A spherical mothball has initial radius 3 cm. Due to evaporation, the radius of the ball reduces to 1 cm in 4 months. In how many months would the mothball evaporate completely if the volume is lost at a rate proportional to the surface area ?

Step-by-step solution

Let V be the volume and S be the surface area of the spherical mothball of radius r . V = 4 3 r^3 S = 4 r^2 Given that the rate of loss of volume is proportional to the surface area: dV dt = -k S Differentiating V with respect to t : dV dt = d dt ( 4 3 r^3 ) = 4 r^2 dr dt = S dr dt Equating the two expressions for dV dt : S dr dt = -k S dr dt = -k Integrating both sides with respect to t : r(t) = -kt + C At t = 0 , r = 3 cm, which gives C = 3 . r(t) = 3 - kt At t = 4 months, r = 1 cm: 1 = 3 - 4k 4k = 2 k = 1 2 The