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QJEEM18Ph2f46: JEE Main 2018 Physics - Class 11 JEEM18Ph2f46 Solution: A thin uniform tube is bent into a circle of radius in the…

A thin uniform tube is bent into a circle of radius rr in the vertical plane. Equal volumes of two immiscible liquids, whose densities are ρ1{\rho _1} and ρ2{\rho _2} \left( {{\rho _1} > {\rho _2}} \right), fill half the circle. The angle θ\theta between the radius vector passing through the common interface and the vertical is :

Multiple Choice Options

  • A θ=tan⁡−1π(ρ1ρ2)\theta = {\tan ^{ - 1}}\pi \left( {{{{\rho _1}} \over {{\rho _2}}}} \right)
  • B θ=tan⁡−1π2(ρ1ρ2)\theta = {\tan ^{ - 1}}{\pi \over 2}\left( {{{{\rho _1}} \over {{\rho _2}}}} \right)
  • C θ=tan⁡−1(ρ1−ρ2ρ1+ρ2)\theta = {\tan ^{ - 1}}\left( {{{{\rho _1} - {\rho _2}} \over {{\rho _1} + {\rho _2}}}} \right)
  • D θ=tan⁡−1π2(ρ1+ρ2ρ1−ρ2)\theta = {\tan ^{ - 1}}{\pi \over 2}\left( {{{{\rho _1} + {\rho _2}} \over {{\rho _1} - {\rho _2}}}} \right)

Step-by-Step Verified Solution

JEE Main 2018 (Online) 15th April Morning Slot Physics - Properties of Matter Question 236 English Explanation

As system is in equilibrium so the pressuse at A from both side of the liquid must be equal .

(r cos θ\theta + r sin θ\theta ) ρ\rho 2g = (r cos θ\theta −- r sin θ\theta ) ρ\rho 1g

⇒\Rightarrow    \,\,\, ρ1ρ2=sin⁡θ+cos⁡θcos⁡θ−sin⁡θ=1+tan⁡θ1−tan⁡θ{{{\rho _1}} \over {{\rho _2}}} = {{\sin \theta + \cos \theta } \over {\cos \theta - \sin \theta }} = {{1 + \tan \theta } \over {1 - \tan \theta }}

⇒\Rightarrow    \,\,\, ρ\rho 1 −- ρ\rho 1 tan θ\theta = ρ\rho 2 + ρ\rho 2 tan θ\theta

⇒\Rightarrow    \,\,\, ( ρ\rho 1 + ρ\rho 2) tan θ\theta = ρ\rho 1 −- ρ\rho 2

⇒\Rightarrow    \,\,\, tan θ\theta = ρ1−ρ2ρ1+ρ2{{{\rho _1} - {\rho _2}} \over {{\rho _1} + {\rho _2}}}

⇒\Rightarrow    \,\,\, θ\theta = tan −- 1 (ρ1−ρ2ρ1+ρ2)\left( {{{{\rho _1} - {\rho _2}} \over {{\rho _1} + {\rho _2}}}} \right)

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