A fluid of density $ρ$ is rotated with a certain angular velocity in a cylindrical vessel as shown:-

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The angular velocity is $ω = \sqrt{\frac{2 g h}{R^{2}}}$

The angular velocity is $ω = \sqrt{\frac{4 g h}{R^{2}}}$

The kinetic energy of fluid is $\frac{16 π ρ g h^{2} R^{2}}{6}$

The kinetic energy of fluid is $\frac{5 π ρ g h^{2} R^{2}}{6}$

answer is A, D.

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Detailed Solution

$2 h = h + \frac{ω^{2} R^{2}}{2 g} \sqrt{\frac{2 g h}{R^{2}}} = ω d K = \frac{1}{2} \times d m ω^{2} x^{2}$
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$= \frac{1}{2} ρ \times 2 π x d x (h + ω^{2} \frac{x^{2}}{2 g}) \times ω^{2} x^{2} K = π ρ ω^{2} [\frac{R^{4}}{2} h + \frac{ω^{2}}{2 g} \times \frac{R^{6}}{6}] K = π ρ ω^{2} [\frac{R^{4}}{2} h + \frac{ω^{2}}{2 g} \times \frac{R^{6}}{6}] = π ρ ω^{2} [\frac{R^{4}}{2} h + \frac{2 g h \times R^{4}}{12 g}] = \frac{5 π ρ ω^{2} R^{4} h}{6}$