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pallavi pradeep bhardwaj Grade: 12
        

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

7 years ago

Answers : (1)

askiitiansexpert soumyajit_iitkanpur
8 Points
										

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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