For rolling down an inclined plane,

\(a_{c m}=\frac{g \sin \theta}{1+\frac{1}{m R^2}}\)

For 5 phere of mass \(m\) and radius \(R, 1=\frac{2}{5} m R^2\)

So, \(\quad a_{c u}=\frac{g \sin 0}{1-\frac{2}{5}\left({m R^2}^2\right)}=\frac{5}{7} g \sin 0\)

As acceleration of centre of mass of rolling body only depends upon angle of inclination,

So, time taken to come down, \(t =\sqrt{\frac{2 L}{a_{c h}}}\)

\(\left(\because L=\frac{1}{2} a_{c n} t^2\right)\)

\(t=\sqrt{\frac{2 L \times 7}{5 g \sin \theta}}=\sqrt{\frac{14 L}{5 g \sin \theta}}\), where \(L=\) length of inclined plane.

Here, \(t\) is independent of mass, radius and density of spheres. Option (d) is correct.