A small spherical droplet of density d is flo | vedclass

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(B) For the droplet to be in equilibrium,the total upward force must equal the downward force.The forces acting on the droplet are:$1$. Buoyant force

Vedclass · Accepted Answer

$r=\sqrt{\frac{3 T}{(2 d-\rho) g}}$

Vedclass · Answer

(B) For the droplet to be in equilibrium,the total upward force must equal the downward force.The forces acting on the droplet are:$1$. Buoyant force $(B)$: $B = V_{	ext{immersed}} ho g = (\frac{1}{2} \cdot \frac{4}{3} \pi r^3) ho g = \frac{2}{3} \pi r^3 ho g$$2$. Surface tension force $(F)$: $F = T \cdot (2 \pi r)$$3$. Weight of the droplet $(mg)$: $mg = (V_{	ext{total}} d) g = (\frac{4}{3} \pi r^3) d g$Equating the forces: $B + F = mg$$\frac{2}{3} \pi r^3 ho g + 2 \pi r T = \frac{4}{3} \pi r^3 d g$Divide by $\pi r$:$\frac{2}{3} r^2 ho g + 2 T = \frac{4}{3} r^2 d g$Rearranging to solve for $r^2$:$2 T = \frac{4}{3} r^2 d g - \frac{2}{3} r^2 ho g$$2 T = \frac{2}{3} r^2 g (2d - ho)$$T = \frac{1}{3} r^2 g (2d - ho)$$r^2 = \frac{3 T}{(2d - ho) g}$$r = \sqrt{\frac{3 T}{(2d - ho) g}}$

$A$ small spherical droplet of density $d$ is floating exactly half immersed in a liquid of density $\rho$ and surface tension $T$. The radius of the droplet is (take note that the surface tension applies an upward force on the droplet).

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