If the work done in blowing a bubble of volume V is W, then the work done in blowing the bubble of volume 2V from the same soap solution will be

To find the work done in blowing a soap bubble, we need to understand the relationship between the work done, the surface tension of the soap solution, and the surface area of the bubble.

A soap bubble has two free surfaces (an inner surface and an outer surface) in contact with air. Therefore, the work done $W$ in blowing a bubble of radius $R$ is equal to the surface tension $T$ multiplied by the total increase in surface area:
$W=2·T·A$
where $A=4\pi R^2$ is the surface area of a sphere of radius $R$.
Substituting $A$ into the equation gives:
$W=8\pi R^{2}T$

Now, let's relate the radius $R$ to the volume $V$ of the bubble. The volume of a spherical bubble is given by:
$V=\frac{4}{3}\pi R^3$
From this, we can express the radius $R$ in terms of the volume $V$:
$R^3=\frac{3V}{4\pi }$
Taking the cube root on both sides:
$R={\left(\frac{3V}{4\pi }\right)}^{\frac{1}{3}}$

Now, substitute this expression for $R$ back into the formula for the work done $W$:
$W=8\pi T{\left[{\left(\frac{3V}{4\pi }\right)}^{\frac{1}{3}}\right]}^2$
$W=8\pi T{\left(\frac{3V}{4\pi }\right)}^{\frac{2}{3}}$
Since the surface tension $T$ of the same soap solution remains constant, we can see that the work done $W$ is directly proportional to $V^{\frac{2}{3}}$:
$W\propto V^{\frac{2}{3}}$

Let $W_1=W$ be the work done to blow a bubble of volume $V_1=V$.
Let $W_2$ be the work done to blow a bubble of volume $V_2=2V$.
Using the proportionality relation:
$\frac{W_2}{W_1}={\left(\frac{V_2}{V_1}\right)}^{\frac{2}{3}}$
Substitute the values of $V_1$ and $V_2$:
$\frac{W_2}{W}={\left(\frac{2V}{V}\right)}^{\frac{2}{3}}$
$\frac{W_2}{W}=2^{\frac{2}{3}}$
$W_2={\left(2^2\right)}^{\frac{1}{3}}W$
$W_2=4^{\frac{1}{3}}W=∛4W$

Therefore, the work done in blowing the bubble of volume 2V is $∛4W$.