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. A sound wave travels in air at a speed of 340 m/s. If the frequency of the sound is 170 Hz, what is the wavelength?
A.
2 m
B.
1 m
C.
0.5 m
D.
0.2 m
Show solution
Solution
Wavelength λ = v/f = 340 m/s / 170 Hz = 2 m.
Correct Answer:
A
— 2 m
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Q. A sound wave travels in air at a speed of 340 m/s. If the frequency of the sound wave is 1700 Hz, what is its wavelength?
A.
0.2 m
B.
0.5 m
C.
2 m
D.
1 m
Show solution
Solution
Wavelength λ = v/f = 340 m/s / 1700 Hz = 0.2 m.
Correct Answer:
B
— 0.5 m
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Q. A sound wave travels through a medium with a speed of 340 m/s and has a frequency of 1700 Hz. What is its wavelength?
A.
0.2 m
B.
0.5 m
C.
2 m
D.
1 m
Show solution
Solution
Wavelength = Speed / Frequency = 340 m/s / 1700 Hz = 0.2 m.
Correct Answer:
B
— 0.5 m
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Q. A sound wave travels through air at 343 m/s. If its frequency is 686 Hz, what is its wavelength? (2023)
A.
0.5 m
B.
1 m
C.
2 m
D.
3 m
Show solution
Solution
Wavelength λ = v/f = 343 m/s / 686 Hz = 0.5 m.
Correct Answer:
A
— 0.5 m
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Q. A sound wave travels through air at a speed of 340 m/s. If its frequency is 1700 Hz, what is its wavelength? (2022)
A.
0.2 m
B.
0.5 m
C.
1 m
D.
2 m
Show solution
Solution
Wavelength (λ) = speed (v) / frequency (f) = 340 m/s / 1700 Hz = 0.2 m.
Correct Answer:
B
— 0.5 m
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Q. A sound wave travels through air at a speed of 340 m/s. If the frequency of the sound is 1700 Hz, what is the wavelength?
A.
0.2 m
B.
0.5 m
C.
1 m
D.
2 m
Show solution
Solution
Using the formula v = fλ, we can rearrange to find λ = v/f. Thus, λ = 340 m/s / 1700 Hz = 0.2 m.
Correct Answer:
A
— 0.2 m
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Q. A sound wave travels through air at a speed of 340 m/s. If the frequency of the sound is 170 Hz, what is the wavelength?
A.
1 m
B.
2 m
C.
3 m
D.
4 m
Show solution
Solution
The wavelength λ is given by λ = v/f. Thus, λ = 340/170 = 2 m.
Correct Answer:
B
— 2 m
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Q. A sound wave travels through air at a speed of 340 m/s. If the wavelength is 0.85 m, what is the frequency? (2022)
A.
400 Hz
B.
300 Hz
C.
250 Hz
D.
200 Hz
Show solution
Solution
Frequency (f) = speed (v) / wavelength (λ) = 340 m/s / 0.85 m ≈ 400 Hz.
Correct Answer:
A
— 400 Hz
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Q. A sound wave travels through air at a speed of 340 m/s. If the wavelength is 0.85 m, what is the frequency of the sound wave? (2022)
A.
400 Hz
B.
300 Hz
C.
250 Hz
D.
500 Hz
Show solution
Solution
The frequency f can be calculated using the formula v = fλ. Rearranging gives f = v/λ = 340 m/s / 0.85 m ≈ 400 Hz.
Correct Answer:
A
— 400 Hz
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Q. A sound wave travels through air at a speed of 340 m/s. What is the wavelength of a sound wave with a frequency of 1700 Hz? (2022)
A.
0.2 m
B.
0.5 m
C.
1.0 m
D.
2.0 m
Show solution
Solution
Using the formula λ = v/f, where v is the speed and f is the frequency, we have λ = 340 m/s / 1700 Hz = 0.2 m.
