Question

Eight drops of water of 0.6 mm radius each merge to form one big drop. If the surface tension of water is $0.072\text{ }N{{m}^{-1}},$ the energy dissipated in the process is

• A. $16~\pi \times {{10}^{-7}}J$
• B. $4.15~\pi \times {{10}^{-7}}J$
• C. $2.0.75~\pi \times {{10}^{-7}}J$
• D. $8~\pi \times {{10}^{-7}}J$

Answer 1

The Correct Option is B

Volume of 8 drops of water = Volume of big drop of water Given that, Radius of small drop r = 0.6 mm $\therefore$ $8\times \frac{4}{3}\pi {{(0.6)}^{3}}=\frac{4}{3}\pi {{R}^{3}}$ or $2\times 0.6=R$ or $R=1.2\,mm$ where R is the radius of big drop. Change in surface area $\Delta A=4\pi {{r}^{2}}\times 8-4\pi {{R}^{2}}$ $=4\pi [{{(0.6)}^{2}}\times 8-{{(1.2)}^{2}}]\times {{10}^{-6}}$ $=4\pi (0.36\times 8-1.44)\times {{10}^{-6}}$ $=4\pi (2.88-1.44)\times {{10}^{-6}}$ $\Delta A=4\pi (1.44)\times {{10}^{-6}}{{m}^{2}}$ Energy dissipated in this process $E=T\times \Delta A$ $=0.072\times 4\pi \times 1.44\times {{10}^{-6}}$ $=28.8\times {{10}^{-2}}\times \pi \times 1.44\times {{10}^{-6}}$ $E=4.15\pi \times {{10}^{-7}}J$