The radius of gyration of a uniform solid sphere of radius R., about an axis passing through a point `R/2` away from the centre of the sphere is:

Options

A. `sqrt((13)/(20))R`
B. `sqrt((23)/(40))R`
C. `sqrt((23)/(20))R`
D. `sqrt((13)/(40))R`

Explanation

To find the radius of gyration of a uniform solid sphere about an axis passing through a point \( \frac{R}{2} \) away from the center of the sphere, we can follow these steps:

### Step 1: Identify the parameters
- Let \( R \) be the radius of the sphere.
- The mass of the sphere is denoted as \( m \).
- The distance from the center of the sphere to the axis of rotation is \( d = \frac{R}{2} \).

### Step 2: Calculate the moment of inertia about the center
The moment of inertia \( I_{CM} \) of a solid sphere about an axis through its center is given by:
\[ I_{CM} = \frac{2}{5} m R^2 \]

### Step 3: Apply the Parallel Axis Theorem
The Parallel Axis Theorem states that:
\[ I = I_{CM} + m d^2 \]
where \( I \) is the moment of inertia about the new axis, \( I_{CM} \) is the moment of inertia about the center of mass, \( m \) is the mass of the sphere, and \( d \) is the distance from the center of mass to the new axis.

Substituting the values:
\[ d = \frac{R}{2} \]
\[ I = \frac{2}{5} m R^2 + m \left(\frac{R}{2}\right)^2 \]
Calculating \( m \left(\frac{R}{2}\right)^2 \):
\[ m \left(\frac{R}{2}\right)^2 = m \cdot \frac{R^2}{4} = \frac{m R^2}{4} \]
Now substituting this back into the equation for \( I \):
\[ I = \frac{2}{5} m R^2 + \frac{m R^2}{4} \]

### Step 4: Find a common denominator and simplify
The common denominator for \( \frac{2}{5} \) and \( \frac{1}{4} \) is 20. Thus, we convert both fractions:
\[ \frac{2}{5} = \frac{8}{20}, \quad \frac{1}{4} = \frac{5}{20} \]
Now substituting these into the moment of inertia equation:
\[ I = \frac{8}{20} m R^2 + \frac{5}{20} m R^2 = \frac{13}{20} m R^2 \]

### Step 5: Calculate the radius of gyration
The radius of gyration \( k \) is given by:
\[ k = \sqrt{\frac{I}{m}} \]
Substituting the value of \( I \):
\[ k = \sqrt{\frac{\frac{13}{20} m R^2}{m}} = \sqrt{\frac{13}{20} R^2} = R \sqrt{\frac{13}{20}} \]

### Final Answer
Thus, the radius of gyration of the uniform solid sphere about the axis passing through a point \( \frac{R}{2} \) away from the center is:
\[ k = \sqrt{\frac{13}{20}} R \]