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 block of mass 10 kg is resting on a horizontal surface. If the coefficient of kinetic friction is 0.3, what is the frictional force acting on the block when it is sliding?
A.
30 N
B.
20 N
C.
10 N
D.
15 N
Solution
Frictional force (f_k) = μ_k * N = μ_k * mg = 0.3 * 10 kg * 9.8 m/s² = 29.4 N, approximately 30 N.
Q. A block of mass 10 kg is resting on a horizontal surface. If the coefficient of static friction is 0.5, what is the maximum static frictional force acting on the block?
A.
25 N
B.
50 N
C.
75 N
D.
100 N
Solution
Maximum static frictional force (Fs) = μs * N = μs * mg = 0.5 * 10 kg * 9.8 m/s² = 49 N, approximately 50 N.
Q. A block of mass 2 kg is pushed along a frictionless surface by a constant force of 10 N. What is the work done by the force when the block moves 5 m?
Q. A block of mass 2 kg is pushed along a horizontal surface with a constant force of 10 N. What is the work done by the force after moving the block 5 m?
Q. A block of mass 2 kg is pushed along a horizontal surface with a force of 10 N. If the block moves a distance of 5 m, what is the work done by the force?
Q. A block of mass 2 kg is pushed along a horizontal surface with a force of 10 N. If the block moves a distance of 5 m, what is the work done on the block?
Q. A block of mass 2 kg is released from a height of 10 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)
A.
14 m/s
B.
20 m/s
C.
10 m/s
D.
5 m/s
Solution
Using conservation of energy, potential energy at height = kinetic energy just before hitting the ground. mgh = 0.5mv². Solving gives v = sqrt(2gh) = sqrt(2 * 9.8 * 10) = 14 m/s.
Q. A block of mass 2 kg is released from a height of 10 m. What is its speed just before it hits the ground?
A.
0 m/s
B.
10 m/s
C.
14 m/s
D.
20 m/s
Solution
Using conservation of energy, potential energy at height = kinetic energy just before hitting the ground. mgh = 0.5mv^2. Solving gives v = sqrt(2gh) = sqrt(2*9.8*10) = 14 m/s.
Q. A block of mass 2 kg is released from a height of 5 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)
A.
5 m/s
B.
10 m/s
C.
15 m/s
D.
20 m/s
Solution
Using conservation of energy, potential energy at height = kinetic energy just before hitting the ground. mgh = 0.5mv². Solving gives v = √(2gh) = √(2 * 9.8 * 5) = 10 m/s.
Q. A block of mass 2 kg is sliding down a frictionless incline of height 10 m. What is its speed at the bottom?
A.
10 m/s
B.
14 m/s
C.
20 m/s
D.
25 m/s
Solution
Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv². Thus, v = sqrt(2gh) = sqrt(2 * 9.8 * 10) = 14 m/s.
Q. A block of mass 2 kg is sliding down a frictionless incline of height 5 m. What is its speed at the bottom of the incline?
A.
10 m/s
B.
5 m/s
C.
20 m/s
D.
15 m/s
Solution
Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv². Thus, v = sqrt(2gh) = sqrt(2 * 9.81 * 5) = 10 m/s.
Q. A block of mass 2 kg is sliding down a frictionless incline of height 5 m. What is its speed at the bottom?
A.
5 m/s
B.
10 m/s
C.
15 m/s
D.
20 m/s
Solution
Using conservation of energy, potential energy at the top (mgh) converts to kinetic energy at the bottom (1/2 mv²). Thus, v = sqrt(2gh) = sqrt(2 * 9.8 * 5) ≈ 10 m/s.
Q. A block of mass 5 kg is pulled along a horizontal surface with a force of 20 N. If the frictional force acting on the block is 5 N, what is the acceleration of the block? (2021)
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
Solution
Net force = Applied force - Frictional force = 20 N - 5 N = 15 N. Using F = ma, a = F/m = 15 N / 5 kg = 3 m/s².
Q. A block of mass 5 kg is pulled on a horizontal surface with a force of 20 N. If the frictional force is 5 N, what is the acceleration of the block? (2020)
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
Solution
Net force = Applied force - Frictional force = 20 N - 5 N = 15 N. Using F = ma, a = F/m = 15 N / 5 kg = 3 m/s².
Q. A block of mass 5 kg is pulled on a horizontal surface with a force of 20 N. If the frictional force acting on the block is 5 N, what is the acceleration of the block? (2021)
A.
1 m/s²
B.
2 m/s²
C.
3 m/s²
D.
4 m/s²
Solution
Net force = Applied force - Frictional force = 20 N - 5 N = 15 N. Acceleration = Net force / Mass = 15 N / 5 kg = 3 m/s².
Q. A block of mass 5 kg is resting on a horizontal surface. If a horizontal force of 20 N is applied, what is the acceleration of the block? (Assume no friction)
A.
2 m/s²
B.
4 m/s²
C.
5 m/s²
D.
10 m/s²
Solution
Using Newton's second law, F = ma, we have a = F/m = 20 N / 5 kg = 4 m/s².
Q. A block of mass 5 kg is resting on a horizontal surface. If a horizontal force of 15 N is applied, what is the acceleration of the block? (Assume no friction) (2021)
A.
1 m/s²
B.
2 m/s²
C.
3 m/s²
D.
4 m/s²
Solution
Using F = ma, acceleration a = F/m = 15 N / 5 kg = 3 m/s².
