A soap bubble of surface tension $0.04 \mathrm{~N} / \mathrm{m}$ is blown to a diameter of 7 cm. If $(15000-\mathrm{x}) \mu \mathrm{J}$ of work is done in blowing it further to make its diameter 14 cm, then the value of x is ____. ( $\pi=22 / 7$ )

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Fluid Mechanics

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

)

Answer: 11304

Solution

A soap film has two surfaces, so blowing the bubble from radius

r_{1}

r_{2}

creates new area

Δ A = 2 (4 π r_{2}^{2} - 4 π r_{1}^{2}) = 8 π (r_{2}^{2} - r_{1}^{2})

and needs work

W = S Δ A

. With

r_{1} = 3.5

cm and

r_{2} = 7

cm,

r_{2}^{2} - r_{1}^{2} = (49 - 12.25) \times 1 0^{- 4} = 36.75 \times 1 0^{- 4} m^{2}

, so

W = 0.04 \times 8 \times \frac{22}{7} \times 36.75 \times 1 0^{- 4} = 3696 \times 1 0^{- 6} J = 3696 μ J

. Setting

15000 - x = 3696

gives

x = 11304

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Integer answer

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