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InterviewSolution
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A liquid drop having surface energy E is spread into 512 droplets of same size. The final surface energy of the droplets isA. 2EB. 4RC. 8ED. 12 |
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Answer» Correct Answer - C According to question, the surface area of the liquid drop is splits into 512 droplets, the surface area becomes `A_(2) = 512 xx 4pi r^(2)` [r = radius of smaller drop] r can be calculated by comparing the total volume of bigger and all smaller droplets. i.e., `(4)/(3)pi R^(3) = 512 xx (4)/(3) pi r^(3) rArr r = (R)/(8)` Hence, total area of smaller droplets is given by `A_(1) = 512 xx 4 xx pi xx ((R)/(8))^(2) = 8A` Change in surface area is given by `A_(2) - A_(1)` `= 4pi ((512 xx R^(2))/(64) - R^(2))` `4pi (8R^(2) - R^(2)) = 7R^(2)` Surface energy, E = AT [ A area, T - tension] So, `(E_(n))/(E_(o)) = (A_(1) xx T)/(A xx T) = (8A)/(A) = 8` `:. E_(n) = 8E` |
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