The amount of work done in blowing a sops bubble such that its diameter increases from ' $d$ ' to…

The amount of work done in blowing a sops bubble such that its diameter increases from ' $d$ ' to $\mathrm{D}$ ' is ( $\mathrm{T}$ = surface tension of solution )
  1. $4 \pi\left(D^2-d^2\right) T$
  2. $8 \pi\left(D^2-d^2\right) T$
  3. $\pi\left(D^2-d^2\right) T$
  4. $2 \pi\left(D^2-d^2\right) T$

Solution

Change in surface area $\begin{aligned} & =2 \times 4 \pi\left[\left(\frac{\mathrm{D}}{2}\right)^2-\left(\frac{\mathrm{d}}{2}\right)^2\right]=2 \pi\left(\mathrm{D}^2-\mathrm{d}^2\right) \\ & \therefore \text { Work done }=\text { surface tension } \times \text { change in area } \\ & =2 \pi\left(\mathrm{D}^2-\mathrm{d}^2\right) \mathrm{T}\end{aligned}$

Asked in: MHT CET 2022 (07 Aug Shift 2)

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