Two loops P and Q are made from a uniform wir | vedclass

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(B) The moment of inertia of a loop (ring) about its central axis is given by $I = Mr^2$. Since the loops are made from the same uniform wire,the mass

Vedclass · Accepted Answer

$4^{1/3}$

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(B) The moment of inertia of a loop (ring) about its central axis is given by $I = Mr^2$. Since the loops are made from the same uniform wire,the mass $M$ is proportional to the circumference,so $M \propto 2\pi r$,which implies $M \propto r$. Substituting $M = kr$ (where $k$ is a constant) into the moment of inertia formula: $I = (kr)r^2 = kr^3$. Therefore,$I \propto r^3$. Given the ratio $I_2/I_1 = 4$,we have $(r_2/r_1)^3 = 4$. Taking the cube root on both sides,we get $r_2/r_1 = 4^{1/3}$.

Two loops $P$ and $Q$ are made from a uniform wire. The radii of $P$ and $Q$ are $r_1$ and $r_2$ respectively. Their moments of inertia are $I_1$ and $I_2$ respectively. If $I_2/I_1 = 4$,then $r_2/r_1 = ?$

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