Let M be the mass and L be the length of a thin uniform rod. In first case, axis of rotation is passing through centre and perpendicular to the length of the rod. In second case, axis of rotation is passing through one end and perpendicular to the length of the rod. The ratio of radius of gyration in first case to second case is

Options

A. 1
B. `(1)/(2)`
C. `(1)/(4)`
D. `(1)/(8)`

Answer

To solve the problem, we need to find the ratio of the radius of gyration of a thin uniform rod when the axis of rotation is at the center versus when it is at one end.

### Step-by-Step Solution:

1. **Define the Radius of Gyration**:
The radius of gyration \( k \) is defined as:
\[ k = \sqrt{\frac{I}{M}} \]
where \( I \) is the moment of inertia and \( M \) is the mass of the rod.

2. **Calculate Moment of Inertia for Case 1** (Axis through the center):
For a thin uniform rod of length \( L \) and mass \( M \), the moment of inertia about an axis through its center and perpendicular to its length is given by:
\[ I_1 = \frac{1}{12} M L^2 \]

3. **Calculate Radius of Gyration for Case 1**:
Using the formula for the radius of gyration:
\[ k_1 = \sqrt{\frac{I_1}{M}} = \sqrt{\frac{\frac{1}{12} M L^2}{M}} = \sqrt{\frac{1}{12} L^2} = \frac{L}{\sqrt{12}} = \frac{L}{2\sqrt{3}} \]

4. **Calculate Moment of Inertia for Case 2** (Axis through one end):
The moment of inertia about an axis through one end and perpendicular to its length is given by:
\[ I_2 = \frac{1}{3} M L^2 \]

5. **Calculate Radius of Gyration for Case 2**:
Using the formula for the radius of gyration:
\[ k_2 = \sqrt{\frac{I_2}{M}} = \sqrt{\frac{\frac{1}{3} M L^2}{M}} = \sqrt{\frac{1}{3} L^2} = \frac{L}{\sqrt{3}} \]

6. **Find the Ratio of Radii of Gyration**:
Now, we calculate the ratio of the radius of gyration from Case 1 to Case 2:
\[ \frac{k_1}{k_2} = \frac{\frac{L}{2\sqrt{3}}}{\frac{L}{\sqrt{3}}} = \frac{1}{2} \]

### Final Answer:
The ratio of the radius of gyration in the first case to the second case is:
\[ \frac{k_1}{k_2} = \frac{1}{2} \]