Physics - Rotation Question with Solution | TestHub

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

Mass of loop P, $M_P=\lambda (2\pi r)$

Mass of loop Q, $M_Q=\lambda (2\pi nr)$

Moment of inertia of a circular loop about its axis is given by $I=M{R}^2$.

Moment of inertia of loop P about its axis:
$I_P=M_P{r}^2=\lambda (2\pi r)r^2=2\pi \lambda r^3$

Moment of inertia of loop Q about its axis:
$I_Q=M_Q(nr{)}^2=\lambda (2\pi nr)(nr{)}^2=2\pi \lambda n^3{r}^3$

Given that $I_Q=4I_P$:
$2\pi \lambda n^3{r}^3=4(2\pi \lambda r^3)$

$n^3=4$

$n=4^{1/3}=(2^2{)}^{1/3}=2^{2/3}$

Answer: $(2{)}^{2/3}$