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Question

Two circular discs A and B are of equal masses and thickness but made of metals with densities dA and dB (dA>dB) . If their moments of inertia about an axis passing through their centres and perpendicular to the circular faces are IA and IB then

A
IA=IB
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B
IA>IB
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C
IA<IB
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D
IAIB
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Solution

The correct option is C IA<IB
Given,
mA=mB=m kg
Let RA and RB be the radii of discs A and B respectively.
Then, m=(πR2At)dA=(πR2Bt)dB
where t thickness
As dA>dB, so R2A<R2B ... (1)

Now,
Moment of inertia of disc A
(IA)=12mR2A
Moment of inertia of disc B
(IB)=12mR2B
IAIB=R2AR2B
Since, R2AR2B<1, we get IA<IB

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