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. If a 15Ω resistor is connected in parallel with a 5Ω resistor, what is the equivalent resistance? (2023)
A.
3.75Ω
B.
4Ω
C.
5Ω
D.
10Ω
Show solution
Solution
1/R_eq = 1/15 + 1/5 = 1/15 + 3/15 = 4/15; R_eq = 15/4 = 3.75Ω.
Correct Answer:
A
— 3.75Ω
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Q. If a 2 kg object is acted upon by a net force of 6 N, what is the object's acceleration?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
Show solution
Solution
Using F = ma, we find a = F/m = 6 N / 2 kg = 3 m/s².
Correct Answer:
B
— 3 m/s²
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Q. If a 2 kg object is dropped from a height of 5 m, what is its speed just before it hits the ground? (g = 10 m/s²)
A.
10 m/s
B.
5 m/s
C.
15 m/s
D.
20 m/s
Show solution
Solution
Using energy conservation: PE_initial = KE_final; m * g * h = 1/2 * m * v^2; v = sqrt(2gh) = sqrt(2 * 10 * 5) = 10 m/s
Correct Answer:
A
— 10 m/s
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Q. If a 2 kg object is dropped from a height of 5 m, what is its velocity just before it hits the ground? (g = 10 m/s²)
A.
10 m/s
B.
5 m/s
C.
15 m/s
D.
20 m/s
Show solution
Solution
Using energy conservation, v = sqrt(2gh) = sqrt(2 * 10 * 5) = 10 m/s
Correct Answer:
A
— 10 m/s
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Q. If a 2 kg object is dropped from a height, what is its velocity just before hitting the ground (g = 10 m/s²)? (2023)
A.
10 m/s
B.
20 m/s
C.
30 m/s
D.
40 m/s
Show solution
Solution
Using v = √(2gh), for h = 10 m, v = √(2 * 10 m/s² * 10 m) = √200 = 20 m/s.
Correct Answer:
B
— 20 m/s
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Q. If a 2 kg object is dropped from a height, what is its velocity just before hitting the ground? (g = 9.8 m/s², h = 20 m) (2023)
A.
14 m/s
B.
19.6 m/s
C.
20 m/s
D.
28 m/s
Show solution
Solution
Using v² = u² + 2gh, where u = 0, we find v = √(2 * 9.8 * 20) = 19.6 m/s.
Correct Answer:
B
— 19.6 m/s
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Q. If a 2 kg object is lifted to a height of 10 m, what is the potential energy gained by the object? (g = 9.8 m/s²)
A.
19.6 J
B.
98 J
C.
39.2 J
D.
78.4 J
Show solution
Solution
Potential Energy (PE) = mgh = 2 kg × 9.8 m/s² × 10 m = 196 J.
Correct Answer:
B
— 98 J
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Q. If a 2 kg object is lifted to a height of 3 m, what is the potential energy gained by the object? (g = 9.8 m/s²)
A.
58.8 J
B.
19.6 J
C.
29.4 J
D.
39.2 J
Show solution
Solution
Potential Energy (PE) = mgh = 2 kg × 9.8 m/s² × 3 m = 58.8 J.
Correct Answer:
A
— 58.8 J
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Q. If a 2 kg object is lifted to a height of 3 m, what is the potential energy gained by the object? (Take g = 9.8 m/s²) (2020)
A.
58.8 J
B.
19.6 J
C.
29.4 J
D.
39.2 J
Show solution
Solution
Potential Energy (PE) = mgh = 2 kg × 9.8 m/s² × 3 m = 58.8 J.
Correct Answer:
A
— 58.8 J
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Q. If a 2 kg object is lifted to a height of 3 m, what is the potential energy gained? (Use g = 9.8 m/s²) (2020)
A.
58.8 J
B.
19.6 J
C.
29.4 J
D.
39.2 J
Show solution
Solution
Potential Energy = mass × g × height = 2 kg × 9.8 m/s² × 3 m = 58.8 J
Correct Answer:
A
— 58.8 J
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Q. If a 2 kg object is lifted to a height of 3 m, what is the potential energy gained by the object? (Use g = 9.8 m/s²)
A.
58.8 J
B.
19.6 J
C.
29.4 J
D.
39.2 J
Show solution
Solution
Potential Energy (PE) = mgh = 2 kg × 9.8 m/s² × 3 m = 58.8 J.
Correct Answer:
A
— 58.8 J
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Q. If a 2 kg object is lifted to a height of 3 m, what is the work done against gravity? (Use g = 9.8 m/s²) (2023)
A.
58.8 J
B.
39.2 J
C.
19.6 J
D.
29.4 J
Show solution
Solution
Work done = mgh = 2 kg × 9.8 m/s² × 3 m = 58.8 J.
Correct Answer:
A
— 58.8 J
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Q. If a 2 kg object is lifted to a height of 5 m, what is the work done against gravity? (g = 10 m/s²) (2019)
A.
20 J
B.
10 J
C.
50 J
D.
30 J
Show solution
Solution
Work Done = mgh = 2 kg * 10 m/s² * 5 m = 100 J
Correct Answer:
A
— 20 J
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Q. If a 2 kg object is moving with a velocity of 15 m/s, what is its kinetic energy?
A.
225 J
B.
150 J
C.
300 J
D.
450 J
Show solution
Solution
Kinetic Energy = 1/2 * m * v^2 = 1/2 * 2 kg * (15 m/s)^2 = 225 J.
Correct Answer:
A
— 225 J
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Q. If a 2 kg object is moving with a velocity of 3 m/s and a force of 6 N is applied in the opposite direction, what will be its final velocity after 2 seconds?
A.
0 m/s
B.
1 m/s
C.
2 m/s
D.
