A solid sphere (mass 2M) and a thin spherical shell (mass M) both of the same size roll down an inclined plane, then:

Options

A. Solid sphere will reach the bottom first
B. Hollow spherical shell will reach the bottom first
C. Both will reach at the same time
D. Cannot he predicted as the data is insufficient

Explanation

To solve the problem of determining which object reaches the bottom of the inclined plane first, we will analyze the motion of both the solid sphere and the thin spherical shell using the principles of rotational motion and acceleration.

### Step-by-Step Solution:

1. **Identify the Objects and Their Properties**:
- We have a solid sphere with mass \(2M\) and a thin spherical shell with mass \(M\). Both objects have the same radius \(R\).

2. **Calculate the Moment of Inertia**:
- The moment of inertia \(I\) for a solid sphere is given by:
\[ I_{\text{solid sphere}} = \frac{2}{5} m r^2 \]
For our solid sphere:
\[ I_1 = \frac{2}{5} (2M) R^2 = \frac{4}{5} M R^2 \]
- The moment of inertia \(I\) for a thin spherical shell is given by:
\[ I_{\text{thin shell}} = \frac{2}{3} m r^2 \]
For our thin spherical shell:
\[ I_2 = \frac{2}{3} M R^2 \]

3. **Determine the Acceleration of Each Object**:
- The formula for the acceleration \(a\) of an object rolling down an incline is:
\[ a = \frac{g \sin \theta}{1 + \frac{I}{m R^2}} \]
- For the solid sphere:
\[ a_1 = \frac{g \sin \theta}{1 + \frac{I_1}{2M R^2}} = \frac{g \sin \theta}{1 + \frac{\frac{4}{5} M R^2}{2M R^2}} = \frac{g \sin \theta}{1 + \frac{2}{5}} = \frac{g \sin \theta}{\frac{7}{5}} = \frac{5g \sin \theta}{7} \]
- For the thin spherical shell:
\[ a_2 = \frac{g \sin \theta}{1 + \frac{I_2}{M R^2}} = \frac{g \sin \theta}{1 + \frac{\frac{2}{3} M R^2}{M R^2}} = \frac{g \sin \theta}{1 + \frac{2}{3}} = \frac{g \sin \theta}{\frac{5}{3}} = \frac{3g \sin \theta}{5} \]

4. **Compare the Accelerations**:
- We have:
\[ a_1 = \frac{5g \sin \theta}{7} \quad \text{and} \quad a_2 = \frac{3g \sin \theta}{5} \]
- To compare \(a_1\) and \(a_2\), we can find a common denominator:
\[ a_1 = \frac{5g \sin \theta}{7} \quad \text{and} \quad a_2 = \frac{21g \sin \theta}{35} \]
\[ a_1 = \frac{25g \sin \theta}{35} \]
- Since \(25g \sin \theta > 21g \sin \theta\), we conclude that:
\[ a_1 > a_2 \]
- Therefore, the solid sphere accelerates faster than the thin spherical shell.

5. **Determine the Time Taken to Reach the Bottom**:
- The time \(t\) taken to reach the bottom can be expressed as:
\[ t \propto \sqrt{\frac{d}{a}} \]
- Since \(a_1 > a_2\), it follows that:
\[ t_1 < t_2 \]
- Thus, the solid sphere will take less time to reach the bottom compared to the thin spherical shell.

### Conclusion:
The solid sphere will reach the bottom of the inclined plane first.

### Final Answer:
**The solid sphere will reach the bottom first.**