A solid cylinder of mass 2 kg and radius 4 cm rotating about its axis at the rate of 3 rpm. The torque required to stop after `2pi` revolutions is :

Options

A. `2xx10^(6)Nm`
B. `2xx10^(-6)Nm`
C. `2xx10^(-3)Nm`
D. `2xx10^(-4)Nm`

Answer

B

Explanation

Initial angular velocity can be written as follows:
`omega_(0)=3r"pm"=3xx(2pi)/(60)rad//sec=(pi)/(10)`
It is to be stopped after `2pi` revolutions. Hence angle rotated `theta` = (number of revolutions) `xx2pi=2pixx2pi`
`omega^(2)=omega_(0)^(2)+2alphatheta`
`0^(2)=((pi)/(10))^(2)+2(alpha)(2pixx2pi)`
`alpha=-(1)/(800)rad//sec^(2)`
`I=(mR^(2))/(2)=((2)((4)/(100))^(2))/(2)=(16)/(10^(4))`
`tau=Ialpha=((16)/(10^(4)))xx(-(1)/(800))=-2xx10^(-6)Nm`
Hence option (b) is correct.