Let $\theta_1, l_1$ be the angle of inclination and distance travelled from top to bottom respectively on plane (1) and $\theta_2, l_2$ be the angle of inclination and distance travelled from top to bottom respectively on plane (2) From diagram $\theta_{1}>\theta_{2}$ i.e. $\sin \theta_{1}>\sin \theta_{2}$ or $\frac{\sin \theta_{1}}{\sin \theta_{2}}>1$...(i) $h =$ height of each inclined plane $= l_1\sin\theta_1 = l_2\sin\theta_2$ (a) Yes. At the top of the planes, the sphere has only P.E = mgh where m = mass of the sphere When the sphere rolls down from the top to the bottom, converted into K.E. ∴ If $v_1 \text{ and } v_2$ be its linear speeds at the bottom of the planes, i.e. (1) and (2) respectively, then $= \left( \frac{1}{2} m v_{1}^{2} + \frac{1}{2} m k^{2} \frac{v_{1}^{2}}{R^{2}} \right)$ or $2 g h = v_{1}^{2} \left(1 + \frac{k^{2}}{R^{2}}\right)$ and $2 g h = v_{2}^{2} \left(1 + \frac{k^{2}}{R^{2}}\right)$ $\therefore v_{1}^{2} = \frac{2 g h}{\left(1 + \frac{k^{2}}{R^{2}}\right)} \dots \text{(ii)}$and $v_{2}^{2} = \frac{2 g h}{\left(1 + \frac{k^{2}}{R^{2}}\right)}$...(iii)where I = MI of the sphere, ω = its angular speed, k is the radius of gyrationFrom (ii) and (iii) it is clear that the sphere reaches the bottom with same speed in each case.(b) To find the time of rolling motion : Yes, it will take longer time down one plane than the other. It will be longer for the plane having smaller angle of inclination.(c) Explanation : Let $t_1 \text{ and } t_2$ be the time taken by the sphere in rolling on plane (1) and (2) respectively.Acceleration of solid sphere on an inclined plane is given by,$a = \frac{g \sin \theta}{\left(1 + \frac{k^{2}}{R^{2}}\right)}$Now for solid sphere $1 + \frac{k^{2}}{R^{2}} = 1 + \frac{\frac{2}{5} R^{2}}{R^{2}} = \frac{7}{5}$If $a_1 \text{ and } a_2$ be the accelerations of the sphere on inclined plane (1) and (2) respectively, then$a_{1} = \frac{g \sin \theta_{1}}{\frac{7}{5}} = \frac{5}{7} g \sin \theta_{1}$Similarly $a_{2} = \frac{5}{7} g \sin \theta_{2}$$S = u t + \frac{1}{2} a t^{2},$ we get $t_{1}^{2} = \frac{2 l_{1}}{a_{1}}$$= \frac{ \frac{2h}{\sin\theta_{1}} }{ \frac{5}{7} g \sin\theta_{1} } = \frac{14 h}{5 g (\sin \theta_{1})^{2}}$... (iv)and $t_{2}^{2} = \frac{2 l_{2}}{a_{2}}$$= \frac{ \frac{2h}{\sin\theta_{2}} }{ \frac{5}{7} g \sin\theta_{2} }$$= \frac{14 h}{5 g (\sin \theta_{2})^{2}}$...(v)It gives $\frac{t_{1}^{2}}{t_{2}^{2}} = \frac{(\sin \theta_{2})^{2}}{(\sin \theta_{1})^{2}}$$\Rightarrow \frac{t_{1}}{t_{2}} = \frac{\sin \theta_{2}}{\sin \theta_{1}}$...(vi)Now from (i) $\frac{\sin \theta_{1}}{\sin \theta_{2}} > 1$ or $\frac{\sin \theta_{2}}{\sin \theta_{1}} < 1$$\therefore \left( \frac{\sin \theta_{2}}{\sin \theta_{1}} \right)^{2} < 1$...(vii)∴ From (vi) and (vii), we get$\frac{t_{1}}{t_{2}} < 1$ or $t_{1} < t_{2}$It is due to the reason that $a \propto \sin \theta$ and t is inversely proportional to $\text{`a'}$ or $\sin\theta$.