A one kg stone attached to the end of a 60 cm chain is revolving at the rate of 3 revolutions/second. If after 30 second, it is making only one revolution per second, find the mean torque

The torque $\mathrm{\tau }$is given by
$\mathrm{\tau }=\frac{\text{ change in momentum }}{\text{ time taken }}$
Here,  ${\mathrm{L}}_1={\mathrm{mr}}^2{\mathrm{\omega }}_1={\mathrm{mr}}^2\times \left(2{\mathrm{\pi v}}_1\right)$
and ${\mathrm{L}}_2={\mathrm{mr}}^2{\mathrm{\omega }}_2={\mathrm{mr}}^2\times \left(2{\mathrm{\pi v}}_2\right)$
so, change in momentum
$\begin{array}{l}={\mathrm{mr}}^2\times 2\mathrm{\pi }\left({\mathrm{v}}_1-{\mathrm{v}}_2\right)\\ \therefore \mathrm{\tau }=\frac{{\mathrm{mr}}^2\times 2\mathrm{\pi }\left({\mathrm{v}}_1-{\mathrm{v}}_2\right)}{\mathrm{t}}\\ =\frac{1(0\cdot 6{)}^2\times 2\mathrm{\pi }(3-1)}{30}\\ =0\cdot 15\mathrm{N}-\mathrm{m}.\end{array}$