A copper wire and an aluminum wire of the same length and cross-sectional area carry the same current. Which wire has greater power dissipation?

The correct answer is B: Aluminum wire. Resistivity of aluminum (2.65 × 10⁻⁸ Ω·m) is greater than copper (1.68 × 10⁻⁸ Ω·m). Since P = I²R and R = ρL/A, aluminum has higher resistance and thus higher power dissipation for same current.

The electric field inside a copper conductor carrying current is approximately:

The correct answer is B: E = ρJ, where ρ is resistivity and J is current density. By Ohm's law in microscopic form: E = ρJ. Inside the conductor, electric field maintains the drift of electrons

The refractive index of a material is found to be 1.5. The speed of light in this material is:

The correct answer is B: 2.0 × 10⁸ m/s. n = c/v. 1.5 = (3 × 10⁸)/v. v = (3 × 10⁸)/1.5 = 2 × 10⁸ m/s

The Doppler effect is observed when:

The correct answer is B: Relative motion exists between source and observer. Doppler effect occurs due to relative motion between source and observer, causing apparent frequency change

Two resistors R₁ and R₂ are connected in series with a battery. If R₁ = 2R₂ and the voltage across R₁ is 8V, the voltage across R₂ is:

The correct answer is A: 4V. In series, current is same. V₁/V₂ = R₁/R₂ = 2R₂/R₂ = 2. Therefore V₂ = V₁/2 = 8/2 = 4V

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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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