Two circular loops P and Q of radii r and nr | vedclass

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Question

(B) The moment of inertia $I$ of a circular loop of mass $M$ and radius $R$ about its central axis is given by $I = MR^2$.Let the linear mass density

Vedclass · Accepted Answer

$(2)^{2/3}$

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(B) The moment of inertia $I$ of a circular loop of mass $M$ and radius $R$ about its central axis is given by $I = MR^2$.Let the linear mass density of the wire be $\lambda$.The mass of loop $P$ is $M_P = \lambda \cdot (2\pi r)$ and its radius is $R_P = r$.Thus,$I_P = M_P R_P^2 = (2\pi r \lambda) r^2 = 2\pi \lambda r^3$.The mass of loop $Q$ is $M_Q = \lambda \cdot (2\pi nr)$ and its radius is $R_Q = nr$.Thus,$I_Q = M_Q R_Q^2 = (2\pi nr \lambda) (nr)^2 = 2\pi \lambda n^3 r^3$.Given that $I_Q = 4 I_P$,we have $2\pi \lambda n^3 r^3 = 4(2\pi \lambda r^3)$.Canceling $2\pi \lambda r^3$ from both sides,we get $n^3 = 4$.Therefore,$n = 4^{1/3} = (2^2)^{1/3} = 2^{2/3}$.

Two circular loops $P$ and $Q$ of radii $r$ and $nr$ are made respectively from a uniform wire. The moment of inertia of loop $Q$ about its axis is $4$ times that of loop $P$ about its axis. The value of $n$ is:

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