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. An object is in equilibrium. If two forces of 10 N and 15 N act on it in opposite directions, what is the net force?
A.
5 N
B.
10 N
C.
15 N
D.
0 N
Show solution
Solution
In equilibrium, the net force is zero. The forces cancel each other out: 15 N - 10 N = 5 N.
Correct Answer:
A
— 5 N
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Q. An object is in equilibrium. What can be said about the net force acting on it?
A.
It is zero
B.
It is equal to its weight
C.
It is equal to the applied force
D.
It is maximum
Show solution
Solution
For an object in equilibrium, the net force acting on it must be zero.
Correct Answer:
A
— It is zero
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Q. An object is in free fall. What is its acceleration?
A.
0 m/s²
B.
9.8 m/s²
C.
10 m/s²
D.
5 m/s²
Show solution
Solution
In free fall, the only force acting on the object is gravity, so its acceleration is approximately 9.8 m/s².
Correct Answer:
B
— 9.8 m/s²
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Q. An object is lifted to a height of 10 m. If the mass of the object is 5 kg, what is the gravitational potential energy gained?
A.
50 J
B.
100 J
C.
150 J
D.
200 J
Show solution
Solution
Potential Energy = mass × g × height = 5 kg × 9.8 m/s² × 10 m = 490 J.
Correct Answer:
B
— 100 J
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Q. An object is lifted to a height of 10 m. If the mass of the object is 5 kg, what is the gravitational potential energy gained? (g = 9.8 m/s²)
A.
49 J
B.
98 J
C.
245 J
D.
490 J
Show solution
Solution
Potential Energy (PE) = mgh = 5 kg * 9.8 m/s² * 10 m = 490 J.
Correct Answer:
B
— 98 J
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Q. An object is lifted vertically 10 m against gravity. If the mass of the object is 5 kg, what is the work done against gravity?
A.
50 J
B.
100 J
C.
150 J
D.
200 J
Show solution
Solution
Work done = mass × g × height = 5 kg × 9.8 m/s² × 10 m = 490 J.
Correct Answer:
B
— 100 J
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Q. An object is lifted vertically 10 m. If the mass of the object is 5 kg, what is the gravitational potential energy gained?
A.
50 J
B.
100 J
C.
150 J
D.
200 J
Show solution
Solution
Potential Energy (PE) = mgh = 5 kg * 9.8 m/s² * 10 m = 490 J.
Correct Answer:
B
— 100 J
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Q. An object is lifted vertically 3 m against gravity. If the mass of the object is 4 kg, what is the work done against gravity? (g = 9.8 m/s²)
A.
117.6 J
B.
39.2 J
C.
29.4 J
D.
19.6 J
Show solution
Solution
Work done = mgh = 4 kg × 9.8 m/s² × 3 m = 117.6 J.
Correct Answer:
A
— 117.6 J
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Q. An object is lifted vertically 4 m against gravity. If the mass of the object is 3 kg, what is the work done against gravity? (g = 9.8 m/s²)
A.
117.6 J
B.
29.4 J
C.
39.2 J
D.
78.4 J
Show solution
Solution
Work done (W) = mgh = 3 kg × 9.8 m/s² × 4 m = 117.6 J.
Correct Answer:
A
— 117.6 J
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Q. An object is lifted vertically 4 m against gravity. If the mass of the object is 10 kg, what is the work done against gravity? (g = 9.8 m/s²)
A.
39.2 J
B.
78.4 J
C.
98 J
D.
196 J
Show solution
Solution
Work done (W) = mgh = 10 kg × 9.8 m/s² × 4 m = 392 J.
Correct Answer:
B
— 78.4 J
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Q. An object is lifted vertically 4 m against gravity. If the mass of the object is 10 kg, what is the work done against gravity? (2021)
A.
40 J
B.
80 J
C.
100 J
D.
200 J
Show solution
Solution
Work Done = m * g * h = 10 kg * 9.8 m/s² * 4 m = 392 J (approx. 400 J).
Correct Answer:
B
— 80 J
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Q. An object is lifted vertically 4 m. If the weight of the object is 50 N, how much work is done against gravity?
A.
100 J
B.
200 J
C.
150 J
D.
50 J
Show solution
Solution
Work done (W) = Force (F) × Distance (d) = 50 N × 4 m = 200 J.
Correct Answer:
B
— 200 J
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Q. An object is lifted vertically to a height of 10 m. If the mass of the object is 5 kg, what is the work done against gravity?
A.
50 J
B.
100 J
C.
150 J
D.
200 J
Show solution
Solution
Work done = mass × g × height = 5 kg × 9.8 m/s² × 10 m = 490 J.
Correct Answer:
B
— 100 J
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Q. An object is moving in a circular path with a constant speed. What can be said about the work done by the centripetal force?
A.
Positive work
B.
Negative work
C.
Zero work
D.
Depends on speed
Show solution
Solution
Centripetal force does no work as it acts perpendicular to the direction of motion.
Correct Answer:
C
— Zero work
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Q. An object is moving in a circular path with a constant speed. What type of energy is primarily involved?
A.
Kinetic Energy
B.
Potential Energy
C.
Mechanical Energy
D.
Thermal Energy
Show solution
Solution
The object has Kinetic Energy due to its motion in the circular path.
Correct Answer:
A
— Kinetic Energy
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Q. An object is moving in a circular path with a radius of 10 m and a speed of 5 m/s. What is the net force acting on the object if its mass is 2 kg?
