Two circular rings A and B of radii nR and R | vedclass

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(D) The moment of inertia of a ring of mass $m$ and radius $r$ about an axis passing through its centre and perpendicular to its plane is $I = mr^2$.S

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$4$

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(D) The moment of inertia of a ring of mass $m$ and radius $r$ about an axis passing through its centre and perpendicular to its plane is $I = mr^2$.Since both rings are made from the same wire,the mass $m$ is proportional to the circumference $(2 \pi r)$,so $m \propto r$.Therefore,$m_A = k(2 \pi nR)$ and $m_B = k(2 \pi R)$,where $k$ is the mass per unit length.Thus,$\frac{m_A}{m_B} = \frac{nR}{R} = n$.The moment of inertia of ring $A$ is $I_A = m_A (nR)^2 = (n m_B) (n^2 R^2) = n^3 (m_B R^2)$.Since $I_B = m_B R^2$,we have $I_A = n^3 I_B$.Given $I_A = 64 I_B$,we get $n^3 = 64$.Therefore,$n = \sqrt[3]{64} = 4$.

Two circular rings $A$ and $B$ of radii $nR$ and $R$ are made from the same wire. The moment of inertia of $A$ about an axis passing through the centre and perpendicular to the plane of $A$ is $64$ times that of the ring $B$. The value of $n$ is:

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