JEE MAIN - Physics (2022 - 25th July Evening Shift - No. 2)
A drop of liquid of density $$\rho$$ is floating half immersed in a liquid of density $${\sigma}$$ and surface tension $$7.5 \times 10^{-4}$$ Ncm$$-$$1. The radius of drop in $$\mathrm{cm}$$ will be :
(g = 10 ms$$-$$2)
$$
\frac{15}{\sqrt{(2 \rho-\sigma)}}
$$
$$\frac{15}{\sqrt{(\rho-\sigma)}}$$
$$\frac{3}{2 \sqrt{(\rho-\sigma)}}$$
$$\frac{3}{20 \sqrt{(2 \rho-\sigma)}}$$
Explanation
_25th_July_Evening_Shift_en_2_1.png)
At equilibrium, forces balance each other
$$ \mathrm{S}(2 \pi \mathrm{r})+\mathrm{F}_{\mathrm{b}}=m g $$
Where $S=$ surface tension
$$ \begin{aligned} & \mathrm{F}_{\mathrm{b}}=\text { buoyant force }=\frac{2}{3} \pi r^3 \sigma g \\\\ & S(2 \pi r)=m g-\mathrm{F}_b=\frac{4}{3} \pi r^3\left(p-\frac{\sigma}{2}\right) g \\\\ & \Rightarrow r^2=\frac{3 S}{(2 p-\sigma) g} \\\\ & \Rightarrow r^2 =\frac{3 \times 7.5 \times 10^{-2}}{(2 p-\sigma) 10} \\\\ & \Rightarrow r^2 =\frac{22.5 \times 10^{-2}}{(2 p-\sigma) 10} \end{aligned} $$
$$ \Rightarrow $$ $$ r=\frac{1.5 \times 10^{-1}}{\sqrt{2 p-\sigma}} \mathrm{m}$$
$$=\frac{15}{\sqrt{2 p-\sigma}} \mathrm{cm} $$
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