A drop of liquid of density $ρ$ is floating half immersed in a liquid of density $σ$ and surface tension $7 .5 \times 10^{- 4} {Ncm}^{- 1}$ . The radius of drop in cm will be : (Take : $g = 10 m / s^{2}$ )

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$\frac{3}{20 \sqrt{2 ρ - σ}}$

$\frac{15}{\sqrt{2 ρ - σ}}$

$\frac{15}{\sqrt{ρ - σ}}$

$\frac{3}{2 \sqrt{ρ - σ}}$

answer is A.

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Detailed Solution

Buoyant force + surface tension = mg

$\begin{array}{l} σ \frac{V}{2} g + 2 πRT = ρVg \\ 2 πRT = \frac{(2 ρ - σ)}{2} \frac{4}{3} {πR}^{3} g; [V = \frac{4}{3} {πR}^{3}] \\ R^{3} = \frac{3 T}{(2 ρ - σ) g} \Rightarrow R = \sqrt{\frac{3 \times 7 .5 \times 10^{- 2} N - m^{- 1}}{(2 ρ - σ) \times 10}} \\ R = \frac{3}{20 \sqrt{(2 ρ - σ)}} m = \frac{15}{\sqrt{2 ρ - σ}} cm \end{array}$