The velocity of a sphere rolling down an inclined plane of height h at an inclination `theta` with the horizontal, will be :

Options

A. `(gh)/(1+k^(2)//r^(2))`
B. `(2gh)/(1+k^(2)//r^(2))`
C. `sqrt((gh)/(1+k^(2)//r^(2)))`
D. `sqrt((2gh)/(1+k^(2)//r^(2)))`

Answer

To find the velocity of a sphere rolling down an inclined plane of height \( h \) at an angle \( \theta \) with the horizontal, we can use the principle of conservation of energy. The potential energy at the top of the incline will be converted into kinetic energy as the sphere rolls down.

### Step-by-Step Solution:

1. **Identify the Potential Energy at Height \( h \)**:
The potential energy (PE) of the sphere at height \( h \) is given by:
\[ PE = mgh \]
where \( m \) is the mass of the sphere, \( g \) is the acceleration due to gravity, and \( h \) is the height.

2. **Identify the Types of Kinetic Energy**:
As the sphere rolls down, its energy is converted into two types of kinetic energy:
- Translational kinetic energy (TKE):
\[ TKE = \frac{1}{2} mv^2 \]
- Rotational kinetic energy (RKE):
\[ RKE = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia of the sphere and \( \omega \) is the angular velocity.

3. **Relate Angular Velocity to Linear Velocity**:
For a sphere rolling without slipping, the relationship between linear velocity \( v \) and angular velocity \( \omega \) is:
\[ \omega = \frac{v}{r} \]
where \( r \) is the radius of the sphere.

4. **Substitute Moment of Inertia for a Solid Sphere**:
The moment of inertia \( I \) of a solid sphere is given by:
\[ I = \frac{2}{5} mr^2 \]

5. **Write the Conservation of Energy Equation**:
The total mechanical energy at the top (potential energy) is equal to the total mechanical energy at the bottom (kinetic energy):
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
Substituting \( I \) and \( \omega \):
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{5} mr^2\right) \left(\frac{v^2}{r^2}\right) \]

6. **Simplify the Equation**:
This simplifies to:
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 \]
Combining the terms on the right:
\[ mgh = \left(\frac{1}{2} + \frac{1}{5}\right) mv^2 \]
The common denominator for \( \frac{1}{2} \) and \( \frac{1}{5} \) is 10:
\[ \frac{1}{2} = \frac{5}{10}, \quad \frac{1}{5} = \frac{2}{10} \quad \Rightarrow \quad \frac{1}{2} + \frac{1}{5} = \frac{7}{10} \]
Thus:
\[ mgh = \frac{7}{10} mv^2 \]

7. **Solve for \( v^2 \)**:
Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gh = \frac{7}{10} v^2 \quad \Rightarrow \quad v^2 = \frac{10gh}{7} \]

8. **Take the Square Root to Find \( v \)**:
\[ v = \sqrt{\frac{10gh}{7}} \]

### Final Answer:
The velocity of the sphere rolling down the inclined plane is:
\[ v = \sqrt{\frac{10gh}{7}} \]