Swirling Disc in Chocolate!

Consider an element - a ring of radius $x$ and $0 < x < r$ and also of thickness $dx$ .

Area of that element would be $A = 2\pi x dx$

A small viscous force $dF$ would act on that element, which is given by $dF = \eta \times A \times \frac{\omega \times x}{h}$

$dF = \eta \times 2\pi x dx \times \frac{\omega \times x}{h}$

Since torque is asked, the torque over the small element would be

$d\tau = dF \times x$

$d\tau = \eta \times 2\pi x dx \times \frac{\omega \times x}{h} \times x$

If we provide this $d\tau$ for that small element, we can keep it at constant angular velocity.

Hence, the net torque to be supplied is:

$\int_0^\tau d\tau = \int_0^r \eta \times 2\pi x dx \times \frac{\omega \times x}{h} \times x$

And hence we get

$\tau = \frac{\eta \times 2\pi \times \omega \times r^{4}}{4 \times h}$

Substituting values, we get

$\tau = 10666.67 S . I . Units$