Viscous liquid of viscosity $η$ is filled between an outer fixed cylinder and an inner cylinder as shown in the figure. The central solid cylinder starts with initial angular velocity $ω_{0}$ . Find out time after which the angular velocity becomes half.

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$\frac{4 π η l R_{1}}{m (R_{2} - R_{1}) \ln 2}$

$\frac{4 π η l R_{2}}{m (R_{2} - R_{1}) \ln 2}$

$\frac{m (R_{2} - R_{1}) \ln 2}{4 π η l R_{1}}$

$\frac{m (R_{2} - R_{1}) \ln 2}{4 π η l R_{2}}$

answer is C.

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Detailed Solution

The velocity of the liquid layer in the contract with outer fixed cylinder will be zero while the speed of the liquid attached with rotating cylinder will be $ω R_{1}$ . The viscous force developed

$| F | = η A \frac{d v}{d z}, w h e r e \frac{d v}{d z} = \frac{ω R_{1} - 0}{R_{2} - R_{1}}$

$F = η (2 π R_{1} l) (\frac{ω R_{1}}{R_{2} - R_{1}})$

The torque of this force about central axes,

$τ = F R_{1} = \frac{2 π η R_{1}^{3} ω l}{R_{2} - R_{1}}$

$I α = \frac{2 π η R_{1}^{3} ω l}{R_{2} - R_{1}} \Rightarrow \frac{M R_{1}^{2}}{2} (- \frac{d ω}{d t}) = \frac{2 π η R_{1}^{3} ω l}{R_{2} - R_{1}}$

$\int_{ω_{0}}^{ω_{0} / 2} - \frac{d ω}{ω} = \frac{4 π ω R_{1} l}{m (R_{2} - R_{1})} \int_{0}^{t} d t \Rightarrow t = \frac{m (R_{2} - R_{1}) \ln 2}{4 π η l R_{1}}$