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Home » A liquid drop having surface energy ‘ ‘ is spread into 512 droplets of same size. The final surface energy of the droplets isA) B) C) D)
A liquid drop having surface energy ‘ E ‘ is spread into 512 droplets of same size. The final surface energy of the droplets is
A) 2 \mathrm{E}
B) 4 \mathrm{E}
C) 8 \mathrm{E}
D) 12 \mathrm{E}
Physics
A liquid drop having surface energy ‘ E ‘ is spread into 512 droplets of same size. The final surface energy of the droplets is
A) 2 \mathrm{E}
B) 4 \mathrm{E}
C) 8 \mathrm{E}
D) 12 \mathrm{E}

Answer is (C)
The surface area of the liquid drop is A=4 \pi R^{2}
Its surface energy is \mathrm{E}
When the drop splits in 512 droplets, the surface area of each droplets is A_{2}=512 \times r^{2}
\therefore=512 \times 4 \pi \mathrm{r}^{2}
The volume of bigger drop is \frac{4}{3} \pi R^{3} and Volume of small droplets is 512 \times \frac{4}{3} \pi r^{3}
\begin{array}{l} \therefore \quad \frac{4}{3} \pi R^{3}=512 \times \frac{4}{3} \pi r^{3} \Rightarrow r=\frac{R}{8} \\ \therefore \quad A_{2}=512 \times 4 \pi r^{2}=512 \times 4 \pi\left(\frac{R}{8}\right)^{2}=8 A_{1} \end{array}
Change in surface area is \mathrm{A}_{2}-\mathrm{A}_{1}=4 \pi\left(\frac{512 \times \mathrm{R}^{2}}{64}-\mathrm{R}^{2}\right)=4 \pi\left(8 \mathrm{R}^{2}-\mathrm{R}^{2}\right)=7 \mathrm{R}^{2}
Surface energy E=A . T (T is surface tension and A is area).
\begin{array}{l} \therefore \frac{E_{a}}{E_{0}}=\frac{A_{2} \cdot T}{A_{1} \cdot T}=\frac{8 \cdot A_{1}}{A_{1}}=8 \\ \therefore E_{n}=8 E \end{array}

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