Correct Answer:
A
— 0.2 m
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Q. A sound wave travels through air at a speed of 343 m/s. If the frequency of the sound is 686 Hz, what is the wavelength? (2022)
A.
0.5 m
B.
1 m
C.
2 m
D.
3 m
Show solution
Solution
Using the wave speed formula v = f * λ, we can rearrange to find λ = v/f. Thus, λ = 343 m/s / 686 Hz = 0.5 m.
Correct Answer:
B
— 1 m
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Q. A sound wave travels through water at a speed of 1500 m/s. If the frequency of the sound wave is 300 Hz, what is the wavelength?
A.
2 m
B.
3 m
C.
4 m
D.
5 m
Show solution
Solution
Wavelength λ = v/f = 1500 m/s / 300 Hz = 5 m.
Correct Answer:
A
— 2 m
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Q. A space probe travels to Mars, which is 225 million km away. If it travels at a speed of 45,000 km/h, how long will it take to reach Mars in days?
A.
125 days
B.
150 days
C.
200 days
D.
180 days
Show solution
Solution
Time = Distance / Speed = 225,000,000 km / 45,000 km/h = 5,000 h. In days: 5,000 h / 24 h/day = 208.33 days.
Correct Answer:
B
— 150 days
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Q. A space shuttle can carry 7 astronauts. If there are 35 astronauts waiting for a mission, how many trips does the shuttle need to make?
Show solution
Solution
Number of trips = Total astronauts / Capacity = 35 / 7 = 5 trips.
Correct Answer:
B
— 5
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Q. A speed is measured as 20 m/s with an uncertainty of ±0.5 m/s. If this speed is used to calculate kinetic energy, what is the percentage error in kinetic energy?
A.
5%
B.
2.5%
C.
1%
D.
10%
Show solution
Solution
K.E. = 0.5 * m * v², percentage error = 2 * (Δv/v) = 2 * (0.5/20) = 0.05 or 5%.
Correct Answer:
B
— 2.5%
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Q. A speed is measured as 20 m/s with an uncertainty of ±0.5 m/s. What is the absolute error in the speed measurement?
A.
0.5 m/s
B.
0.25 m/s
C.
1 m/s
D.
0.1 m/s
Show solution
Solution
The absolute error is given directly as ±0.5 m/s.
Correct Answer:
A
— 0.5 m/s
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Q. A speed is recorded as 60 km/h with an error of 2 km/h. What is the percentage error?
A.
3.33%
B.
2.5%
C.
1.67%
D.
4%
Show solution
Solution
Percentage error = (Absolute error / Measured value) * 100 = (2 / 60) * 100 = 3.33%
Correct Answer:
A
— 3.33%
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Q. A speed of 30 m/s is measured with an uncertainty of ±0.5 m/s. What is the total uncertainty if this speed is used to calculate kinetic energy?
A.
0.25 J
B.
0.5 J
C.
1 J
D.
2 J
Show solution
Solution
Kinetic energy = 0.5 * m * v²; uncertainty in KE = 2 * v * uncertainty in v = 2 * 30 * 0.5 = 30 J.
Correct Answer:
C
— 1 J
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Q. A sphere has a diameter of 10 cm. What is its surface area?
A.
314.16 cm²
B.
250.00 cm²
C.
200.00 cm²
D.
150.00 cm²
Show solution
Solution
Surface Area = 4πr² = 4π(5)² = 100π ≈ 314.16 cm²
Correct Answer:
A
— 314.16 cm²
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Q. A sphere has a radius of 4 cm. What is its volume?
A.
32π cm³
B.
64π cm³
C.
48π cm³
D.
16π cm³
Show solution
Solution
Volume = (4/3)πr³ = (4/3)π × (4 cm)³ = 64π cm³.
Correct Answer:
A
— 32π cm³
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Q. A sphere has a radius of 5 cm. What is its volume?
A.
