Question

Work done in splitting a drop of water of $1\, mm$ radius into $ {{10}^{6}} $ droplets is (surface tension of water $ =72\times {{10}^{-3}}J/{{m}^{2}} $ )

• A. $ 89.5\times {{10}^{-5}}J $
• B. 0.895 J
• C. $ 895\times {{10}^{-5}}J $
• D. None of these

Answer 1

Answer: (B)

The volume of big drop $ ={{10}^{6}}\times $ volume of small drop $ \Rightarrow $ $ \frac{4}{3}\pi {{R}^{3}}={{10}^{6}}\times \frac{4}{3}\pi {{r}^{3}} $ $ \Rightarrow $ $ R=100r $ Therefore, work done $ W=T.\Delta A $ $ =T[n.4\pi {{r}^{2}}-4\pi {{R}^{2}}] $ $ =4\pi T[{{10}^{6}}{{r}^{2}}-{{10}^{4}}{{r}^{2}}] $ $ =4\pi \times 72\times {{10}^{-3}}\times 990000\times {{10}^{-6}} $ $ =0.895J $