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A drop of water is divided in 27 equal droplets. If $Δ P_{1} and Δ P_{2}$ be the excess pressures in the big drop and smaller droplet respectively, then $Δ P_{1} / Δ P_{2}$ is equal to

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answer is D.

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Detailed Solution

$\begin{array}{l} \frac{4}{3} π r^{3} \times 27 = \frac{4}{3} \times R^{3} \\ \Rightarrow r = \frac{R}{3} \end{array}$

Now $Δ P_{1} = \frac{2 S}{R} a n d Δ P_{2} = \frac{2 S}{r} = \frac{2 S}{R / 3} = 3 Δ P_{2}$

$∴ \frac{Δ P_{1}}{Δ P_{2}} = \frac{1}{3}$

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