A solid sphere is roilling down on inclined plane from rest and rectangular block of same mass is also slipping down simultaneously from rest on the same plane. Then:

Options

A. both of them will reach the bottom simultaneously
B. the sphere will reach the bottom first
C. the rectangular block will reach the bottom first.
D. depends on density of material

Answer

To solve the problem of a solid sphere rolling down an inclined plane and a rectangular block slipping down the same plane, we need to analyze the motion of both objects. We will derive their respective accelerations and compare them to determine which one reaches the bottom first.

### Step-by-Step Solution:

1. **Identify the Forces Acting on Each Object**:
- For the **solid sphere**, the forces acting on it are:
- Gravitational force: \( mg \)
- Normal force: \( N \)
- For the **rectangular block**, the forces acting on it are:
- Gravitational force: \( mg \)
- Normal force: \( N \)

2. **Determine the Acceleration of the Rectangular Block**:
- The block is sliding down without rolling, so the only force causing it to accelerate is the component of gravitational force along the incline.
- The acceleration \( a_{\text{block}} \) can be given by:
\[ a_{\text{block}} = g \sin \theta \]

3. **Determine the Acceleration of the Solid Sphere**:
- The solid sphere rolls down the incline, which means it has both translational and rotational motion.
- The net force acting on the sphere along the incline is:
\[ F_{\text{net}} = mg \sin \theta - f \]
where \( f \) is the frictional force.
- The moment of inertia \( I \) for a solid sphere is given by:
\[ I = \frac{2}{5} m r^2 \]
- The frictional force provides the torque for the sphere:
\[ f \cdot r = I \cdot \alpha \]
where \( \alpha \) is the angular acceleration. Since \( \alpha = \frac{a}{r} \), we can substitute:
\[ f \cdot r = \frac{2}{5} m r^2 \cdot \frac{a}{r} \]
Simplifying gives:
\[ f = \frac{2}{5} m a \]

4. **Apply Newton's Second Law for the Sphere**:
- The total force equation for the sphere becomes:
\[ mg \sin \theta - \frac{2}{5} m a = m a \]
- Rearranging gives:
\[ mg \sin \theta = m a + \frac{2}{5} m a = \frac{7}{5} m a \]
- Thus, the acceleration \( a_{\text{sphere}} \) is:
\[ a_{\text{sphere}} = \frac{5}{7} g \sin \theta \]

5. **Compare the Accelerations**:
- The acceleration of the rectangular block is:
\[ a_{\text{block}} = g \sin \theta \]
- The acceleration of the solid sphere is:
\[ a_{\text{sphere}} = \frac{5}{7} g \sin \theta \]
- Since \( g \sin \theta > \frac{5}{7} g \sin \theta \), it follows that:
\[ a_{\text{block}} > a_{\text{sphere}} \]

6. **Conclusion**:
- The rectangular block will reach the bottom of the incline first, while the solid sphere will take longer due to its lower acceleration.