Physics - Rotation Question with Solution | TestHub

Let $\lambda$ be the linear mass density of the uniform metal wire.

The mass of a circular loop of radius $R$ is given by $M=2\pi R\lambda$.

The moment of inertia of a circular loop about its central axis is $I=M{R}^2$.

Substituting the expression for mass, we get:

$I=(2\pi R\lambda )R^2=2\pi \lambda R^3$

This shows that the moment of inertia is directly proportional to the cube of the radius, $I\propto R^3$.

For loops P and Q, the ratio of their moments of inertia is:

$\frac{I_P}{I_Q}={\left(\frac{R_1}{R_2}\right)}^3$

Given $\frac{I_P}{I_Q}=27$, we have:

${\left(\frac{R_1}{R_2}\right)}^3=27$

Taking the cube root on both sides:

$\frac{R_1}{R_2}=3$

The ratio $R_1:R_2$ is $3:1$.

Answer: $3:1$