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A drop of liquid of diametre 2.8mm breaks up inti 125 identical drops The change in energy is nearly
let the radius of each droplet be r m Volume of bigger drop = total volume of 125 droplets (4/3)* pi * (1.4*10-3)3 = 125 * (4/3) * pi * r3 => r = (1.4/5)* 10-3 = 0.28*10-3 change in surface area = ds = 4*pi* [125* (0.28*10-3) 2 - (1.4*10-3)2] m = 0.0984 m if surface tension of the liquid ne T Nm-1 then work done W = Tds = 0.098T J.
let the radius of each droplet be r m
Volume of bigger drop = total volume of 125 droplets
(4/3)* pi * (1.4*10-3)3 = 125 * (4/3) * pi * r3
=> r = (1.4/5)* 10-3 = 0.28*10-3
change in surface area = ds = 4*pi* [125* (0.28*10-3) 2 - (1.4*10-3)2] m
= 0.0984 m
if surface tension of the liquid ne T Nm-1 then work done W = Tds = 0.098T J.
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