Let r be the radius, V be the volume and S be the surface area of the spherical raindrop at time t. Then, V=
4
3
πr3 and S=4πr2 The rate at which the raindrop evaporates is
dV
dt
which is proportional to the surface area. ∴
dV
dt
∝S⇒
dV
dt
=−kS, where k>0. . . (i) Now, V=
4
3
πr3 and S=4πr2 ∴
dV
dt
=
4π
3
×3r2
dr
dt
=4πr2
dr
dt
4πr2
dr
dt
=−k(4πr2) [from Eq. (i)]
dr
dt
=−k⇒dr=−kdt On integrating, we get ∫dr=−k∫dt+C ∴r=−kt+C Initially, i.e. when t=0,r=3 ∴3=−k×0+C ∴C=3 ∴r=−kt+3 When t=1,r=2 ∴2=−k×1+3 ∴k=1 ∴r=−t+3 ∴r=3−t, where 0≤t≤3 This is the required expression for the radius of the raindrop at any time t.