A sherical raindrop evaporate at a rate proportional to its surface area at any instant t.
The differential equation giving the rate of change of radius of raindrop has degree?
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For a sherical raindrop, the rate of evaporation is proportional to its surface area.
Hence, !!(dV)/dt prop surface area!!
!!(dV)/dt = lambda4pir^2!!........(i)
Volume of sphere = !!(4/3)pir^3!!
Putting this in equation (i), we get
!!(d(4/3 pir^3))/dt = lambda4pir^2!!
!!=> (3r^2)/3(dr)/dt = lambdar^2!!
!!=> (dr)/dt = lambda!!
So !!degree = 1!!