1.

What will happen if n drops of a liquid each has surface energy E, combine to form a single drop.A. No energy will be released in the precessB. Some energy will be absorbed in the processC. Energy released or absorbed will be `E(n-n^(2//3))`D. Energy released or absorbed will be `nE(n-n^(2//3-1))`

Answer» Let radius of smalll drop =r
Radius of single drop =R
Surface tension S= surface energy per unit area
`nxx(4)/(3)xxpir^(3)=(4)/(3)piR^(2)`
`R=n^(1//3) r`
Initial surface energy `=E_(i)=nxx4pir^(2)xxS=nE`
Final surface energy `=E_(f)=4piR^(2)S`
`=4pir^(2)n^(2..3)S=n^(2//3)E`
Energy released `=E_(i)-E_(f)=E(n-n^(2//3))`


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