JEE Mains · Physics · STD 11 - Fluid mechanics
- JEE Mains 2026
- Hard
- MCQ
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)
Pick an option to check your answer.
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 \([ PM =\rho RT ]\)
\(dP =\left(\frac{ PM }{ RT }\right) \omega^2 x dx\)
\(\int_{P_A}^{P_B} \frac{d P}{P}=\frac{\omega^2 M}{R T} \int_0^{\ell} x d x\)
\(\ell n\left(\frac{P_B}{P_A}\right)=\frac{\omega^2 \ell^2 M }{2 RT }\)
\(P_B=P_A e^{\frac{M \omega^2 \ell^2}{2 R T}}\)
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