 N
B.
30 N
C.
40 N
D.
50 N
Show solution
Solution
Force = Torque / Distance = 12 Nm / 0.4 m = 30 N.
Correct Answer:
A
— 20 N
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Q. A torque of 12 Nm is produced by a force acting at a distance of 4 m from the pivot. What is the magnitude of the force?
A.
2 N
B.
3 N
C.
4 N
D.
5 N
Show solution
Solution
Force = Torque / Distance = 12 Nm / 4 m = 3 N.
Correct Answer:
B
— 3 N
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Q. A torque of 15 N·m is applied to a wheel with a moment of inertia of 3 kg·m². What is the angular acceleration? (2023)
A.
3 rad/s²
B.
5 rad/s²
C.
10 rad/s²
D.
15 rad/s²
Show solution
Solution
Using τ = Iα, we have α = τ/I = 15/3 = 5 rad/s².
Correct Answer:
B
— 5 rad/s²
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Q. A torque of 15 N·m is applied to a wheel with a radius of 0.3 m. What is the force applied tangentially to the wheel?
A.
25 N
B.
50 N
C.
45 N
D.
30 N
Show solution
Solution
Torque (τ) = Force (F) × Radius (r) => F = τ / r = 15 N·m / 0.3 m = 50 N.
Correct Answer:
B
— 50 N
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Q. A torque of 25 Nm is applied to a wheel with a radius of 0.5 m. What is the force applied at the edge of the wheel?
A.
50 N
B.
25 N
C.
75 N
D.
100 N
Show solution
Solution
Force = Torque / Radius = 25 Nm / 0.5 m = 50 N.
Correct Answer:
A
— 50 N
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Q. A torque of 25 Nm is applied to a wheel. If the radius of the wheel is 0.5 m, what is the force applied tangentially at the edge of the wheel?
A.
10 N
B.
25 N
C.
50 N
D.
5 N
Show solution
Solution
Torque (τ) = Force (F) × Radius (r) => F = τ / r = 25 Nm / 0.5 m = 50 N.
Correct Answer:
A
— 10 N
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Q. A torque of 30 Nm is applied to a wheel with a radius of 0.5 m. What is the force applied at the edge of the wheel?
A.
60 N
B.
30 N
C.
15 N
D.
75 N
Show solution
Solution
Torque (τ) = Force (F) × Radius (r) => F = τ / r = 30 Nm / 0.5 m = 60 N.
Correct Answer:
A
— 60 N
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Q. A torque of 30 Nm is applied to a wheel. If the radius of the wheel is 0.5 m, what is the force applied tangentially?
A.
15 N
B.
30 N
C.
60 N
D.
75 N
Show solution
Solution
Torque (τ) = Force (F) × Radius (r) => F = τ / r = 30 Nm / 0.5 m = 60 N.
Correct Answer:
C
— 60 N
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Q. A torque of 30 Nm is applied to a wheel. If the radius of the wheel is 0.5 m, what is the force applied tangentially at the edge of the wheel?
A.
15 N
B.
30 N
C.
60 N
D.
75 N
Show solution
Solution
Torque (τ) = F × r, thus F = τ / r = 30 Nm / 0.5 m = 60 N.
Correct Answer:
C
— 60 N
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Q. A torque of 40 Nm is required to rotate a wheel. If the radius of the wheel is 0.4 m, what is the force applied tangentially?
A.
100 N
B.
80 N
C.
60 N
D.
40 N
Show solution
Solution
Force = Torque / Radius = 40 Nm / 0.4 m = 100 N.
Correct Answer:
B
— 80 N
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Q. A torque of 5 Nm is applied to a wheel. If the radius of the wheel is 0.25 m, what is the force applied tangentially?
A.
10 N
B.
20 N
C.
5 N
D.
15 N
Show solution
Solution
Torque (τ) = F × r, thus F = τ / r = 5 Nm / 0.25 m = 20 N.
Correct Answer:
A
— 10 N
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Q. A torque of 5 N·m is applied to a wheel with a moment of inertia of 2 kg·m². What is the angular acceleration of the wheel?
A.
2.5 rad/s²
B.
5 rad/s²
C.
10 rad/s²
D.
1 rad/s²
Show solution
Solution
Angular acceleration α = Torque/I = 5 N·m / 2 kg·m² = 2.5 rad/s².
Correct Answer:
A
— 2.5 rad/s²
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Q. A torque of 50 Nm is applied to a wheel with a radius of 0.25 m. What is the force applied at the edge of the wheel?
A.
100 N
B.
200 N
C.
250 N
D.
300 N
Show solution
Solution
Force = Torque / Radius = 50 Nm / 0.25 m = 200 N.
Correct Answer:
B
— 200 N
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Q. A torque of 50 Nm is applied to a wheel with a radius of 0.5 m. What is the force applied tangentially to the wheel?
A.
100 N
B.
50 N
C.
25 N
D.
75 N
Show solution
Solution
Torque (τ) = Force (F) × Radius (r) => F = τ / r = 50 Nm / 0.5 m = 100 N.
Correct Answer:
A
— 100 N
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Q. A torque of 50 Nm is created by a force acting at a distance of 2 m. What is the force applied?
A.
20 N
B.
25 N
C.
30 N
D.
35 N
Show solution
Solution
Force = Torque / Distance = 50 Nm / 2 m = 25 N.
Correct Answer:
B
— 25 N
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