The speed of a longitudinal wave in a metalli | vedclass

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(D) The speed of a longitudinal wave in a metallic bar is given by $V = \sqrt{\frac{Y}{\rho}}$, where $Y$ is Young's modulus and $\rho$ is the density

Vedclass · Accepted Answer

$401$

Vedclass · Answer

(D) The speed of a longitudinal wave in a metallic bar is given by $V = \sqrt{\frac{Y}{ho}}$, where $Y$ is Young's modulus and $ho$ is the density.Taking the logarithmic derivative, we get $\frac{\Delta V}{V} = \frac{1}{2} \frac{\Delta Y}{Y} - \frac{1}{2} \frac{\Delta ho}{ho}$.Given $\frac{\Delta Y}{Y} 	imes 100 = 1\%$ and $\frac{\Delta ho}{ho} 	imes 100 = 0.5\%$.Substituting these values, the percentage change in speed is $\frac{\Delta V}{V} 	imes 100 = \frac{1}{2}(1\%) - \frac{1}{2}(0.5\%) = 0.5\% - 0.25\% = 0.25\%$.The change in speed $\Delta V = \frac{0.25}{100} 	imes 400 \ m/s = 1 \ m/s$.The final speed $V_{final} = V + \Delta V = 400 + 1 = 401 \ m/s$.

$A$. 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

The speed of a longitudinal wave in a metallic bar is $400 \ m/s$. If the density and Young's modulus of the bar material are increased by $0.5\%$ and $1\%$ respectively, then the speed of the wave is changed approximately to . . . . . . $m/s$.

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