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Question

Physics Question on System of Particles & Rotational Motion

Two circular loops A and B or radii rAr_A and rB 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 IBIA=8,then(rBrA)\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

2

Explanation

Solution

Ratio of moment of inertia of a loop about axis passing through their center and perpendicular to their plane is
IBIA=8\frac{I_B}{I_A}=8
=12mrB212mrA2=8\frac{\frac{1}{2}mr^2_B}{\frac{1}{2}mr^2_A}=8
orrB2rA2=82=4orrBrA=2\frac{r^2_B}{r^2_A}=\frac{8}{2}=4 \,or\,\frac{r_B}{r_A}=2
so,rBrA=21\frac{r_B}{r_A}=\frac{2}{1}