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.

Explanation

`(dm)/(dt) = K.4pir^(2)`
Where K is a constant and .r. is the radius of the drop at any instant .
But ` m=(4)/(m) pir^(3) rho=(4)/(3) pi r^(3)`
since ` rho`= density of water is 1 gram/c.c
`(dm)/(dt) =(4)/(3) pi. 3r^(2) (dr)/(dt) = 4pir^(2) (dr)/(dt) = K. 4 pir^(3)" ":. K=(dr)/(dt)`
`:. ` r=Kt ( since it starts with zero radius )
Since ` m(dv)/(dt) + v(dm)/(dt) = mg `
`(4)/(3) pir^(3) (dv)/(dt) + v.k 4pir^(2)=(4)/(3) pir^(2) g , (dv)/(dt) + vK(3)/(r) =g `
But `(dv)/(dt) =a , v = at `
`:. a + at. K. (3)/(Kt) =g ` becomes `4a=g,a =` acceleration of drop =g/4