Major Competitive Exams

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Major Competitive Exams

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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!

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Q. If the magnetic field strength is doubled, what happens to the force on a current-carrying conductor in that field? (2022)

A. It doubles
B. It halves
C. It remains the same
D. It quadruples

Solution

The force on a current-carrying conductor is directly proportional to the magnetic field strength, so if the field strength is doubled, the force also doubles.

Correct Answer: A — It doubles

Q. If the magnetic field strength is doubled, what happens to the induced EMF in a coil with a constant number of turns and area?

A. It doubles
B. It halves
C. It remains the same
D. It quadruples

Solution

According to Faraday's law, the induced EMF is directly proportional to the rate of change of magnetic flux, which depends on the magnetic field strength.

Correct Answer: A — It doubles

Q. If the magnetic field strength is doubled, what happens to the magnetic force on a charged particle moving perpendicular to the field?

A. Doubles
B. Halves
C. Remains the same
D. Quadruples

Solution

The magnetic force on a charged particle is given by F = qvB sin(θ). If the magnetic field strength B is doubled, the force F also doubles, assuming charge q and velocity v remain constant.

Correct Answer: A — Doubles

Q. If the magnetic field through a loop is doubled while the area remains constant, what happens to the magnetic flux?

A. Magnetic flux doubles
B. Magnetic flux halves
C. Magnetic flux remains the same
D. Magnetic flux becomes zero

Solution

Magnetic flux is given by the product of magnetic field strength and area. If the magnetic field is doubled, the magnetic flux also doubles.

Correct Answer: A — Magnetic flux doubles

Q. If the magnetic field through a loop is increased uniformly, what happens to the induced current in the loop?

A. It flows in the direction of the magnetic field
B. It flows in the opposite direction to the magnetic field
C. It remains constant
D. It stops flowing

Solution

According to Lenz's law, the induced current will flow in the opposite direction to oppose the increase in magnetic flux.

Correct Answer: B — It flows in the opposite direction to the magnetic field

Q. If the magnetic field through a loop is increasing at a constant rate, what can be said about the induced current?

A. It is constant
B. It is increasing
C. It is decreasing
D. It is zero

Solution

If the magnetic field is increasing at a constant rate, the induced EMF is also increasing, which means the induced current is increasing.

Correct Answer: B — It is increasing

Q. If the magnetic flux through a coil changes from 0.2 Wb to 0.5 Wb in 2 seconds, what is the average induced EMF? (2021)

A. 150 V
B. 100 V
C. 75 V
D. 50 V

Solution

Average induced EMF = -ΔΦ/Δt = -(0.5 Wb - 0.2 Wb) / 2 s = -0.15 Wb/s = 150 V.

Correct Answer: B — 100 V

Q. If the magnetic flux through a coil changes from 0.2 Wb to 0.5 Wb in 2 seconds, what is the average induced EMF in the coil? (2021)

A. 150 V
B. 100 V
C. 75 V
D. 50 V

Solution

Average induced EMF (ε) = -ΔΦ/Δt = -(0.5 Wb - 0.2 Wb) / 2 s = -0.3 Wb / 2 s = -0.15 V. The absolute value is 0.15 V or 150 mV.

Correct Answer: B — 100 V

Q. If the magnetic flux through a coil changes from 0.5 Wb to 0.2 Wb in 2 seconds, what is the average induced emf? (2023)

A. 150 V
B. 100 V
C. 75 V
D. 50 V

Solution

Average induced emf = -ΔΦ/Δt = -(0.2 - 0.5)/2 = 0.15 V = 150 V.

Correct Answer: B — 100 V

Q. If the magnetic flux through a coil changes, what happens to the induced EMF? (2023)

A. It remains constant
B. It increases
C. It decreases
D. It is induced

Solution

An induced EMF is generated when there is a change in magnetic flux through a coil.

Correct Answer: D — It is induced

Q. If the mass of a satellite is doubled while keeping its orbital radius constant, what happens to the gravitational force acting on it?

A. It doubles.
B. It remains the same.
C. It halves.
D. It quadruples.

Solution

The gravitational force acting on the satellite will double if its mass is doubled, as gravitational force is directly proportional to mass.

Correct Answer: A — It doubles.

Q. If the mass of a simple harmonic oscillator is doubled while keeping the spring constant the same, how does the period change?

A. Increases
B. Decreases
C. Remains the same
D. Cannot be determined

Solution

The period T = 2π√(m/k). If m is doubled, T increases.

Correct Answer: A — Increases

Q. If the mass of an object is 10 kg and it is located at a point where the gravitational potential is -30 J/kg, what is the gravitational potential energy of the object?

A. -300 J
B. -30 J
C. 300 J
D. 30 J

Solution

Gravitational potential energy U = m * V = 10 * (-30) = -300 J.

Correct Answer: A — -300 J

Q. If the mass of an object is 2 kg, what is its energy equivalent according to Einstein's equation E=mc^2? (2022)

A. 1.8 x 10^16 J
B. 1.8 x 10^14 J
C. 1.8 x 10^12 J
D. 1.8 x 10^10 J

Solution

E = mc^2 = 2 kg * (3 x 10^8 m/s)^2 = 1.8 x 10^16 J.

Correct Answer: A — 1.8 x 10^16 J

Q. If the mass of an object is 5 kg and it is at a height of 10 m above the ground, what is its gravitational potential energy?

A. 50 J
B. 100 J
C. 500 J
D. 0 J

Solution

Gravitational potential energy U = mgh = 5 * 10 * 10 = 500 J.

