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JEE Main 20252 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
2 April
Shift
Evening Shift
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Main 2025 · 2 April · Evening Shift · Physics · Mechanical Properties of Fluids

Two water drops each of radius 'r' coalesce to from a bigger drop....

Two water drops each of radius 'r' coalesce to from a bigger drop. If ' T ' is the surface tension, the surface energy released in this process is :

Options

  1. A4πr2 T[2−223]4 \pi \mathrm{r}^2 \mathrm{~T}\left[2-2^{\frac{2}{3}}\right]
  2. B4πr2 T[2−213]4 \pi r^2 \mathrm{~T}\left[2-2^{\frac{1}{3}}\right]
  3. C4πr2 T[1+2]4 \pi r^2 \mathrm{~T}[1+\sqrt{2}]
  4. D4πr2T[2−1]4 \pi r^2 T[\sqrt{2}-1]

Correct answer

A. 4πr2 T[2−223]4 \pi \mathrm{r}^2 \mathrm{~T}\left[2-2^{\frac{2}{3}}\right]

Step-by-step solution

2×43πR3=43πr3⇒r=21/3RUi=2×4πR2TUf=4πr2T=4πR2T22/3∴ Heat lost =ui−uf=4πR2T[2−22/3]\begin{aligned} & 2 \times \frac{4}{3} \pi \mathrm{R}^3=\frac{4}{3} \pi \mathrm{r}^3 \Rightarrow \mathrm{r}=2^{1 / 3} \mathrm{R} \\ & \mathrm{U}_{\mathrm{i}}=2 \times 4 \pi \mathrm{R}^2 \mathrm{ T} \\ & \mathrm{U}_{\mathrm{f}}=4 \pi \mathrm{r}^2 \mathrm{ T}=4 \pi \mathrm{R}^2 \mathrm{ T} 2^{2 / 3} \\ & \therefore \text{ Heat lost }=\mathrm{u}_{\mathrm{i}}-\mathrm{u}_{\mathrm{f}}=4 \pi \mathrm{R}^2 \mathrm{ T}\left[2-2^{2 / 3}\right]\end{aligned}$$
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