InterviewSolution
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InterviewSolution
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A liquid drop of diameter 4 mm breaks into 1000 droplets of equal size. Calculate the resultant change in surface energy, the surface tension of the liquid is `0.07 Nm^(-1)` |
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Answer» Correct Answer - `3.168 xx 10^(-5) J` Here, `R = 2xx10^(-3)m` , `n = 1000`. Let r be the radius of each small droplet formed. As, volume of big drop `= 1000 xx` volume of small droplet `(4)/(3) pi R^3 = 1000 xx(4)/(3) pi r^3` or `r= (R )/(10) = (2xx10^(-3))/(10) = 2xx10^(-4)m` Increase in surface area, `Delta A = 1000 4 pi r^2 - 4pi R^2 = 4pi (1000 r - R^2)` ` =4xx(22)/(7) [1000 (2xx10^(-4))^2 - (2xx 10^(-3))^2]` `= 4xx(22)/(7) [4 xx 10^(-5) - 0.4xx10^(-5)]` `=4xx(22)/(7) xx 3.6 xx10^(-5) m^2` Resultant increase in surface energy `= S.Txx Delta A = S xx Delta A` ` = 0.07 xx 4xx(22)/(7) xx3.6 xx 10^(-5)` `=3.168xx 10^(-5) J` |
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