Question

If the work done in blowing a bubble of volume $V$ is $W$, then the work done in blowing a soap bubble of volume $2\, V$ will be :

• A. $W$
• B. $2\, W$
• C. $\sqrt{2}$ W
• D. ${{4}^{1/3}}W$

Answer 1

Answer: (D)

${{4}^{1/3}}W$

Answer 2

From the definition of surface tension $(T)$, the surface tension of a liquid is equal to the work (W) required to increase the surface area $(A)$ of the liquid film by unity at constant temperature.
$\therefore W=T \times \Delta A$
Since, surface area of a sphere is $4 \pi R^{2}$ and there are two free surfaces, we have
$W=T \times 8 \pi R^{2}$ ...(i)
and volume of sphere $=\frac{4}{3} \pi R^{3}$
i.e, $V=\frac{4}{3} \pi R^{3}$
$\Rightarrow R=\left(\frac{3 V}{4 \pi}\right)^{1 / 3}$ ...(ii)
From Eqs. (i) and (ii), we get
$\Rightarrow W \propto V^{2 / 3}$
$W=T \times 8 \pi \times\left(\frac{3 V}{4 \pi}\right)^{2 / 3}$
$\therefore W_{1} \propto V_{1}^{2 / 3}$ and $W_{2} \propto V_{2}^{2 / 3}$
$\therefore \frac{W_{2}}{W_{1}}=\left(\frac{2 V_{1}}{V_{1}}\right)^{2 / 3}$
$\Rightarrow W_{2}=2^{2 / 3} W_{1}=4^{1 / 3} W$