The radius of gyration of a solid sphere of radius R about its tangent 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 \( k \) of a solid sphere of radius \( R \) about its tangent, we can follow these steps:

### Step 1: Understand the Concept of Radius of Gyration
The radius of gyration \( k \) is defined in relation to the moment of inertia \( I \) and mass \( m \) of the object. The relationship is given by:
\[ I = m k^2 \]
Thus, we can express the radius of gyration as:
\[ k = \sqrt{\frac{I}{m}} \]

### Step 2: Determine the Moment of Inertia about the Center of Mass
For a solid sphere, the moment of inertia about an axis through its center of mass (denoted as \( I_{AB} \)) is given by:
\[ I_{AB} = \frac{2}{5} m R^2 \]

### Step 3: Apply the Parallel Axis Theorem
To find the moment of inertia about the tangent axis (denoted as \( I_{XY} \)), we apply the parallel axis theorem, which states:
\[ I_{XY} = I_{AB} + m d^2 \]
where \( d \) is the distance between the two axes. In this case, since the tangent is at a distance \( R \) from the center of the sphere, we have \( d = R \).

### Step 4: Calculate the Moment of Inertia about the Tangent Axis
Substituting the values into the equation:
\[ I_{XY} = \frac{2}{5} m R^2 + m R^2 \]
This simplifies to:
\[ I_{XY} = \frac{2}{5} m R^2 + \frac{5}{5} m R^2 = \frac{7}{5} m R^2 \]

### Step 5: Calculate the Radius of Gyration about the Tangent Axis
Now, we can find the radius of gyration \( k_{XY} \):
\[ k_{XY} = \sqrt{\frac{I_{XY}}{m}} = \sqrt{\frac{\frac{7}{5} m R^2}{m}} = \sqrt{\frac{7}{5} R^2} \]
This simplifies to:
\[ k_{XY} = R \sqrt{\frac{7}{5}} \]

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