Question

Two circular loops P and Q are made from a uniform wire. The radii of P and Q are $R_1$ and $R_2$ respectively. The moments of inertia about their own central axis are $I_P$ and $I_Q$ respectively. If $\frac{I_P}{I_Q} = \frac{1}{8}$ then the ratio $\frac{R_2}{R_1}$ is

• A. $4$
• B. $3$
• C. $2$
• D. $5$

Hint:

Be careful! For a solid disc or a pre-existing object where mass is held constant, $I \propto R^2$. But when loops are bent from a continuous wire, increasing the radius automatically adds more wire length (and thus more mass), causing the relationship to change to $I \propto R^3$.

Answer 1

The Correct Option is C

Step 1: State what we want.
Two rings P and Q are bent from the same uniform wire, with radii $R_1$ and $R_2$. Given $\dfrac{I_P}{I_Q} = \dfrac{1}{8}$, find $\dfrac{R_2}{R_1}$.
Step 2: Moment of inertia of a ring.
About its central axis, a ring of mass $M$ and radius $R$ has $I = M R^2$.
Step 3: Express the mass through the wire.
The same wire means the same mass per unit length $\lambda$. A ring of radius $R$ uses length $2\pi R$, so its mass is $M = \lambda (2\pi R)$.
Step 4: Get $I$ in terms of $R$ only.
Then $I = \lambda(2\pi R) R^2 = 2\pi\lambda R^3$, so $I \propto R^3$.
Step 5: Form the ratio.
$\dfrac{I_P}{I_Q} = \left(\dfrac{R_1}{R_2}\right)^3 = \dfrac{1}{8}$. Taking the cube root, $\dfrac{R_1}{R_2} = \dfrac{1}{2}$.
Step 6: Answer the asked ratio.
Inverting, $\dfrac{R_2}{R_1} = 2$, which is option (3). Because mass itself grows with radius, $I$ scales as $R^3$ rather than $R^2$ here.
\[ \boxed{\dfrac{R_2}{R_1} = 2} \]