A spherical liquid drop of radius R is divided into 8 equal droplets. If surface tension is S , then the work done in this process will be

Step-by-step solution

Work done, W = S A Where, A is a change in surface area. Since, the radius of the big drop be R A _ initial =4 R ^2 Let the radius of the small drops be r A _ final =8 4 r ^2 The volume is constant array ll & 4 3 R ^3=8 4 3 r ^3 & R ^3=8 r ^3 & R =2 r array Substituting the values in the formula. aligned & W = S A & W = S ( A _ final - A _ initial ) aligned aligned & W = S (8 4 r ^2-4 R ^2 ) & W = S (32 ( R 2 )^2-4 R ^2 ) & W = S (8 R ^2-4 R ^2 ) & W =4 R ^2 ~S aligned