Sound waves travel from a location A at absolute temperature $T_{1}$ to location B at absolute temperature $T_{2}$ in a time t . The air temperature increase linearly from $T_{1}$ to $T_{2}$ . The speed of sound varies with absolute temperature T as $V = K \sqrt{T}$ where K is a positive constant. Then

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The distance average velocity over the distance AB is $\frac{K (\sqrt{T_{1}} + \sqrt{T_{2}})}{2}$

The distance between A and B is $\frac{Kt}{2} (\sqrt{T_{1}} + \sqrt{T_{2}})$

The distance average velocity over the distance AB is $\frac{2 K}{3} \frac{(T_{2}^{3 / 2} - T_{1}^{3 / 2})}{(T_{2} - T_{1})}$

The distance between A and B is $Kt (\sqrt{T_{1}} + \sqrt{T_{2}})$

answer is A, B.

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Detailed Solution

$\begin{array}{l} \frac{dT}{dx} = (\frac{T_{2} - T_{1}}{L}) \Rightarrow T (x) = T_{1} + (\frac{T_{2} - T_{1}}{L}) x \frac{dx}{dt} = K \sqrt{T} = K \sqrt{T_{1} + (\frac{T_{2} - T_{1}}{L})} x \\ t = \int_{0}^{L} \frac{dx}{K \sqrt{T_{1} + (\frac{T_{2} - T_{1}}{L})}} = (\frac{L}{T_{2} - T_{1}}) \frac{1}{K} [2 \sqrt{T_{2}} - 2 \sqrt{T_{1}}] \\ = \frac{2 L}{K (\sqrt{T_{2}} + \sqrt{T_{1}})} L = \frac{kt}{2} (\sqrt{T_{2}} + \sqrt{T_{1}}) \\ V_{avg} = \frac{\int_{0}^{2} vdx}{L} = \frac{\int_{0}^{2} K \sqrt{T_{1} + (\frac{T_{2} - T_{1}}{L}) xdx}}{L} = \frac{2 k}{3} \frac{(T_{2}^{3 / 2} - T_{1}^{3 / 2})}{(T_{2} - T_{1})} \end{array}$