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From a circular disc of radius R and mas...

From a circular disc of radius R and mass 9M, a small disc of radius R/3 is removed from the disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is

A

`4M^(2)`

B

`40/9MR^(2)`

C

`10MR^(2)`

D

`37/9MR^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
Mass per unit area of disc`=(9M)/(piR^(2))`
`therefore` Mass of removed portion `=(9M)/(piR^(2))xxpi(R/3)^(2)=M`
Let moment of inertia of removed portion = `I_(1)`
`thereforeI_(1)=M/2(R/3)^(2)+M((2R)/3)^(2)`, by theorem of parallel or `I_(1)=(MR^(2))/2`
Let `I_(2)` = Moment of inertia of the whole disc
`I_(2)=(9MR^(2))/2`
`therefore` Let I = Moment of inertia of remaining disc
`thereforeI=I_(2)-I_(1)`
or `I=(9MR^(2))/2-(MR^(2))/2=(8MR^(2))/2=4MR^(2)`
or `I=4MR^2`
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Knowledge Check

  • From a circular disc of radius R and mass 9M, a small disc of radius R/3 is removed as shown in figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is

    A
    `4 MR^(2)`
    B
    `40/9 MR^(2)`
    C
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    D
    `37/9 MR^(2)`

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    `40/9 MR^(2)`
    B
    `MR^(2)`
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    `4MR^(2)`
    D
    `4/9MR^(2)`

  • From a circular disc of radius R and mass 9 M , a small disc of mass M and radius (R)/(3) is removed concentrically . The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is

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