1. Work done in increasing the size of a soap bubble

For a soap bubble, there are two surfaces, so the increase in surface energy is:

$W = 2T\,\Delta A$

Since surface area of a sphere is

$A = 4\pi r^2$

for two surfaces, total area is

$A_{\text{total}} = 2(4\pi r^2)=8\pi r^2$

Hence, work done in expanding the bubble from radius $r_1$ to $r_2$ is

$W = T\,(8\pi r_2^2-8\pi r_1^2)=8\pi T(r_2^2-r_1^2)$

2. Given data

• Initial diameter $=7\text{ cm}$
• Initial radius:

$r_1=\frac{7}{2}=3.5\text{ cm}$

• Work done:

$W=36960\text{ erg}$

• Surface tension:

$T=40\text{ dyne/cm}$

Also, note that in CGS units:

$1\text{ erg} = 1\text{ dyne-cm}$

So units are consistent.

3. Substitute into the formula

$36960=8\pi(40)(r_2^2-r_1^2)$

Using $\pi=\frac{22}{7}$:

$36960=8\times \frac{22}{7}\times 40\,(r_2^2-r_1^2)$

$36960=\frac{7040}{7}(r_2^2-r_1^2)$

So,

$r_2^2-r_1^2=\frac{36960\times 7}{7040}$

$r_2^2-r_1^2=36.75$

Now,

$r_1^2=(3.5)^2=12.25$

Thus,

$r_2^2=36.75+12.25=49$

$r_2=7\text{ cm}$

4. Final answer

The new radius is

$\boxed{7\text{ cm}}$