JEE Mains · Physics · STD 11 - 9.2 surface tension
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 :
- A \(4 \pi \mathrm{r}^2 \mathrm{~T}\left[2-2^{\frac{2}{3}}\right]\)
- B \(4 \pi r^2 \mathrm{~T}\left[2-2^{\frac{1}{3}}\right]\)
- C \(4 \pi r^2 \mathrm{~T}[1+\sqrt{2}]\)
- D \(4 \pi r^2 T[\sqrt{2}-1]\)
Answer & Solution
Correct Answer
(A) \(4 \pi \mathrm{r}^2 \mathrm{~T}\left[2-2^{\frac{2}{3}}\right]\)
Step-by-step Solution
Detailed explanation
\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…
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