JEE Main 2026 — Apr 4, Shift 1 — Physics solution

JEE Main2026Apr 4, Shift 1Physics

The surface tension of a soap solution is $3.5 \times 1 0^{- 2}$ N/m. The work required to increase the radius of a soap bubble from $1$ cm to $2$ cm is $α \times 1 0^{- 6}$ J. The value of $α$ is ______. ( $π = 22/7$ )

Your answer

Solution

A soap bubble has two surfaces (inner and outer). Work required $=$ surface tension $\times$ change in total surface area:

$W = T \cdot 2 \cdot 4 π (R_{2}^{2} - R_{1}^{2}) = 8 π T (R_{2}^{2} - R_{1}^{2}) .$

$R_{2}^{2} - R_{1}^{2} = (2^{2} - 1^{2}) \times 1 0^{- 4} = 3 \times 1 0^{- 4}$ m $^{2}$ .

$W = 8 \times \frac{22}{7} \times 3.5 \times 1 0^{- 2} \times 3 \times 1 0^{- 4} = 8 \times 11 \times 3 \times 1 0^{- 6} = 264 \times 1 0^{- 6}$ J.

$∴ α = 264.$

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