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A stationary horizontal disc is free to ...

A stationary horizontal disc is free to rotate about its axis. When a torque is applied on it, its kinetic energy as a function of `theta`, where `theta` is the angle by which it has rotated, is given as `ktheta^(2)` If its moment of inertia is I then the angular acceleration of the disc is

A

`(k)/(4I)theta`

B

`(k)/(2I)theta`

C

`(k)/(I)theta`

D

`(2k)/(I)theta`

Text Solution

Verified by Experts

The correct Answer is:
D
`K.E. = ktheta^(2)`
`(1)/(2)I omega^(2)=ktheta^(2)`
`{:(omega^(2)=(2ktheta^(2))/(I),implies2omega(domega)/(d theta)=(4ktheta)/(I)):}`
implies `omega(domega)/(d theta)=(2ktheta)/(I)implies alpha=(2ktheta)/(I)`
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