A certain number of spherical drops of a liquid of radius 'f coalesce to form a single drop of radius 'R' and volume 'V'. If 'T' is the surface tension of the liquid, then :
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a
energy =4VT1r-1Ris released.
b
energy =3VT1r-1Ris absorbed
c
energy =3VT1r-1Ris released
d
energy is neither released nor absorbed
answer is C.
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Detailed Solution
As surface area decreases so energy is released Released energy =4πR2T[n13-1]Where, R=n13r =4πR3T1r-1R=3VT1r-1R