A disc and a solid sphere of same radius but different masses roll off on two inclined planes of the same
altitude and length. Which one of the two objects gets to the bottom of the plane first? [NEET 2016] 

Acceleration of an object rolling down an inclined plane is given by

$a=\frac{g\sin \theta }{1+I/m{r}^2}$

where,  $\theta$ = angle of inclination of the inclined plane,

m = mass of the object
and I = moment of inertia about the axis through centre of mass.
Type of sphere is not mentioned in the question. Therefore, we will assume the given sphere as solid sphere.

For disc,    $\frac{I}{m{r}^2}=\frac{1/2m{r}^2}{m{r}^2}=\frac{1}{2}$

For solid sphere,   $\frac{I}{m{r}^2}=\frac{2/5m{r}^2}{m{r}^2}=\frac{2}{5}$

$\therefore a_{\text{disc }}=\frac{g\sin \theta }{1+\frac{1}{2}}=\frac{2}{3}g\sin \theta =0.66g\sin \theta$

$a_{\text{solid sphere }}=\frac{g\sin \theta }{1+\frac{2}{5}}=\frac{5}{7}g\sin \theta =0.71g\sin \theta$

Clearly,  $a_{\text{solid sphere }}>a_{\text{disc }}$

Hence, sphere gets to the bottom of the plane first.