Q160 / 180 · COMEDK 2015 · Physics

A soap bubble of radius $r$ is blown up to form a bubble of radius $3 r$ under isothermal conditions. If $\sigma$ is the surface tension of soap solution, then the energy spent in blowing is

  • A
    $3 \pi \sigma r^{2}$
  • B
    $6 \pi \sigma r^{2}$
  • C
    $12 \pi \sigma r^{2}$
  • Correct answer
    D
    $64 \pi \sigma r^{2}$

Solution

Surface area of bubble of radius, $r=4 \pi r^{2}$ Surface area of bubble of radius, $3 r=4 \pi(3 r)^{2}=36 \pi r^{2}$ Therefore, increase in surface area $$ =36 \pi r^{2}-4 \pi r^{2}=32 \pi r^{2} $$ Since, a bubble has two surfaces, the total increase in surface area $=2 \times 32 \pi r^{2}=64 \pi r^{2}$ $\therefore$ Energy spent $=$ Work done $=$ Surface tension $\times$ Total increase in surface area $=64 \pi \sigma r^{2}$

Hence, the correct answer is option: 'D'.

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