If a drop of liquid breaks into smaller droplets, it result in lowering of temperature of the droplets. Let a drop of radius R, breaks into N small droplets each of radius r. Estimate the drop in temperature.

Options

A. `(3T)/(rho S) [(1)/(r )- (1)/(R )]`
B. `-(2T)/(rho S) [(1)/(r )- (1)/(R )]`
C. `(2R)/(rho S) [(1)/(R )- (1)/(r )]`
D. `(3T)/(rho S)[(1)/(R )- (1)/(r )]`

Answer

D

Explanation

Since, volume remains unchanged, during this phenomenon, so `(4)/(3) pi R^(3) = N xx (4)/(3) pi r^(3)`
`N = (R^(3))/(r^(3))`
Now, change in surface area `=4piR^(2)-N4pi r^(2)`
`=4pi (R^(2)-Nr^(2))`
Energy released `(Delta U)= Txx` change in surface area `=T xx 4pi [R^(2)-Nr^(2)]`
Here, all this energy released is at the cost of lowering the temperature and mass of the big drop of liquid `=(4)/(3) pi R^(2)rho`. Now, change in temperature, `Delta theta =(Delta U)/(ms)=(T xx 4pi (R^(2)-Nr^(2)))/(((4)/(3) pi R^(3) rho)S)`
`=(3T)/(rho S) ((1)/(R )- (Nr^(2))/(R^(3)))`
`=(3T)/(rho S) ((1)/(R )- (R^(3) xx r^(2))/(r^(3) xx R^(3)))`
`=(3T)/(rho S) ((1)/(R )-(1)/(r ))`