Two circular loops P and Q of radii ' r 'and ' nr ' are made respectively from a uniform wire. Moment of…
Two circular loops P and Q of radii ' r 'and ' nr ' are made respectively from a uniform wire. Moment of inertia of loop Q about its axis is four times that of loop P about its axis. The value of ' $n$ ' is
$(2)^{1 / 3}$
$(2)^{2 / 3}$
$\quad(2)^{3 / 4}$
$(2)^{1 / 4}$
Solution
The two loops are formed from the same wire. Hence, the linear density will remain constant.
$\therefore \quad \lambda=\frac{\mathrm{M}_{\mathrm{P}}}{\mathrm{~L}_{\mathrm{p}}}=\frac{\mathrm{M}_{\mathrm{Q}}}{\mathrm{~L}_{\mathrm{Q}}}$
$\begin{array}{ll}\therefore \quad & \mathrm{M}_{\mathrm{P}}=\lambda \times \mathrm{L}_{\mathrm{P}}=\lambda \times\left(2 \pi \mathrm{R}_{\mathrm{P}}\right)=2 \pi \mathrm{r} \lambda \\ & \text { Similarly, } \\ & \mathrm{M}_{\mathrm{Q}}=\lambda \times \mathrm{L}_{\mathrm{Q}}=\lambda \times\left(2 \pi \mathrm{R}_{\mathrm{Q}}\right)=2 \pi(\mathrm{nr}) \lambda \\ & \text { Also, M.I. of circular loop } \mathrm{P} \text { is, } \\ & \mathrm{I}_{\mathrm{P}}=\mathrm{M}_{\mathrm{P}} \mathrm{r}^2=2 \pi \mathrm{r} \lambda(\mathrm{r})^2=2 \pi \mathrm{r}^3 \lambda \\ & \text { and that of loop } \mathrm{Q}, \\ & \mathrm{I}_{\mathrm{Q}}=\mathrm{M}_{\mathrm{Q}}(\mathrm{nr})^2=2 \pi(\mathrm{nr}) \lambda \times(\mathrm{nr})^2=2 \pi \mathrm{n}^3 \mathrm{r}^3 \lambda \\ & \mathrm{Given}^2 \text { that, } \mathrm{I}_{\mathrm{Q}}=4 \mathrm{I}_{\mathrm{P}} \\ \therefore \quad & 2 \pi \mathrm{n}^3 \mathrm{r}^3 \lambda=4 \times 2 \pi \mathrm{r}^3 \lambda \\ \therefore \quad \mathrm{n}^3=4 \\ \therefore \quad & \mathrm{n}=(4)^{\frac{1}{3}}=(2)^{\frac{2}{3}}\end{array}$