JEE Main 2026 — Apr 2, Shift 2 — Physics solution

JEE Main2026Apr 2, Shift 2Physics

The surface tension of a soap bubble is $0.03$ N/m. The work done in increasing the diameter of bubble from $2$ cm to $6$ cm is $α π \times 1 0^{- 4}$ J. The value of $α$ is_______. (Take $π = 3.14$ )

A
$0.86$
B
$0.64$
C
$1.92$
D
$7.68$

Solution

A soap bubble has two surfaces, so the work done equals the change in surface energy: $W = T \cdot 2 (4 π R_{2}^{2} - 4 π R_{1}^{2})$ .

With $R_{1} = 1$ cm $= 1 0^{- 2}$ m and $R_{2} = 3$ cm $= 3 \times 1 0^{- 2}$ m:

$W = 0.03 \cdot 8 π (9 \times 1 0^{- 4} - 1 0^{- 4}) = 0.03 \cdot 8 π \cdot 8 \times 1 0^{- 4} = 1.92 π \times 1 0^{- 4}$ J.

$∴ α = 1.92$ .

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