Two loops P and Q are made from a uniform wir | vedclass

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(B) The moment of inertia of a loop of mass $M$ and radius $R$ about its central axis is given by $I = MR^2$.Since the loops are made from the same un

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$4^{1/3}$

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(B) The moment of inertia of a loop of mass $M$ and radius $R$ about its central axis is given by $I = MR^2$.Since the loops are made from the same uniform wire,the mass $M$ is proportional to the circumference,so $M = \lambda (2\pi r)$,where $\lambda$ is the linear mass density.Thus,$I = (\lambda 2\pi r) r^2 = 2\pi \lambda r^3$.Therefore,the ratio of moments of inertia is $\frac{I_1}{I_2} = \frac{r_1^3}{r_2^3}$.Given $\frac{I_2}{I_1} = 4$,we have $\frac{I_1}{I_2} = \frac{1}{4}$.So,$\frac{r_1^3}{r_2^3} = \frac{1}{4}$,which implies $\left(\frac{r_2}{r_1}\right)^3 = 4$.Taking the cube root on both sides,we get $\frac{r_2}{r_1} = 4^{1/3}$.

List-$I$	List-$II$
$(i) \ a \bar{a}$	$(A) -\frac{\pi}{3}$
$(ii) \ \arg \left(\frac{1}{\bar{a}}\right)$	$(B) -i \sqrt{3}$
$(iii) \ a-\bar{a}$	$(C) \frac{2i}{\sqrt{3}}$
$(iv) \ \operatorname{Im}\left(\frac{4}{3a}\right)$	$(D) 1$
	$(E) \frac{\pi}{3}$
	$(F) \frac{2}{\sqrt{3}}$

Two loops $P$ and $Q$ are made from a uniform wire. The radii of $P$ and $Q$ are $r_1$ and $r_2$ respectively,and their moments of inertia are $I_1$ and $I_2$ respectively. If $\frac{I_2}{I_1} = 4$,then $\frac{r_2}{r_1}$ equals:

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