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Question

(A) The speed of sound in a gas is given by $v = \sqrt{\frac{\gamma RT}{M}}$,which implies $v \propto \sqrt{T}$,where $T$ is the absolute temperature

Vedclass · Accepted Answer

$1.9$

Vedclass · Answer

(A) The speed of sound in a gas is given by $v = \sqrt{\frac{\gamma RT}{M}}$,which implies $v \propto \sqrt{T}$,where $T$ is the absolute temperature in Kelvin.Let $d$ be the distance between the two fixed points.The time taken is $t = \frac{d}{v}$.Therefore,$\frac{t_1}{t_2} = \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.Given $T_1 = 10^{\circ}C = 10 + 273 = 283 \ K$ and $T_2 = 30^{\circ}C = 30 + 273 = 303 \ K$.Given $t_1 = 2.0 \ s$.Substituting the values: $\frac{2.0}{t_2} = \sqrt{\frac{303}{283}} \approx \sqrt{1.0706} \approx 1.0347$.$t_2 = \frac{2.0}{1.0347} \approx 1.93 \ s$.Rounding to the nearest given option,the correct answer is $1.9 \ s$.

$A$. 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

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 ...... $s$.

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