JEE Advanced · Physics · 8. Rotational Motion

Four solid spheres each of diameter $\sqrt{5} \mathrm{~cm}$ and mass $0.5 \mathrm{~kg}$ ar placed with their centres at the corners of a square of side $4 \mathrm{~cm}$. The moment of inertia of the system about the diagonal of the square is $N \times 10^{-4} \mathrm{~kg}-\mathrm{m}^2$, then $N$ is

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Answer & Solution

Correct Answer

(D) 9

Step-by-step Solution

Detailed explanation

$r =\frac{d}{2}=\frac{\sqrt{5}}{2} \mathrm{~cm} $
$ =\frac{\sqrt{5}}{2} \times 10^{-2} \mathrm{~m} $
$ m =0.5 \mathrm{~kg} $
$ a =4 \mathrm{~cm} $
$ =4 \times 10^{-2} \mathrm{~m} $
$ I_{X X} =I_1+I_2+I_3+I_4$

$= {\left[\frac{2}{5} m r^2+m\left(\frac{a}{\sqrt{2}}\right)^2\right]+\frac{2}{5} m r^2 } $ $ +\left[\frac{2}{5} m r^2+m\left(\frac{a}{\sqrt{2}}\right)^2\right]+\frac{2}{5} m r^2$
Substituting the values, we get
$I_{X X} =9 \times 10^{-4} \mathrm{Kgm}^{-2} $
$ \therefore N =9$
Answer is 9.
Analysis of Question
(i) Question is simple.
(ii) Only theorem of parallel axes is to be used properly.

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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