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Question

A recording disc rotates steadily at 45 rpm on a table. When a small mass of 0.02 kg is dropped gently on the disc at a distance of 0.04 m from its axis, it gets stuck to the disc and the rate of revolution falls to 36 rpm. The moment of inertia of the disc about its centre is

A
1.28×104 kgm2
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B
1.28×105 kgm2
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C
1.28×103 kgm2
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D
1.28×102 kgm2
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Solution

The correct option is A 1.28×104 kgm2
By conservation of angular momentum,
I1ω1=I2ω2 .....(i)
where,
I1= Moment of Inertia of disc before mass gets stuck
I2= Moment of Inertia of disc after mass gets stuck
ω1= Angular velocity of disc before mass gets stuck
ω2 =Angular velocity of disc after mass gets stuck

using eq. (i),
I×(2π×4560)=[I+0.02×(0.04)2]×(2π×3660)
I=1.28×104 kgm2

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