The moment of inertia of a solid sphere of mass M and radius R, about its diameter is `(2//5)MR^(2)` . Its M.I. about parallel axis passing through a point at a distance (R/2) from its centre is

Options

A. `(15)/(20)MR^(2)`
B. `(7)/(5)MR^(2)`
C. `(13)/(20)MR^(2)`
D. `(8)/(15)MR^(2)`

Explanation

To find the moment of inertia of a solid sphere about a parallel axis that is a distance \( \frac{R}{2} \) from its center, we can use the parallel axis theorem. The parallel axis theorem states that the moment of inertia \( I \) about any axis parallel to an axis through the center of mass is given by:

\[ I = I_{cm} + Md^2 \]

where:
- \( I_{cm} \) is the moment of inertia about the center of mass axis,
- \( M \) is the mass of the object,
- \( d \) is the distance between the two axes.

### Step 1: Identify the given values
- The moment of inertia of the solid sphere about its diameter (center of mass axis) is given as:
\[ I_{cm} = \frac{2}{5}MR^2 \]
- The mass \( M \) of the sphere is \( M \).
- The distance \( d \) from the center to the new axis is:
\[ d = \frac{R}{2} \]

### Step 2: Apply the parallel axis theorem
Using the parallel axis theorem:
\[ I = I_{cm} + Md^2 \]

Substituting the known values:
\[ I = \frac{2}{5}MR^2 + M\left(\frac{R}{2}\right)^2 \]

### Step 3: Calculate \( Md^2 \)
Calculate \( Md^2 \):
\[ Md^2 = M\left(\frac{R}{2}\right)^2 = M\left(\frac{R^2}{4}\right) = \frac{MR^2}{4} \]

### Step 4: Substitute back into the equation
Now substitute \( Md^2 \) back into the equation for \( I \):
\[ I = \frac{2}{5}MR^2 + \frac{MR^2}{4} \]

### Step 5: Find a common denominator and combine the terms
To combine the fractions, find a common denominator (which is 20):
\[ I = \frac{2}{5}MR^2 + \frac{MR^2}{4} = \frac{8}{20}MR^2 + \frac{5}{20}MR^2 = \frac{13}{20}MR^2 \]

### Final Answer
Thus, the moment of inertia of the solid sphere about the parallel axis passing through a point at a distance \( \frac{R}{2} \) from its center is:
\[ I = \frac{13}{20}MR^2 \]

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