MHT CET · Physics · Rotational Motion

Two discs having moment of inertia \(\mathrm{l}_{1}\) and \(\mathrm{I}_{2}\) are made from same material have
same mass. Their thickness and radii are \(\mathrm{t}_{1}, \mathrm{t}_{2}\) and \(\mathrm{R}_{1}, \mathrm{R}_{2}\) respectively. The relation
between moment of inertia of each disc about an axis passing through its centre
and perpendicular to its plane and its thickness is

A \(\mathrm{I}_{1} \mathrm{t}_{1}=\mathrm{I}_{2} \mathrm{t}_{2}\)
B \(\mathrm{I}_{1} \mathrm{t}_{2}^{2}=\mathrm{I}_{2} \mathrm{t}_{1}^{2}\)
C \(\mathrm{I}_{1} \mathrm{t}_{2}=\mathrm{I}_{2} \mathrm{t}_{1}\)
D \(\mathrm{I}_{1} \mathrm{t}_{1}^{2}=\mathrm{I}_{2} \mathrm{t}_{2}^{2}\)

Verified Solution

Answer & Solution

Correct Answer

(A) \(\mathrm{I}_{1} \mathrm{t}_{1}=\mathrm{I}_{2} \mathrm{t}_{2}\)

Step-by-step Solution

Detailed explanation

\(I_{1}=\frac{m R_{1}^{2}}{2}\)
\(I_{2}=\frac{m R_{2}^{2}}{2}\)
\(\frac{I_{1}}{I_{2}}=\frac{R_{1}^{2}}{R_{2}^{2}}=\frac{t_{2}}{t_{1}}\)
\(\therefore \mathrm{I}_{1} \mathrm{t}_{1}=\mathrm{I}_{2} \mathrm{t}_{2}\)
\(\mathrm{m}_{1}=\pi \mathrm{R}_{1}^{2} \mathrm{t}_{1}\)
\(\mathrm{~m}_{2}=\pi \mathrm{R}_{2}^{2} \mathrm{t}_{2}\)\
\(\mathrm{m}_{1}=\mathrm{m}_{2}\)
\(\therefore \mathrm{R}_{1}^{2} \mathrm{t}_{1}=\mathrm{R}_{2}^{2} \mathrm{t}_{2}\)

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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