A system absorbs 500 J of heat and performs 200 J of work on the surroundings. What is the change in internal energy of the system?

The correct answer is A: 300 J. By first law of thermodynamics: ΔU = Q − W = 500 − 200 = 300 J

In a MOSFET, the threshold voltage (Vth) increases when:

The correct answer is B: Oxide thickness increases. Vth = Vfb + 2φF + (√(2εsqNa(2φF))/Cox). Threshold voltage increases with oxide thickness (Cox decreases) and increases with substrate doping. Temperature has weak effect.

A transverse wave travels along a stretched string with equation y = 0.02sin(50πt - 4πx), where x and y are in meters and t is in seconds. What is the wave velocity and wavelength of this wave?

The correct answer is A: v = 12.5 m/s, λ = 0.5 m. From y = 0.02sin(50πt - 4πx), comparing with y = Asin(ωt - kx): ω = 50π rad/s, k = 4π rad/m. Wavelength λ = 2π/k = 2π/(4π) = 0.5 m. Wave velocity v = ω/k = 50π/(4π) = 12.5 m/s.

In X-ray production, the minimum wavelength of continuous X-ray spectrum depends on:

The correct answer is B: Accelerating voltage only. Minimum wavelength λ_min = hc/eV, depends only on accelerating voltage V. Higher voltage produces shorter wavelength X-rays.

Compared to a p-n junction diode, a Schottky diode has:

The correct answer is B: Lower forward voltage drop and faster switching. Schottky diodes have lower barrier height (typically 0.3-0.5V vs 0.7V for Si), resulting in lower forward voltage. Faster switching due to majority carrier conduction (no minority carrier storage).

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JEE Physics
Waves

Physics questions for JEE Main — Mechanics, Electrostatics, Optics, Modern Physics.

16 Q 9 Topics Take Mock Test

Difficulty: All Easy Medium Hard 11–16 of 16

Topics in JEE Physics

All Mechanics 100 Thermodynamics 100 Electrostatics 100 Current Electricity 100 Magnetism 100 Optics 100 Modern Physics 100 Semiconductors 100 Waves 100

Q.11 Hard Waves

Two coherent sources with intensity ratio 4:1 interfere. The maximum intensity in the interference pattern is:

A 5I (where I is smaller intensity)

B 9I

C 12I

D 16I

Correct Answer: B. 9I

EXPLANATION

Let intensities be I and 4I. Amplitudes are A and 2A. Max intensity = (A + 2A)² = 9A² = 9I

Test

Q.12 Hard Waves

A string of mass 100 g and length 2 m is fixed at both ends. If tension is 200 N, what is the frequency of the fundamental mode?

A 22.4 Hz

B 31.6 Hz

C 44.7 Hz

D 50 Hz

Correct Answer: C. 44.7 Hz

EXPLANATION

v = √(T/μ) = √(200/(0.1/2)) = √4000 = 63.2 m/s. f = v/(2L) = 63.2/4 = 15.8 Hz... Recalculating: f₁ = √(T/μ)/(2L) where μ = 0.05 kg/m, so f₁ ≈ 44.7 Hz

Test

Q.13 Hard Waves

Two waves with intensities 4I and I interfere. The maximum intensity is:

A 5I

B 9I

C 4I

D (√5 + 1)I

Correct Answer: B. 9I

EXPLANATION

I_max = (√I₁ + √I₂)² = (√(4I) + √I)² = (2√I + √I)² = (3√I)² = 9I

Test

Q.14 Hard Waves

In a resonance tube experiment, the first resonance length is l₁ and second is l₂. The end correction is approximately:

A (l₂ - l₁)/2

B (l₂ - l₁)/3

C (l₂ + l₁)/2

D l₂ - l₁

Correct Answer: A. (l₂ - l₁)/2

EXPLANATION

For successive resonances in a resonance tube, l₂ - l₁ = λ/2, and end correction e ≈ (l₂ - l₁)/2 = λ/4

Test

Q.15 Hard Waves

A progressive wave y = 10sin(100πt - 0.01πx) cm is given. The wavelength is:

A 100 cm

B 200 cm

C 50 cm

D 400 cm

Correct Answer: B. 200 cm

EXPLANATION

Comparing with y = Asin(2πft - 2πx/λ), we have 2π/λ = 0.01π, so λ = 200 cm

Test

Q.16 Hard Waves

A transverse wave on a string has amplitude A, wavelength λ, and speed v. The maximum velocity of a particle is:

A v

B 2πAv/λ

C Av/λ

D A/v

Correct Answer: B. 2πAv/λ

EXPLANATION

Maximum particle velocity = ωA = 2πfA = 2πA(v/λ) = 2πAv/λ

Test

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