3 m/s
Show solution
Solution
Net force = -6 N, acceleration = F/m = -6 N / 2 kg = -3 m/s². Final velocity = initial velocity + at = 3 m/s + (-3 m/s² * 2 s) = 0 m/s.
Correct Answer:
B
— 1 m/s
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Q. If a 2 kg object is moving with a velocity of 3 m/s, what is its kinetic energy?
A.
9 J
B.
6 J
C.
3 J
D.
12 J
Show solution
Solution
Kinetic Energy = 0.5 * m * v² = 0.5 * 2 kg * (3 m/s)² = 9 J.
Correct Answer:
A
— 9 J
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Q. If a 2 kg object is pulled with a force of 10 N and experiences a frictional force of 4 N, what is its acceleration?
A.
3 m/s²
B.
5 m/s²
C.
2 m/s²
D.
1 m/s²
Show solution
Solution
Net force = applied force - friction = 10 N - 4 N = 6 N. Acceleration a = F/m = 6 N / 2 kg = 3 m/s².
Correct Answer:
A
— 3 m/s²
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Q. If a 2 kg object is subjected to a net force of 10 N, what will be its acceleration? (2022)
A.
2 m/s²
B.
5 m/s²
C.
10 m/s²
D.
20 m/s²
Show solution
Solution
Using F = ma, a = F/m = 10 N / 2 kg = 5 m/s².
Correct Answer:
B
— 5 m/s²
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Q. If a 2 kg object is subjected to a net force of 6 N, what is its acceleration?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
Show solution
Solution
Using F = ma, we have a = F/m = 6 N / 2 kg = 3 m/s².
Correct Answer:
B
— 3 m/s²
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Q. If a 3 kg object is in free fall, what is the force acting on it due to gravity?
A.
3 N
B.
9 N
C.
30 N
D.
None of the above
Show solution
Solution
Force due to gravity F = mg = 3 kg * 9.8 m/s² = 29.4 N, approximately 9 N for simplification.
Correct Answer:
B
— 9 N
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Q. If a 3 kg object is moving with a speed of 4 m/s and comes to a stop, what is the work done by friction?
A.
-24 J
B.
-48 J
C.
0 J
D.
24 J
Show solution
Solution
Initial kinetic energy = 0.5 × m × v² = 0.5 × 3 kg × (4 m/s)² = 24 J. Work done by friction = -24 J.
Correct Answer:
B
— -48 J
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Q. If a 3 kg object is moving with a speed of 4 m/s and comes to a stop, what is the work done by the friction force?
A.
-24 J
B.
-48 J
C.
-72 J
D.
-96 J
Show solution
Solution
Initial kinetic energy = 0.5 × 3 kg × (4 m/s)² = 24 J. Work done by friction = -24 J.
Correct Answer:
B
— -48 J
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Q. If a 3 kg object is moving with a speed of 4 m/s and comes to a stop, what is the work done by the friction?
A.
-24 J
B.
-48 J
C.
-12 J
D.
-36 J
Show solution
Solution
Initial kinetic energy = 0.5 × 3 kg × (4 m/s)² = 24 J. Work done by friction = -24 J.
Correct Answer:
B
— -48 J
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Q. If a 3 kg object is moving with a speed of 4 m/s, what is its kinetic energy?
A.
12 J
B.
24 J
C.
36 J
D.
48 J
Show solution
Solution
KE = 0.5 × m × v² = 0.5 × 3 kg × (4 m/s)² = 0.5 × 3 × 16 = 24 J.
Correct Answer:
B
— 24 J
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Q. If a 3 kg object is moving with a speed of 4 m/s, what is the total mechanical energy if it is at a height of 2 m?
A.
30 J
B.
40 J
C.
50 J
D.
60 J
Show solution
Solution
Total mechanical energy = K.E + P.E = 0.5 × 3 kg × (4 m/s)² + 3 kg × 9.8 m/s² × 2 m = 24 J + 58.8 J = 82.8 J.
Correct Answer:
C
— 50 J
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Q. If a 3 kg object is moving with a velocity of 4 m/s, what is its momentum?
A.
12 kg m/s
B.
6 kg m/s
C.
8 kg m/s
D.
10 kg m/s
Show solution
Solution
Momentum (p) = m * v = 3 kg * 4 m/s = 12 kg m/s
Correct Answer:
A
— 12 kg m/s
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Q. If a 3 kg object is subjected to a net force of 12 N, what is its acceleration? (2023)
A.
4 m/s²
B.
3 m/s²
C.
2 m/s²
D.
1 m/s²
Show solution
Solution
Acceleration a = F/m = 12 N / 3 kg = 4 m/s².
Correct Answer:
A
— 4 m/s²
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Q. If a 4 kg object is acted upon by a net force of 16 N, what is its acceleration?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
Show solution
Solution
Using F = ma, we have a = F/m = 16 N / 4 kg = 4 m/s².
Correct Answer:
C
— 4 m/s²
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Q. If a 4 kg object is at rest and a net force of 8 N is applied, what will be its velocity after 2 seconds?
A.
4 m/s
B.
2 m/s
C.
1 m/s
D.
0 m/s
Show solution
Solution
Using F = ma, a = F/m = 8 N / 4 kg = 2 m/s². Velocity after 2 seconds = a * t = 2 m/s² * 2 s = 4 m/s.
Correct Answer:
B
— 2 m/s
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Q. If a 4 kg object is lifted to a height of 2 m, what is the work done against gravity? (2021)
A.
40 J
B.
80 J
C.
20 J
D.
60 J
Show solution
Solution
Work done = mass × g × height = 4 kg × 10 m/s² × 2 m = 80 J
Correct Answer:
B
— 80 J
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