1. Pressure inside a soap bubble

For a soap bubble of radius $r$, the excess pressure inside it is

$\Delta P = \frac{4T}{r}$

where $T$ is the surface tension.

Hence, pressure inside the bubble is

$P_{\text{in}} = P_0 + \frac{4T}{r}$

where $P_0$ is atmospheric pressure.

2. Volume increases at a constant rate

Given that volume increases linearly with time,

$\frac{dV}{dt} = \text{constant}$

So,

$V = kt + V_0$

for some constant $k$.

Now for a spherical bubble,

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

Thus,

$r \propto V^{1/3} \propto (kt+V_0)^{1/3}$

Therefore,

$P_{\text{in}} = P_0 + \frac{4T}{r} = P_0 + \frac{4T}{(kt+V_0)^{1/3}}\times (\text{constant factor})$

So the excess pressure decreases with time.

3. Shape of the graph

Since

$P_{\text{in}} - P_0 \propto (kt+V_0)^{-1/3}$

this means:

• pressure decreases with time,
• but not linearly,
• the decrease is rapid initially and then slower later.

So the graph is a decreasing curve that flattens with time, approaching atmospheric pressure asymptotically.

4. Correct option

Thus the correct graph is the one showing a nonlinear decreasing curve approaching a constant value.

So, the correct answer is:

$\boxed{\text{B}}$

5. Comparison with stored answer

Stored correct answer: $\text{B}$

Our derived answer: $\text{B}$

They match.