Physics Q33 – JEE-MAIN 2026

Eight mercury drops, each of radius

r

, coalesce to form a bigger drop. The surface energy released in this process is _______ . (

S

is the surface tension of mercury)

When multiple liquid drops coalesce to form a single larger drop, the total volume of the liquid remains constant.

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

When 8 small mercury drops, each of radius $r$ , coalesce to form a bigger drop of radius $R$ , the total volume of mercury is conserved.

8 \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}

Simplifying this equation, we get:

8 r^{3} = R^{3} ⟹ R = (8)^{1 / 3} r = 2 r

Step 2: Calculate the change in surface area○ Expand

The initial total surface area of the 8 small drops is:

A_{i n i t i a l} = 8 \times (4 π r^{2}) = 32 π r^{2}

The final surface area of the single large drop is:

A_{f i n a l} = 4 π R^{2} = 4 π (2 r)^{2} = 4 π (4 r^{2}) = 16 π r^{2}

The decrease in surface area is:

Δ A = A_{i n i t i a l} - A_{f i n a l} = 32 π r^{2} - 16 π r^{2} = 16 π r^{2}

Step 3: Calculate the surface energy released○ Expand

The surface energy released during the coalescence process is the product of the surface tension $S$ and the decrease in surface area $Δ A$ .

E_{r e l e a s e d} = S \times Δ A = S \times 16 π r^{2} = 16 π r^{2} S

This matches option 2.

💡 Teacher's Secret Hint

Remember that energy is released when the total surface area decreases, as the system moves to a lower energy state.

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Physics Question 33 – JEE-MAIN 2026