A.
1 N
B.
2 N
C.
5 N
D.
10 N
Show solution
Solution
Centripetal force (F_c) = mv²/r = 2 kg * (5 m/s)² / (10 m) = 5 N.
Correct Answer:
D
— 10 N
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Q. An object is moving in a circular path with a radius of 10 m at a speed of 5 m/s. What is the period of the motion?
A.
2π s
B.
4π s
C.
10 s
D.
20 s
Show solution
Solution
Period (T) = 2πr/v = 2π(10)/5 = 4π s.
Correct Answer:
A
— 2π s
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Q. An object is moving in a circular path with a radius of 3 m and an angular speed of 2 rad/s. What is the linear speed of the object?
A.
3 m/s
B.
6 m/s
C.
9 m/s
D.
12 m/s
Show solution
Solution
Linear speed (v) = rω = 3 m * 2 rad/s = 6 m/s.
Correct Answer:
B
— 6 m/s
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Q. An object is moving with a speed of 15 m/s and has a mass of 3 kg. What is its kinetic energy?
A.
67.5 J
B.
45 J
C.
90 J
D.
135 J
Show solution
Solution
Kinetic Energy (KE) = 0.5 × m × v² = 0.5 × 3 kg × (15 m/s)² = 0.5 × 3 × 225 = 337.5 J.
Correct Answer:
C
— 90 J
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Q. An object is moving with a velocity of 10 m/s. What is its kinetic energy if its mass is 2 kg? (2022)
A.
10 J
B.
20 J
C.
30 J
D.
40 J
Show solution
Solution
Kinetic Energy = (1/2)mv² = (1/2) × 2 kg × (10 m/s)² = 100 J
Correct Answer:
B
— 20 J
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Q. An object is placed 10 cm from a convex lens with a focal length of 5 cm. What is the nature of the image formed? (2022)
A.
Real and inverted
B.
Virtual and erect
C.
Real and erect
D.
Virtual and inverted
Show solution
Solution
Since the object distance is less than the focal length, the image formed is virtual and erect.
Correct Answer:
B
— Virtual and erect
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Q. An object is placed 10 cm in front of a convex mirror with a focal length of 5 cm. What is the nature of the image formed?
A.
Real and inverted
B.
Virtual and erect
C.
Real and erect
D.
Virtual and inverted
Show solution
Solution
Convex mirrors always produce virtual and erect images.
Correct Answer:
B
— Virtual and erect
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Q. An object is placed 15 cm in front of a convex lens of focal length 10 cm. What is the position of the image formed? (2020)
A.
5 cm
B.
10 cm
C.
30 cm
D.
15 cm
Show solution
Solution
Using the lens formula: 1/f = 1/v - 1/u, where f = 10 cm and u = -15 cm, we find v = 30 cm.
Correct Answer:
C
— 30 cm
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Q. An object is placed 25 cm from a convex lens of focal length 10 cm. Where is the image formed?
A.
10 cm
B.
15 cm
C.
20 cm
D.
30 cm
Show solution
Solution
Using the lens formula 1/f = 1/v - 1/u, we find v = 20 cm.
Correct Answer:
C
— 20 cm
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Q. An object is placed 30 cm from a concave lens of focal length 15 cm. What is the nature of the image formed?
A.
Real and inverted
B.
Virtual and erect
C.
Real and erect
D.
Virtual and inverted
Show solution
Solution
For a concave lens, the image formed is virtual and erect when the object is placed in front of it.
Correct Answer:
B
— Virtual and erect
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Q. An object is placed 30 cm in front of a convex lens of focal length 15 cm. What is the position of the image formed? (2020)
A.
10 cm
B.
15 cm
C.
20 cm
D.
25 cm
Show solution
Solution
Using the lens formula 1/f = 1/v - 1/u, where u = -30 cm and f = 15 cm, we find v = 10 cm.
Correct Answer:
D
— 25 cm
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Q. An object is placed 30 cm in front of a convex lens of focal length 15 cm. What is the image distance?
A.
10 cm
B.
15 cm
C.
20 cm
D.
25 cm
Show solution
Solution
Using the lens formula 1/f = 1/v - 1/u, where u = -30 cm and f = 15 cm, we find v = 10 cm.
Correct Answer:
A
— 10 cm
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Q. An object is placed 40 cm from a convex lens with a focal length of 10 cm. Where is the image formed?
A.
At 10 cm
B.
At 20 cm
C.
At 30 cm
D.
At 40 cm
Show solution
Solution
Using the lens formula, the image is formed at 30 cm on the opposite side.
Correct Answer:
C
— At 30 cm
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Q. An object is placed 50 cm from a converging lens of focal length 25 cm. Where will the image be formed?
A.
16.67 cm
B.
33.33 cm
C.
25 cm
D.
20 cm
Show solution
Solution
Using the lens formula 1/f = 1/v - 1/u, we find v = 33.33 cm.
Correct Answer:
B
— 33.33 cm
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Q. An object is placed at a distance of 15 cm from a convex lens of focal length 10 cm. Where is the image formed?
A.
5 cm
B.
10 cm
C.
15 cm
D.
20 cm
Show solution
Solution
Using the lens formula, 1/f = 1/v - 1/u, we find v = 30 cm, meaning the image is formed 30 cm on the opposite side.
Correct Answer:
D
— 20 cm
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