Correct Answer: B — 100 J

Q. If the mass of an object is doubled, how does its gravitational potential energy change at a constant height?

A. It remains the same
B. It doubles
C. It halves
D. It quadruples

Solution

The gravitational potential energy U = mgh will double if the mass m is doubled at constant height h.

Correct Answer: B — It doubles

Q. If the mass of an object is halved and the distance from the center of the Earth is doubled, what happens to the gravitational potential energy?

A. It remains the same.
B. It doubles.
C. It halves.
D. It quadruples.

Solution

Gravitational potential energy is directly proportional to mass and inversely proportional to distance, so it halves.

Correct Answer: C — It halves.

Q. If the mass of an object is halved, how does its gravitational potential energy change at a constant height?

A. It doubles
B. It halves
C. It remains the same
D. It becomes zero

Solution

The gravitational potential energy U = mgh is directly proportional to the mass, so if the mass is halved, the potential energy also halves.

Correct Answer: B — It halves

Q. If the mass of an object is measured as 10.0 kg with an uncertainty of ±0.2 kg, what is the percentage error?

A. 2%
B. 0.2%
C. 0.02%
D. 0.5%

Solution

Percentage error = (absolute error / measured value) * 100 = (0.2 / 10.0) * 100 = 2%.

Correct Answer: A — 2%

Q. If the mass of an object is measured as 10.0 kg with an uncertainty of ±0.2 kg, what is the percentage error in the mass measurement?

A. 2%
B. 0.2%
C. 0.5%
D. 1%

Solution

Percentage error = (absolute error / measured value) * 100 = (0.2 / 10.0) * 100 = 2%.

Correct Answer: D — 1%

Q. If the mass of an object is measured as 50 kg with an uncertainty of ±0.5 kg, what is the percentage error in the mass measurement?

A. 1%
B. 0.1%
C. 0.5%
D. 2%

Solution

Percentage error = (absolute error / measured value) * 100 = (0.5 / 50) * 100 = 1%.

Correct Answer: A — 1%

Q. If the mass of an object is measured in kilograms and its volume in cubic meters, what is the unit of density? (2022)

A. kg/m
B. kg/m^3
C. m^3/kg
D. m/kg

Solution

Density is defined as mass per unit volume, so the unit of density is kg/m³.

Correct Answer: B — kg/m^3

Q. If the mass of an object is tripled and the distance from the center of the Earth is halved, how does the gravitational force change?

A. It triples
B. It increases by a factor of 6
C. It increases by a factor of 9
D. It decreases

Solution

F ∝ m/r²; if m is tripled and r is halved, F = 3m/(1/4) = 12m, so it increases by a factor of 9.

Correct Answer: C — It increases by a factor of 9

Q. If the mass of an object is tripled, how does its gravitational potential energy change at a constant height?

A. It triples
B. It doubles
C. It remains the same
D. It increases by a factor of nine

Solution

Gravitational potential energy U = mgh, so if mass is tripled, U also triples.

Correct Answer: A — It triples

Q. If the mass of an object is tripled, how does the gravitational potential energy change for a constant height?

A. It triples
B. It doubles
C. It halves
D. It remains the same

Solution

Gravitational potential energy is directly proportional to mass, so it triples.

Correct Answer: A — It triples

Q. If the mass of the Earth is 6 x 10^24 kg and the radius is 6.4 x 10^6 m, what is the gravitational force on a 1 kg mass at the surface?

A. 9.8 N
B. 6.67 N
C. 10 N
D. 8.9 N

Solution

Using F = G * (m1 * m2) / r^2, F = (6.67 x 10^-11) * (6 x 10^24 * 1) / (6.4 x 10^6)^2 = 9.8 N.

Correct Answer: A — 9.8 N

Q. If the mass of the Earth is 6 x 10^24 kg and the radius is 6.4 x 10^6 m, what is the gravitational acceleration at the surface of the Earth?

A. 9.8 m/s²
B. 10 m/s²
C. 9.81 m/s²
D. 11 m/s²

Solution

Using g = G * M / R², where G = 6.67 x 10^-11 N m²/kg², M = 6 x 10^24 kg, and R = 6.4 x 10^6 m, we find g ≈ 9.8 m/s².

Correct Answer: A — 9.8 m/s²

Q. If the mass of the Earth is 6 × 10^24 kg and its radius is 6.4 × 10^6 m, what is the acceleration due to gravity at its surface?

A. 9.8 m/s²
B. 10 m/s²
C. 9.81 m/s²
D. 8.9 m/s²

Solution

g = G * M / R² = (6.67 × 10^-11) * (6 × 10^24) / (6.4 × 10^6)² = 9.81 m/s²

Correct Answer: C — 9.81 m/s²

Q. If the mass of the Earth is M and its radius is R, what is the gravitational acceleration at the surface of the Earth?

A. g = GM/R^2
B. g = GMR^2
C. g = G/MR^2
D. g = MR/G

Solution

The gravitational acceleration at the surface of the Earth is given by g = GM/R^2.

Correct Answer: A — g = GM/R^2

Q. If the mass of the Earth is M and the radius is R, what is the gravitational acceleration at the surface of the Earth?

A. g = GM/R^2
B. g = GMR^2
C. g = G/R
D. g = M/R^2

Solution

The gravitational acceleration at the surface of the Earth is given by g = GM/R^2.

Correct Answer: A — g = GM/R^2

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