When a 12000 joule of work is done on a flywheel, its frequency of rotation increases from 10 Hz to 20 Hz. The moment of inertia of flywheel about its axis of rotation is `(pi^(2) = 10)`

Options

A. `1 kgm^(2)`
B. `2 kgm ^(2)`
C. `1.688 kgm^(2)`
D. `1.5 kgm^(2)`

Answer

B

Explanation

Given , work done W = 12000J ,
Initial frequnecy , `f_(1) = 10 Hz`
and final frequnecy `f_(2) = 20 Hz`
Angular velcocity for rotational motion is given kinetic energy . ltbr ` omega = 2pif`
`therefore " " omega_(1) = 2pif_(1) = 2pi xx 10 = 2o pi rad//s`
and `omega_(2) = 2pif_(2) = 2pi xx 20 = 40 pi rads//s`
According to work -energy theorem, work done in rotation = change in rotational kinetic energy
`rArr " "W = (1)/(2) /Omega_(2)^(2) - (1)/(2)lomega_(1)^(2) " "[because pi^(2) = 10]`
`= (1)/(2)l(omega_(2)^(2) - omega_(1)^(2))" "......(i)`
where , I = moment of inertia of the flywheel.
Substituing given value of Eq.(i) ew get
`12000 =(1)/(2)l(1600pi^(2) - 400pi^(2))`
`rArr = (1)/(2)l(1200 xx 10)" "[because pi^(2) = 10]`
`rArr" "l = (12000xx2)/(12000) = 2 kgm^(2)`