Question

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

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

Answer 1

Answer: (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 $R=1.2mm$ 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 $E=T\times \Delta A$ $=0.072\times 4\pi \times 1.44\times {10}^{-6}$ $=28.8\times {10}^{-2}\times \pi \times 1.44\times {10}^{-6}$ $E=4.15\pi \times {10}^{-7}J$