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Work done in spliitting a drop of water of 1 mm radius into 10^6 droplets is
Let the radius of each droplet be r m and all the 10^6 droplets are identical. Volume of bigger drop = Total volume of droplets (4/3)* pi* (1* 10-3 )3 = 106 * (4/3)*pi * r3 => r = 10-5 Change in surface area = ds = 4* pi * [106 * (10-5)2 - (10-3)2] m2 = 4* pi * 99* 10-6 m2 If the surface tension of water be T Nm-1 then work done W = T.ds J
Let the radius of each droplet be r m and all the 10^6 droplets are identical.
Volume of bigger drop = Total volume of droplets
(4/3)* pi* (1* 10-3 )3 = 106 * (4/3)*pi * r3
=> r = 10-5
Change in surface area = ds = 4* pi * [106 * (10-5)2 - (10-3)2] m2 = 4* pi * 99* 10-6 m2
If the surface tension of water be T Nm-1 then work done W = T.ds J
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