A drop of mercury of radius 0.2 cm is broken into 8 identical droplets. Find the work done if the surface tens

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Q. A drop of mercury of radius 0.2 cm is broken into 8 identical droplets. Find the work done if the surface tension of mercury is 435.5 dyne/cm. [Ans. 2.189\times10⁻⁵ J]

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Concept understanding — Surface Energy

Surface Energy: The Cost of Making a Surface

Imagine you are blowing a soap bubble. You have to keep blowing — pushing air in — even after the bubble is fully formed. Why? Because every time the bubble gets bigger, you are creating new surface. That surface is not free. The liquid film resists being stretched, and you have to do work against that resistance.

That resistance is surface tension. But the work you do — the energy you spend — to create that new surface is surface energy.

The Intuition: Molecules at the Edge

Inside a liquid, a molecule is surrounded by neighbours on all sides. It feels a net pull of zero — it is happy. But a molecule on the surface has neighbours only below and to the sides, not above. It is pulled inward. That means every molecule on the surface is in a higher-energy state than one in the bulk. To bring a molecule from the inside to the surface, you must do work against this inward pull.

So a surface is like a stretched membrane that wants to shrink. Creating more surface means pulling more molecules up from the bulk — and that costs energy.

The Precise Definition

Surface energy is the work done to increase the surface area of a liquid by one unit.

If you increase the area by $\Delta A$ , and the work required is $W$ , then the surface energy $E$ per unit area is:

E = \frac{W}{\Delta A}

For a liquid, this quantity is numerically equal to the surface tension $\sigma$ (or $T$ ). Surface tension is force per unit length ( $\text{N/m}$ ), while surface energy is energy per unit area ( $\text{J/m}^2$ ). But:

1\ \text{J/m}^2 = 1\ \frac{\text{N·m}}{\text{m}^2} = 1\ \text{N/m}

So they are the same number, just expressed in different units. Surface tension is the force that resists stretching; surface energy is the work you do when you stretch.

\text{Surface energy} = \frac{\text{Work done}}{\text{Increase in area}} = \sigma

A Concrete Example: The Soap Film

Take a U-shaped wire with a sliding wire across it, forming a soap film. The film has two surfaces (top and bottom). If you pull the slider by a distance $x$ , you increase the area by $2lx$ (two sides, each of length $l$ times $x$ ). The force you apply is $F = 2\sigma l$ (surface tension acts along both sides). The work done is:

W = F \cdot x = 2\sigma l x = \sigma \cdot (2lx) = \sigma \cdot \Delta A

So the work per unit area is exactly $\sigma$ .

Tip

For a liquid film with two surfaces, always remember to double the area when relating work to surface tension.

Why This Matters

Surface energy explains why: …

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Step 1. $R = 0.2$ cm $= 2×10^{-3}$ m breaks into 8 droplets, so $8r^3 = R^3 \Rightarrow r = R/2 = 1×10^{-3}$ m (since $8 = 2^3$ ).

Step 2. Initial area: $4\pi R^2 = 4\pi(2×10^{-3})^2 = 4\pi×4×10^{-6} = 5.0265×10^{-5}$ m². Final area (8 droplets): $8×4\pi r^2 = 32\pi(10^{-3})^2 = 32\pi×10^{-6} = 1.00531×10^{-4}$ m² — exactly twice the initial area. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$Q. When a number of droplets coalesce to form a single drop, the total surface area of the drop ______.$
(a) decreases
(b) becomes zero
(c) remains same
(d) increases
›Reveal solutionSolution

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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CBSE 2024Set ANNUAL1 markMCQ
$Q. The work done to make a mercury drop of radius 4 cm will be equal to (Surface tension of mercury = 0.465 N/m) (A) 7.03 \times 10^-3 joule (B) 10 \times 10^-2 joule (C) 9.35 \times 10^-3 joule (D) 18.68 \times 10^-3 joule$
›Reveal solutionSolution

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

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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CBSE 2024Set ANNUAL1 markMCQ
$Q. The unit of surface tension is same as that of$
(a) surface energy per unit volume
(b) force per unit area
(c) surface energy per unit area
(d) surface energy per unit length
›Reveal solutionSolution

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 …

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CBSE 2023Set ANNUAL1 markMCQ
$Q. When many small drops of mercury make a big drop, then it's temperature:$
(a) decreases
(b) increases
(c) no change
(d) none of these
›Reveal solutionSolution

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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CBSE 2022Set ANNUAL1 mark
$Q. Which quantity is equal to the potential energy stored per unit area of a liquid?$
›Reveal solutionSolution

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 …

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CBSE 2020Set ANNUAL1 markMCQ
$Q. Two droplets coalesce in a single drop. In this process _______. (A) energy is liberated (B) energy is obsorbed (C) energy does not change (D) some mass is converted into energy$
›Reveal solutionSolution

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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Q.A drop of mercury of radius 0.2 cm is broken into 8 identical droplets. Find the work done if the surface tension of mercury is 435.5 dyne/cm. [Ans. 2.189×10⁻⁵ J]

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