Terminal velocity of a raindrop 
I have solved the question and got the answer , but I do not understand why will the raindrop attain a terminal velocity.Without any resistive forces why does it attain a terminal velocity?
 A: You can actually think of the "little rain drops" that you are picking up as providing some resistance.
When you travel at velocity $v$, and have area $A$, you are "picking up" all the material in a cylinder with volume $V=vA$ per unit time. That volume of material needs to be accelerated to velocity $v$, requiring a force $F\Delta t \propto m \Delta v = Av^2$.
So you have $v^2$ type drag - the same as you would have if you were a drop of constant size, moving through air (where drag is usually given to $F=\frac12 \rho A v^2 C_D$). The "approximation" comes from the fact that $C_D$, the drag factor, is a weak function of velocity.
The equation you were given is actually slightly different - it says to make an assumption about the rate of growth of the drop as being proportional to the mass. That may be mathematically easier (I didn't try it) but doesn't really make sense from a physics perspective. But either way, "directionally" the two equations will get you to the same thing - as the drop goes faster, it picks up more material that needs accelerating and that results in additional force. Eventually that force will limit the velocity - work done by gravity will be proportional to $mv$ but the force experienced by the drop increases faster than that. That is why there will be a terminal velocity.
