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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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Verified Answer
The correct answer is:
$\sqrt{\frac{3 \sigma}{g(2 \rho-d)}}$
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,
$$
r=\sqrt{\frac{3 \sigma}{g(2 \rho-d)}}
$$
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