Moment of inertia of a thin rod of a mass M and length L about an axis passing through its centre is`(ML^(2))/(12)`. Its moment of inertia about a parallel axis at a distance of `(L)/(4) from this acis is given by

Options

A. `(ML^(2))/(48)`
B. `(ML^(3))/(48)`
C. `(ML^(2))/(12)`
D. `(7ML^(2))/(48)`

Explanation

To find the moment of inertia of a thin rod of mass \( M \) and length \( L \) about a parallel axis at a distance of \( \frac{L}{4} \) from its center, we will use the Parallel Axis Theorem. Let's solve this step by step.

### Step 1: Understand the Given Information
We know:
- The moment of inertia of the thin rod about its center of mass (CM) is given by:
\[ I_{\text{CM}} = \frac{ML^2}{12} \]
- The distance \( x \) from the center of mass to the new axis is:
\[ x = \frac{L}{4} \]

### Step 2: Apply the Parallel Axis Theorem
The Parallel Axis Theorem states that the moment of inertia \( I \) about any axis parallel to the axis through the center of mass is given by:
\[ I = I_{\text{CM}} + Mx^2 \]
where \( x \) is the distance between the two axes.

### Step 3: Substitute the Values
Now, substituting the known values into the equation:
\[ I = \frac{ML^2}{12} + M\left(\frac{L}{4}\right)^2 \]

### Step 4: Calculate \( M\left(\frac{L}{4}\right)^2 \)
Calculating \( M\left(\frac{L}{4}\right)^2 \):
\[ M\left(\frac{L}{4}\right)^2 = M \cdot \frac{L^2}{16} = \frac{ML^2}{16} \]

### Step 5: Combine the Terms
Now, we can combine the two terms:
\[ I = \frac{ML^2}{12} + \frac{ML^2}{16} \]

### Step 6: Find a Common Denominator
To add these fractions, we need a common denominator. The least common multiple of 12 and 16 is 48. Thus, we rewrite the fractions:
\[ \frac{ML^2}{12} = \frac{4ML^2}{48} \]
\[ \frac{ML^2}{16} = \frac{3ML^2}{48} \]

### Step 7: Add the Fractions
Now, we can add the two fractions:
\[ I = \frac{4ML^2}{48} + \frac{3ML^2}{48} = \frac{(4 + 3)ML^2}{48} = \frac{7ML^2}{48} \]

### Final Answer
Thus, the moment of inertia of the thin rod about the parallel axis is:
\[ I = \frac{7ML^2}{48} \]