MHT CET · Physics · Rotational Motion

Let $M$ be the mass and $L$ be the length of a thin uniform rod. In first case, axis of rotation is passing through center 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

A $1$
B $\frac{1}{2}$
C $\frac{1}{4}$
D $\frac{1}{8}$

Verified Solution

Answer & Solution

Correct Answer

(B) $\frac{1}{2}$

Step-by-step Solution

Detailed explanation

Moment of inertia of rod whose axis of rotation is passing through center and perpendicular to the plane of rod is

I = \frac{M L^{2}}{12} a n d I = M K_{1}^{2}

(where

K_{1}

radius of gravitation)

\Rightarrow K_{1} = \frac{L}{2 \sqrt{3}}

..... (i)
When axis of rotation of rod is passing through one end of rod, then

I = M K_{2}^{2} = \frac{M L^{2}}{3} \Rightarrow K_{2} = \frac{L}{\sqrt{3}}

..... (ii)
Taking ratios of (i) and (ii) we get

\frac{K_{1}}{K_{2}} = \frac{L}{2 \sqrt{3}} \times \frac{\sqrt{3}}{L} = \frac{1}{2}

\Rightarrow \frac{K_{1}}{K_{2}} = \frac{1}{2}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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