JEE PYQ: Mechanical Properties Of Fluids Question 22
Question 22 - 2020 (09 Jan 2020 Shift 2)
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):
(1) $r = \sqrt{\frac{2T}{3(d+\rho)g}}$
(2) $r = \sqrt{\frac{T}{(d-\rho)g}}$
(3) $r = \sqrt{\frac{T}{(d+\rho)g}}$
(4) $r = \sqrt{\frac{3T}{(2d-\rho)g}}$
Show Answer
Answer: (4) $r = \sqrt{\frac{3T}{(2d-\rho)g}}$
Solution
For the drops to be in equilibrium, upward force on drop = downward force on drop $T \cdot 2\pi R = \frac{4}{3}\pi R^3 dg - \frac{2}{3}\pi R^3 \rho g$ $\Rightarrow T(2\pi R) = \frac{2}{3}\pi R^3(2d - \rho)g$ $\Rightarrow T = \frac{R^2}{3}(2d - \rho)g \Rightarrow R = \sqrt{\frac{3T}{(2d - \rho)g}}$
BETA
