MHTCET 16 May 2024 Shift 1 Paper

Concept:For a circular loop, the moment of inertia is $\mathrm{I} = \mathrm{MR}^2$, where $\mathrm{M}$ is the mass of the loop and $\mathrm{R}$ is its radius.Explanation:The mass of the loop is given by $\mathrm{M} = \sigma \cdot 2\pi \mathrm{R}$, where $\sigma$ is the mass per unit length.Since both loops are made from the same uniform wire, $\sigma$ is the same for both.Substituting $\mathrm{M}$ into $\mathrm{I} = \mathrm{MR}^2$ gives:$$\mathrm{I} = \sigma \cdot 2\pi \mathrm{R}^3$$Therefore, the ratio of moments of inertia is:$$\frac{\mathrm{I}_P}{\mathrm{I}_Q} = \frac{\sigma \cdot 2\pi \mathrm{R}_1^3}{\sigma \cdot 2\pi \mathrm{R}_2^3} = \left(\frac{\mathrm{R}_1}{\mathrm{R}_2}\right)^3$$Given $\frac{\mathrm{I}_P}{\mathrm{I}_Q} = 27$, we get:$$27 = \left(\frac{\mathrm{R}_1}{\mathrm{R}_2}\right)^3$$Taking the cube root on both sides:$$\frac{\mathrm{R}_1}{\mathrm{R}_2} = 3$$Thus, the ratio is $3 : 1$.Answer:$\frac{\mathrm{R}_1}{\mathrm{R}_2} = 3 : 1$, which corresponds to option B.