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Question

The diameter of one drop of water is 0.2 cm. The work done in breaking one drop into 1000 equal droplets will be :-(surface tension of water =7×10−2N/m)

A
7.9×106J
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B
5.92×106J
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C
2.92×106J
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D
1.92×106J
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Solution

The correct option is A 7.9×10−6JCurrent radius of a drop = 0.1 cmOn breaking it into 1000 droplets, the volume would remain constant. Therefore, radius of each small drop,1.33πR3=1000×1.33πr3Or r=R/10=0.01 cmChange in surface areaδA=1000×4πr2−4πR2= 4π(10R2−R2)= 36πR2Therefore, the change in surface energy,δU=TδAδU=7×10−2×36π×10−6δU=7.91×10−6J

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