MHT CET · Physics · Rotational Motion

Three identical rods each of mass ‘M’ and length ‘L’ are joined to form a symbol ‘H’. The moment of inertia of the system about one of the sides of ‘H’ is

A $\frac{2 M L^{2}}{3}$
B $\frac{M L^{2}}{2}$
C $\frac{M L^{2}}{6}$
D $\frac{4 M L^{2}}{3}$

Verified Solution

Answer & Solution

Correct Answer

(D) $\frac{4 M L^{2}}{3}$

Step-by-step Solution

Detailed explanation

The given situation can be shown as

Let us take the moment of inertia of the system about rod

R_{1}

then total moment of inertia is
For rod

R_{1} l_{1} = 0

For rod

R_{2}

, using perpendicular axis theorem,

l_{2} = \frac{M L^{2}}{3}

For rod

R_{3}

, using parallel axis theorem,

l_{3} = l_{C M} + l_{(a t L)} = 0 + M L^{2} = M L^{2}

Now, putting the values of

l_{1}, l_{2}

and

l_{3}

in Eq. (i), we get

l_{τ} = 0 + \frac{M L^{2}}{3} + M L^{2} \Rightarrow l_{τ} = \frac{4 M L^{2}}{3}

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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