The radius of gyration of a uniform rod of length l , about an axis passing through a point `(l)/(8)`away form the centre of the rod , and perpendicular to it is:

Options

A. `(l)/(4)`
B. `(l)/(64)`
C. `sqrt((19)/(192))l`
D. `sqrt((7)/(48))l`

Explanation

To find the radius of gyration of a uniform rod of length \( l \) about an axis passing through a point \( \frac{l}{8} \) away from the center of the rod and perpendicular to it, we can follow these steps:

### Step 1: Identify the Moment of Inertia about the Center of Mass
The moment of inertia \( I_{CM} \) of a uniform rod of length \( l \) about an axis passing through its center and perpendicular to its length is given by the formula:
\[ I_{CM} = \frac{1}{12} ml^2 \]
where \( m \) is the mass of the rod.

### Step 2: Use the Parallel Axis Theorem
To find the moment of inertia \( I \) about the new axis (which is \( \frac{l}{8} \) away from the center), we apply the Parallel Axis Theorem:
\[ I = I_{CM} + md^2 \]
where \( d \) is the distance from the center of mass to the new axis. Here, \( d = \frac{l}{8} \).

### Step 3: Substitute the Values
Substituting the values into the formula:
\[ I = \frac{1}{12} ml^2 + m\left(\frac{l}{8}\right)^2 \]
Calculating \( \left(\frac{l}{8}\right)^2 \):
\[ \left(\frac{l}{8}\right)^2 = \frac{l^2}{64} \]
So, substituting this back into the equation:
\[ I = \frac{1}{12} ml^2 + m\left(\frac{l^2}{64}\right) \]

### Step 4: Find a Common Denominator
To combine the two terms, we need a common denominator:
\[ I = \frac{16}{192} ml^2 + \frac{3}{192} ml^2 = \frac{19}{192} ml^2 \]

### Step 5: Calculate the Radius of Gyration
The radius of gyration \( k \) is defined as:
\[ I = mk^2 \]
Substituting for \( I \):
\[ mk^2 = \frac{19}{192} ml^2 \]
Dividing both sides by \( m \):
\[ k^2 = \frac{19}{192} l^2 \]
Taking the square root gives:
\[ k = \sqrt{\frac{19}{192}} l \]

### Final Answer
Thus, the radius of gyration of the uniform rod about the specified axis is:
\[ k = \sqrt{\frac{19}{192}} l \]