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Question

# Work done in splitting a drop of water of 1 mm radius into 106 droplets is (Surface tension of water = 72 x 10âˆ’3J/m2)

A
9.58 x 105J
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B
8.95 x 105J
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C
5.89 x 105J
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D
5.98 x 106J
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Solution

## The correct option is B 8.95 x 10−5Jlet R the radius of the big drop R= 10−3m r is the radius of small drops since volume remains constant ⟹43πR3 = n(43πr2)⟹ n = (Rr)3intial area of the big drop = 4πR2final drops (small) area = n×4πr2change in area = 4π(nr2−R2)work done W =4πR3T(1r−1R) =4πR2T[n13−1] =4π×(10−3)2×72×10−3×[1063−1] = 8.95×10−5J

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