Question

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

• A. \(\frac{1}{4}\)
• B. \(\frac{1}{2}\)
• C. \(\frac{1}{8}\)
• D. \(\frac{1}{6}\)

Hint:

Use I = ML^2/12 about the centre and ML^2/3 about the end, with I = M k^2.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The radius of gyration \(k\) is defined by \(I=Mk^2\), so \(k=\sqrt{I/M}\).

Step 2: Moment of inertia in each case:
Centre axis: \(I_1=\dfrac{ML^2}{12}\). End axis: \(I_2=\dfrac{ML^2}{3}\).

Step 3: Find the radii of gyration:
\(k_1=\dfrac{L}{\sqrt{12}}\) and \(k_2=\dfrac{L}{\sqrt3}\).

Step 4: Take the ratio:
\[ \dfrac{k_1}{k_2}=\sqrt{\dfrac{3}{12}}=\sqrt{\dfrac14}=\dfrac12 \]
Option B.

Step 5: Why the other options are wrong.
\(\dfrac14\) is the ratio of the moments of inertia, \(I_1/I_2\), not the radii. \(\dfrac18\) and \(\dfrac16\) do not follow from the formulas.

Final Answer:
The ratio of radii of gyration is 1/2. \[ \boxed{\text{(B) }\dfrac12} \]