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(B) The relationship between velocity $(v)$,frequency $(n)$,and wavelength $(\lambda)$ is given by $v = n \lambda$.Since both sound sources are at the

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$303$

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(B) The relationship between velocity $(v)$,frequency $(n)$,and wavelength $(\lambda)$ is given by $v = n \lambda$.Since both sound sources are at the same temperature,the velocity of sound $(v)$ is the same for both waves.Let the frequencies be $n_1$ and $n_2$ corresponding to wavelengths $\lambda_1 = 50 \ cm = 0.5 \ m$ and $\lambda_2 = 50.5 \ cm = 0.505 \ m$.We have $v = n_1 \lambda_1 = n_2 \lambda_2$.Thus,$n_1 (0.5) = n_2 (0.505)$,which implies $\frac{n_1}{n_2} = \frac{0.505}{0.5} = 1.01$.Given $|n_1 - n_2| = 6 \ Hz$,we can write $n_1 = n_2 + 6$.Substituting this into the ratio: $\frac{n_2 + 6}{n_2} = 1.01$.$1 + \frac{6}{n_2} = 1.01 \implies \frac{6}{n_2} = 0.01$.$n_2 = \frac{6}{0.01} = 600 \ Hz$.Now,calculate the velocity: $v = n_2 \lambda_2 = 600 	imes 0.505 \ m = 303 \ m/s$.

$A$. Temperature of gas is made $4$ times and pressure $2$ times	$P$. Speed becomes $2\sqrt{2}$ times
$B$. Only pressure is made $4$ times without change in temperature	$Q$. Speed becomes $2$ times
$C$. Only temperature is changed to $4$ times	$R$. Speed remains unchanged
$D$. Molecular mass of the gas is made $4$ times	$S$. Speed becomes half

Two sounding sources send waves at a certain temperature in air of wavelength $50 \ cm$ and $50.5 \ cm$ respectively. The frequency of the sources differs by $6 \ Hz$. The velocity of sound in air at the same temperature is (in $m/s$)

Similar Questions

Which one of the following statements is true?

The velocity of sound in air is:

It takes $2.0 \ s$ for a sound wave to travel between two fixed points when the day temperature is $10^{\circ}C$. If the temperature rises to $30^{\circ}C$,the sound wave travels between the same fixed points in ...... $s$.

$Assertion :$ The change in air pressure affects the speed of sound.
$Reason :$ The speed of sound in gases is proportional to the square of pressure.

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