InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Two mercury droplets of radii 0.1 cm and 0.2 cm collapse into one single drop. What amount of energy is released? The surface tension of mercury `T = 435. 5xx10^(-3) Nm^(-1)` |
|
Answer» Let R be the radius of big single drop formed when two small drops of radii `r_1 and r_2` collapse together. Then volume of bid drop = volume of two small drops `:. (4)/(3) pi R^1 = (4)/(3) pi r_1^3 + (4)/(3) pi r_2^3 or R^3 = r_1^3 + r_2^3 = (0.1)^3 + (0.2)^3 = 0.001 + 0.008 = 0.009 cm^3` `:. R = 0.21 cm` Change in surface area ` = 4pi R^2 - (4pi r_1^2 + 4pi_2^2)` Energy relased ` =ST xx` change in surface area `= T xx [ 4pi R^2 - (4pi r_1^2 + 4pi r_2^2)] = 4pi T[R^2 - (r_1^2 + r_2^2)]` `4xx3.142 xx 435.5 xx10^(-3) [(0.21)^2 - {(0.1)^2 + (0.2)^2}] = - 32.33xx 10^(-7)J` |
|