Q.The work done to make a mercury drop of radius 4 cm will be equal to (Surface tension of mercury = 0.465 N/m) (A) 7.03 × 10^-3 joule
(B) 10 × 10^-2 joule
(C) 9.35 × 10^-3 joule
(D) 18.68 × 10^-3 joule

When droplets coalesce, total surface area decreases because the same total volume is now enclosed by less surface than before.

When $n$ small droplets, each of radius $r$, coalesce to form a single big drop of radius $R$, the volume is conserved:

$n \cdot \tfrac{4}{3}\pi r^3 = \tfrac{4}{3}\pi R^3 \implies R = n^{1/3} r$

Total surface area before coalescence: $A_i = n \cdot 4\pi r^2$.

Surface area after coalescence: $A_f = 4\pi R^2 = 4\pi n^{2/3} r^2$.

$\frac{A_f}{A_i} = \frac{4\pi n^{2/3} r^2}{4\pi n r^2} = n^{-1/3} < 1 \;(\text{for } n>1)$

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$W=T\cdot4\pi r^2\approx9.35\times10^{-3}\,\text{J}$.

A drop has a single free surface, so the work done to create it (against surface tension) equals $T$ times the total surface area formed: $W=T\times4\pi r^2$.

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Surface tension (force per unit length, N/m) and surface energy per unit area (J/m^2) are dimensionally and numerically the same quantity, just viewed two different ways.

Surface tension is usually defined as force per unit length: T = F/L, with SI unit N/m.

Surface energy per unit area has units: J/m^2 = (N·m)/m^2 = N/m — exactly the same unit as surface tension. This is not a coincidence: the work needed to increase a liquid surface's area by dA equals T*dA, so T is numerically equal to the surface energy per unit area as well as …

Coalescing drops release surface energy, so the temperature increases.

Each drop has surface energy = surface tension × surface area. When many small drops combine into one big drop, the total surface area decreases sharply (a single big drop has far less area than the many small ones combined).

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Surface tension T is numerically equal to the potential (surface) energy stored per unit area of a liquid surface, T = surface energy / area, unit J/m² (= N/m).

A liquid surface behaves like a stretched elastic membrane because molecules at the surface have fewer neighbours than those in the bulk, giving them extra potential energy. Increasing the surface area requires work done against this imbalance, and that work per unit increase in area defines the surface energy per unit area — which is nu …

Coalescing reduces total surface area for the same volume, so surface energy decreases and the excess is released (mostly as heat).

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