In the figure shown, is given a flat plate and a rotating cone with a very acute angle $θ$ (less than 5⁰). The apex of cone just touches the plate surface. A liquid with coefficient of viscosity $η$ fills the narrow gap formed by the cone and plate. Cone is rotated with angular velocity $ω$ as shown. Neglect end effects and assume a linear velocity profile :-

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The torque on the driven cone is equal to $(\frac{1}{3} π R^{3} η) \frac{ω}{θ}$

Velocity gradient in the liquid is $θ$

Velocity gradient in the liquid is $\frac{ω}{θ}$

The torque on the driven cone is equal to $(\frac{2}{3} π R^{3} η) \frac{ω}{θ}$

answer is A, C.

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Detailed Solution

Velocity gradient
$\frac{d v}{d y} = \frac{ω r \cos θ - 0}{r \sin θ} = \frac{ω}{\tan θ} = \frac{ω}{θ}$
Question Image
(as  is small)
Considering elemental ring of radius x and thickness dx torque on this elemental ring,
$d τ = [(2 π xdx) \frac{ω}{θ}] d η$

$\Rightarrow Net torque, τ = \frac{2}{3} π R^{3} η \frac{ω}{θ}$