Fluid Mechanics

JEE Main level, numerical answer, medium

88 of 192

Question 88 of 192

The surface tension of a soap solution is 3.5×10−23.5 \times 10^{-2} N/m. The work required to increase the radius of a soap bubble from 11 cm to 22 cm is α×10−6\alpha \times 10^{-6} J. The value of α\alpha is _____. (π=22/7\pi = 22/7)

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Answer: 264

The work done to increase the radius of a soap bubble is given by W=T×ΔAW = T \times \Delta A

Since a soap bubble has two free surfaces, the change in surface area is ΔA=2×4π(r22−r12)\Delta A = 2 \times 4\pi (r_2^2 - r_1^2)

Substituting the given values:

W=3.5×10−2×8π×((2×10−2)2−(1×10−2)2)W = 3.5 \times 10^{-2} \times 8\pi \times ((2 \times 10^{-2})^2 - (1 \times 10^{-2})^2)

W=3.5×10−2×8π×(4×10−4−1×10−4)W = 3.5 \times 10^{-2} \times 8\pi \times (4 \times 10^{-4} - 1 \times 10^{-4})

W=3.5×10−2×8π×3×10−4W = 3.5 \times 10^{-2} \times 8\pi \times 3 \times 10^{-4}

W=84π×10−6W = 84\pi \times 10^{-6}

Using π=227\pi = \dfrac{22}{7}:

W=84×227×10−6=264×10−6W = 84 \times \dfrac{22}{7} \times 10^{-6} = 264 \times 10^{-6} J

Comparing with α×10−6\alpha \times 10^{-6} J, we get α=264\alpha = 264

Answer: 264264

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