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 a uniform w

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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 a uniform wire,the mass $m$ is proportional to the circumference,so $m = \lambda (2\pi r)$,where $\lambda$ is the linear mass density.Thus,$I = (\lambda 2\pi r) r^2 = 2\pi\lambda r^3$.Given the ratio of moments of inertia: $\frac{I_2}{I_1} = \frac{2\pi\lambda r_2^3}{2\pi\lambda r_1^3} = \left(\frac{r_2}{r_1}\right)^3$.We are given $\frac{I_2}{I_1} = 4$.Therefore,$\left(\frac{r_2}{r_1}\right)^3 = 4$.Taking the cube root on both sides,we get $\frac{r_2}{r_1} = 4^{1/3}$.

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 $I_2/I_1 = 4$,then $\frac{r_2}{r_1}$ equals:

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