The radius of gyration of a solid sphere of radius R about its tangential is

Options

A. `sqrt((7)/(5))R`
B. `sqrt((2)/(5))R`
C. `sqrt((5)/(7))`
D. `R`

Explanation

To find the radius of gyration of a solid sphere of radius \( R \) about its tangential edge, we will follow these steps:

### Step 1: Determine the Moment of Inertia about the Center
The moment of inertia \( I_C \) of a solid sphere about its center is given by the formula:
\[ I_C = \frac{2}{5} m R^2 \]
where \( m \) is the mass of the sphere and \( R \) is its radius.

### Step 2: Apply the Parallel Axis Theorem
To find the moment of inertia about the tangential edge, we will use the parallel axis theorem, which states:
\[ I = I_C + m d^2 \]
where \( d \) is the distance from the center of mass to the new axis. For a solid sphere, when we move from the center to the tangential edge, the distance \( d \) is equal to the radius \( R \).

### Step 3: Substitute the Values
Substituting the values into the parallel axis theorem:
\[ I = I_C + m R^2 \]
\[ I = \frac{2}{5} m R^2 + m R^2 \]
\[ I = \frac{2}{5} m R^2 + \frac{5}{5} m R^2 = \frac{7}{5} m R^2 \]

### Step 4: Relate Moment of Inertia to Radius of Gyration
The moment of inertia can also be expressed in terms of the radius of gyration \( k \):
\[ I = m k^2 \]
Setting the two expressions for \( I \) equal gives:
\[ m k^2 = \frac{7}{5} m R^2 \]

### Step 5: Solve for the Radius of Gyration
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ k^2 = \frac{7}{5} R^2 \]
Taking the square root of both sides gives:
\[ k = R \sqrt{\frac{7}{5}} \]

### Final Answer
Thus, the radius of gyration of the solid sphere about its tangential edge is:
\[ k = R \sqrt{\frac{7}{5}} \]