The radius of gyration of a sphere of radius `R` about a tangent is.

Options

A. `(sqrt(2))/(3) R`
B. `(sqrt(2))/(5) R`
C. `sqrt((5)/(3)) R`
D. `sqrt((7)/(5)) R`

Explanation

To find the radius of gyration of a sphere of radius \( R \) about a tangent, we can follow these steps:

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

### Step 2: Use the Parallel Axis Theorem
To find the moment of inertia about a tangent to the sphere, we can use the parallel axis theorem. The parallel axis theorem states that:
\[ I = I_{\text{center}} + md^2 \]
where \( d \) is the distance from the center of mass to the new axis. For a tangent to the sphere, \( d = R \).

Thus, we can write:
\[ I_{\text{tangent}} = I_{\text{center}} + mR^2 \]
Substituting the value of \( I_{\text{center}} \):
\[ I_{\text{tangent}} = \frac{2}{5} m R^2 + m R^2 \]

### Step 3: Simplify the Moment of Inertia
Now, combine the terms:
\[ I_{\text{tangent}} = \frac{2}{5} m R^2 + \frac{5}{5} m R^2 = \frac{7}{5} m R^2 \]

### Step 4: Relate Moment of Inertia to Radius of Gyration
The radius of gyration \( k \) is related to the moment of inertia by the equation:
\[ I = m k^2 \]
Substituting the expression for \( I_{\text{tangent}} \):
\[ \frac{7}{5} m R^2 = m k^2 \]

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

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