Major Competitive Exams

My Learning Cart

Major Competitive Exams

Download Q&A

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!

Civil Services Defence Exams Engineering & Architecture Admissions Government Jobs Management Admissions Undergraduate

Q. A plane mirror is placed at an angle of 60 degrees to the incident ray. What is the angle of reflection?

A. 30 degrees
B. 60 degrees
C. 90 degrees
D. 120 degrees

Solution

According to the law of reflection, the angle of reflection is equal to the angle of incidence. Therefore, the angle of reflection is also 60 degrees.

Correct Answer: B — 60 degrees

Q. A plane mirror produces an image that is: (2023)

A. Real and inverted
B. Virtual and erect
C. Real and erect
D. Virtual and inverted

Solution

A plane mirror always produces a virtual and erect image, which is the same size as the object.

Correct Answer: B — Virtual and erect

Q. A planet orbits a star in an elliptical path. What remains constant throughout its orbit?

A. Angular momentum
B. Kinetic energy
C. Potential energy
D. Total energy

Solution

Angular momentum is conserved in the absence of external torques, even in elliptical orbits.

Correct Answer: A — Angular momentum

Q. A planet orbits the sun in a circular path. If the radius of the orbit is doubled, what happens to the angular momentum of the planet if its speed remains constant?

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

Solution

Angular momentum L = mvr, so if the radius is doubled and speed remains constant, angular momentum doubles.

Correct Answer: A — Doubles

Q. A planet orbits the sun in a circular path. If the radius of the orbit is halved, what happens to the angular momentum of the planet?

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

Solution

Angular momentum L = mvr; if the radius is halved and speed remains constant, angular momentum halves.

Correct Answer: B — It halves

Q. A planet orbits the sun in an elliptical path. What remains constant for the planet as it moves in its orbit?

A. Kinetic energy
B. Potential energy
C. Angular momentum
D. Total energy

Solution

Angular momentum remains constant for a planet in orbit due to the conservation of angular momentum.

Correct Answer: C — Angular momentum

Q. A planet orbits the sun in an elliptical path. Which of the following statements is true regarding its angular momentum?

A. It is constant.
B. It varies with distance from the sun.
C. It is zero at perihelion.
D. It is maximum at aphelion.

Solution

The angular momentum of a planet about the sun is conserved as there is no external torque acting on it.

Correct Answer: A — It is constant.

Q. A player has a batting average of 45 runs per match. If he plays 10 matches, what is his total expected score? (2020)

A. 400
B. 450
C. 500
D. 550

Solution

Total expected score = batting average * number of matches = 45 runs/match * 10 matches = 450 runs.

Correct Answer: B — 450

Q. A player has a batting average of 50 runs after 10 innings. If he scores 60 runs in his next innings, what will be his new average? (2022)

A. 52
B. 55
C. 54
D. 53

Solution

Total runs after 10 innings = 50 * 10 = 500. New total runs = 500 + 60 = 560. New average = 560 / 11 = 50.91, which rounds to 55.

Correct Answer: B — 55

Q. A player has a batting average of 50 runs per match. If he has played 20 matches, how many runs has he scored in total? (2020)

A. 800
B. 1000
C. 900
D. 950

Solution

Total runs = Batting average * Number of matches = 50 * 20 = 1000 runs.

Correct Answer: B — 1000

Q. A player has a batting average of 50 runs. If he has played 20 matches, how many runs has he scored in total? (2019)

A. 800
B. 1000
C. 900
D. 700

Solution

Total runs scored = Batting average * Number of matches = 50 * 20 = 1000 runs.

Correct Answer: A — 800

Q. A player hits a ball at an angle of 30 degrees with the horizontal. If the initial speed is 20 m/s, what is the vertical component of the velocity? (2022)

A. 10 m/s
B. 15 m/s
C. 20 m/s
D. 5 m/s

Solution

Vertical component = Initial speed × sin(angle) = 20 m/s × sin(30°) = 20 m/s × 0.5 = 10 m/s.

Correct Answer: A — 10 m/s

Q. A player hits a baseball at an angle of 30 degrees with an initial speed of 40 m/s. What is the horizontal component of the velocity? (2020)

A. 20
B. 34.64
C. 30
D. 35

Solution

Horizontal component = initial speed × cos(angle) = 40 m/s × cos(30°) = 40 × √3/2 ≈ 34.64 m/s.

Correct Answer: B — 34.64

Q. A player hits a baseball at an angle of 30 degrees with an initial speed of 40 m/s. What is the vertical component of the velocity? (2020)

A. 20 m/s
B. 30 m/s
C. 35 m/s
D. 25 m/s

Solution

Vertical component = initial speed × sin(angle) = 40 m/s × sin(30°) = 40 m/s × 0.5 = 20 m/s.

Correct Answer: A — 20 m/s

Q. A player runs 100 meters in 12 seconds. What is his average speed in m/s? (2019)

A. 8.33
B. 9.00
C. 7.50
D. 8.00

Solution

Average speed = Distance / Time = 100 meters / 12 seconds = 8.33 m/s.

