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JEE Main 2025 · Physics · Fluid Mechanics, Elasticity and Surface Tension · medium

One thousand identical spherical water droplets, each of radius 1 mm, coalesce into a single large drop. The surface tension of water is 0.07 N m^-1.…

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One thousand identical spherical water droplets, each of radius 1 mm1\ \mathrm{mm}, coalesce into a single large drop. The surface tension of water is 0.07 N m−10.07\ \mathrm{N\,m^{-1}}. The surface energy released is Nπ×10−6 JN\pi\times10^{-6}\ \mathrm{J}, where NN is ______.
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Answer: 252

Volume conservation: 43πR3=1000⋅43πr3\tfrac43\pi R^3 = 1000\cdot\tfrac43\pi r^3 gives R=10r=10 mmR=10r=10\ \mathrm{mm}. Initial surface area =1000×4πr2=4000π mm2=1000\times4\pi r^2 = 4000\pi\ \mathrm{mm^2}; final area =4πR2=400π mm2=4\pi R^2 = 400\pi\ \mathrm{mm^2}. Decrease in area =3600π mm2=3600π×10−6 m2=3600\pi\ \mathrm{mm^2} = 3600\pi\times10^{-6}\ \mathrm{m^2}. Energy released =T ΔA=0.07×3600π×10−6=252π×10−6 J=T\,\Delta A = 0.07\times3600\pi\times10^{-6} = 252\pi\times10^{-6}\ \mathrm{J}. Hence N=252N=252.
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