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A drop of liquid of diameter 2.8 mm split into 125 identical drops .The change in energy is …..(Surface tension T=75dyne/cm) |
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Answer» zero No. of drops N=125 Change in energy `DeltaU`=(?) Surface energy of larger drop of radius R `U=TA` `thereforeU=T(4piR^(2))` …(1) Surface energy of smaller drop of radius r, `thereforeU.=T(4pir^(2))`...(2) Volume of n small drops = volume of large drop `n((4)/(3)PIR^(3))=(4)/(3)piR^(3)` `thereforenr^(3)=R^(3)` `thereforen^((1)/(3))r=R` `thereforer=(R)/(n^((1)/(3)))` `thereforer^(2)=(R^(2))/(n^((2)/(3)))`... (3) Equation (2) obtain as below, `U.=T((4piR^(2))/(n^((2)/(3))))` `n=125` `thereforen^((2)/(3))=(5^(3))^((2)/(3))=5^(2)=25` `U.=T((4piR)/(25))^(2)`....(4) Change in energy , `DeltaU=U-U`. `DeltaU=4piR^(2)T-(4piR^(2)T)/(25)` `=4piR^(2)T*[1-(1)/(25)]` `=4piR^(2)T((24)/(25))` `DeltaU=4xx3.14xx(2.8xx10^(-1))^(2)(75)((24)/(25))` `=7.89` `DeltaU=71erg` |
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