A water drop of radius 1 cm is broken into 1000equal droplets. What would be the gain in surfaceenergy if the surface tension of water is0.075 Nm?\lbrack UK PMT 2015\rbrack(b) 8.48 x 10(d) 2.12x 10(a) 16.96 x 10 J(c) 4.24 x 10 JJJ

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10^{6}

10^{- 3} J / m^{2}

1 m m

10^{6}

= 72 \times 10^{- 3} J / m^{2}

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

1 m m

64

72 \times 10^{- 3} J / m^{2}

1 m m

10^{6}

72 \times 10^{- 3} N / M

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