Question

A liquid drop of radius R breaks into 27 small drops. What will be the change in surface energy if the surface tension of the liquid is $ T $ ?

• A. $ 8\pi {{R}^{2}}T $
• B. $ 12\pi {{R}^{2}}T $
• C. $ 8{{\pi }^{2}}{{R}^{2}}T $
• D. $ 12{{\pi }^{2}}{{R}^{2}}T $

Answer 1

Answer: (A)

: Change in surface energy $ =T\times $ change in area. Volume of one big drop = Volume of 27 small drops $ \therefore $ $ \frac{4}{3}\pi {{R}^{3}}=27\times \frac{4}{3}\pi {{r}^{3}} $ or $ R=3r $ .... (i) Surface area of 27 drops $ =27\times 4\pi {{r}^{2}} $ Surface area of one bigger drop $ =4\pi {{R}^{2}} $ $ \therefore $ change in area $ =27\times 4\pi {{r}^{2}}-4\pi {{R}^{2}} $ $ =4\pi [27{{r}^{2}}-{{R}^{2}}]=4\pi \left[ 27\times \frac{{{R}^{2}}}{9}-{{R}^{2}} \right] $ $ =4\pi \times 2{{R}^{2}}=8\pi {{R}^{2}} $ . $ \therefore $ Change in surface energy $ =8\pi {{R}^{2}}T $ .