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(C) The speed of sound in a liquid is given by the formula $v = \sqrt{\frac{1}{K \rho}}$, where $K$ is the compressibility and $\rho$ is the density o

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$2.5$

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(C) The speed of sound in a liquid is given by the formula $v = \sqrt{\frac{1}{K ho}}$, where $K$ is the compressibility and $ho$ is the density of the medium.Given: $K = 4.84 	imes 10^{-10} \,m^2 \,N^{-1}$, $ho = 1024 \,kg \,m^{-3}$.Substituting the values: $v = \sqrt{\frac{1}{4.84 	imes 10^{-10} 	imes 1024}} = \sqrt{\frac{1}{4956.16 	imes 10^{-10}}} = \sqrt{\frac{10^{10}}{4956.16}} \approx \sqrt{2017700} \approx 1420.46 \,m/s$.The total time taken for the echo is $t = 3.52 \,s$. The time taken for the sound to reach the bottom is $t' = \frac{t}{2} = \frac{3.52}{2} = 1.76 \,s$.The depth $d$ is given by $d = v 	imes t' = 1420.46 	imes 1.76 \approx 2500 \,m = 2.5 \,km$.

$A$. 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

The depth of the sea is measured by means of an echo. If the time between the moment the sound is produced and received is $3.52 \,s$, then the depth of the sea is (The compressibility of sea water is $4.84 \times 10^{-10} \,m^2 \,N^{-1}$ and its density is $1024 \,kg \,m^{-3}$): (in $km$)

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