Question

The absolute temperature of air in a region linearly increases from $T_1$ to $T_2$ in a space of width d. Find the time taken by a sound wave to go through the region in terms of $T_1,T_2,d$ and the speed v of sound at 273 K.

• A. $\frac{d}{v(\sqrt{T_2}+\sqrt{T_1})}$
• B. $\frac{d\sqrt{273}}{v(\sqrt{T_2}+\sqrt{T_1})}$
• C. $\frac{d}{2v}\frac{\sqrt{273}}{v(\sqrt{T_2}+\sqrt{T_1})}$
• D. None of these

Answer 1

The correct option is C

$\frac{d}{2v}\frac{\sqrt{273}}{v(\sqrt{T_2}+\sqrt{T_1})}$

If temperature increases from $T{­}_{1}to{T}_2$ linearly with distance.Then the equation of temperature (T) with Distance (x) will be

$T-T_1=\left(\frac{T_2-T_1}{d}\right)(n-0)$

$T==(\frac{(T_2-T_1)}{d}x+T_1$

$T=\left(\frac{T_2-T_1}{d}\right)x+T_1$

At an arbitary point x,

Speed of sound =$\sqrt{\frac{\gamma RT}{M}}$

$=\sqrt{\frac{\gamma R}{d}[\left(\frac{T_2-T_1}{d}\right)x+T_1]}$

On small displacment dx the time taken

$dt=\frac{dx}{\sqrt{\frac{\gamma R}{M}\frac{(T_2-T_1)}{d}x+\frac{\gamma R{T}_1}{M}}}$

$t=\int dt=\underset{0}{\overset{d}{\int }}\frac{dx}{\sqrt{\frac{\gamma R}{M}\frac{(T_2-T_1)}{d}x+\frac{\gamma R{T}_1}{M}}}$

$=\frac{\frac{1}{2}\sqrt{\frac{\gamma R}{M}\frac{(T_2-T_1)}{d}x+\frac{\gamma R{T}_1}{M}}}{\frac{\gamma R(T_2-T_1)}{Md}}{|}_0^d$

$=\frac{Md}{2\gamma R(T_2-T_1)}[\sqrt{\frac{\gamma R{T}_2}{M}}-\sqrt{\frac{\gamma R{T}_1}{M}})$

$=\frac{d}{2}\sqrt{\frac{M}{\gamma R}}x\frac{1}{(T_2-T_1)}[\sqrt{T_2}-\sqrt{T_1}]$

as $\sqrt{\frac{\gamma R\times 273}{M}}=v$

$\sqrt{\frac{\gamma R}{M}}=\frac{v}{\sqrt{273}}\Rightarrow \sqrt{\frac{M}{\gamma R}}=\frac{\sqrt{273}}{v}$

$=\frac{d}{2}\frac{\sqrt{273}}{v}\frac{1}{(\sqrt{T_2}+\sqrt{T_1})}$