Surface tension of a soap bubble is $2.0 \times 10^{-2} \mathrm{Nm}^{-1}$. Work done to increase the radius of soap bubble from 3.5 cm to 7 cm will be : [Take $\pi=\frac{22}{7}$]

(C) $18.48 \times 10^{-4} \mathrm{~J}$

(A) $0.72 \times 10^{-4} \mathrm{~J}$

(B) $5.76 \times 10^{-4} \mathrm{~J}$

(D) $9.24 \times 10^{-4} \mathrm{~J}$

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Fluid Mechanics

Surface tension of a soap bubble is

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

. Work done to increase the radius of soap bubble from 3.5 cm to 7 cm will be : [Take

π = \frac{22}{7}

]

Answer: (C)

A soap bubble has two surfaces, so its surface area is

8 π R^{2}

and the work done in enlarging it equals the gain in surface energy,

W = T_{S} (8 π) (R_{2}^{2} - R_{1}^{2})

. With

T_{S} = 2 \times 1 0^{- 2}

N m

^{- 1}

R_{1} = 0.035

R_{2} = 0.07

m and

π = \frac{22}{7}

R_{2}^{2} - R_{1}^{2} = (0.0 7^{2} - 0.03 5^{2}) = 3.675 \times 1 0^{- 3} m^{2}

, giving

W = 2 \times 1 0^{- 2} \times 8 \times \frac{22}{7} \times 3.675 \times 1 0^{- 3}

. The work done is

18.48 \times 1 0^{- 4} J

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Single correct option