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The moment of inertia of the disc used i...

The moment of inertia of the disc used in a torsional pendulum about the suspension wire is `0.2kg-m^2`. It oscillates with a period of 2s. Asnother disc is placed over the first one and time period of the system becomes 2.5s. Find the moment of inertia of the second disc about the wire.

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Let the torsioN/Al constant of the wire be k. The moment of inertia of the first disc about the wire is `0.2kgm^2`. Hence the time period is
`2s=2pisqrt(I/K)`
`=2pi(sqrt(0.2kg-m^2)/k)`……….i
When the sencond disc having moment of inertia `I_1` about the wire is added time period is
`2.5s=2pi (sqrt(0.2kg-m^2+I_1)/k)`......ii
From i and iij `3.25/4=(0.2kg-m^2+I_1)/(0.2kgm^2)`
this gives `I_1=0.11 kg-m^2`
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