JEE Mains · Physics · STD 11 - 9.1 fluid mechanics
A cylindrical vessel containing a liquid is rotated about its axis so that the liquid rises at its sides as shown in the figure. The radius of vessel is \(5\, cm\) and the angular speed of rotation is \(\omega\; rad \,s^{-1}\). The difference in the height, \(h(\) in \(cm )\) of liquid at the centre of vessel and at the side will be
- A \(\frac{25 \omega^{2}}{2 g}\)
- B \(\frac{2 \omega^{2}}{5 g}\)
- C \(\frac{5 \omega^{2}}{2 g}\)
- D \(\frac{2 \omega^{2}}{25 g}\)
Answer & Solution
Correct Answer
(A) \(\frac{25 \omega^{2}}{2 g}\)
Step-by-step Solution
Detailed explanation
Applying pressure equation from \(A\) to \(B\) \(P_{0}+\rho \cdot \frac{R \omega^{2}}{2} \cdot R-\rho g h=P_{0}\) \(\frac{\rho R ^{2} \omega^{2}}{2}=\rho gh\) \(h =\frac{ R ^{2} \omega^{2}}{2 g }=(5)^{2} \frac{\omega^{2}}{2 g }=\frac{25}{2} \frac{\omega^{2}}{ g }\)
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