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'm' grams of a gas of a molecular weight M is flowing in an isolated tube with velocity 2V. If the gas flow is suddenly stopped the rise in its temperature is; ( $γ$ = ratio of specific heats; R = universal gas constant; J = Mechanical equivalent of heat)

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$\frac{2 {Mv}^{2} (γ - 1)}{RJ}$

$\frac{{mv}^{2} (γ - 1)}{M 2 RJ}$

$\frac{{mv}^{2} γ}{2 RJ}$

$\frac{{Mv}^{2} γ}{2 RJ}$

answer is A.

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Detailed Solution

U = 2V; V = 0, dQ = 0 (Suddenly stopped)
$ΔKE$ = Wadia
$\begin{array}{l} \frac{1}{2} m (2 v)^{2} = J \frac{m}{M} \times \frac{R}{γ - 1} \times dT \Rightarrow 2 V^{2} = \frac{JRdT}{M (γ - 1)} \\ dT = \frac{2 V^{2} M (γ - 1)}{JR} \end{array}$

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