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  4. Two circular loops A and B or radii rA and rB respectively are made from a uniform wire. The ratio of their moments of inertia about axes passing through their centers and perpendicular to their planes is (IB/IA)=8, then ((rB/rA)) equal to

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Q. Two circular loops A and B or radii $r_A$ and $ r_B$ respectively are made from a uniform wire. The ratio of their moments of inertia about axes passing through their centers and perpendicular to their planes is $\frac{I_B}{I_A}=8,\,then\,\left(\frac{r_B}{r_A}\right) $ equal to

EAMCETEAMCET 2003System of Particles and Rotational Motion

Solution:

Ratio of moment of inertia of a loop about axis passing through their center and perpendicular to their plane is
$\frac{I_B}{I_A}=8$
=$\frac{\frac{1}{2}mr^2_B}{\frac{1}{2}mr^2_A}=8$
or$\frac{r^2_B}{r^2_A}=\frac{8}{2}=4 \,or\,\frac{r_B}{r_A}=2$
so,$\frac{r_B}{r_A}=\frac{2}{1}$