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 force of 100 N is applied to a 50 kg object. What is the resulting acceleration? (2023)
A.
1 m/s²
B.
2 m/s²
C.
3 m/s²
D.
4 m/s²
Show solution
Solution
Using F = ma, a = F/m = 100 N / 50 kg = 2 m/s².
Correct Answer:
B
— 2 m/s²
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Q. A force of 15 N is applied at an angle of 30° to the horizontal while moving an object 4 m. What is the work done by the force?
A.
30 J
B.
60 J
C.
90 J
D.
120 J
Show solution
Solution
Work done = F × d × cos(θ) = 15 N × 4 m × cos(30°) = 15 N × 4 m × (√3/2) = 60 J.
Correct Answer:
C
— 90 J
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Q. A force of 15 N is applied at an angle of 30° to the horizontal. What is the horizontal component of the force? (2020)
A.
7.5 N
B.
12.99 N
C.
15 N
D.
10 N
Show solution
Solution
Horizontal component = F × cos(θ) = 15 N × cos(30°) = 15 N × √3/2 ≈ 12.99 N.
Correct Answer:
B
— 12.99 N
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Q. A force of 15 N is applied at an angle of 60 degrees to a lever arm of 1 m. What is the torque?
A.
7.5 Nm
B.
12.5 Nm
C.
15 Nm
D.
25 Nm
Show solution
Solution
Torque = Force × Distance × sin(θ) = 15 N × 1 m × sin(60°) = 15 N × 1 m × (√3/2) = 12.5 Nm.
Correct Answer:
B
— 12.5 Nm
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Q. A force of 15 N is applied at an angle of 60 degrees to the horizontal while moving an object 4 m. What is the work done?
A.
30 J
B.
60 J
C.
45 J
D.
75 J
Show solution
Solution
Work Done = F * d * cos(θ) = 15 N * 4 m * cos(60°) = 30 J
Correct Answer:
C
— 45 J
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Q. A force of 15 N is applied at an angle of 60° to the horizontal while moving an object 4 m. What is the work done by the force?
A.
30 J
B.
60 J
C.
45 J
D.
20 J
Show solution
Solution
Work done (W) = F × d × cos(θ) = 15 N × 4 m × cos(60°) = 15 N × 4 m × 0.5 = 30 J.
Correct Answer:
C
— 45 J
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Q. A force of 15 N is applied to a 3 kg mass. What is the resulting acceleration? (2023)
A.
3 m/s²
B.
5 m/s²
C.
7 m/s²
D.
10 m/s²
Show solution
Solution
Using F = ma, acceleration a = F/m = 15 N / 3 kg = 5 m/s².
Correct Answer:
B
— 5 m/s²
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Q. A force of 15 N is applied to a 3 kg object. What is the net force acting on the object if it is also experiencing a frictional force of 5 N?
A.
10 N
B.
15 N
C.
20 N
D.
5 N
Show solution
Solution
Net force = applied force - frictional force = 15 N - 5 N = 10 N.
Correct Answer:
A
— 10 N
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Q. A force of 15 N is applied to a 3 kg object. What is the object's acceleration?
A.
3 m/s²
B.
5 m/s²
C.
7 m/s²
D.
10 m/s²
Show solution
Solution
Using F = ma, acceleration a = F/m = 15 N / 3 kg = 5 m/s².
Correct Answer:
B
— 5 m/s²
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Q. A force of 15 N is applied to a 3 kg object. What is the resulting acceleration?
A.
3 m/s²
B.
5 m/s²
C.
7 m/s²
D.
10 m/s²
Show solution
Solution
Using F = ma, we find a = F/m = 15 N / 3 kg = 5 m/s².
Correct Answer:
B
— 5 m/s²
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Q. A force of 15 N is applied to a 5 kg object. What is the object's acceleration?
A.
3 m/s²
B.
2 m/s²
C.
1 m/s²
D.
