Q116 / 150 · MHT CET 2011 · Physics

The radius $R$ of the soap bubble is doubled under isothermal condition. If $T$ be the surface tension of soap bubble. The work done in doing so it given by

  • Correct answer
    A
    $32 \pi R{ }^{2} T$
  • B
    $24 \pi R^{2} T$
  • C
    $8 \pi R^{2} T$
  • D
    $4 \pi R^{2} T$

Solution

Surface energy of a bubble $$ \begin{array}{l} E=\text { Tension } \times \text { surface area } \\ E=T \times A \end{array} $$ For soap bubble $$ \begin{array}{l} E=2 T \times A \\ E=2 T \times 4 \pi R^{2} \end{array} $$ Here, $\quad R=$ radius of bubble If radius is doubled then we have $$ \begin{array}{l} E^{\prime}=2 T \times 4 \pi R^{2} \\ E^{\prime}=2 T \times 4 \pi(2 R)^{2} \\ E^{\prime}=32 \pi R^{2} T \end{array} $$

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

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