Correct Answer: A — 8.33

Q. A player runs 5 km in 25 minutes. What is his speed in km/h? (2023)

A. 10
B. 12
C. 14
D. 15

Solution

Total time in hours = 25 minutes / 60 = 25/60 hours. Speed = distance/time = 5 km / (25/60) h = 12 km/h.

Correct Answer: A — 10

Q. A player runs around a circular track of radius 10 meters. What is the circumference of the track? (2022)

A. 62.83 m
B. 31.42 m
C. 20 m
D. 40 m

Solution

Circumference = 2 * π * radius = 2 * π * 10 = 62.83 m.

Correct Answer: A — 62.83 m

Q. A player runs around a circular track of radius 50 meters. What is the distance covered in one complete round? (2019)

A. 100π
B. 200π
C. 50π
D. 25π

Solution

Circumference = 2πr = 2π * 50 = 100π meters.

Correct Answer: A — 100π

Q. A player scores 10 goals in 5 matches. What is the average number of goals scored per match? (2019)

A. 1.5
B. 2
C. 2.5
D. 3

Solution

Average goals per match = total goals / total matches = 10 / 5 = 2.

Correct Answer: B — 2

Q. A player scores 15 points in a game. If he plays 10 games, what is the total number of points scored? (2023)

A. 150
B. 120
C. 130
D. 140

Solution

Total points scored = 15 points/game * 10 games = 150 points.

Correct Answer: A — 150

Q. A player scores 5 goals in a match. If each goal is worth 2 points, how many points does the player score? (2023)

A. 10
B. 15
C. 5
D. 20

Solution

Total points = number of goals × points per goal = 5 × 2 = 10 points.

Correct Answer: A — 10

Q. A player scores 60% of the total points in a game. If the total points are 150, how many points did the player score? (2022)

A. 90
B. 80
C. 70
D. 60

Solution

Points scored = 60% of 150 = 0.6 × 150 = 90 points.

Correct Answer: A — 90

Q. A point charge +Q is placed at the center of a spherical Gaussian surface of radius R. What is the electric flux through the surface?

A. 0
B. Q/ε₀
C. Q/(4πε₀R²)
D. Q/(4πε₀R)

Solution

The electric flux through the Gaussian surface is given by Φ = Q/ε₀, where Q is the charge enclosed.

Correct Answer: B — Q/ε₀

Q. A point charge +Q is placed at the center of a spherical Gaussian surface. What is the total electric flux through the surface?

A. 0
B. Q/ε₀
C. Q/4πε₀
D. 4πQ/ε₀

Solution

The total electric flux through the surface is given by Gauss's law as Φ = Q/ε₀, and for a point charge at the center, it results in 4πQ/ε₀.

Correct Answer: D — 4πQ/ε₀

Q. A point charge of +5 µC is placed at the origin. What is the electric potential at a point 2 m away from the charge?

A. 1125 V
B. 450 V
C. 225 V
D. 0 V

Solution

The electric potential V at a distance r from a point charge Q is given by V = kQ/r. Here, V = (9 x 10^9 Nm²/C²)(5 x 10^-6 C)/(2 m) = 1125 V.

Correct Answer: A — 1125 V

Q. A point charge of +Q is placed at the center of a spherical Gaussian surface. What is the total electric flux through the surface?

A. 0
B. Q/ε₀
C. Q/2ε₀
D. 4πQ/ε₀

Solution

According to Gauss's law, the total electric flux through a closed surface is Φ = Q_enc/ε₀. Here, Q_enc = Q, so Φ = Q/ε₀.

Correct Answer: D — 4πQ/ε₀

Q. A point charge of +Q is placed at the center of a spherical shell of radius R with surface charge density σ. What is the electric field inside the shell?

A. 0
B. Q/(4πε₀R²)
C. σ/ε₀
D. Q/(4πε₀R)

Solution

According to Gauss's law, the electric field inside a conductor in electrostatic equilibrium is zero.

Correct Answer: A — 0

Q. A point charge Q is placed at the center of a cube. What is the electric flux through one face of the cube?

A. Q/ε₀
B. Q/6ε₀
C. Q/4ε₀
D. 0

Solution

The total flux through the cube is Q/ε₀. Since there are 6 faces, the flux through one face is Q/(6ε₀).

Correct Answer: B — Q/6ε₀

Q. A point charge Q is placed at the center of a spherical Gaussian surface. What is the electric flux through the surface?

A. 0
B. Q/ε₀
C. Q/4πε₀
D. Q²/ε₀

Solution

According to Gauss's law, the electric flux Φ = Q/ε₀ when a point charge Q is at the center of a spherical surface.

Correct Answer: B — Q/ε₀

Q. A policy affects 1,000 households. If 70% of them report satisfaction, how many households are satisfied?

A. 600
B. 700
C. 800
D. 900

Solution

70% of 1,000 = 0.70 * 1,000 = 700

Correct Answer: B — 700

Showing 2641 to 2670 of 31669 (1056 Pages)

School

College

Degree

Competitve

Skills

Soulshift Feedback ×

On a scale of 0–10, how likely are you to recommend The Soulshift Academy?

Not likely Very likely