A drop of liquid of density rho is floating h | vedclass

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(D) The drop is in equilibrium under the action of the following forces:Weight of the drop,$W = Mg = \frac{4}{3} \pi r^3 \rho g$ (downwards).Upthrust

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$\sqrt{\frac{12 T}{g(2 \rho-d)}}$

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(D) The drop is in equilibrium under the action of the following forces:Weight of the drop,$W = Mg = \frac{4}{3} \pi r^3 \rho g$ (downwards).Upthrust (buoyant force) = weight of the liquid displaced = $\frac{2}{3} \pi r^3 d g$ (upwards).Force due to surface tension,$F = 2 \pi r T$ (upwards).For equilibrium,the downward force must equal the sum of the upward forces:$Mg = F + F_{t}$$F = Mg - F_{t}$$2 \pi r T = \frac{4}{3} \pi r^3 \rho g - \frac{2}{3} \pi r^3 d g$$2 \pi r T = \frac{2}{3} \pi r^3 (2 \rho - d) g$$T = \frac{1}{3} r^2 (2 \rho - d) g$$r^2 = \frac{3 T}{g(2 \rho - d)}$$r = \sqrt{\frac{3 T}{g(2 \rho - d)}}$The diameter $D = 2r = 2 \sqrt{\frac{3 T}{g(2 \rho - d)}} = \sqrt{\frac{12 T}{g(2 \rho - d)}}$.

$A$ drop of liquid of density $\rho$ is floating half-immersed in a liquid of density $d$. If $T$ is the surface tension,then the diameter of the drop is:

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