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A horizontal tube of length $l$ closed at both ends, contains an ideal gas of molecular weight $M$. The tube is rotated at a constant angular velocity $\omega$ about a vertical axis passing through an end. Assuming the temperature to be uniform and constant. If $p_1$ and $p_2$ denote the pressure at free and the fixed end respectively, then choose the correct relation.
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The correct answer is:
$\frac{p_2}{p_1}=e^{\frac{M \omega^2 l^2}{2 R T}}$
Consider the diagram
Consider the elemetary part of thickness $d x$ at a distance $x$ from axis of rotation, then force on this part
$A d p=(d m) \omega^2 x$ $\ldots$ (i)
where, $d m=$ mass of element Now, from ideal gas equation $p V=n R T$ we get
$p A d x=\frac{d m}{M} R T$
$\Rightarrow \quad d m=\frac{M p A}{R T} d x$ ...(ii)
From Eqs. (i) and (ii)
$\begin{aligned} & A d p=\frac{M p A}{R T} \omega^2 x d x \\ & \int_{p_1}^{p_2} \frac{d p}{p}=\int_0^1 \frac{M \omega^2}{R T} x d x \\ & \Rightarrow \quad \ln \frac{p_2}{p_1}=\frac{M \omega^2 I^2}{2 R T} \\ & \Rightarrow \quad \frac{p_2}{p_1}=e^{\frac{M \omega^2 I^2}{2 R T}} \\ & \end{aligned}$
Consider the elemetary part of thickness $d x$ at a distance $x$ from axis of rotation, then force on this part
$A d p=(d m) \omega^2 x$ $\ldots$ (i)
where, $d m=$ mass of element Now, from ideal gas equation $p V=n R T$ we get
$p A d x=\frac{d m}{M} R T$
$\Rightarrow \quad d m=\frac{M p A}{R T} d x$ ...(ii)
From Eqs. (i) and (ii)
$\begin{aligned} & A d p=\frac{M p A}{R T} \omega^2 x d x \\ & \int_{p_1}^{p_2} \frac{d p}{p}=\int_0^1 \frac{M \omega^2}{R T} x d x \\ & \Rightarrow \quad \ln \frac{p_2}{p_1}=\frac{M \omega^2 I^2}{2 R T} \\ & \Rightarrow \quad \frac{p_2}{p_1}=e^{\frac{M \omega^2 I^2}{2 R T}} \\ & \end{aligned}$
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