A solid sphere slides down a smooth inclined plane of inclination θ with acceleration a 1 . A disc…

Acceleration of solid sphere sliding down an inclined plane, $a_1=g\sin \theta$

For disc plane is rough, so it will roll down the incline and its acceleration in rolling is given as, $a_2=\frac{g\sin \theta }{1+{\displaystyle \frac{I}{M{R}^2}}}=\frac{g\sin \theta }{1+{\displaystyle \frac{\left({\displaystyle \frac{M{R}^2}{2}}\right)}{M{R}^2}}}=\frac{2g\sin \theta }{3}$

$\Rightarrow \frac{a_1}{a_2}=\frac{g\sin \theta }{\left({\displaystyle \frac{2g\sin \theta }{3}}\right)}=\frac{3}{2}$