JEE Physics - MCQ Practice Questions

First resonance in pipe closed at one end: L₁ = λ/4. So λ/4 = 0.16 m, λ = 0.64 m. v = fλ = 512 × 0.64 = 327.68 ≈ 328 m/s

Beat frequency = |f₁ - f₂| = 4 Hz only when waves overlap in the same region of space

At boundary from denser to less dense medium, reflected pulse is NOT inverted but has smaller amplitude due to partial reflection

Using v = fλ, f = v/λ = $\frac{340}{0}$.85 = 400 Hz

The fundamental wave equation is v = fλ, where v is velocity, f is frequency, and λ is wavelength

In the nth harmonic, the number of antinodes equals n. For the third harmonic (n=3), there are 3 antinodes

Intensity is proportional to the square of amplitude (I ∝ A²). When A becomes 2A, I becomes 4I

Using Doppler formula: f' = f(v + v_observer)/v = 500(340 + 20)/340 = 529.4 Hz

At equidistant points, path difference = 0, hence phase difference = 0°. This leads to constructive interference

Using Doppler formula: f' = f × v/(v + v_source) = 600 × 340/(340 + 30) = 547 Hz

f₁ = ($\frac{1}{2}$L)√(T/μ) where μ = 0.$\frac{01}{1}$ = 0.01 kg/m. f₁ = 0.5 × √($\frac{40}{0}$.01) = 31.6 Hz

For constructive interference, A_resultant = A₁ + A₂ = 3 + 4 = 7 cm

Diffraction is more prominent when wavelength is comparable to or larger than the aperture size

From y = A sin(2πx/λ - 2πt/T), we get λ = 4 m and T = 0.5 s. v = λ/T = $\frac{4}{0}$.5 = 8 m/s

In the second harmonic (n=2), f₂ = 2f₁. Therefore, f₁ = f₂/2 = $\frac{200}{2}$ = 100 Hz

Beat frequency = |f₁ - f₂| = |250 - 254| = 4 Hz

Using Snell's law: n₁sin(θ₁) = n₂sin(θ₂). 1.5 × sin(30°) = 1 × sin(θ₂). θ₂ = 48.75°

From y = A cos(kx) sin(ωt), we have k = π, so λ = 2π/k = 2π/π = 2 cm

# Solution: Organ Pipe Resonance Frequencies

For a pipe closed at one end, only odd harmonics can resonate due to boundary conditions at the closed end.

Step 1: Identify the Resonance Pattern for Closed Pipe

A pipe closed at one end has a node at the closed end and an antinode at the open end. This allows only odd multiples of the fundamental frequency to resonate.

Step 2: Calculate the Next Possible Resonance

The fundamental frequency (n = 1) is given as 200 Hz. The next possible resonance occurs at n = 2, which is the 3rd harmonic.

The frequency of the next possible resonance is 600 Hz.

The answer is (D) 600 Hz.

Sound intensity decreases with the square of the distance from a point source due to spherical spreading of the wavefront.

Step 1: Identify the Inverse Square Law

For a point source radiating uniformly in all directions, the intensity at distance r is inversely proportional to r². The power remains constant, but spreads over an increasing spherical surface area.

Step 2: Apply the Relationship Between Two Distances

We know intensity at r₁ = 1 m is I₁ = 10⁻⁸ W/m², and we need intensity at r₂ = 10 m.

The intensity at 10 m distance is 10⁻¹⁰ W/m². Answer: (A)