A liquid drop of density ' $\rho$ ' is floating half immersed in a liquid of density ' $d$ '. If ' $T$ ' is…

A liquid drop of density ' $\rho$ ' is floating half immersed in a liquid of density ' $d$ '. If ' $T$ ' is the surface tension then the diameter of the liquid drop is ( $\mathrm{g}=$ acceleration due to gravity)
  1. $\left[\frac{6 \mathrm{~T}}{\mathrm{~g}(2 \rho-\mathrm{d})}\right]^{\frac{1}{2}}$
  2. $\left[\frac{8 \mathrm{~T}}{3 \mathrm{~g}(2 \rho-\mathrm{d})}\right]^{\frac{1}{2}}$
  3. $\left[\frac{12 \mathrm{~T}}{\mathrm{~g}(2 \rho-\mathrm{d})}\right]^{\frac{1}{2}}$
  4. $\left[\frac{3 \mathrm{~T}}{\mathrm{~g}(2 \rho-\mathrm{d})}\right]^{\frac{\mathrm{t}}{2}}$

Solution

$\begin{aligned} & \text { Surface Tension }=\mathrm{T} \times 2 \pi \mathrm{r} \\ & \frac{4}{3} \pi \mathrm{r}^3 \rho \mathrm{~g}=2 \pi \mathrm{rT}+\frac{1}{2} \times \frac{4}{3} \pi \mathrm{r}^3 \mathrm{dg}\end{aligned}$ $\begin{array}{ll}\therefore \quad & 2 \pi r T=\frac{4}{3} \pi r^3 \rho g-\left(\frac{4}{3} \pi r^3 d g\right) \times \frac{1}{2} \\ & 2 \pi T=\frac{4}{3} \pi r^2 g\left(\rho-\frac{d}{2}\right) \\ \therefore & r^2=\frac{2 \pi T}{\frac{4}{3} \pi g\left(\rho-\frac{d}{2}\right)} \\ \therefore & r^2=\frac{3 T}{g(2 \rho-d)} \Rightarrow r=\sqrt{\frac{3 T}{g(2 \rho-d)}} \\ & \text { Diameter }=2 r=\sqrt{\frac{12 T}{g(2 \rho-d)}}\end{array}$

Asked in: MHT CET 2024 (03 May Shift 2)

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