Question

A uniform disc of mass $M$ and radius $R$ rolls down an inclined (angle of inclination=$\theta$) plane without slipping then magnitude of torque on the disk is given by $\frac{kMRgsin\theta }{3}$,then the value of $k$,is

Answer 1

$a=\frac{mg\sin \theta -f}{m}\alpha =\frac{2fr}{m{r}^2}=\frac{2f}{mr}r\alpha =a\Rightarrow \frac{2f}{m}=g\sin \theta -\frac{f}{m}\frac{f}{m}=\frac{1}{3}g\sin \theta r=\frac{mg\sin \theta }{3}r=\frac{kmgr\sin \theta }{3}\Rightarrow k=1$

Hence the correct answer is $1$