A solid sphere and a hollow sphere of the same material and of a same size can be distinguished without weighing

• by determining their moments of inertia about their coaxial axe

• by rolling them simultaneously on an inclined plane

• by rotating them about a common axis of rotation

• by applying equal torque on them

B.

by rolling them simultaneously on an inclined plane

The acceleration of a body rolling down the plane $= \mathrm{\alpha }$

∴                $\mathrm{\alpha } = \frac{\mathrm{g} \mathrm{sin\theta }}{1 + {\displaystyle \frac{{\mathrm{K}}^2}{{\mathrm{R}}^2}}}$

where K is radius of gyration and R the radius of sphere.

For solid sphere,

                      $\frac{{\mathrm{K}}^2}{{\mathrm{R}}^2} = \frac{2}{5}$

                       $\mathrm{\alpha } = \frac{\mathrm{g} \mathrm{sin\theta }}{1 + {\displaystyle \frac{2}{5}}}$

                       $\mathrm{\alpha } = \frac{\mathrm{g} \sin \mathrm{\theta }}{{\displaystyle \frac{7}{5}}}$                         

 ∴                       $\mathrm{\alpha } = \frac{5}{7}\mathrm{g} \mathrm{sin\theta }$

For hollow sphere,

                        $\frac{{\mathrm{K}}^2}{{\mathrm{R}}^2} = \frac{2}{3}$

                         $\mathrm{\alpha } = \frac{\mathrm{g} \sin \mathrm{\theta }}{1 + {\displaystyle \frac{2}{3}}}$

                          $\mathrm{\alpha } = \frac{3}{5} \mathrm{g} \sin \mathrm{\theta }$

                           $\mathrm{\alpha } = 0.6 \left(\mathrm{g} \mathrm{sin\theta } \right)$

Since, acceleration of solid sphere is more than of hollow sphere, it rolls faster, and reaches the
bottom of the inclined plane earlier. Hence, solid sphere and hollow sphere can be distinguished by rolling them simultaneously on an inclined plane.