$A$ solid sphere rolls down two different inclined planes of the same heights but different angles of inclination.
$(a)$ Will it reach the bottom with the same speed in each case?
$(b)$ Will it take longer to roll down one plane than the other?
$(c)$ If so,which one and why?

(A) Yes. $(b)$ Yes. $(c)$ On the smaller inclination.
$(a)$ Let the mass of the sphere be $m$,height of the plane be $h$,and velocity at the bottom be $v$. By the law of conservation of energy,the potential energy at the top equals the total kinetic energy (translational + rotational) at the bottom:
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
For a solid sphere,$I = \frac{2}{5}mr^2$ and $\omega = v/r$. Substituting these:
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(v/r)^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$
$v = \sqrt{\frac{10}{7}gh}$. Since $v$ depends only on $h$ and $g$,the speed at the bottom is the same for both planes.
$(b)$ and $(c)$ The acceleration of a rolling sphere on an incline $\theta$ is $a = \frac{g \sin \theta}{1 + I/mr^2} = \frac{g \sin \theta}{1 + 2/5} = \frac{5}{7}g \sin \theta$. Since $a \propto \sin \theta$,the plane with the smaller angle $\theta$ has smaller acceleration. Using $v = u + at$ with $u=0$,$t = v/a$. Since $v$ is constant and $a$ is smaller for the smaller angle,the time $t$ taken to reach the bottom will be longer for the plane with the smaller inclination.