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

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(D) Since the wire is uniform,the mass of the loop is proportional to its circumference (length of the wire).Let the mass of loop $P$ be $m_P = m$. Th

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(D) Since the wire is uniform,the mass of the loop is proportional to its circumference (length of the wire).Let the mass of loop $P$ be $m_P = m$. The circumference of $P$ is $2\pi r$.The circumference of loop $Q$ is $2\pi(nr) = n(2\pi r)$.Therefore,the mass of loop $Q$ is $m_Q = n 	imes m_P = nm$.The moment of inertia of a circular ring of mass $M$ and radius $R$ about an axis passing through its centre and perpendicular to its plane is $I = MR^2$.For loop $P$,$I_P = m_P r^2 = mr^2$.For loop $Q$,$I_Q = m_Q (nr)^2 = (nm)(n^2r^2) = n^3 mr^2$.Given that $I_Q = 8 I_P$,we have:$n^3 mr^2 = 8(mr^2)$$n^3 = 8$$n = \sqrt[3]{8} = 2$.

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