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(C) Let $\eta$ be the coefficient of viscosity and $x$ be the thickness of the oil film.Consider an elemental ring of radius $r$ and width $dr$ on the

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$R^4$

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(C) Let $\eta$ be the coefficient of viscosity and $x$ be the thickness of the oil film.Consider an elemental ring of radius $r$ and width $dr$ on the disc.The velocity of the disc at radius $r$ is $v = \omega r$.The velocity gradient across the film is $\frac{dv}{dx} = \frac{\omega r - 0}{x} = \frac{\omega r}{x}$.The viscous force $dF$ acting on this elemental ring is given by $dF = \eta A \frac{dv}{dx}$,where $A = 2\pi r dr$.$dF = \eta (2\pi r dr) \frac{\omega r}{x} = \frac{2\pi \eta \omega}{x} r^2 dr$.The torque $d	au$ required to overcome this viscous force is $d	au = (dF) r$.$d	au = \left( \frac{2\pi \eta \omega}{x} r^2 dr ight) r = \frac{2\pi \eta \omega}{x} r^3 dr$.Integrating from $r = 0$ to $r = R$ to find the total torque $	au$:$	au = \int_{0}^{R} \frac{2\pi \eta \omega}{x} r^3 dr = \frac{2\pi \eta \omega}{x} \left[ \frac{r^4}{4} ight]_{0}^{R} = \frac{\pi \eta \omega}{2x} R^4$.Thus,the torque $	au$ is proportional to $R^4$.

$A$ plastic circular disc of radius $R$ is placed on a thin oil film,spread over a flat horizontal surface. The torque required to spin the disc about its central vertical axis with a constant angular velocity is proportional to

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