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Question

A horizontal disc rotating freely about a vertical axis makes 200 rpm. A piece of wax of mass 20 g falls verticaly on the disc and adheres to it at a distance of 8 cm from the axis. If the number of revolutions per minute is thereby reduced to 120 rpm, then calculate the moment of inertia of disc.

A
3.8×104 kg-m2
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B
1.92×104 kg-m2
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C
1.28×104 kg-m2
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D
2.56×104 kg-m2
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Solution

The correct option is B 1.92×104 kg-m2
Let I1 is the moment of inertia of disc.
Here, system is disc and a piece of wax.
Since, there is no external torque acting on the system. Therefore, the angular momentum of the system will be conserved.
Li=Lf
I1ω1=I2ω2
where I2 be the moment of inertia of disc after wax adheres to it.

(ω=2πN60)
N= no. of revolutions per minute.
I1(2π×20060)=(I1+20×82)(2π×12060)
I1×200=(I1+1280)×120
80I1=1280×120
I1=1920 g-cm2
I1=1.92×104 kg-m2

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