JEE MAIN - Physics (2020 - 9th January Evening Slot - No. 21)
A small spherical droplet of density d is floating
exactly half immersed in a liquid of density $$\rho $$
and surface tension T. The radius of the droplet
is (take note that the surface tension applies an
upward force on the droplet) :
$$r = \sqrt {{T \over {\left( {d - \rho } \right)g}}} $$
$$r = \sqrt {{{2T} \over {3\left( {d + \rho } \right)g}}} $$
$$r = \sqrt {{T \over {\left( {d + \rho } \right)g}}} $$
$$r = \sqrt {{{3T} \over {\left( {2d - \rho } \right)g}}} $$
Explanation
_9th_January_Evening_Slot_en_21_1.png)
$$T.2\pi r + {2 \over 3}\pi {r^3}\rho g = {4 \over 3}\pi {r^3}dg$$
$$ \Rightarrow $$ T = $${{{r^2}} \over 3}\left( {2d - \rho } \right)g$$
$$ \Rightarrow $$ r = $$\sqrt {{{3T} \over {\left( {2d - \rho } \right)g}}} $$
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