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Question

# A certain number of spherical drops of a liquid of radius 'r' coalesce to form a single drop of radius 'R' and volume 'V'. If 'T' is the surface tension of the liquid then:

A
Energy=4VT(1r1R) is released
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B
Energy=3VT(1r+1R) is released
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C
Energy=3VT(1r1R) is released
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D
Energy is neither released nor absorbed
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Solution

## The correct option is C Energy=3VT(1r−1R) is releasedΔU=(T)(ΔA)A(initial)=(4πr2)nA(final)=4πR2ΔA=(4πr2)n−4πR2(43πr3)n=43πR3n=R3r3ΔA=4π[R3r3⋅r2−R2]=4π[R3r−R3R]=(4πR33)3[1r−1R]ΔA=3V[1r−1R]ΔU=3VT[1r−1R]

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