A person pulls along a rope wound up around a pulley with a constant force `F` for a time interval of `t` seconds. If `a` and `b` are the radii of the inner and the outer circumference `(a lt b)`, then find the ratio of work done by the person in the two cases shown in the figuer is `W_(1)//W_(2)`.
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Answer

`a^(2)//b^(2)`

Explanation

case `- 1`
`F xx a = I alpha rArr alpha = (Fa)/(I)`
As `alpha` is constant `omega = alpha t, omega = ((Fa)/(I)) alpha`
`W_(1) = Delta KE rArr WD_(1) =(1)/(2) I omega^(2) rArr WD_(1) = (1)/(2) (I) (alpha^(2) t^(2))`
`WD_(1) =(1)/(2)I((Fa)/(I))^(2) t^(2)`.....(1)
similarly in case II `WD_(2) = (1)/(2) I ((Fb)/(I))^(2) t^(2)` ....(2)
from (1) & (2)
`(WD_(1))/(WD_(2)) = (a^(2))/(b^(2))`.