JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Fluid Mechanics

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

)

Answer: 264

Solution

A soap bubble has two surfaces, so the work equals

2 T

times the increase in area:

W = 2 T (4 π R_{2}^{2} - 4 π R_{1}^{2}) = 8 π T (R_{2}^{2} - R_{1}^{2})

. With

T = 3.5 \times 1 0^{- 2}

N/m,

R_{1} = 1 0^{- 2}

R_{2} = 2 \times 1 0^{- 2}

m and

π = \frac{22}{7}

W = 8 \times \frac{22}{7} \times 3.5 \times 1 0^{- 2} \times (4 - 1) \times 1 0^{- 4} = 264 \times 1 0^{- 6}

J, so

α = 264

JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Integer answer

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