4 m/s²
Show solution
Solution
Using F = ma, acceleration a = F/m = 15 N / 5 kg = 3 m/s².
Correct Answer:
A
— 3 m/s²
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Q. A force of 15 N is applied to move an object 3 m at an angle of 60 degrees to the horizontal. What is the work done by the force?
A.
15 J
B.
20 J
C.
30 J
D.
45 J
Show solution
Solution
Work done = F × d × cos(θ) = 15 N × 3 m × cos(60°) = 15 N × 3 m × 0.5 = 22.5 J.
Correct Answer:
C
— 30 J
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Q. A force of 15 N is applied to move an object 4 m in the direction of the force. What is the work done?
A.
30 J
B.
60 J
C.
75 J
D.
90 J
Show solution
Solution
Work done = Force × Distance = 15 N × 4 m = 60 J.
Correct Answer:
D
— 90 J
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Q. A force of 20 N is applied at an angle of 30 degrees to the lever arm of 1 m. What is the torque about the pivot?
A.
10 Nm
B.
17.32 Nm
C.
20 Nm
D.
5 Nm
Show solution
Solution
Torque = Force × Distance × sin(30°) = 20 N × 1 m × 0.5 = 10 Nm.
Correct Answer:
B
— 17.32 Nm
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Q. A force of 20 N is applied at an angle of 60 degrees to a lever arm of 2 m. What is the torque about the pivot?
A.
10 Nm
B.
20 Nm
C.
30 Nm
D.
40 Nm
Show solution
Solution
Torque = Force × Distance × sin(θ) = 20 N × 2 m × sin(60°) = 20 N × 2 m × (√3/2) = 20√3 Nm.
Correct Answer:
B
— 20 Nm
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Q. A force of 20 N is applied at an angle of 60 degrees to a lever arm of length 0.5 m. What is the torque about the pivot?
A.
5 Nm
B.
10 Nm
C.
8.66 Nm
D.
17.32 Nm
Show solution
Solution
Torque (τ) = F × r × sin(θ) = 20 N × 0.5 m × sin(60°) = 20 N × 0.5 m × (√3/2) = 8.66 Nm.
Correct Answer:
C
— 8.66 Nm
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Q. A force of 20 N is applied at an angle of 60 degrees to the lever arm of length 2 m. What is the torque about the pivot?
A.
10 Nm
B.
20 Nm
C.
17.32 Nm
D.
34.64 Nm
Show solution
Solution
Torque (τ) = F × r × sin(θ) = 20 N × 2 m × sin(60°) = 20 × 2 × (√3/2) = 20√3 Nm ≈ 34.64 Nm.
Correct Answer:
C
— 17.32 Nm
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Q. A force of 20 N is applied at an angle of 60° to the horizontal while moving an object 4 m. What is the work done?
A.
40 J
B.
80 J
C.
20 J
D.
60 J
Show solution
Solution
Work done (W) = F × d × cos(θ) = 20 N × 4 m × cos(60°) = 20 × 4 × 0.5 = 40 J.
Correct Answer:
B
— 80 J
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Q. A force of 25 N is applied at an angle of 30 degrees to the horizontal while moving an object 10 m. What is the work done?
A.
100 J
B.
125 J
C.
150 J
D.
175 J
Show solution
Solution
Work done = Force × Distance × cos(θ) = 25 N × 10 m × cos(30°) = 216.5 J.
Correct Answer:
B
— 125 J
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Q. A force of 25 N is applied at an angle of 60 degrees to the horizontal while moving an object 3 m. What is the work done?
A.
37.5 J
B.
50 J
C.
75 J
D.
100 J
Show solution
Solution
Work done = Force × Distance × cos(θ) = 25 N × 3 m × cos(60°) = 37.5 J.
Correct Answer:
A
— 37.5 J
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Q. A force of 30 N is applied at an angle of 60 degrees to a lever arm of length 2 m. What is the torque about the pivot?
A.
15 Nm
B.
