Question

Two loops P and Q are made from a uniform wire. The radii of P and Q are $r_1$ and $r_2$ respectively, and their moments of inertia about their own axises are $I_1$ and $I_2$ respectively. If $I_2=4I_1$, the $r_1/r_2$ equals:

• A. $4^{2/3}$
• B. $4^{1/3}$
• C. $4^{7/3}$
• D. $4^{-1/3}$

Answer 1

The correct option is D $4^{-1/3}$

$\begin{array}{c}\text{Given - * Two loops made from same wire pound of}\\ \text{of radius re, and}{\mu }_2\text{and moment of}\\ \text{Inertia I, and}{I}_2\end{array}$

$\begin{array}{c}\begin{array}{cc}I=M{R}^2& M\to \text{Mass of the loop}\\ & R\to \text{Radius}\\ M=\sigma \cdot 2\pi R& \sigma \to \text{linear density of wire.}\end{array}\end{array}$

$\begin{array}{c}\Rightarrow I=\sigma 2\pi R^3\\ \Rightarrow \frac{I_1}{I_2}=\frac{\sigma 2\pi r_1^3}{\sigma 2\pi r_2^3}=\frac{r_1^3}{r_2^3}\\ \text{we have}{I}_2=4I_1\\ \Rightarrow \frac{r_1^3}{r_2^3}=\frac{1}{4}\Rightarrow \frac{r_{a1}}{r_2}=\frac{1}{4^{1/3}}=4^{-1/3}\end{array}$