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A thin horizontal disc of radius R is located within a cylindrical cavity filled with a liquid whose viscosity is $η$ . The distance between the disc and horizontal planes of cavity is h.The power developed by the viscous forces acting on the disc when it rotates with angular velocity $2 ω$ (The end effects are to be neglected)

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$\frac{π η ω^{2} R^{4}}{4 h}$

$\frac{π η ω^{2} R^{4}}{h}$

$\frac{4 π η ω^{2} R^{4}}{h}$

$\frac{π η ω^{2} R^{4}}{2 h}$

answer is A.

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Detailed Solution

$τ = \int [\frac{2 η r^{2} 2 ω (2 π r d r)}{h}] = \frac{2 η π r^{4} ω^{}}{h} \Rightarrow P = τ 2 ω = \frac{4 η π r^{4} ω^{2}}{h}$

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