A solid sphere and a hollow sphere of the same mass and of same radius are rolled on an inclined plane. Let the time taken to reach the bottom by the solid sphere and the hollow sphere be $$t_1 \text{ and } t_2$$, respectively, then

Consider an inclined plane of length $$s$$ and angle of inclination $$\theta$$. Both spheres are released from rest and roll down without slipping.

For any rigid body rolling without slipping, the linear acceleration of the centre of mass is derived from Newton’s 2nd law plus the rotational equation:

$$m a = m g \sin\theta - f$$ $$-(1)$$
$$f R = I \alpha$$ and $$\alpha = \frac{a}{R}$$ $$-(2)$$

Eliminating the static friction $$f$$ using $$(2)$$ and substituting into $$(1)$$ gives the standard result:

$$a = \frac{g \sin\theta}{1 + \dfrac{I}{m R^{2}}}$$ $$-(3)$$

Write $$I = m k^{2}$$, where $$k$$ is the radius of gyration. Then

$$a = \frac{g \sin\theta}{1 + \dfrac{k^{2}}{R^{2}}}$$ $$-(4)$$

Case 1: Solid sphere (uniform density)

Moment of inertia: $$I = \frac{2}{5} m R^{2} \Rightarrow k^{2} = \frac{2}{5} R^{2}$$.

Put this in $$(4)$$:

$$a_{1} = \frac{g \sin\theta}{1 + \dfrac{2}{5}} = \frac{g \sin\theta}{\dfrac{7}{5}} = \frac{5}{7} g \sin\theta$$ $$-(5)$$

Case 2: Hollow sphere (thin spherical shell)

Moment of inertia: $$I = \frac{2}{3} m R^{2} \Rightarrow k^{2} = \frac{2}{3} R^{2}$$.

Put this in $$(4)$$:

$$a_{2} = \frac{g \sin\theta}{1 + \dfrac{2}{3}} = \frac{g \sin\theta}{\dfrac{5}{3}} = \frac{3}{5} g \sin\theta$$ $$-(6)$$

Numerical comparison:

$$a_{1} = \frac{5}{7} g \sin\theta \approx 0.714 \, g \sin\theta$$
$$a_{2} = \frac{3}{5} g \sin\theta = 0.6 \, g \sin\theta$$.

Thus $$a_{1} \gt a_{2}$$. The solid sphere gains more acceleration.

The time taken to travel the same distance $$s$$ with constant acceleration is

$$t = \sqrt{\frac{2 s}{a}}$$ $$-(7)$$

Therefore

$$t_{1} = \sqrt{\frac{2 s}{a_{1}}}, \qquad t_{2} = \sqrt{\frac{2 s}{a_{2}}}$$.

Since $$a_{1} \gt a_{2}$$, the denominator of $$t_{1}$$ is larger, making $$t_{1}$$ smaller:

$$t_{1} \lt t_{2}$$.

Hence the correct relation is $$t_{1} \lt t_{2}$$, corresponding to Option C.