Physics Q31 – JEE-MAIN 2025

Home Physics Mechanical Properties of FluidsQuestion #270

Two water drops each of radius '

r

' coalesce to form a bigger drop. If '

T

' is the surface tension, the surface energy released in this process is :

Clue 1 of 3: Conservation Principle

When drops coalesce, the total volume of the liquid is conserved.

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Ninja Strategy•Volume Conservation and Area Change

Recognize that coalescence conserves volume, leading to a specific relationship between radii, and that energy is released due to a decrease in total surface area.

🧠 Full Solution Path

Step 1: Determine the radius of the larger drop✦ Active

Assume the drops are spherical and volume is conserved during coalescence. If two small drops of radius $r$ combine to form a single large drop of radius $R$ , then the total volume before and after must be equal.

2 \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3} ⟹ R^{3} = 2 r^{3} ⟹ R = 2^{1 / 3} r

Step 2: Calculate the initial and final surface areas○ Expand

The initial total surface area of the two small drops is $A_{i n i t i a l} = 2 \times (4 π r^{2}) = 8 π r^{2}$ . The final surface area of the single large drop is $A_{f i n a l} = 4 π R^{2}$ .

A_{f i n a l} = 4 π (2^{1 / 3} r)^{2} = 4 π (2^{2 / 3} r^{2}) = 4 π r^{2} 2^{2 / 3}

Step 3: Calculate the surface energy released○ Expand

The surface energy released is the product of the surface tension $T$ and the decrease in surface area ( $Δ A = A_{i n i t i a l} - A_{f i n a l}$ ). Energy is released because the total surface area decreases.

E_{r e l e a s e d} = T \times (A_{i n i t i a l} - A_{f i n a l}) = T \times (8 π r^{2} - 4 π r^{2} 2^{2 / 3})

E_{r e l e a s e d} = 4 π r^{2} T (2 - 2^{2 / 3})

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Physics Question 31 – JEE-MAIN 2025

Clue 1 of 3: Conservation Principle

🧠 Full Solution Path