A mercury drop of radius 1 ~cm is sprayed into 10^6 drops of equal size. The energy expended in joules is (Surface tension of mercury is 460 10⁻³ N / m )

Step-by-step solution

R=1 ~cm , n=10^6 Total volume of small drops = volume of big drop aligned 10^6 4 3 r^3 & = 4 3 R^3 10^6 r^3 & =R^3 r & = R 10^2 aligned Energy expended = work done aligned & = A T & = (n 4 r^2-4 R^2 ) T & =4 (n r^2-R^2 ) T & =4 (n ( R 10^2 )^2-R^2 ) T & =4 R^2 [n 1 10^4 -1 ] T & =4 3.14 (1 10⁻² )^2 & [10^6 1 10^4 -1 ] 460 10⁻³ & =4 3.14 10⁻⁴ (10^2-1 ) 460 10⁻³ & =4 3.14 10⁻⁴ 99 460 10⁻³ & =0.057 ~J aligned