Two mercury drops of radii r and 2 r merge to form a bigger drop. The surface energy released in the process is nearly (Surface tension mercury is S and take 9^ 2 / 3 =4.326 )

Step-by-step solution

Radii of drop, r₁=r and r₂=2 r Sum of volume of two drops = Volume of the bigger drop aligned & 4 3 r^3+ 4 3 (2 r)^3= 4 3 R^3 & r^3+8 r^3=R^3 & R^3=9 r^3=(9)^ 1 / 3 r aligned Surface energy, E=4 R^2 S aligned & =4 (9)^ 2 / 3 r ^2 ~S & =4 4.326 r ^2 ~S & =17.1 r ^2 ~S aligned