Question

Let \( M \) and \( L \) be the mass and length of a thin uniform rod respectively. In the first case, the axis of rotation passes through the centre and is perpendicular to its length. In the second case, the axis of rotation passes through one end and is perpendicular to its length. The ratio of radii of gyration in first case to second case is

• A. \( 3 : 1 \)
• B. \( 1 : 2 \)
• C. \( 2 : 1 \)
• D. \( 1 : 3 \)

Hint:

Radius of gyration compares mass distribution relative to axis of rotation.

Answer 1

The Correct Option is B

Step 1: Moment of inertia about centre.
\[ I_1 = \frac{1}{12} ML^2 \]
Step 2: Moment of inertia about one end.
\[ I_2 = \frac{1}{3} ML^2 \]
Step 3: Radius of gyration formula.
\[ k = \sqrt{\frac{I}{M}} \]
Step 4: Calculate radii of gyration.
\[ k_1 = \frac{L}{\sqrt{12}}, \quad k_2 = \frac{L}{\sqrt{3}} \]
Step 5: Find ratio.
\[ \frac{k_1}{k_2} = \frac{1/\sqrt{12}}{1/\sqrt{3}} = \frac{1}{2} \]
Step 6: Conclusion.
The required ratio is \( 1 : 2 \).