Multiple choice

The work done in blowing a soap bubble of radius $0.2 m$, given that the surface tension of soap solution is $\displaystyle 60\times { 10 }^{ -3 }{ N }/{ m }$ is :

  1. $\displaystyle 24\pi \times { 10 }^{ -2 }J$
  2. $\displaystyle 4\pi \times { 10 }^{ -4 }J$
  3. $\displaystyle 96\pi \times { 10 }^{ -4 }J$
  4. $\displaystyle 1.92\pi \times { 10 }^{ -2 }J$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Work done in blowing a soap bubble is W = 2 * (Surface Area * Surface Tension) because a bubble has two surfaces. W = 2 * (4 * pi * r^2) * T. W = 8 * pi * (0.2)^2 * 60 * 10^-3 = 8 * pi * 0.04 * 60 * 10^-3 = 19.2 * pi * 10^-2 J.

AI explanation

The work done against surface tension in blowing a soap bubble is given by the formula W = T * A, where T is the surface tension and A is the total surface area of the two surfaces. Substituting the given values, the total area is 2 * 4 * pi * (0.2)^2, which equals 0.32*pi m^2. Multiplying this area by the surface tension of 0.06 N/m results in 0.0192*pi J, which is the same as 1.92*pi * 10^-2 J. The result is 1.92*pi * 10^-2 J.

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