A thin wire of length ' $L$ ' and uniform linear mass density ' $\mathrm{m}$ ' is bent into a circular coil.…

A thin wire of length ' $L$ ' and uniform linear mass density ' $\mathrm{m}$ ' is bent into a circular coil. The moment of inertia of this coil about tangential axis and in plane of the coil is
  1. $\frac{3 \mathrm{~mL}^2}{5 \pi^2}$
  2. $\frac{3 m L^3}{8 \pi^2}$
  3. $\frac{3 \mathrm{~mL}^3}{4 \pi^2}$
  4. $\frac{3 \mathrm{~mL}^2}{7 \pi^2}$

Solution

$\therefore \quad$ Moment of inertia of thin wire: $\begin{aligned} \mathrm{I} & =\frac{\mathrm{MR}^2}{2} \\ \mathrm{M} & =\mathrm{V} \times \mathrm{m} \text { and } \mathrm{L}=2 \pi \mathrm{R} \\ \mathrm{R} & =\frac{\mathrm{L}}{2 \pi} \\ \therefore \quad \mathrm{I} & =\frac{\mathrm{Lm}}{2}\left(\frac{\mathrm{L}}{2 \pi}\right)^2 \\ \mathrm{I} & =\frac{\mathrm{mL}^3}{8 \pi^2} \end{aligned}$ $\therefore \quad$ Using Parallel axis theorem: $\begin{aligned} & I^{\prime}=I+M R^2 \\ & I=\frac{m L^3}{8 \pi^2}+L m\left(\frac{L}{2 \pi}\right)^2 \\ & I=\frac{3 m L^3}{8 \pi^2} \end{aligned}$ ~

Asked in: MHT CET 2023 (12 May Shift 2)

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