1. Acceleration of a body rolling without slipping on an incline

For a rigid body rolling down an incline of angle $\theta$ without slipping,

$a = \frac{g\sin\theta}{1 + \frac{I}{mR^2}}$

where:

• $I$ = moment of inertia about the center,
• $m$ = mass,
• $R$ = radius.

So, the body with smaller $\frac{I}{mR^2}$ gets larger acceleration, and hence reaches the bottom in less time.

2. For the solid sphere

Moment of inertia of a solid sphere:

$I_{\text{solid}} = \frac{2}{5}mR^2$

Thus,

$a_1 = \frac{g\sin\theta}{1 + \frac{2}{5}} = \frac{g\sin\theta}{\frac{7}{5}} = \frac{5}{7}g\sin\theta$

3. For the hollow sphere

Moment of inertia of a hollow sphere (thin spherical shell):

$I_{\text{hollow}} = \frac{2}{3}mR^2$

Thus,

$a_2 = \frac{g\sin\theta}{1 + \frac{2}{3}} = \frac{g\sin\theta}{\frac{5}{3}} = \frac{3}{5}g\sin\theta$

4. Compare accelerations

$a_1 = \frac{5}{7}g\sin\theta, \qquad a_2 = \frac{3}{5}g\sin\theta$

Now,

$\frac{5}{7} > \frac{3}{5}$

since

$25 > 21$

Therefore,

$a_1 > a_2$

So the solid sphere has greater acceleration and reaches the bottom earlier.

5. Compare times

If both start from rest and travel the same distance $s$ down the incline,

$s = \frac{1}{2}at^2 \implies t = \sqrt{\frac{2s}{a}}$

Larger acceleration means smaller time. Hence,

$t_1 < t_2$

6. Option check

• A: $t_1 < t_2$ ✅ Correct
• B: $t_1 = 2t_2$ ❌ No such ratio follows
• C: $t_1 > t_2$ ❌ Opposite
• D: $t_1 = t_2$ ❌ Incorrect

Therefore, the correct option is A.