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Q.

A liquid drop of density $ρ$ is floating half immersed in a liquid of surface tension ' $S$ ' and density $\frac{ρ}{2}$ . If the surface tension ' $S$ ' of the liquid is numerically equal to 10 times of acceleration due to gravity, then the diameter of the drop is $\sqrt{\frac{2 X}{ρ}},$ then $X = ?$

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answer is 40.

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Detailed Solution

$F_{S} = S \times π D = 10 g (π D) W = F_{S} + F_{b}, F_{S} = W - F_{b} \Rightarrow F_{S} = ρ v g - \frac{ρ}{2} \frac{v}{2} g = \frac{3}{4} ρ v g \Rightarrow 10 g π D = \frac{3}{4} ρ g [{\frac{4 π}{3} (\frac{D}{2})}^{3}] (∵ V = \frac{4}{3} π r^{3}) D = \sqrt{\frac{80}{ρ}} m, = \sqrt{\frac{2 x}{ρ}}, X = 40$

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