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AIIMS 201925 May
Exam
AIIMS
Year
2019
Date
25 May
Subject
Physics
Chapter
Mechanical Properties of Fluids

AIIMS 2019 · 25 May · Physics · Mechanical Properties of Fluids

If T is the surface tension of the liquid, the change in surface...

If TT is the surface tension of the liquid, the change in surface energy by splitting a liquid drop of radius RR into 125 droplets of radius rr is

Options

  1. A16πR2T16 \pi R^2 T
  2. B32πR2T32 \pi R^2 T
  3. C18πR2T18 \pi R^2 T
  4. D36πR2T36 \pi R^2 T

Correct answer

A. 16πR2T16 \pi R^2 T

Step-by-step solution

As volume of 125 droplets = Volume of the drop ∴12543πr3=43πR3\therefore \quad 125 \frac{4}{3} \pi r^3=\frac{4}{3} \pi R^3 or 125r3=R3orr=R5125 r^3=R^3 or r=\frac{R}{5} The surface area of the drop =4πR2=4 \pi R^2 and the surface area of 125 droplets =125(4πr2)=125(4π(R5)2)=20πR2=125\left(4 \pi r^2\right)=125\left(4 \pi\left(\frac{R}{5}\right)^2\right)=20 \pi R^2∴ The increase in surface area =20πR2−4πR2=16πR2\begin{aligned} \therefore \quad \text{ The increase in surface area } & =20 \pi R^2-4 \pi R^2 \\ & =16 \pi R^2\end{aligned}$Thechangeinsurfaceenergy=SurfacetensionThe change in surface energy = Surface tension\timesincreaseinsurfaceareaincrease in surface area=T \times 16 \pi R^2=16 \pi R^2 T$
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