MHT CET · Physics · Rotational Motion
Two circular loops \(\mathrm{P}\) and \(\mathrm{Q}\) of radii \(r\) and \(n r\) respectively are made from a uniform wire. Moment of inertia of \(Q\) about its axis is four times that of \(\mathrm{P}\) about its axis. The value of \(n\) is
- A \((2)^{\frac{2}{3}}\)
- B \((2)^{\frac{1}{3}}\)
- C \(2\)
- D \(\sqrt{2}\)
Answer & Solution
Correct Answer
(A) \((2)^{\frac{2}{3}}\)
Step-by-step Solution
Detailed explanation
Given, \(I_Q=4 I_P\).
Let \(m\) be the mass per unit length of wire.
\(\therefore m \times 2 \pi n r(n r)^2=4\left[(m 2 \pi r) r^2\right]\)
Implies \(n^3=4\)
\(\Rightarrow n=4^{1 / 3}\)
Let \(m\) be the mass per unit length of wire.
\(\therefore m \times 2 \pi n r(n r)^2=4\left[(m 2 \pi r) r^2\right]\)
Implies \(n^3=4\)
\(\Rightarrow n=4^{1 / 3}\)
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