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A drop of liquid of radius $R = 10^{- 2} m$ having surface tension $S = \frac{0.1}{4 π} N m^{- 1}$ divides itself into $K$ identical drops. In this process the total change in the surface energy $Δ U = 10^{- 3} J$ . If $K = 10^{α}$ then the value of $α$ is

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answer is F.

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Detailed Solution

Let radius of small drops $= r$ .

$\frac{4}{3} π R^{3} = K \frac{4}{3} π r^{3}$

$R = K^{\frac{1}{3}} r$ … (i)

$S (K 4 π r^{2} - 4 π R^{2}) = 10^{- 3}$

$\frac{0.1}{4 π} 4 π (k \frac{R^{2}}{K^{\frac{1}{3}}} - R^{2}) = 10^{- 3}$

$R^{2} (K^{\frac{1}{3}} - 1) = 10^{- 2}$

$K^{\frac{1}{3}} - 1 = 100$

$K^{\frac{1}{3}} = 101$

$10^{\frac{α}{3}} = 101$

$α \approx 6$

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