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A spherical liquid drop of radius `R` is divided into eight equal droplets. If the surface tension is `T`, then the work done in this process will beA. `2piR^2T`B. `3piR^2T`C. `4piR^2T`D. `2piRT^2` |
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Answer» Correct Answer - C Radius of the larger drop `=R` Suppose radius of the droplets `=r` Since, volume will be remain constant, `(4)/(3)piR^3=8xx(4)/(3)pir^2` [`because` No. of droplets`=8pixx(0.01)xx75` work done`=`(increase in surface area)`xx`Surface tension `=[84pi((R )/(2))^2-4piR^2]xxT` `=(8piR^2-4piR^2)xxT=4piR^2T` |
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