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The surface tension of water = 72 xx10^...

The surface tension of water ` = 72 xx10^(-3) jm^(-2)` .The work done in splitting a drop of water of 1 mm radius into 64 droplets is

A

`2xx10^(-6)`J

B

`2.7 xx10^(-6)`J

C

`4xx10^(-6)`J

D

`6.4 xx10^(-6)`J

Text Solution

Verified by Experts

The correct Answer is:
B
` 4/3 piR^(3) = 64 xx4/3 pir^(3)`
` :. r^(3) = (R^(3))/64 " " r = R/4 = 1/4 ` mm
Initial surface energy `E_(1) = 4piT(R^(2))`
Final surface energy `E_(2) = 64 xx 4piT(r^(2))`
`:. E_(2) -E_(1) = 4piT (64r^(2)-R^(2))`
` :. ` Work done `= 4piT [64r^(2)-R^(2)]`
` = 4piT [4 xx 10^(-6)-10^(-6)]` ltbr. ` = 12piT xx 10^(-6) = 2.7 xx 10^(-6)` J
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