Two loops $P$ and $Q$ of radii $R_1$ and $R_2$ are made from uniform metal wire of same material.…
Two loops $P$ and $Q$ of radii $R_1$ and $R_2$ are made from uniform metal wire of same material. $\mathrm{I}_{\mathrm{P}}$ and $\mathrm{I}_{\mathrm{Q}}$ be the moment of inertia of loop P and Q respectively then ratio $R_1 / R_2$ is $\left(\right.$ Given $\left.I_P / I_Q=27\right)$
$4: 1$
$3: 1$
$9: 1$
$6: 1$
Solution
Two loops made from same material
$\mathrm{I}=\mathrm{MR}^2...(i)$
$\mathrm{M}=\sigma .2 \pi \mathrm{R}$ Substituting in (i),
...( $\sigma$ is mass per unit length)
$\begin{aligned}
& I=\sigma \cdot 2 \pi \mathrm{R}^3 \\
\therefore \quad & \frac{\mathrm{I}_{\mathrm{P}}}{\mathrm{I}_{\mathrm{Q}}}= \\
& \frac{\sigma \cdot 2 \pi \mathrm{R}_1^3}{\sigma \cdot 2 \pi \mathrm{R}_2^3} \\
& \frac{\mathrm{I}_{\mathrm{P}}}{\mathrm{I}_{\mathrm{Q}}}=\left(\frac{\mathrm{R}_1}{\mathrm{R}_2}\right)^3 \\
& \frac{27}{1}=\left(\frac{\mathrm{R}_1}{\mathrm{R}_2}\right)^3 \quad \ldots(\text { (ii) } \\
\therefore \quad & \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{3}{1}
\end{aligned}$