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
B
8.95 x 105J
C
5.89 x 105J
D
5.98 x 106J

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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