When 8 droplets of water of radius 1 mm each combine into one, then the energy evolved is approximately equal to 7 xx 10^(-6)J. Find the value of x.

When 8 droplets of water of radius 1 mm each combine into one, then th

1.	When 8 droplets of water of radius 1 mm each combine into one, then the energy evolved is approximately equal to 7 xx 10^(-6)J. Find the value of x.
Answer» Solution :`S=0.07 N//m, r=1 mm =1 XX 10^(-3)m` Let R=radius of big drop `4/3 pi R^(3)=8 xx 4/3 PIR^(3)` `R=2r =2 xx 10^(-3)m` Surface area of bigger drop `=4piR^(2)` `=4pi xx (2 xx 10^(-3))^(2)` `=4pi xx 4 xx 10^(-6)m^(2)` `=16 pi xx 10^(-6) m^(2)` Surface area of 8 small drops `=8 xx 4pi xx (1 xx 10^(-3))^(2)` `=32 pi xx 10^(-6) m^(2)` `triangleA=` decreases in surface area `=(32 pi xx 10^(-6) -16pi xx 10^(-6))` `=16 pi xx 10^(-6) m^(2)` Energy evolved `="Surface Tesnion "xx " decreases in surface area"` `=0.06 xx 16pi xx 10^(-6)` `=0.96 pi xx 10^(-6)J=1pi xx 10^(-6) H`