If a body starts rotating from rest because of a torque of 2 Nm, then its kinetic energy after 20 revolutions will be
A. `60 pi J`
B. `80 pi J`
C. `70 pi J`
D. `40 pi J`
A. `60 pi J`
B. `80 pi J`
C. `70 pi J`
D. `40 pi J`
Correct Answer – B
Change in K.E. `=(1)/(2)I(omega_(2)^(2)-omega_(1)^(2))`
`=(1)/(2)I(omega_(2)+omega_(1))((omega_(2)-omega_(1)))/(t)t`
`=(1)/(2)I alpha (omega_(2)+omega_(1))t = tau theta`
In one rev …….. `2 pi` rad
`therefore` 20 rev …… ?
`theta = 20xx2pi = 40 pi`