JEE MAIN - Physics (2018 - 15th April Morning Slot - No. 4)
A thin uniform tube is bent into a circle of radius $$r$$ in the vertical 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 :
$$\theta = {\tan ^{ - 1}}\pi \left( {{{{\rho _1}} \over {{\rho _2}}}} \right)$$
$$\theta = {\tan ^{ - 1}}{\pi \over 2}\left( {{{{\rho _1}} \over {{\rho _2}}}} \right)$$
$$\theta = {\tan ^{ - 1}}\left( {{{{\rho _1} - {\rho _2}} \over {{\rho _1} + {\rho _2}}}} \right)$$
$$\theta = {\tan ^{ - 1}}{\pi \over 2}\left( {{{{\rho _1} + {\rho _2}} \over {{\rho _1} - {\rho _2}}}} \right)$$
Explanation
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As system is in equilibrium so the pressuse at A from both side of the liquid must be equal .
(r cos $$\theta $$ + r sin $$\theta $$) $$\rho $$2g = (r cos $$\theta $$ $$-$$ r sin $$\theta $$) $$\rho $$1g
$$ \Rightarrow $$$$\,\,\,$$ $${{{\rho _1}} \over {{\rho _2}}} = {{\sin \theta + \cos \theta } \over {\cos \theta - \sin \theta }} = {{1 + \tan \theta } \over {1 - \tan \theta }}$$
$$ \Rightarrow $$$$\,\,\,$$ $$\rho $$1 $$-$$ $$\rho $$1 tan$$\theta $$ = $$\rho $$2 + $$\rho $$2 tan$$\theta $$
$$ \Rightarrow $$$$\,\,\,$$ ($$\rho $$1 + $$\rho $$2) tan$$\theta $$ = $$\rho $$1 $$-$$ $$\rho $$2
$$ \Rightarrow $$$$\,\,\,$$ tan$$\theta $$ = $${{{\rho _1} - {\rho _2}} \over {{\rho _1} + {\rho _2}}}$$
$$ \Rightarrow $$$$\,\,\,$$ $$\theta $$ = tan$$-$$1 $$\left( {{{{\rho _1} - {\rho _2}} \over {{\rho _1} + {\rho _2}}}} \right)$$
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