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Question

Work done in splitting a drop of water of 1 mm radius into $$10^6$$ droplets is (Surface tension of water = 72 x $$10^{-3}J / m^2$$)


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 $$10^{-5}J$$
let R the radius of the big drop  R= $$10^{-3}$$m
 r is the radius of small drops 
since volume remains constant $$\implies \dfrac{4}{3} \pi R^3 $$ = n($$\dfrac{4}{3} \pi r^2$$)
$$\implies$$ n = $$(\dfrac{R}{r})^3$$
intial area of the big drop = $$ 4 \pi R^2 $$
final drops (small) area = $$ n \times 4 \pi r^2 $$
change in area = $$ 4 \pi (nr^2 - R^2)$$
work done W =$$ 4 \pi R^3 T(\dfrac{1}{r} -\dfrac{1}{R})$$ 
         
             =$$ 4 \pi R^2T[n^{\dfrac{1}{3}}-1]$$

               =$$ 4 \pi \times (10^{-3})^2 \times 72 \times 10^{-3} \times [10^{\dfrac{6}{3}}-1]$$
  = $$ 8.95 \times 10^{-5} J$$

Physics

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