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. A wave has a frequency of 500 Hz and a wavelength of 0.68 m. What is the speed of the wave? (2021)

A. 340 m/s
B. 500 m/s
C. 680 m/s
D. 250 m/s

Solution

The speed of the wave can be calculated using the formula v = fλ. Thus, v = 500 Hz * 0.68 m = 340 m/s.

Correct Answer: A — 340 m/s

Q. A wave has a frequency of 60 Hz and a wavelength of 1.5 m. What is its speed? (2022)

A. 90 m/s
B. 60 m/s
C. 40 m/s
D. 30 m/s

Solution

Speed (v) = frequency (f) × wavelength (λ) = 60 Hz × 1.5 m = 90 m/s.

Correct Answer: A — 90 m/s

Q. A wave has a frequency of 60 Hz and a wavelength of 3 m. What is its speed? (2016)

A. 180 m/s
B. 120 m/s
C. 60 m/s
D. 90 m/s

Solution

Speed v = f × λ = 60 Hz × 3 m = 180 m/s.

Correct Answer: A — 180 m/s

Q. A wave has a frequency of 60 Hz and a wavelength of 3 m. What is the speed of the wave?

A. 180 m/s
B. 120 m/s
C. 60 m/s
D. 30 m/s

Solution

The speed v of a wave is given by v = fλ. Thus, v = 60 * 3 = 180 m/s.

Correct Answer: B — 120 m/s

Q. A wave has a speed of 300 m/s and a frequency of 150 Hz. What is its wavelength? (2022)

A. 1 m
B. 2 m
C. 3 m
D. 4 m

Solution

Wavelength λ can be calculated using the formula λ = v/f. Therefore, λ = 300 m/s / 150 Hz = 2 m.

Correct Answer: B — 2 m

Q. A wave has a wavelength of 0.5 m and a frequency of 300 Hz. What is its speed? (2020)

A. 150 m/s
B. 300 m/s
C. 600 m/s
D. 1200 m/s

Solution

Speed (v) = frequency (f) × wavelength (λ) = 300 Hz × 0.5 m = 150 m/s.

Correct Answer: A — 150 m/s

Q. A wave has a wavelength of 0.5 m and a frequency of 600 Hz. What is the speed of the wave?

A. 300 m/s
B. 600 m/s
C. 1200 m/s
D. 1500 m/s

Solution

The speed v of a wave is given by v = fλ. Here, v = 600 Hz * 0.5 m = 300 m/s.

Correct Answer: A — 300 m/s

Q. A wave has an amplitude of 0.5 m and a frequency of 10 Hz. What is the maximum speed of a particle in the wave?

A. 3.14 m/s
B. 6.28 m/s
C. 9.42 m/s
D. 12.56 m/s

Solution

The maximum speed v_max is given by v_max = ωA, where ω = 2πf. Thus, ω = 2π(10) = 20π and v_max = 20π(0.5) = 10π ≈ 31.42 m/s.

Correct Answer: B — 6.28 m/s

Q. A wave is described by the equation y(x, t) = 0.2 cos(4x - 5t). What is the amplitude of the wave?

A. 0.1 m
B. 0.2 m
C. 0.3 m
D. 0.4 m

Solution

The amplitude of the wave is the coefficient of the cosine function, which is 0.2 m.

Correct Answer: B — 0.2 m

Q. A wave on a string has a frequency of 10 Hz and a wavelength of 2 m. What is the speed of the wave? (2022)

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

Solution

Wave speed v is given by v = fλ. Thus, v = 10 Hz * 2 m = 20 m/s.

Correct Answer: B — 10 m/s

Q. A wave on a string has a frequency of 60 Hz and a wavelength of 0.5 m. What is the speed of the wave? (2021)

A. 30 m/s
B. 60 m/s
C. 120 m/s
D. 15 m/s

Solution

Speed (v) = Frequency (f) × Wavelength (λ) = 60 Hz × 0.5 m = 30 m/s.

Correct Answer: C — 120 m/s

Q. A wave on a string has a speed of 50 m/s and a wavelength of 2 m. What is its frequency? (2018)

A. 25 Hz
B. 50 Hz
C. 100 Hz
D. 75 Hz

Solution

Frequency f = v/λ = 50 m/s / 2 m = 25 Hz.

Correct Answer: A — 25 Hz

Q. A wave on a string is described by the equation y(x, t) = 0.05 cos(4x - 3t). What is the amplitude of the wave?

A. 0.05 m
B. 0.1 m
C. 0.2 m
D. 0.3 m

Solution

The amplitude of the wave is the coefficient of the cosine function, which is 0.05 m.

Correct Answer: A — 0.05 m

Q. A wave on a string is described by the equation y(x, t) = 0.05 sin(2π(0.1x - 5t)). What is the amplitude of the wave?

A. 0.01 m
B. 0.05 m
C. 0.1 m
D. 0.2 m

Solution

The amplitude of the wave is the coefficient in front of the sine function. Here, the amplitude is 0.05 m.

Correct Answer: B — 0.05 m

Q. A wave on a string is described by the equation y(x, t) = 0.1 sin(2π(0.5x - 10t)). What is the speed of the wave?

A. 5 m/s
B. 10 m/s
C. 20 m/s
D. 25 m/s

Solution

The speed of the wave v is given by v = fλ. Here, f = 10 Hz and λ = 2 m (from k = 2π/λ). Thus, v = 10 * 2 = 20 m/s.

