Question

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

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

Answer: (A)

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