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

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energy =4VT1r−1R is released.

energy =3VT1r+1Ris absorbed.

energy =3VT1r−1R is released.

energy is neither released nor absorbed

answer is C.

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Detailed Solution

Let n droplets each of radius r coalesce to form a big drop of radius R.∴ Volume of n droplets = Volume of big dropn×43πr3=43πR3 n=R3r3 ... (I) Volume of big drop, V=43πR3 ....(II)Initial surface area of n droplets. Ai=n×4πr2=R3r3×4πr2 (Using .... (i)) =4πR3r=43πR33r=3Vr (Using.....( (ii) ) Final surface area of big dropAf=4πR2=43πR33R=3VR (Usin⁡g….(iii)Decrease in surface areaΔA=Ai−Af=3Vr−3VR=3V1r−1R∴ Energy released = Surface tension × Decrease in surface area=T×ΔA =3VT1r−1R

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