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 solid sphere and a hollow sphere of the same mass and radius are released from rest at the same height. Which one will have a greater translational speed when they reach the ground?
A.
Solid sphere
B.
Hollow sphere
C.
Both will have the same speed
D.
Depends on the mass
Solution
The solid sphere will have a greater translational speed because it has a smaller moment of inertia.
Q. A solid sphere of mass M and radius R is rolling without slipping on a horizontal surface. What is the expression for its total angular momentum about its center of mass?
A.
(2/5)MR^2ω
B.
MR^2ω
C.
MR^2
D.
0
Solution
Total angular momentum L = Iω, where I = (2/5)MR^2 for a solid sphere.
Q. A solid sphere of mass m and radius r rolls without slipping down an inclined plane of angle θ. What is the acceleration of the center of mass of the sphere?
A.
g sin(θ)
B.
g sin(θ)/2
C.
g sin(θ)/3
D.
g sin(θ)/4
Solution
The acceleration of the center of mass of a rolling object is given by a = g sin(θ) / (1 + k^2/r^2). For a solid sphere, k^2/r^2 = 2/5, thus a = g sin(θ) / (1 + 2/5) = g sin(θ) / (7/5) = (5/7)g sin(θ).
Q. A solid sphere of mass M and radius R rolls without slipping down an inclined plane of height h. What is the speed of the center of mass of the sphere when it reaches the bottom? (2021)
A.
√(2gh)
B.
√(5gh/7)
C.
√(3gh/5)
D.
√(gh)
Solution
Using conservation of energy, potential energy at the top = kinetic energy at the bottom. The total kinetic energy is the sum of translational and rotational kinetic energy. Thus, v = √(5gh/7).
Q. A solid sphere of mass M and radius R rolls without slipping down an inclined plane of height h. What is the speed of the center of mass of the sphere at the bottom of the incline? (2021)
A.
√(2gh)
B.
√(3gh/2)
C.
√(gh)
D.
√(4gh/3)
Solution
Using conservation of energy, potential energy at the top = kinetic energy at the bottom. The total kinetic energy is the sum of translational and rotational kinetic energy. Thus, mgh = (1/2)mv^2 + (1/5)mv^2, leading to v = √(10gh/7).
Q. A solid sphere of radius R rolls without slipping down an incline of height h. What is its speed at the bottom of the incline? (2021)
A.
√(2gh)
B.
√(3gh)
C.
√(4gh)
D.
√(5gh)
Solution
Using conservation of energy, potential energy at the top = kinetic energy at the bottom. For a solid sphere, v = √(5gh/7). Thus, speed at the bottom is √(3gh).
Q. A solid sphere of radius R rolls without slipping down an inclined plane of angle θ. What is the acceleration of the center of mass of the sphere?
A.
g sin(θ)
B.
g sin(θ)/2
C.
g sin(θ)/3
D.
g sin(θ)/4
Solution
The acceleration of the center of mass of a solid sphere rolling down an incline is given by a = g sin(θ) / (1 + (2/5)) = g sin(θ) / (7/5) = (5/7) g sin(θ).
Q. A solid sphere of radius R rolls without slipping down an inclined plane of height h. What is the speed of the center of mass at the bottom of the incline? (2021)
A.
√(2gh)
B.
√(3gh)
C.
√(4gh)
D.
√(5gh)
Solution
Using conservation of energy, potential energy at height h = kinetic energy at the bottom. For a solid sphere, v = √(3gh).
Q. A solid sphere rolls down a hill without slipping. If the height of the hill is h, what is the speed of the sphere at the bottom of the hill?
A.
√(2gh)
B.
√(3gh)
C.
√(4gh)
D.
√(5gh)
Solution
Using conservation of energy, potential energy at the top (mgh) converts to kinetic energy (1/2 mv^2 + 1/2 Iω^2). For a solid sphere, I = (2/5)mr^2 and ω = v/r. Solving gives v = √(2gh).
Q. A solid sphere rolls down an inclined plane without slipping. What is the ratio of its translational kinetic energy to its total kinetic energy at the bottom of the incline?
A.
1:2
B.
2:3
C.
1:3
D.
1:1
Solution
The total kinetic energy is the sum of translational and rotational kinetic energy. For a solid sphere, the ratio of translational to total kinetic energy is 2:5, which simplifies to 2:3.
Q. A solid sphere rolls down an inclined plane without slipping. What is the ratio of its translational kinetic energy to its total kinetic energy at the bottom?
A.
1:2
B.
2:3
C.
1:3
D.
1:1
Solution
The total kinetic energy is the sum of translational and rotational kinetic energy. For a solid sphere, the ratio of translational to total kinetic energy is 2:3.
Q. A solid sphere rolls without slipping down an incline. What is the ratio of its translational kinetic energy to its total kinetic energy at the bottom?
A.
1:2
B.
2:3
C.
1:1
D.
1:3
Solution
For a solid sphere, the ratio of translational kinetic energy to total kinetic energy is 2:3.
Q. A solution contains 20% sugar. If 5 liters of this solution is diluted with 10 liters of water, what is the new percentage of sugar in the solution?
A.
10%
B.
15%
C.
20%
D.
25%
Solution
Initial sugar = 20% of 5 liters = 1 liter. Total volume after dilution = 5 + 10 = 15 liters. New percentage = (1/15) * 100 = 6.67%.
Q. A solution contains 25% sugar. If 10 liters of water is added to 30 liters of this solution, what is the new concentration of sugar in the solution?
A.
15%
B.
20%
C.
25%
D.
30%
Solution
Initial sugar = 0.25 * 30 = 7.5 liters. New total volume = 30 + 10 = 40 liters. New concentration = (7.5/40) * 100 = 18.75%.
Q. A solution contains 25% sugar. If 8 liters of this solution is diluted with 4 liters of water, what is the new concentration of sugar in the solution?
Q. A solution contains 40% sugar. If 10 liters of this solution is mixed with 5 liters of pure sugar, what is the percentage of sugar in the new solution?
A.
50%
B.
60%
C.
70%
D.
80%
Solution
Sugar in 10L = 40% of 10L = 4L. Total sugar = 4L + 5L = 9L. Total volume = 10L + 5L = 15L. Percentage = (9/15) * 100 = 60%.
