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Question: Answered & Verified by Expert
Two circular loops $\mathrm{A}$ and $\mathrm{B}$ of radii ${ }^{\prime} \mathrm{R}^{\prime}$ and ' $\mathrm{NR}^{\prime}$ respectively are made from a
uniform wire. Moment of inertia of B about its axis is 3 times that of A about its
axis. The value of $\mathrm{N}$ is
PhysicsRotational MotionMHT CETMHT CET 2020 (19 Oct Shift 2)
Options:
  • A $[5]^{\frac{1}{3}}$
  • B $[3]^{\frac{1}{3}}$
  • C $[4]^{\frac{1}{3}}$
  • D $[2]^{\frac{1}{3}}$
Solution:
2487 Upvotes Verified Answer
The correct answer is: $[3]^{\frac{1}{3}}$
(D)
If $m$ is the mass per unit length of the wire, then mass of loop $\mathrm{A} \quad=\mathrm{M}_{\mathrm{A}}=2 \pi \mathrm{Rm}$ mass of loop $\mathrm{B} \quad=\mathrm{M}_{\mathrm{B}}=2 \pi \mathrm{NRm}$ $\therefore \frac{\mathrm{M}_{\mathrm{B}}}{\mathrm{M}_{\mathrm{A}}}=\mathrm{N}$
$\mathrm{I}_{\mathrm{A}}=\mathrm{M}_{\mathrm{A}} \mathrm{R}^{2}, \mathrm{I}_{\mathrm{B}}=\mathrm{M}_{\mathrm{B}}(\mathrm{NR})^{2}=\mathrm{M}_{\mathrm{B}} \mathrm{N}^{2} \mathrm{R}^{2}$
$\therefore \frac{\mathrm{I}_{\mathrm{B}}}{\mathrm{I}_{\mathrm{A}}}=\frac{\mathrm{M}_{\mathrm{B}} \mathrm{N}^{2}}{\mathrm{M}_{\mathrm{A}}}=\mathrm{N}^{3}$
$\therefore \quad 3=\mathrm{N}^{3}$
$\therefore \quad \mathrm{N}=(3)^{\frac{1}{3}}$

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