A spherical body of density $\rho$ is floating half immersed in a liquid of density $d$. If $\sigma$ is the surface tension of the liquid, then the diameter of the body is

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Aicharya · Accepted Answer

Weight of body $=$ Buoyant force + Force of surface tension $\begin{aligned} \frac{4}{3} \pi r^3 \rho \times g & =\frac{2}{3} \pi r^3 d g+2 \pi r \sigma \\ \frac{2}{3} \pi r^3 g(2 \rho-d) & =2 \pi r \sigma \end{aligned}$ So, $r^2=\frac{3 \sigma}{g(2 \rho-d)}$ So, $r=\sqrt{\frac{3 \sigma}{g(2 \rho-d)}}$

A spherical body of density $\rho$ is floating half immersed in a liquid of density $d$. If $\sigma$ is the…

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