On an inclined plane, two spheres of similar size, one solid while the other hollow, start rolling from the position of rest. The time taken by the hollow sphere, in order to cover the same distance, in comparison with solid one, will be:

Options

A. more
B. less
C. equal
D. data is incomplete

Answer

To solve the problem of comparing the time taken by a solid sphere and a hollow sphere rolling down an inclined plane, we can follow these steps:

### Step 1: Understand the Problem
We have two spheres of the same size: one is solid and the other is hollow. Both spheres start rolling from rest down an inclined plane. We need to determine which sphere takes more time to cover the same distance.

### Step 2: Identify the Relevant Equations
For rolling objects, the acceleration \( a \) of the center of mass can be expressed as:
\[ a = \frac{g \sin \theta}{1 + \frac{I}{mR^2}} \]
where:
- \( g \) is the acceleration due to gravity,
- \( \theta \) is the angle of the incline,
- \( I \) is the moment of inertia,
- \( m \) is the mass of the sphere,
- \( R \) is the radius of the sphere.

### Step 3: Calculate the Moment of Inertia
1. **Solid Sphere**: The moment of inertia \( I \) for a solid sphere is given by:
\[ I_{\text{solid}} = \frac{2}{5} m R^2 \]
2. **Hollow Sphere**: The moment of inertia \( I \) for a hollow sphere is given by:
\[ I_{\text{hollow}} = \frac{2}{3} m R^2 \]

### Step 4: Calculate the Acceleration for Each Sphere
1. **Acceleration of Solid Sphere**:
\[ a_{\text{solid}} = \frac{g \sin \theta}{1 + \frac{I_{\text{solid}}}{mR^2}} = \frac{g \sin \theta}{1 + \frac{2/5 m R^2}{m R^2}} = \frac{g \sin \theta}{1 + \frac{2}{5}} = \frac{g \sin \theta}{\frac{7}{5}} = \frac{5g \sin \theta}{7} \]

2. **Acceleration of Hollow Sphere**:
\[ a_{\text{hollow}} = \frac{g \sin \theta}{1 + \frac{I_{\text{hollow}}}{mR^2}} = \frac{g \sin \theta}{1 + \frac{2/3 m R^2}{m R^2}} = \frac{g \sin \theta}{1 + \frac{2}{3}} = \frac{g \sin \theta}{\frac{5}{3}} = \frac{3g \sin \theta}{5} \]

### Step 5: Compare the Accelerations
From the calculations:
- \( a_{\text{solid}} = \frac{5g \sin \theta}{7} \)
- \( a_{\text{hollow}} = \frac{3g \sin \theta}{5} \)

To compare, we can see that:
\[ \frac{5g \sin \theta}{7} > \frac{3g \sin \theta}{5} \]
This indicates that the solid sphere has a greater acceleration than the hollow sphere.

### Step 6: Relate Acceleration to Time
Since both spheres start from rest and roll down the same distance \( s \), we can use the kinematic equation:
\[ s = \frac{1}{2} a t^2 \]
From this, we can express time \( t \) as:
\[ t = \sqrt{\frac{2s}{a}} \]
Since the solid sphere has a greater acceleration, it will take less time to cover the same distance compared to the hollow sphere.

### Conclusion
Thus, the time taken by the hollow sphere to cover the same distance is greater than that taken by the solid sphere.

### Final Answer
The time taken by the hollow sphere is greater than that taken by the solid sphere.

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