A liquid drop of diameter 2 mm breaks into 512 droplets. The change in surface energy is α× 10^-6 J. The value of α is _______. (Take surface tension of liquid = 0.08 N/m)

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q45 – JEE-MAIN 2026

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Home Physics Mechanical Properties of FluidsQuestion #254

A liquid drop of diameter 2 mm breaks into 512 droplets. The change in surface energy is

α \times 10^{- 6}

J. The value of

α

is _______. (Take surface tension of liquid = 0.08 N/m)

Clue 1 of 3: Conservation of Volume

When a large liquid drop breaks into smaller droplets, the total volume of the liquid remains constant.

🧠 Full Solution Path

Step 1: Relate radii using volume conservation✦ Active

Let $R$ be the radius of the large drop and $r$ be the radius of each small droplet. The volume of the large drop is $V_{l a r g e} = \frac{4}{3} π R^{3}$ . If it breaks into $N$ smaller droplets, the total volume of the small droplets is $V_{t o t a l, s m a l l} = N \times \frac{4}{3} π r^{3}$ . By conservation of volume, $V_{l a r g e} = V_{t o t a l, s m a l l}$ :

\frac{4}{3} π R^{3} = N \times \frac{4}{3} π r^{3} ⟹ R^{3} = N r^{3} ⟹ r = \frac{R}{N^{1 / 3}}

Given $N = 512$ , we find $N^{1 / 3} = (512)^{1 / 3} = 8$ . Therefore, $r = \frac{R}{8}$ .

Step 2: Calculate initial and final surface areas○ Expand

The initial surface area of the large drop is $A_{i n i t i a l} = 4 π R^{2}$ . The final total surface area of $N$ small droplets is $A_{f i n a l} = N \times 4 π r^{2}$ . Substituting $r = R / 8$ :

A_{f i n a l} = N \times 4 π {(\frac{R}{8})}^{2} = 512 \times 4 π \frac{R^{2}}{64} = 8 \times 4 π R^{2}

Step 3: Calculate the change in surface energy○ Expand

The change in surface energy is $Δ E = σ (A_{f i n a l} - A_{i n i t i a l})$ . Substituting the expressions for areas:

Δ E = σ (8 \times 4 π R^{2} - 4 π R^{2}) = σ (7 \times 4 π R^{2}) = 28 π σ R^{2}

Given diameter $D = 2$ mm, so radius $R = 1$ mm $= 1 \times 10^{- 3}$ m. Surface tension $σ = 0.08$ N/m. Substitute these values:

Δ E = 28 π (0.08) (1 \times 10^{- 3})^{2} = 28 π \times 0.08 \times 10^{- 6}

Calculate the numerical value:

Δ E = (2.24 π) \times 10^{- 6} \approx (2.24 \times 3.14159) \times 10^{- 6} \approx 7.037 \times 10^{- 6} J

The problem states the change in surface energy is $α \times 10^{- 6}$ J. Comparing this with our result, $α \approx 7.037$ . The closest integer value for $α$ among the options is 7.

💡 Teacher's Secret Hint

Remember that surface energy is released when drops combine, and energy is absorbed (or work is done) when a large drop breaks into smaller ones, hence the positive change in surface energy.

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Physics Question 45 – JEE-MAIN 2026

Clue 1 of 3: Conservation of Volume

🧠 Full Solution Path