JEE Main 2026 Jan 24 Shift 2, Physics Q25: Properties of Fluids

JEE Main2026Jan 24, Shift 2Physics

A soap bubble of surface tension $0.04$ N/m is blown to a diameter of $7$ cm. If $(15000 - x) μ J$ of work is done in blowing it further to make its diameter $14$ cm, then the value of $x$ is ____.

$(π = 22/7)$

Your answer

Solution

A soap bubble has two surfaces (inner and outer), so the change in surface area is $Δ A = 2 (4 π r_{2}^{2} - 4 π r_{1}^{2}) = 8 π (r_{2}^{2} - r_{1}^{2})$ .

Radii: $r_{1} = 3.5 cm = 0.035$ m, $r_{2} = 7 cm = 0.07$ m.

$r_{2}^{2} - r_{1}^{2} = 0.0049 - 0.001225 = 0.003675 m^{2}$ .

Work done: $W = S \cdot Δ A = 8 π S (r_{2}^{2} - r_{1}^{2})$

$= 8 \times \frac{22}{7} \times 0.04 \times 0.003675 \approx 3.696 \times 1 0^{- 3} J = 3696 μ J$ .

Given $W = (15000 - x) μ J$ : $3696 = 15000 - x \Rightarrow x = 11304$ .

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JEE Main 2026 Jan 24 Shift 2, Physics Q25: Properties of Fluids

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