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A solid cylinder of radius 0.5 m and mas...

A solid cylinder of radius 0.5 m and mass 50 kg is rotating at 300 rpm. Then the torque which will bring it to rest in 5 seconds is

A

`25 pi Nm`

B

`(25 pi)/(2)Nm`

C

`50 pi Nm`

D

`100 pi Nm`

Text Solution

Verified by Experts

The correct Answer is:
B
`tau = I alpha =(I(omega_(2)-omega_(1)))/(t)`
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