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Q.

Drops of liquid of density d are floating half immersed in a liquid of density $ρ$ . If the surface tension of liquid is T then the radius of the drop will be (d= density of liquid drop)

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a

$\sqrt{\frac{3 T}{g (2 d - ρ)}}$

b

$\sqrt{\frac{6 T}{g (2 d - ρ)}}$

c

$\sqrt{\frac{2 T}{g (2 d - ρ)}}$

d

$\sqrt{\frac{3 T}{g (4 d - 3 ρ)}}$

answer is A.

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Detailed Solution

$\begin{array}{l} mg = V_{in} ρg + 2 πrT; vdg - \frac{V}{2} ρgg = 2 πrT \\ \frac{Vg}{2} (2 d - ρ) = 2 πrT \\ \frac{1}{2} \frac{4}{3} {πr}^{3} g (2 d - ρ) = 2 πr T \\ \Rightarrow r^{2} = \frac{3 T}{g (2 d - ρ)} \Rightarrow r = \sqrt{\frac{3 T}{g (2 d - ρ)}} \end{array}$

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