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QJEEM20Phba9f: JEE Main 2020 Physics - Class 11 JEEM20Phba9f Solution: A small spherical droplet of density d is floating exactly half …

A small spherical droplet of density d is floating exactly half immersed in a liquid of density ρ\rho and surface tension T. The radius of the droplet is (take note that the surface tension applies an upward force on the droplet) :

Multiple Choice Options

  • A r=T(d−ρ)gr = \sqrt {{T \over {\left( {d - \rho } \right)g}}}
  • B r=2T3(d+ρ)gr = \sqrt {{{2T} \over {3\left( {d + \rho } \right)g}}}
  • C r=T(d+ρ)gr = \sqrt {{T \over {\left( {d + \rho } \right)g}}}
  • D r=3T(2d−ρ)gr = \sqrt {{{3T} \over {\left( {2d - \rho } \right)g}}}

Step-by-Step Verified Solution

JEE Main 2020 (Online) 9th January Evening Slot Physics - Properties of Matter Question 198 English Explanation

T.2πr+23πr3ρg=43πr3dgT.2\pi r + {2 \over 3}\pi {r^3}\rho g = {4 \over 3}\pi {r^3}dg

⇒\Rightarrow T = r23(2d−ρ)g{{{r^2}} \over 3}\left( {2d - \rho } \right)g

⇒\Rightarrow r = 3T(2d−ρ)g\sqrt {{{3T} \over {\left( {2d - \rho } \right)g}}}

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