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(B) The moment of inertia $I$ of a ring about its central axis is given by $I = mR^2$,where $m$ is the mass and $R$ is the radius.Since the diameter $

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$1:2$

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(B) The moment of inertia $I$ of a ring about its central axis is given by $I = mR^2$,where $m$ is the mass and $R$ is the radius.Since the diameter $D = 2R$,we can write $R = D/2$.Substituting this into the formula,we get $I = m(D/2)^2 = m(D^2/4)$.This implies $I \propto mD^2$,or $m \propto I/D^2$.Given the ratios $I_1/I_2 = 2/1$ and $D_1/D_2 = 2/1$,we find the ratio of masses as:$m_1/m_2 = (I_1/I_2) 	imes (D_2/D_1)^2$$m_1/m_2 = (2/1) 	imes (1/2)^2 = 2 	imes (1/4) = 2/4 = 1/2$.Therefore,the ratio of their masses is $1:2$.

Two rings have their moments of inertia in the ratio $2:1$ and their diameters are in the ratio $2:1$. The ratio of their masses is

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