The speed of a homogeneous solid sphere after rolling down an inclined plane of vertical height `h` from rest without slipping will be.

Options

A. `sqrt((10 gh)/(7))`
B. `sqrt(gh)`
C. `sqrt((6)/(5)gh)`
D. `sqrt((4)/(3)gh)`

Explanation

To find the speed of a homogeneous solid sphere after rolling down an inclined plane of vertical height \( h \) from rest without slipping, we can use the principles of conservation of energy.

### Step-by-Step Solution:

1. **Identify the Initial and Final States**:
- Initial state: The sphere is at rest at a height \( h \).
- Final state: The sphere has rolled down the incline and is moving with speed \( v \).

2. **Apply Conservation of Energy**:
- The potential energy (PE) at the top will convert into kinetic energy (KE) at the bottom.
- The potential energy at height \( h \) is given by:
\[ PE = mgh \]
- The kinetic energy when the sphere is rolling consists of two parts: translational kinetic energy and rotational kinetic energy.
- The translational kinetic energy (TKE) is:
\[ TKE = \frac{1}{2} mv^2 \]
- The rotational kinetic energy (RKE) for a solid sphere is:
\[ RKE = \frac{1}{2} I \omega^2 \]
- For a solid sphere, the moment of inertia \( I \) is given by:
\[ I = \frac{2}{5} m r^2 \]
- The relationship between linear speed \( v \) and angular speed \( \omega \) for rolling without slipping is:
\[ v = r \omega \implies \omega = \frac{v}{r} \]

3. **Substituting for Rotational Kinetic Energy**:
- Substitute \( \omega \) in the RKE:
\[ RKE = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v}{r}\right)^2 = \frac{1}{5} mv^2 \]

4. **Total Kinetic Energy**:
- The total kinetic energy (KE) at the bottom is:
\[ KE = TKE + RKE = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \left(\frac{5}{10} + \frac{2}{10}\right) mv^2 = \frac{7}{10} mv^2 \]

5. **Setting Up the Energy Equation**:
- According to the conservation of energy:
\[ mgh = \frac{7}{10} mv^2 \]
- We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gh = \frac{7}{10} v^2 \]

6. **Solving for \( v \)**:
- Rearranging gives:
\[ v^2 = \frac{10}{7} gh \]
- Taking the square root:
\[ v = \sqrt{\frac{10}{7} gh} \]

### Final Answer:
The speed of the homogeneous solid sphere after rolling down an inclined plane of vertical height \( h \) from rest without slipping will be:
\[ v = \sqrt{\frac{10}{7} gh} \]