JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Fluid Mechanics

The surface tension of soap solution is

3.5 \times 1 0^{- 2} Nm^{- 1}

. The amount of work done required to increase the radius of soap bubble from 10 cm to 20 cm is

____

\times 1 0^{- 4} J

Answer: 264

Solution

A soap bubble has two surfaces (inner and outer), so the work done is

W = T \times Δ A = T \times 2 \times 4 π (r_{2}^{2} - r_{1}^{2}) = 8 π T (r_{2}^{2} - r_{1}^{2})

. With

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

N/m,

r_{1} = 0.1

r_{2} = 0.2

W = 8 π \times 3.5 \times 1 0^{- 2} \times (0.04 - 0.01) = 8 π \times 3.5 \times 1 0^{- 2} \times 0.03 = 8 \times 3.14 \times 1.05 \times 1 0^{- 3} \approx 2.64 \times 1 0^{- 2}

= 264 \times 1 0^{- 4}

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Integer answer

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