When an air bubble of radius r rises from the bottom to the surface of a lake, its radius becomes $$\frac{5r}{4}$$. Taking the atmospheric pressure to be equal to 10 m height of water column, the depth of the lake would approximately be (ignore the surface tension and the effect of temperature):

The bubble is a pocket of gas. During its slow ascent we can safely assume that the temperature of this gas does not change, so the gas obeys Boyle’s law, which states first: $$P_1V_1 = P_2V_2,$$ where $$P_1,\,V_1$$ are the pressure and volume at the lake bottom and $$P_2,\,V_2$$ are the corresponding quantities at the surface.

We denote by $$h$$ the depth of the lake. Let $$\rho$$ be the density of water and $$g$$ the acceleration due to gravity. At the bottom, the total pressure acting on the bubble is the sum of the atmospheric pressure and the hydrostatic pressure of the water column above it. Hence we have

$$P_1 = P_{\text{atm}} + \rho g h.$$

At the surface the bubble is exposed only to the atmospheric pressure, so

$$P_2 = P_{\text{atm}}.$$

The problem tells us that the atmospheric pressure is equivalent to the pressure of a 10 m column of water, that is

$$P_{\text{atm}} = \rho g \times 10\;\text{m} = 10\rho g.$$

Now we examine the volumes. The volume of a sphere varies as the cube of its radius. If the initial radius is $$r$$ and the final radius is $$\dfrac{5r}{4},$$ then

$$\frac{V_2}{V_1} = \left(\frac{5r/4}{\,r}\right)^3 = \left(\frac{5}{4}\right)^3 = \frac{125}{64}.$$

With Boyle’s law $$P_1V_1 = P_2V_2,$$ we substitute the pressures and the volume ratio:

$$\bigl(P_{\text{atm}} + \rho g h\bigr)\,V_1 \;=\; P_{\text{atm}}\,V_2 \;=\; P_{\text{atm}}\left(\frac{125}{64}V_1\right).$$

The factor $$V_1$$ appears on both sides, so it cancels out, leaving

$$P_{\text{atm}} + \rho g h = P_{\text{atm}}\left(\frac{125}{64}\right).$$

We now isolate $$\rho g h$$:

$$\rho g h = P_{\text{atm}}\left(\frac{125}{64} - 1\right) = P_{\text{atm}}\left(\frac{125 - 64}{64}\right) = P_{\text{atm}}\left(\frac{61}{64}\right).$$

Because $$P_{\text{atm}} = 10\rho g,$$ we substitute this value:

$$\rho g h = 10\rho g \times \frac{61}{64}.$$

Both $$\rho$$ and $$g$$ cancel out, leaving a simple numerical result:

$$h = 10 \times \frac{61}{64} = \frac{610}{64}\,\text{m} \approx 9.53\,\text{m}.$$

The depth of the lake is therefore very close to 9.5 m, which matches the numerical choice in the options.

Hence, the correct answer is Option 4.