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

  • [JEE MAIN 2018]
  • A

    $\theta  = {\tan ^{ - 1}}\left[ {\frac{\pi }{2}\left( {\frac{{{\rho _1} - {\rho _2}}}{{{\rho _1} + {\rho _2}}}} \right)} \right]$

  • B

    $\theta  = {\tan ^{ - 1}}\frac{\pi }{2}\left( {\frac{{{\rho _1} + {\rho _2}}}{{{\rho _1} - {\rho _2}}}} \right)$

  • C

    $\theta  = {\tan ^{ - 1}}\pi \left( {\frac{{{\rho _1}}}{{{\rho _2}}}} \right)$

  • D

    $\theta  = {\tan ^{ - 1}}\frac{\pi }{2}\left( {\frac{{{\rho _2}}}{{{\rho _1}}}} \right)$

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$1.$  The piston is now pulled out slowly and held at a distance $2 \mathrm{~L}$ from the top. The pressure in the cylinder between its top and the piston will then be

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