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A mercury drop of radius R splits up into 1000 droplets equal radii. What is the change in surface area? |
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Answer» Solution :Volume of ONE drop = Volume of 1000 DROPLETS `THEREFORE 4/3 piR^3 = 1000 xx 4/3 pir^3` `therefore R^3 = 1000r^3` `therefore r = R/10` Change in surface area = `1000 xx 4pir^2 - 4piR^2` `4pi [ 1000 xx R^2/100 - R^2]` dA = `4pi (10R^2 -R^2) = 36 pi R^2` |
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