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Work done in spliitting a drop of water of 1 mm radius into 10^6 droplets is

7 years ago


Answers : (1)


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 * [10* (10-5)2 - (10-3)2] m= 4* pi * 99* 10-6 m2

If the surface tension of water be T Nm-1 then work done W = T.ds J

7 years ago

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