JEE Mains · Physics · STD 11 - 9.1 fluid mechanics
A cylindrical tube AB of length \(l\), closed at both ends contains an ideal gas of 1 mol having molecular weight M. The tube is rotated in a horizontal plane with constant angular velocity \( \omega \) about an axis perpendicular to AB and passing through the edge at end A. If \( P_{A} \) and \( P_{B} \) are the pressures at A and B respectively, then (Consider the temperature is same at all points in the tube)
- A \( P_{B}=P_{A}exp(M\omega^{2}l^{2}/2RT) \)
- B \( P_{B}=P_{A} \)
- C \( P_{B}=P_{A}exp(M\omega^{2}l^{2}/3RT) \)
- D \( P_{B}=P_{A}exp(M\omega^{2}l^{2}/RT) \)
Answer & Solution
Correct Answer
(A) \( P_{B}=P_{A}exp(M\omega^{2}l^{2}/2RT) \)
Step-by-step Solution
Detailed explanation
\(A[(P+d P)-P]=(d m)\left(\omega^2 x\right)\) \(dP =\frac{( dm )}{ A } \omega^2 x\) \(dP =\frac{(\rho)( A )( dx ) \omega^2 x }{ A }\) also \([ PM =\rho RT ]\) \(dP =\left(\frac{ PM }{ RT }\right) \omega^2 x dx\)…
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