If a number of litle of droplets of water of surface tension S, all of the same radius r combine to from a single drop of radius R and the energy relased is converted into kinetic energy. Find the velocity acquired by the bigger drop. If the energy relased in converted inot heat, find the rise in temperature.

Answer

`sqrt(6S(R -r)/(r R)), (3S)/(J)[(1)/(r )- (1)/(R )]`

Explanation

Volume of bigger drop = nxx volume of a smaller drop `So, (4)/(3)pi R^3 = nxx(4)/(3)pi r^3 or n = R^3//r^3….(i)`
Mass of the bigger drop. `m = (4)/(3)pi R^3 xx 1 = (4)/(3)pi R^3…… (ii)`
The energy released, `Delta W = 4pi S (n r^2 - R^2)`
`= 4pi S[(R^3)/(r^3) r^2-R^2] = 4pi SR^3[(1)/(r ) - (1)/(R )]`
` =3S xx(4)/(3)pi R^3[(1)/(r ) - (1)/(R )] = 3Sm((1)/(r )- (1)/(R ))`
As per equation, `Delta W = (1)/(2)m upsilon^2 = 3Sm((1)/(r ) - (1)/(R ))`
`or upsilon = sqrt(6S((1)/(r ) -(1)/(R ))) = sqrt((6S(R_r))/(r R))` Quantity of heat produced, `dH = (dW)/(J) = (3S)/(J)m ((1)/(r ) - (1)/(R ))` Heat taken by water, dH = mass xx specific heat xx rise in temp. ` =mxx1 xx delta theta`
` :. m Delta theta = (3Sm)/(J)((1)/(r ) - (1)/(R ))`
or `Delta theta = (3S)/(J)((1)/(r ) - (1)/(R ))`