MHT CET · Physics · Rotational Motion
Two circular rings 'A' and 'B' of radii 'nR' and 'R' are made from the same wire. The
moment of inertia of 'A' about an axis passing through the centre and
perpendicular to the plane of 'A' is 64 times that of the ring 'B'. The value of ' \(n\) ' is
- A 8
- B 3
- C 6
- D 4
Answer & Solution
Correct Answer
(D) 4
Step-by-step Solution
Detailed explanation
\(\frac{\mathrm{I}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{B}}}=\frac{64}{1}\)
\(64=\frac{m_{A} r_{A}^{2}}{m_{B} r_{B}^{2}}=\frac{2 \pi r_{A}}{2 \pi r_{B}} \frac{r_{A}^{2}}{r_{B}^{2}}=\frac{r_{A}^{3}}{r_{B}^{3}}=n^{3}\)
\(\therefore n=4\)
\(64=\frac{m_{A} r_{A}^{2}}{m_{B} r_{B}^{2}}=\frac{2 \pi r_{A}}{2 \pi r_{B}} \frac{r_{A}^{2}}{r_{B}^{2}}=\frac{r_{A}^{3}}{r_{B}^{3}}=n^{3}\)
\(\therefore n=4\)
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