From a uniform wire,two circular loops are ma | vedclass

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(D) The moment of inertia of a circular loop of mass $m$ and radius $R$ about an axis passing through its centre and perpendicular to its plane is $I

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$2$

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(D) The moment of inertia of a circular loop of mass $m$ and radius $R$ about an axis passing through its centre and perpendicular to its plane is $I = mR^2$.Let the mass of loop $P$ be $m_P$ and its radius be $r_P = r$. The mass of loop $Q$ is $m_Q$ and its radius is $r_Q = nr$.Since the wire is uniform,the mass is proportional to the circumference $(m \propto 2\pi R)$. Thus,$m_Q = n m_P$.The moment of inertia of $P$ is $I_P = m_P r^2$.The moment of inertia of $Q$ is $I_Q = m_Q (nr)^2 = (n m_P) (n^2 r^2) = n^3 m_P r^2$.Given that $I_Q = 8 I_P$,we have $n^3 m_P r^2 = 8 m_P r^2$.Therefore,$n^3 = 8$,which gives $n = 2$.

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