Energy needed in breaking a liquid drop of radius $R$, into $n$ smaller drops each of radius $r$, is [T-…
Energy needed in breaking a liquid drop of radius $R$, into $n$ smaller drops each of radius $r$, is [T- Surface tension of the liquid]
- $\left(4 \pi r^2 n-4 \pi R^2\right) T$
- $\left(\frac{4}{3} \pi r^3 n-\frac{4}{3} \pi R^3\right) T$
- $\left(4 \pi R^2-4 \pi r^2\right) n T$
- $\left(4 \pi R^2-n 4 \pi r^2\right) / T$
Solution
We have $\Delta \mathrm{U}=$ (surface energy of bigger drop) - (surface energy of lower drop)
$\begin{aligned} & \Delta \mathrm{U}=\left(\mathrm{T} \times 4 \pi \mathrm{R}^2\right)-\left(\mathrm{nT} \times 4 \pi \mathrm{r}^2\right) \\ & |\Delta \mathrm{U}|=\left(4 \pi \mathrm{r}^2 \mathrm{n}-4 \pi \mathrm{R}^2\right) \mathrm{T}\end{aligned}$
Asked in: AP EAMCET 2022 (05 Jul Shift 2)
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