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A recording disc rotates steadily at n(1...

A recording disc rotates steadily at `n_(1)` rps on a table. When a small mass m is dropped gently on the disc at a distance x from its axis, it sticks to the disc, the rate of revolution falls to `n_(2)` rps. The original moment of inertia of the disc about a perpendicular axis through its centre is

A

`I = (mx^(2))/(n_(1) - n_(2))`

B

`I = (n_(1) mx^(2))/(n_(1) - n_(2))`

C

`I = (n_(2)mx^(2))/(n_(1) - n_(2))`

D

`I = (n_(2) mx^(2))/(n_(1))`

Text Solution

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The correct Answer is:
C
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