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The moment of inertia of a disc rotating...

The moment of inertia of a disc rotating about an axis passing through its centre and perpendicular to its axis is `20 kgm^(2)` . If its rotational K.E. is 10 J, then its angular momentum will be

A

`"10 kg m"^(2)//s`

B

`"15 kg m"^(2)//s`

C

`"20 kg m"^(2)//s`

D

`"25 kg m"^(2)//s`

Text Solution

Verified by Experts

The correct Answer is:
C
`E=(1)/(2)Iomega^(2)=(1)/(2)(I^(2)omega^(2))/(I)=(1)/(2)(L^(2))/(I)`
`therefore L^(2)=2IE`
`therefore L=sqrt(2IE)=sqrt(2xx20xx10)="20 kg m"^(2)//s`
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