Fluid Mechanics

JEE Main level, single correct, medium

89 of 192

Question 89 of 192

The surface tension of a soap bubble is 0.030.03 N/m. The work done in increasing the diameter of bubble from 22 cm to 66 cm is απ×10−4\alpha \pi \times 10^{-4} J. The value of α\alpha is _______. (Take π=3.14\pi = 3.14)

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Answer: (C) 1.921.92

Initial radius r1=22=1r_1 = \dfrac{2}{2} = 1 cm =10−2= 10^{-2} m

Final radius r2=62=3r_2 = \dfrac{6}{2} = 3 cm =3×10−2= 3 \times 10^{-2} m

Surface tension T=0.03T = 0.03 N/m

A soap bubble has two free surfaces. The change in total surface area is given by:

ΔA=2×4π(r22−r12)\Delta A = 2 \times 4\pi (r_2^2 - r_1^2)

ΔA=8π((3×10−2)2−(10−2)2)\Delta A = 8\pi ((3 \times 10^{-2})^2 - (10^{-2})^2)

ΔA=8π(9×10−4−10−4)=64π×10−4\Delta A = 8\pi (9 \times 10^{-4} - 10^{-4}) = 64\pi \times 10^{-4} m2^2

Work done W=TΔAW = T \Delta A

W=0.03×64π×10−4W = 0.03 \times 64\pi \times 10^{-4}

W=1.92π×10−4W = 1.92\pi \times 10^{-4} J

Comparing with απ×10−4\alpha \pi \times 10^{-4} J, we get α=1.92\alpha = 1.92.

Answer: 1.921.92

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