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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 103J/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 105J
let R the radius of the big drop R= 103m
r is the radius of small drops
since volume remains constant 43πR3 = n(43πr2)
n = (Rr)3
intial area of the big drop = 4πR2
final drops (small) area = n×4πr2
change in area = 4π(nr2R2)
work done W =4πR3T(1r1R)
=4πR2T[n131]

=4π×(103)2×72×103×[10631]
= 8.95×105J

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