A drop of liquid of diameter 2.8 mm split into 125 identical drops .The change in energy is …..(Surface tension T=75dyne/cm)

Options

A. zero
B. 19 erg
C. 46 erg
D. 71 erg

Answer

d

Explanation

Radius of large drop `R=2.8mm=2.8xx10^(-1)`cm
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`