A spherical rain drop, falling in a constant gravitational field , grows by absorption of moisture from the surroundings at a rate proportional to its surface area. If it starts with zero radius, find its acceleration .

A spherical rain drop, falling in a constant gravitational field , gro

1.	A spherical rain drop, falling in a constant gravitational field , grows by absorption of moisture from the surroundings at a rate proportional to its surface area. If it starts with zero radius, find its acceleration .
Answer» <P> Solution :`(dm)/(dt)=K.4pir^(2)` Where K is a constant and .r. is the radius of the drop at any instant . But `m=4/3pir^(3)P=4/3pir^(3)` SINCE P density of water is `1 gram//c.c` `(dm)/(dt)=4/3pi.3r^(2)(dr)/(dt) =4pir^(2)(dr)/(dt)=K.4pir^(2)``:. K=(dr)/(dt)` `:. r=Kt `(since it STARS with zero radius) Since `m(dv)/(dt) +V(dm)/(dt) =mg` `4/3pir^(3)(dv)/(dt)+v.k4pir^(2)=4/3pir^(2)g, (dv)/(dt)+vK3/r=g` But`(dv)/(dt)=a, v=at` `:. a+at.K.3/(Kt)=g` becomes `4a=g, a`=ACCELERATION of drop=`g//4`