Fluid Mechanics

JEE Main level, numerical answer, medium

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Question 91 of 192

A soap bubble of surface tension 0.04 N/m0.04 \mathrm{~N} / \mathrm{m} is blown to a diameter of 7 cm. If (15000−x)μJ(15000-x) \mu \mathrm{J} of work is done in blowing it further to make its diameter 14 cm, then the value of xx is ____\_\_\_\_. (π=22/7\pi=22 / 7)

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Answer: 11304

Soap bubble has two surfaces. Work done: W=ΔA×σW = \Delta A \times \sigma
Initial radius: ri=3.5r_i = 3.5 cm = 0.035 m, Final radius: rf=7r_f = 7 cm = 0.07 m
Change in surface area (both surfaces): ΔA=8π(rf2−ri2)=8π(0.072−0.0352)\Delta A = 8\pi(r_f^2 - r_i^2) = 8\pi(0.07^2 - 0.035^2)
ΔA=8π(0.0049−0.001225)=8π×0.003675=0.0294π\Delta A = 8\pi(0.0049 - 0.001225) = 8\pi \times 0.003675 = 0.0294\pi m²
W=0.0294π×0.04=0.001176πW = 0.0294\pi \times 0.04 = 0.001176\pi J = 0.001176×227=0.0036960.001176 \times \frac{22}{7} = 0.003696 J = 3696 μJ
Therefore: x=15000−3696=11304x = 15000 - 3696 = 11304

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