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Rotational Motion question

2021 · 18 Mar · Shift 2 · Q51
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  5. /2021 · 18 Mar · Shift 2 · Q51

Rotational Motion question

2021 · 18 Mar · Shift 2 · Q51

JEE MainPhysicsRotational MotionMCQ+4 / −1
Consider a uniform wire of mass M and length L. It is bent into a semicircle. Its moment of inertia about a line perpendicular to the plane of the wire passing through the center is :
  1. A
    14ML2π2{1 \over 4}{{M{L^2}} \over {{\pi ^2}}}41​π2ML2​
  2. B
    12ML2π2{1 \over 2}{{M{L^2}} \over {{\pi ^2}}}21​π2ML2​
  3. C
    25ML2π2{2 \over 5}{{M{L^2}} \over {{\pi ^2}}}52​π2ML2​
  4. D
    ML2π2{{M{L^2}} \over {{\pi ^2}}}π2ML2​
View written solutionFree

Correct answer: D

  1. Relate the wire length to the radius of the semicircle

If the wire of length LLL is bent into a semicircle of radius RRR, then its arc length is

L=πRL = \pi RL=πR

So,

R=LπR = \frac{L}{\pi}R=πL​

  1. Moment of inertia of the semicircular wire about the given axis

The axis is perpendicular to the plane of the semicircle and passes through the center of the circle.

Every mass element dmdmdm of the wire lies at the same distance RRR from this axis.

Hence,

dI=R2 dmdI = R^2\,dmdI=R2dm

Integrating over the whole wire,

I=∫R2 dm=R2∫dm=MR2I = \int R^2\,dm = R^2\int dm = MR^2I=∫R2dm=R2∫dm=MR2

  1. Substitute R=LπR = \dfrac{L}{\pi}R=πL​

I=M(Lπ)2=ML2π2I = M\left(\frac{L}{\pi}\right)^2 = \frac{ML^2}{\pi^2}I=M(πL​)2=π2ML2​

  1. Match with the options

I=ML2π2I = \frac{ML^2}{\pi^2}I=π2ML2​

This corresponds to Option D.

  1. Comparison with stored correct answer

Stored correct answer: D

Derived answer: D

So, the derived answer agrees with the stored correct answer.

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