If a body starts rotating from rest because of a torque of 2 Nm, then its kinetic energy after 20 revolutions will beA. `60 pi J`B. `80 pi J`C. `70 pi J`D. `40 pi J`

If a body starts rotating from rest because of a torque of 2 Nm, then

1.	If a body starts rotating from rest because of a torque of 2 Nm, then its kinetic energy after 20 revolutions will beA. `60 pi J`B. `80 pi J`C. `70 pi J`D. `40 pi J`
Answer» 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`

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If a body starts rotating from rest because of a torque of 2 Nm, then its kinetic energy after 20 revolutions will beA. `60 pi J`B. `80 pi J`C. `70 pi J`D. `40 pi J`

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