100π/3 cm³
B.
125π/3 cm³
C.
150π/3 cm³
D.
75π/3 cm³
Show solution
Solution
Volume = (4/3)πr³ = (4/3)π × (5 cm)³ = (4/3)π × 125 cm³ = 125π/3 cm³.
Correct Answer:
B
— 125π/3 cm³
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Q. A sphere has a radius of 7 cm. What is its surface area?
A.
154 cm²
B.
196 cm²
C.
308 cm²
D.
616 cm²
Show solution
Solution
Surface area of a sphere = 4πr² = 4π(7)² = 4π(49) = 196 cm².
Correct Answer:
B
— 196 cm²
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Q. A sphere rolls down a ramp of height h. What is the total mechanical energy at the top?
A.
mgh
B.
1/2 mv^2
C.
mgh + 1/2 mv^2
D.
0
Show solution
Solution
The total mechanical energy at the top is purely potential energy, which is mgh.
Correct Answer:
A
— mgh
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Q. A sphere rolls down a ramp. If the height of the ramp is h, what is the speed of the sphere at the bottom assuming no energy loss?
A.
√(2gh)
B.
√(3gh)
C.
√(4gh)
D.
√(gh)
Show solution
Solution
Using conservation of energy, the potential energy at height h converts to kinetic energy at the bottom, giving speed √(2gh).
Correct Answer:
A
— √(2gh)
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Q. A sphere rolls on a flat surface with a speed v. What is the kinetic energy of the sphere?
A.
(1/2)mv^2
B.
(1/2)mv^2 + (1/5)mv^2
C.
(1/2)mv^2 + (2/5)mv^2
D.
(1/2)mv^2 + (3/5)mv^2
Show solution
Solution
The total kinetic energy of a rolling sphere is the sum of translational and rotational kinetic energy, which is (1/2)mv^2 + (2/5)mv^2.
Correct Answer:
C
— (1/2)mv^2 + (2/5)mv^2
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Q. A sphere rolls without slipping on a flat surface. If it has a radius R and rolls with a speed v, what is its angular speed?
A.
v/R
B.
2v/R
C.
v/2R
D.
v²/R
Show solution
Solution
The angular speed ω of a rolling sphere is given by ω = v/R.
Correct Answer:
A
— v/R
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Q. A spherical conductor has a charge Q. What is the electric potential inside the conductor?
A.
0
B.
Q/(4πε₀r)
C.
Q/(4πε₀R)
D.
Constant throughout
Show solution
Solution
The electric potential inside a charged spherical conductor is constant and equal to the potential on its surface, which is Q/(4πε₀R).
Correct Answer:
D
— Constant throughout
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Q. A spherical conductor has a radius R and carries a charge Q. What is the electric potential on its surface?
A.
kQ/R
B.
kQ/2R
C.
0
D.
kQ/R²
Show solution
Solution
The electric potential V on the surface of a charged spherical conductor is given by V = kQ/R.
Correct Answer:
A
— kQ/R
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Q. A spherical Gaussian surface of radius R encloses a charge Q. What is the electric field at a distance 2R from the center?
A.
Q/4πε₀R²
B.
Q/4πε₀(2R)²
C.
0
D.
Q/ε₀(2R)²
Show solution
Solution
The electric field outside a spherical charge distribution is given by E = Q/4πε₀r². At 2R, it becomes Q/4πε₀(2R)².
Correct Answer:
B
— Q/4πε₀(2R)²
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Q. A spherical shell of radius R carries a total charge Q. What is the electric field at a point outside the shell?
A.
0
B.
Q/(4πε₀R²)
C.
Q/(4πε₀R)
D.
Q/(4πε₀R³)
Show solution
Solution
For a spherical shell, the electric field outside the shell behaves as if all the charge were concentrated at the center, so E = Q/(4πε₀R²).
Correct Answer:
B
— Q/(4πε₀R²)
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