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 \(\left.I_P / I_Q=27\right)\)

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)
\(\begin{aligned}
& I=\sigma \cdot 2 \pi \mathrm{R}^3 \\
\therefore \quad & \frac{\mathrm{I}_{\mathrm{P}}}{\mathrm{I}_{\mathrm{Q}}}= \\
& \frac{\sigma \cdot 2 \pi \mathrm{R}_1^3}{\sigma \cdot 2 \pi \mathrm{R}_2^3} \\
& \frac{\mathrm{I}_{\mathrm{P}}}{\mathrm{I}_{\mathrm{Q}}}=\left(\frac{\mathrm{R}_1}{\mathrm{R}_2}\right)^3 \\
& \frac{27}{1}=\left(\frac{\mathrm{R}_1}{\mathrm{R}_2}\right)^3 \quad \ldots(\text { (ii) } \\
\therefore \quad & \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{3}{1}
\end{aligned}\)