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JEE Main 20209 JanuaryEvening Shift
Exam
JEE Main
Year
2020
Date
9 January
Shift
Evening Shift
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Main 2020 · 9 January · Evening Shift · Physics · Mechanical Properties of Fluids

A small spherical droplet of density d is floating exactly half...

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

Options

  1. Ar=2T3(d+ρ)gr = \sqrt{\dfrac{2T}{3 ( d + ρ ) g}}
  2. Br=T(d−ρ)gr = \sqrt{\dfrac{T}{( d - ρ ) g}}
  3. Cr=T(d+ρ)gr = \sqrt{\dfrac{T}{( d + ρ ) g}}
  4. Dr=3T(2d−ρ)gr = \sqrt{\dfrac{3T}{( 2d - ρ ) g}}

Correct answer

D. r=3T(2d−ρ)gr = \sqrt{\dfrac{3T}{( 2d - ρ ) g}}

Step-by-step solution

A small spherical droplet of density d is floating exactly half immersed in a liquid of density ρ and surface tension T
dVg=pdVg = p\left(\dfrac{V}{2}\right)g + T(2\pir)T\left(2\pir\right)
⇒d.43πr3g=p.23πr3g+2\pirT⇒ d . \dfrac{4}{3}\pi{r}^{3}g = p . \dfrac{2}{3}\pi{r}^{3}g + 2\pirT
⇒23r2g(2d−p)=2T⇒ \dfrac{2}{3}{r}^{2}g\left(2d - p\right) = 2T
⇒r=3T(2d−p)g⇒ r = \sqrt{\dfrac{3T}{\left(2d - p\right)g}}
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