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QJEEM22Phc892: JEE Main 2022 Physics - Class 11 JEEM22Phc892 Solution: A drop of liquid of density is floating half immersed in a…

A drop of liquid of density ρ\rho is floating half immersed in a liquid of density σ{\sigma} and surface tension 7.5×10−47.5 \times 10^{-4} Ncm −- 1. The radius of drop in cm\mathrm{cm} will be :

(g = 10 ms −- 2)

Multiple Choice Options

  • A 15(2ρ−σ)\frac{15}{\sqrt{(2 \rho-\sigma)}}
  • B 15(ρ−σ)\frac{15}{\sqrt{(\rho-\sigma)}}
  • C 32(ρ−σ)\frac{3}{2 \sqrt{(\rho-\sigma)}}
  • D 320(2ρ−σ)\frac{3}{20 \sqrt{(2 \rho-\sigma)}}

Step-by-Step Verified Solution

JEE Main 2022 (Online) 25th July Evening Shift Physics - Properties of Matter Question 128 English Explanation
At equilibrium, forces balance each other

S(2πr)+Fb=mg\mathrm{S}(2 \pi \mathrm{r})+\mathrm{F}_{\mathrm{b}}=m g

Where S=S= surface tension

amp;Fb= buoyant force =23πr3σgamp;S(2πr)=mg−Fb=43πr3(p−σ2)gamp;⇒r2=3S(2p−σ)gamp;⇒r2=3×7.5×10−2(2p−σ)10amp;⇒r2=22.5×10−2(2p−σ)10\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=1.5×10−12p−σmr=\frac{1.5 \times 10^{-1}}{\sqrt{2 p-\sigma}} \mathrm{m}

=152p−σcm=\frac{15}{\sqrt{2 p-\sigma}} \mathrm{cm}

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