Physics - W.E.P. Question with Solution | TestHub

The work done in blowing a soap bubble of radius $R$ and surface tension $T$ is given by:

$W=8\pi R^{2}T$

The volume of the soap bubble is:

$V=\frac{4}{3}\pi R^3$

From this, the radius $R$ in terms of volume $V$ is:

$R={\left(\frac{3V}{4\pi }\right)}^{\frac{1}{3}}$

Substituting $R$ into the expression for work done:

$W=8\pi T{\left(\frac{3V}{4\pi }\right)}^{\frac{2}{3}}$

This implies that the work done is directly proportional to $V^{\frac{2}{3}}$:

$W\propto V^{\frac{2}{3}}$

For a new volume $V^{\prime }=2V$, the new work done $W^{\prime }$ is:

$\frac{W^{\prime }}{W}={\left(\frac{2V}{V}\right)}^{\frac{2}{3}}=2^{\frac{2}{3}}=(2^2{)}^{\frac{1}{3}}=4^{\frac{1}{3}}$

$W^{\prime }=W(4{)}^{\frac{1}{3}}$

Answer: $\mathrm{W}(4{)}^{\frac{1}{3}}$