A thick walled hollow sphere has outer radius `R`. It down an inclined plane without slipping and its -speed the bottom is `v`. If the inclined plane is frictionless and the sphere slides down without rolling, its speed at the bottom `5v//4`. What is the radius of gyration of the sphere?

Options

A. `R/(sqrt(2))`
B. `R//2`
C. `3R//4`
D. `(sqrt3R)/4`

Explanation

To solve the problem, we will analyze the two cases of the thick-walled hollow sphere as it moves down the inclined plane: one where it rolls without slipping and another where it slides without rolling. We will derive the radius of gyration from the given information.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a thick-walled hollow sphere with an outer radius \( R \).
- In the first case, it rolls down the incline without slipping, reaching a speed \( v \) at the bottom.
- In the second case, it slides down the incline (without rolling) and reaches a speed of \( \frac{5v}{4} \) at the bottom.
- We need to find the radius of gyration \( k \) of the sphere.

2. **Case 1: Rolling Without Slipping**:
- When the sphere rolls down, the potential energy at height \( h \) converts to kinetic energy.
- The potential energy lost is \( mgh \).
- The kinetic energy at the bottom consists of translational and rotational parts:
\[ \text{K.E.} = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
- For a rolling object, \( \omega = \frac{v}{R} \) and \( I \) is the moment of inertia about the center.
- Thus, we can express the kinetic energy as:
\[ \frac{1}{2} mv^2 + \frac{1}{2} I \left(\frac{v}{R}\right)^2 \]
- Setting the potential energy equal to the kinetic energy gives:
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} I \frac{v^2}{R^2} \]
- Rearranging, we get:
\[ 2mgh = mv^2 + \frac{Iv^2}{R^2} \]
- Dividing through by \( m \):
\[ 2gh = v^2 + \frac{Iv^2}{mR^2} \]
- This can be rearranged to:
\[ 2gh - v^2 = \frac{Iv^2}{mR^2} \quad \text{(Equation 1)} \]

3. **Case 2: Sliding Without Rolling**:
- In this case, the potential energy converts only to translational kinetic energy:
\[ mgh = \frac{1}{2} m \left(\frac{5v}{4}\right)^2 \]
- Calculating the kinetic energy:
\[ mgh = \frac{1}{2} m \cdot \frac{25v^2}{16} \]
- Rearranging gives:
\[ 2gh = \frac{25v^2}{16} \]
- Thus:
\[ v^2 = \frac{32gh}{25} \quad \text{(Equation 2)} \]

4. **Substituting Equation 2 into Equation 1**:
- Substitute \( v^2 \) from Equation 2 into Equation 1:
\[ 2gh - \frac{32gh}{25} = \frac{Iv^2}{mR^2} \]
- Finding a common denominator:
\[ \frac{50gh - 32gh}{25} = \frac{Iv^2}{mR^2} \]
- Simplifying gives:
\[ \frac{18gh}{25} = \frac{Iv^2}{mR^2} \]
- Now substituting \( v^2 \) from Equation 2:
\[ \frac{18gh}{25} = \frac{I \cdot \frac{32gh}{25}}{mR^2} \]
- Canceling \( gh \) and simplifying:
\[ 18 = \frac{32I}{mR^2} \]
- Rearranging gives:
\[ I = \frac{18mR^2}{32} = \frac{9mR^2}{16} \]

5. **Finding the Radius of Gyration**:
- The radius of gyration \( k \) is defined as:
\[ I = mk^2 \]
- Setting the two expressions for \( I \) equal:
\[ mk^2 = \frac{9mR^2}{16} \]
- Canceling \( m \):
\[ k^2 = \frac{9R^2}{16} \]
- Taking the square root gives:
\[ k = \frac{3R}{4} \]

### Final Answer:
The radius of gyration \( k \) of the sphere is \( \frac{3R}{4} \).