InterviewSolution
Saved Bookmarks
| 1. |
A mercury drop of radius 1 cm is broken into `10^6` droplets of equal size. The work done is `(T=35xx10^-2(N)/(m)`)A. `4.35xx10^-2J`B. `4.35xx10^-3J`C. `4.35xx10^-6J`D. `4.35xx10^-8J` |
|
Answer» Correct Answer - A Initial radius of drop, `R=1cm1xx10^-2m` suppose, `r` is the radius of droplets since, volume will be remain constant. `(4)/(3)piR^3=10^6xx(4)/(3)pir^3` [since, number of droplets`=10^6`] `r=(1xx10^-2)/(10^2)` Increase in surface area, `triangleS=10^6xx4pir^2-4piR^2` `=10^6xx4xxpixx10^-8-4pixx10^-4` `=4xx9.9xxpixx10^-3` Work done `=`increase in surface area `xx` Surface tension. `=4xx9.9xxpixx10^-3xx35xx10^-2` `=4.35xx10^-2J` |
|