Question 126 of 192

A drop of liquid of density $\rho$ is floating half immersed in a liquid of density $\sigma$ and surface tension $7.5\times {10}^{-4}N{cm}^{-1}$. The radius of drop in $cm$ will be : (Take : $g=10{ms}^{-2}$)

A$\dfrac{15}{\sqrt{2\rho -\sigma }}$B$\dfrac{15}{\sqrt{\rho -\sigma }}$C$\dfrac{3}{2\sqrt{\rho -\sigma }}$D$\dfrac{3}{20\sqrt{2\rho -\sigma }}$

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Answer: (A) $\dfrac{15}{\sqrt{2\rho -\sigma }}$

Balancing the forces on drop
Buoyant force $+$ surface tension $=mg$
$\sigma \dfrac{V}{2}g+2\pi RT=\rho Vg$
$2\pi RT=\dfrac{\left(2\rho -\sigma \right)}{2}\cdot \dfrac{4}{3}\pi {R}^{3}g;\left[\text{As}V=\dfrac{4}{3}\pi {R}^{3}\right]$
$\Rightarrow {R}^{3}=\dfrac{3T}{\left(2\rho -\sigma \right)g}\Rightarrow R=\sqrt{\dfrac{3\times 7.5\times {10}^{-2}N-{m}^{-1}}{\left(2\rho -\sigma \right)\times 10}}$
$R=\dfrac{3}{20\sqrt{\left(2\rho -\sigma \right)}}m=\dfrac{15}{\sqrt{2\rho -\sigma }}cm$