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Properties of Matter question

2020 · 9 Jan · Shift 2 · Q61
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Properties of Matter question

2020 · 9 Jan · Shift 2 · Q61

JEE MainPhysicsProperties of MatterMCQ+4 / −1
A small spherical droplet of density d is floating exactly half immersed in a liquid of density ρ\rhoρ and surface tension T. The radius of the droplet is (take note that the surface tension applies an upward force on the droplet) :
  1. A
    r=T(d−ρ)gr = \sqrt {{T \over {\left( {d - \rho } \right)g}}}r=(d−ρ)gT​​
  2. B
    r=2T3(d+ρ)gr = \sqrt {{{2T} \over {3\left( {d + \rho } \right)g}}}r=3(d+ρ)g2T​​
  3. C
    r=T(d+ρ)gr = \sqrt {{T \over {\left( {d + \rho } \right)g}}}r=(d+ρ)gT​​
  4. D
    r=3T(2d−ρ)gr = \sqrt {{{3T} \over {\left( {2d - \rho } \right)g}}}r=(2d−ρ)g3T​​
View written solutionFree

Correct answer: D

  1. Forces acting on the droplet

A small spherical droplet of radius rrr and density ddd floats with half of its volume immersed in a liquid of density ρ\rhoρ.

The forces on the droplet are:

  • Weight of droplet (downward)
  • Buoyant force due to displaced liquid (upward)
  • Surface tension force (upward, as stated)

Since the droplet is floating in equilibrium,

Weight=Buoyant force+Surface tension force\text{Weight} = \text{Buoyant force} + \text{Surface tension force}Weight=Buoyant force+Surface tension force
  1. Weight of the spherical droplet

Volume of sphere:

V=43πr3V = \frac{4}{3}\pi r^3V=34​πr3

So weight is

W=(43πr3)dgW = \left(\frac{4}{3}\pi r^3\right) d gW=(34​πr3)dg
  1. Buoyant force

Since the droplet is half immersed, submerged volume is

Vsub=12⋅43πr3=23πr3V_{\text{sub}} = \frac{1}{2}\cdot \frac{4}{3}\pi r^3 = \frac{2}{3}\pi r^3Vsub​=21​⋅34​πr3=32​πr3

Hence buoyant force is

Fb=ρg(23πr3)F_b = \rho g \left(\frac{2}{3}\pi r^3\right)Fb​=ρg(32​πr3)
  1. Surface tension force

The droplet is half immersed, so the contact circle is a great circle of radius rrr.

Surface tension acts along the circumference:

Circumference=2πr\text{Circumference} = 2\pi rCircumference=2πr

Thus upward force due to surface tension is

FT=2πrTF_T = 2\pi r TFT​=2πrT
  1. Apply equilibrium condition
43πr3dg=23πr3ρg+2πrT\frac{4}{3}\pi r^3 d g = \frac{2}{3}\pi r^3 \rho g + 2\pi r T34​πr3dg=32​πr3ρg+2πrT

Divide throughout by πr\pi rπr:

43r2dg=23r2ρg+2T\frac{4}{3}r^2 d g = \frac{2}{3}r^2 \rho g + 2T34​r2dg=32​r2ρg+2T

Bring buoyancy term to the left:

r2g(4d−2ρ3)=2Tr^2 g\left(\frac{4d-2\rho}{3}\right) = 2Tr2g(34d−2ρ​)=2T r2g⋅2(2d−ρ)3=2Tr^2 g\cdot \frac{2(2d-\rho)}{3} = 2Tr2g⋅32(2d−ρ)​=2T

Cancel 2:

r2g⋅2d−ρ3=Tr^2 g\cdot \frac{2d-\rho}{3} = Tr2g⋅32d−ρ​=T

So,

r2=3T(2d−ρ)gr^2 = \frac{3T}{(2d-\rho)g}r2=(2d−ρ)g3T​

Therefore,

r=3T(2d−ρ)g\boxed{r = \sqrt{\frac{3T}{(2d-\rho)g}}}r=(2d−ρ)g3T​​​
  1. Compare with options

This matches:

Option D\boxed{\text{Option D}}Option D​
  1. Comparison with stored correct answer

Stored correct answer: D

Our derived answer: D

So they agree.

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