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Rotational Motion question

2019 · 12 Jan · Shift 1 · Q60
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  5. /2019 · 12 Jan · Shift 1 · Q60

Rotational Motion question

2019 · 12 Jan · Shift 1 · Q60

JEE MainPhysicsRotational MotionMCQ+4 / −1
Let the moment of inertia of a hollow cylinder of length 30 cm (inner radius 10 cm and outer radius 20 cm), about its axis be I. The radius of a thin cylinder of the same mass such that its moment of inertia about its axis is also I, is :
  1. A
    16 cm
  2. B
    12 cm
  3. C
    14 cm
  4. D
    18 cm
View written solutionFree

Correct answer: A

  1. Moment of inertia of the hollow cylinder

For a hollow cylinder of mass MMM, inner radius R1R_1R1​, and outer radius R2R_2R2​, the moment of inertia about its own axis is

I=12M(R12+R22).I=\frac{1}{2}M\left(R_1^2+R_2^2\right).I=21​M(R12​+R22​).

Given:

R1=10 cm,R2=20 cm.R_1=10\text{ cm},\qquad R_2=20\text{ cm}.R1​=10 cm,R2​=20 cm.

So,

I=12M(102+202)=12M(100+400)=12M(500)=250M.I=\frac{1}{2}M(10^2+20^2)=\frac{1}{2}M(100+400)=\frac{1}{2}M(500)=250M.I=21​M(102+202)=21​M(100+400)=21​M(500)=250M.

  1. Moment of inertia of the thin cylinder

A thin cylinder (thin-walled hollow cylinder) of radius rrr and same mass MMM has moment of inertia about its axis:

I=Mr2.I=Mr^2.I=Mr2.

Since this is equal to the given III,

Mr2=250M.Mr^2=250M.Mr2=250M.

Cancelling MMM,

r2=250.r^2=250.r2=250.

Thus,

r=250=510≈15.8 cm.r=\sqrt{250}=5\sqrt{10}\approx 15.8\text{ cm}.r=250​=510​≈15.8 cm.

  1. Choose the nearest option

r≈16 cm.r\approx 16\text{ cm}.r≈16 cm.

So the correct option is A.

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