Q.How much amount of work is done in forming a soap bubble of radius r?

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

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

So $A_f < A_i$ — the total surface area decreases. Physically, this is why the process releases energy: the excess surface energy ($T \times \Delta A$) is given out as heat, which is why coalescing droplets warm up slightly.

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

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: $W=T\times4\pi r^2$.

Here $T=0.465\,\text{N/m}$, $r=4\,\text{cm}=0.04\,\text{m}$: $4\pi r^2=4\pi(0.04)^2=4\pi(0.0016)=0.02011\,\text{m}^2$.

$W=0.465\times0.02011\approx9.35\times10^{-3}\,\text{J}$.

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 the force per unit length.

'Force per unit area' (option b) is actually the unit of pressure (N/m^2, i.e. Pascal), a different unit altogether — not the same as surface tension's N/m.

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).

The lost surface energy is converted into heat, which the mercury absorbs, so its temperature increases.

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 numerically the same as the surface tension (force per unit length). Both are measured in J/m² or equivalently N/m.

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

Two droplets of radius $r$ have combined surface area $2(4\pi r^2) = 8\pi r^2$. On coalescing into one drop of radius $R$ (with $R^3 = 2r^3$ by volume conservation), the new surface area $4\pi R^2$ is smaller than $8\pi r^2$. Since surface energy $= T \times (\text{surface area})$, the surface area decrease means surface energy decreases; this excess energy is liberated (chiefly as heat, slightly warming the drop).