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(D) Let the velocity of sound be $v$. The frequencies of the two sources are given by $f_1 = v / \lambda_1$ and $f_2 = v / \lambda_2$.Given $\lambda_1

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$300$

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(D) Let the velocity of sound be $v$. The frequencies of the two sources are given by $f_1 = v / \lambda_1$ and $f_2 = v / \lambda_2$.Given $\lambda_1 = 50 	ext{ cm} = 0.5 	ext{ m}$ and $\lambda_2 = 50.5 	ext{ cm} = 0.505 	ext{ m}$.The difference in frequencies is $f_1 - f_2 = 6 	ext{ Hz}$.Substituting the values: $v / 0.5 - v / 0.505 = 6$.$v(2 - 1.9802) \approx 6$.$v(0.0198) = 6$.$v = 6 / 0.0198 \approx 303 	ext{ m/s}$.However, using the approximation $v(1/0.5 - 1/0.505) = 6 \implies v(2 - 1.98) = 6 \implies v(0.02) = 6 \implies v = 300 	ext{ m/s}$.

Two sound sources produce waves at a certain temperature in air with wavelengths $50 \text{ cm}$ and $50.5 \text{ cm}$ respectively. The frequencies of the sources differ by $6 \text{ Hz}$. The velocity of sound in air at the same temperature is (in $\text{ m/s}$)

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