30 Nm
C.
60 Nm
D.
52 Nm
Show solution
Solution
Torque (τ) = F × r × sin(θ) = 30 N × 2 m × sin(60°) = 30 × 2 × (√3/2) = 30√3 Nm ≈ 51.96 Nm.
Correct Answer:
D
— 52 Nm
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Q. A force of 30 N is applied to a 3 kg object. What is the resulting acceleration? (2020)
A.
5 m/s²
B.
10 m/s²
C.
15 m/s²
D.
20 m/s²
Show solution
Solution
Using F = ma, a = F/m = 30 N / 3 kg = 10 m/s².
Correct Answer:
B
— 10 m/s²
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Q. A force of 30 N is applied to a 5 kg object. What is the object's acceleration?
A.
3 m/s²
B.
6 m/s²
C.
9 m/s²
D.
12 m/s²
Show solution
Solution
Using F = ma, a = F/m = 30 N / 5 kg = 6 m/s².
Correct Answer:
B
— 6 m/s²
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Q. A force of 30 N is applied to a 6 kg object. What is the object's acceleration? (2023)
A.
3 m/s²
B.
4 m/s²
C.
5 m/s²
D.
6 m/s²
Show solution
Solution
Using F = ma, we find a = F/m = 30 N / 6 kg = 5 m/s².
Correct Answer:
B
— 4 m/s²
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Q. A force of 40 N is applied at a distance of 0.5 m from the pivot. What is the torque?
A.
10 Nm
B.
15 Nm
C.
20 Nm
D.
25 Nm
Show solution
Solution
Torque = Force × Distance = 40 N × 0.5 m = 20 Nm.
Correct Answer:
C
— 20 Nm
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Q. A force of 40 N is applied at an angle of 60 degrees to the lever arm of length 2 m. What is the torque about the pivot?
A.
20 Nm
B.
40 Nm
C.
34.64 Nm
D.
69.28 Nm
Show solution
Solution
Torque (τ) = F × r × sin(θ) = 40 N × 2 m × sin(60°) = 40 × 2 × (√3/2) = 40√3 Nm ≈ 34.64 Nm.
Correct Answer:
C
— 34.64 Nm
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Q. A force of 50 N is applied at a distance of 0.5 m from the pivot at an angle of 60 degrees. What is the torque?
A.
25 Nm
B.
43.3 Nm
C.
50 Nm
D.
0 Nm
Show solution
Solution
Torque (τ) = F × r × sin(θ) = 50 N × 0.5 m × sin(60°) = 50 N × 0.5 m × (√3/2) = 43.3 Nm.
Correct Answer:
B
— 43.3 Nm
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Q. A force of 50 N is applied at an angle of 30 degrees to the lever arm of length 1 m. What is the torque about the pivot?
A.
25 Nm
B.
43.3 Nm
C.
50 Nm
D.
86.6 Nm
Show solution
Solution
Torque (τ) = F × d × sin(θ) = 50 N × 1 m × sin(30°) = 50 N × 1 m × 0.5 = 25 Nm.
Correct Answer:
B
— 43.3 Nm
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Q. A force of 50 N is applied at an angle of 30 degrees to the lever arm of length 2 m. What is the torque about the pivot?
A.
25 N·m
B.
50 N·m
C.
86.6 N·m
D.
100 N·m
Show solution
Solution
Torque (τ) = F × r × sin(θ) = 50 N × 2 m × sin(30°) = 50 N × 2 m × 0.5 = 50 N·m.
Correct Answer:
C
— 86.6 N·m
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Q. A force of 50 N is applied at an angle of 30° to the horizontal. What is the horizontal component of the force? (2020)
A.
25 N
B.
43.3 N
C.
50 N
D.
35 N
Show solution
Solution
Horizontal component = Fcos(θ) = 50 N × cos(30°) = 50 N × (√3/2) ≈ 43.3 N.
Correct Answer:
B
— 43.3 N
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