A small steel ball falls through a syrup at a constant speed of `10cms^-1`. If the steel ball is pulled upwards with a force equal to twice its effective weight, how fast will it move upwards?

Explanation

`6pietarv=mg^(1)` (effective weight)
Let `W_(e)=` effective weight and `6pietar=K`
In equilibrium `w_(e)=kv` ..(1)
Again `2w_(e)-w_(e)=kv^(1)` ..(2)
From Eqs. (i) and (ii) `v^(1)=v=1.0m//s`