Two uniform solid spheres possessing unequal masses and unequal...

Let $M$ be the mass and $R$ be the radius of the solid sphere.

The equation of motion along the inclined plane is $Mg \sin \theta - f = Ma$.

The torque equation about the center of mass is $fR = I\alpha$.

For pure rolling, $a = R\alpha$, which gives $f = \dfrac{Ia}{R^2}$.

Substituting $f$ in the first equation, we get $Mg \sin \theta - \dfrac{Ia}{R^2} = Ma$.

$a = \dfrac{Mg \sin \theta}{M + \dfrac{I}{R^2}}$

For a solid sphere, the moment of inertia is $I = \dfrac{2}{5}MR^2$.

$a = \dfrac{Mg \sin \theta}{M + \dfrac{2}{5}M} = \dfrac{5}{7}g \sin \theta$

The acceleration of a solid sphere rolling down an incline is independent of its mass and radius. Since both spheres are released from rest at the same height, they travel the same distance with the same acceleration.

Therefore, both spheres reach the bottom at the same time.