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

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=1.2\,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 $