Correct Answer: B — 10 m/s

Q. A wave on a string is described by the equation y(x, t) = 0.1 sin(2π(0.5x - 2t)). What is the speed of the wave?

A. 1 m/s
B. 2 m/s
C. 0.5 m/s
D. 4 m/s

Solution

The speed of the wave v is given by v = ω/k. Here, ω = 2π(2) = 4π rad/s and k = 2π(0.5) = π rad/m. Thus, v = (4π)/(π) = 4 m/s.

Correct Answer: B — 2 m/s

Q. A wave on a string is described by the equation y(x, t) = 0.1 sin(2πx - 4πt). What is the amplitude of the wave?

A. 0.1 m
B. 0.2 m
C. 0.4 m
D. 1 m

Solution

The amplitude is the coefficient of the sine function, which is 0.1 m.

Correct Answer: A — 0.1 m

Q. A wave on a string is described by the equation y(x, t) = A sin(kx - ωt). What does 'A' represent?

A. Wavelength
B. Frequency
C. Amplitude
D. Wave number

Solution

'A' represents the amplitude of the wave, which is the maximum displacement from the equilibrium position.

Correct Answer: C — Amplitude

Q. A wave on a string is described by the equation y(x, t) = A sin(kx - ωt). What does 'k' represent?

A. Angular frequency
B. Wave number
C. Amplitude
D. Phase constant

Solution

'k' is the wave number, which is related to the wavelength of the wave.

Correct Answer: B — Wave number

Q. A wave pulse travels along a string and reflects off a fixed end. What happens to the wave upon reflection? (2021)

A. It inverts
B. It remains unchanged
C. It doubles in amplitude
D. It disappears

Solution

When a wave pulse reflects off a fixed end, it inverts due to the boundary condition at the fixed end.

Correct Answer: A — It inverts

Q. A wave traveling along a string has a frequency of 50 Hz and a wavelength of 2 m. What is the speed of the wave? (2020)

A. 25 m/s
B. 50 m/s
C. 100 m/s
D. 75 m/s

Solution

Speed (v) = frequency (f) × wavelength (λ) = 50 Hz × 2 m = 100 m/s.

Correct Answer: C — 100 m/s

Q. A wave traveling along a string is described by the equation y(x, t) = 0.1 sin(2π(0.5x - 2t)). What is the speed of the wave?

A. 1 m/s
B. 2 m/s
C. 3 m/s
D. 4 m/s

Solution

The speed of the wave v is given by v = fλ. Here, f = 2 Hz and λ = 1 m (from the wave equation). Thus, v = 2 * 1 = 2 m/s.

Correct Answer: B — 2 m/s

Q. A wave traveling along a string is described by the equation y(x, t) = 0.1 sin(2π(0.5x - 4t)). What is the wave speed?

A. 2 m/s
B. 4 m/s
C. 8 m/s
D. 1 m/s

Solution

The wave speed v can be found from the equation v = fλ. Here, f = 4 Hz and λ = 2 m (from k = 2π/λ, k = 1 m⁻¹). Thus, v = 4 Hz * 2 m = 8 m/s.

Correct Answer: B — 4 m/s

Q. A wave traveling along a string is described by the equation y(x, t) = A sin(kx - ωt). What does 'k' represent?

A. Amplitude
B. Wave number
C. Frequency
D. Angular frequency

Solution

'k' is the wave number, which is related to the wavelength λ by the equation k = 2π/λ.

Correct Answer: B — Wave number

Q. A wave traveling along a string is described by the equation y(x, t) = A sin(kx - ωt). What does 'A' represent?

A. Wavelength
B. Amplitude
C. Frequency
D. Speed

Solution

'A' represents the amplitude of the wave, which is the maximum displacement from the equilibrium position.

Correct Answer: B — Amplitude

Q. A wave traveling along a string is described by the equation y(x, t) = A sin(kx - ωt). What does the parameter A represent?

A. Wavelength
B. Amplitude
C. Frequency
D. Speed

Solution

In the wave equation y(x, t) = A sin(kx - ωt), A represents the amplitude of the wave, which is the maximum displacement from the equilibrium position.

Correct Answer: B — Amplitude

Q. A wave traveling along a string is described by the equation y(x, t) = A sin(kx - ωt). What is the phase velocity of the wave?

A. A/k
B. ω/k
C. k/ω
D. Aω

Solution

The phase velocity v is given by v = ω/k.

Correct Answer: B — ω/k

Q. A wave traveling along a string is described by the equation y(x, t) = A sin(kx - ωt). If A = 2 m, k = 3 rad/m, and ω = 6 rad/s, what is the amplitude of the wave?

A. 1 m
B. 2 m
C. 3 m
D. 4 m

Solution

The amplitude of the wave is given directly by A in the wave equation. Here, A = 2 m.

Correct Answer: B — 2 m

Q. A wave travels along a string with a speed of 300 m/s and has a frequency of 150 Hz. What is the wavelength of the wave?

A. 1.0 m
B. 2.0 m
C. 3.0 m
D. 4.0 m

Solution

Wavelength λ is given by λ = v/f. Here, v = 300 m/s and f = 150 Hz. Thus, λ = 300/150 = 2.0 m.

Correct Answer: B — 2.0 m

Q. A wave travels at a speed of 300 m/s and has a frequency of 150 Hz. What is its wavelength?

A. 2 m
B. 1.5 m
C. 3 m
D. 0.5 m

Solution

The wavelength λ can be calculated using the formula v = fλ. Thus, λ = v/f = 300 m/s / 150 Hz = 2 m.

Correct Answer: B — 1.5 m

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