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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 6.

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Detailed Solution

$\frac{4}{3} π R^{3} = \frac{4}{3} π r^{3} \times K$ (Conservation of volume)

R=radius of bigger drop

r=radius of smaller drop

$R^{3} = r^{3} \times K$

$U_{i} = S (4 π R^{2})$

$U_{f} = K S (4 π r^{2})$

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

$4 π S [K r^{2} - R^{2}] = 10^{- 3}$

$K^{1 / 3} R^{2} - R^{2} = 10^{- 2}$

$10^{a / 3} - 1 = 100$

$10^{a / 3} = 101$

$a \approx 6$

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