Rotation 1 Question 16

19. A thin wire of length L and uniform linear mass density ρ is bent into a circular loop with centre at O as shown. The moment of inertia of the loop about the axis XX is (2000)

(a) ρL38π2

(b) ρL316π2

(c) 5ρL316π2

(d) 3ρL38π2

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Answer:

Correct Answer: 19. (d)

Solution:

  1. Mass of the ringM=ρL. Let R be the radius of the ring, then

or

L=2πRR=L2π

Moment of inertia about an axis passing through O and parallel to XX will be

I0=12MR2

Therefore, moment of inertia about XX (from parallel axis theorem) will be given by

IXX=12MR2+MR2=32MR2

Substituting values of M and R

IXX=32(ρL)L24π2=3ρL38π2



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