A spherical body of density rho is floating h | vedclass

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(A) For the body to be in equilibrium,the downward force (weight) must be balanced by the upward forces (buoyant force and surface tension force).Weig

Vedclass · Accepted Answer

$2 \sqrt{\frac{3 \sigma}{g(2 \rho-d)}}$

Vedclass · Answer

(A) For the body to be in equilibrium,the downward force (weight) must be balanced by the upward forces (buoyant force and surface tension force).Weight of the body = $W = \frac{4}{3} \pi r^3 ho g$Buoyant force = $F_B = 	ext{Volume immersed} 	imes d 	imes g = \frac{2}{3} \pi r^3 d g$Surface tension force = $F_S = 2 \pi r \sigma$Equating the forces: $W = F_B + F_S$$\frac{4}{3} \pi r^3 ho g = \frac{2}{3} \pi r^3 d g + 2 \pi r \sigma$$\frac{2}{3} \pi r^3 g (2 ho - d) = 2 \pi r \sigma$$r^2 = \frac{3 \sigma}{g(2 ho - d)}$$r = \sqrt{\frac{3 \sigma}{g(2 ho - d)}}$The diameter $D = 2r = 2 \sqrt{\frac{3 \sigma}{g(2 ho - d)}}$.

$A$ 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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