Two sounding sources send waves at certain temperature in air of wavelength 50 cm and 50.5 cm respectively. The frequency of sources differ by 6 Hz. The velocity of sound in air at same temperature is (a) 300 m/s (b) 303 m/s (c) 313 m/s (d) 330 m/s

Answer: (b). n₁ - n₂ = (v)/(λ₁) - (v)/(λ₂) = 6 with λ₁ = 0.50 m, λ₂ = 0.505 m. v · (0.005)/(0.50 × 0.505) = 6 v = (6 × 0.2525)/(0.005) = 303 m/s So the answer is (b) 303 m/s.

Two sounding sources send waves at certain temperature in air of… | MHT CET 2023 (12 May, Shift 1)

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Two sounding sources send waves at certain temperature in air of wavelength $50\text{ cm}$ and $50.5\text{ cm}$ respectively. The frequency of sources differ by $6\text{ Hz}$ . The velocity of sound in air at same temperature is

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Answer: (b)

$n_1 - n_2 = \dfrac{v}{\lambda_1} - \dfrac{v}{\lambda_2} = 6$ with $\lambda_1 = 0.50\text{ m}$ , $\lambda_2 = 0.505\text{ m}$ .
$v \cdot \dfrac{0.005}{0.50 \times 0.505} = 6$
$v = \dfrac{6 \times 0.2525}{0.005} = 303\text{ m/s}$
So the answer is (b) $303\text{ m/s}$ .

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