Q135 / 160 · TS EAMCET 2022 (20 Jul Shift 1) · Physics

A soap bubble of initial radius $R$ is to be blown up. The surface tension of the soap film is $\mathrm{T}$. The surface energy needed to double the diameter of the bubble is

  • A
    $12 \pi R^2 T$
  • B
    $4 \pi \mathrm{R}^2 \mathrm{~T}$
  • C
    $16 \pi R^2 T$
  • Correct answer
    D
    $24 \pi \mathrm{R}^2 \mathrm{~T}$

Solution

$$ \begin{aligned} & \Delta \mathrm{U}=\mathrm{U}_f-\mathrm{U}_{\mathrm{i}} \\ & =\left[\mathrm{T}\left[4 \pi(2 \mathrm{R})^2\right]-\mathrm{T}\left[4 \pi \mathrm{R}^2\right]\right] \times 2 \\ & =32 \pi \mathrm{R}^2 \mathrm{~T}-8 \pi \mathrm{R}^2 \mathrm{~T}=24 \pi \mathrm{R}^2 \mathrm{~T} \end{aligned} $$ Note:- As soap bubble is thick, so we have to take care of both inner and outer layer $$ \text { So, } U=\left(T \times 4 \pi R^2\right) \times 2 $$

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

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