A solid sphere and a disc of same radii are rolling  down an inclined plane without slip. One reaches earlier than the other due to

THE TIME TAKEN TO ROLL DOWN IS GIVEN BY  $T=\sqrt{\frac{2L}{a}}$

$\mathrm{a} = \frac{\mathrm{g} \sin \mathrm{\theta }}{1+{\displaystyle \frac{\mathrm{I}}{{\mathrm{MR}}^2}}} = \frac{\mathrm{g} \sin \mathrm{\theta }}{1+{\displaystyle \frac{{\mathrm{K}}^{2^{ }}}{{\mathrm{R}}^2}}} [\mathrm{I} = {\mathrm{MK}}^2]$

Clearly one reaches earlier than other due to different radius of gyration.