A certain number of spherical drops of a liquid of radius r each, coalesce to form a single drop of radius R and volume V. If T is the surface tension of the liquid, then
A
Energy =3VT(1r+1R) is released
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B
Energy =3VT(1r+1R) is absorbed
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C
Energy =3VT(1r−1R) is released
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D
Energy =3VT(1r−1R) is absorbed
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Solution
The correct option is C Energy =3VT(1r−1R) is released Given,
Radius of smaller liquid drop =r
Radius of larger drop =R
Volume of larger drop =V
Surface tension of the liquid =T
Let us suppose n drops coaslece to form a bigger drop.
As volume of liquid will be same,
Volume of n smaller drops = volume of single bigger drop ⇒n×43πr3=43πR3=V ⇒n×4πr3=4πR3=3V ........(1)
We know, surface energy = surface tension × change in surface area ⇒E=TΔA =T[4πR2−n×4πr2] =T[4πR3R−n×4πr3r] =T[3VR−3Vr] ......[ from (1) ] =3VT[1R−1r]
As R>r⇒1R<1r ∴E is negative.
Hence, 3VT[1r−1R] energy will be released.