Assume that the displacement(s) of air is proportional to the pressure difference ( Δ p) created by a sound wave. Displacement (s) further depends on the speed of sound (v), density of air ( ρ) and the frequency (f). If Δ p ~ 10 Pa, v ~ 300 m/s, ρ ~ 1 kg/m3 and f ~ 1000 Hz, then s will be of the order of (take the multiplicative constant to be 1): (a) 1 mm (b) 3/100 mm (c) 10 mm (d) 1/10 mm

Answer: (b). Write s = Δ p va ρb fc; [s] = L and [Δ p] = M L ⁻¹ T ⁻². Matching M, L, T gives b = -1, a = -1, c = -1: s = (Δ p)/(ρ v f). s = (10)/(1 × 300 × 1000) = 3.3 × 10⁻⁵ m ≈ 3/100 mm So the answer is (b) 3/100 mm.

Assume that the displacement(s) of air is proportional to the… | JEE Main 2020 (5 Sep, Shift 1)

Q29JEE Main 2020 · 5 SEP · SHIFT 1Medium

Wave Basics & Properties of Sound

Assume that the displacement(s) of air is proportional to the pressure difference ( $\Delta$ p) created by a sound wave. Displacement (s) further depends on the speed of sound (v), density of air ( $\rho$ ) and the frequency (f). If $\Delta$ p ~ 10 Pa, v ~ 300 m/s, $\rho$ ~ 1 kg/m3 and f ~ 1000 Hz, then s will be of the order of (take the multiplicative constant to be 1):

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Answer: (b)

Write $s = \Delta p\, v^a \rho^b f^c$ ; $[s] =$ L and $[\Delta p] =$ M L $^{-1}$ T $^{-2}$ .
Matching M, L, T gives $b = -1$ , $a = -1$ , $c = -1$ : $s = \frac{\Delta p}{\rho v f}$ .
$s = \frac{10}{1 \times 300 \times 1000} = 3.3 \times 10^{-5}$ m $\approx \frac{3}{100}$ mm
So the answer is (b) $\frac{3}{100}$ mm.

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Wave Basics & Properties of Sound question from JEE Main 2020 (5 Sep, Shift 1)

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