1. Use Boyle’s law

Since temperature is constant and surface tension is ignored, $P_1 V_1 = P_2 V_2$

For a spherical bubble, $V \propto r^3$

Given that the radius changes from $r$ to $\frac{5r}{4}$, $\frac{V_2}{V_1} = \left(\frac{5r/4}{r}\right)^3 = \left(\frac{5}{4}\right)^3 = \frac{125}{64}$

So, $P_1 V_1 = P_2 V_2 \implies P_1 = P_2 \cdot \frac{V_2}{V_1} = P_2 \cdot \frac{125}{64}$

2. Pressures at bottom and surface

Let the depth of the lake be $h$ m.

• At the surface, pressure on the bubble is atmospheric pressure: $P_2 = P_{\text{atm}}$
• At the bottom, pressure is: $P_1 = P_{\text{atm}} + \rho g h$

Given atmospheric pressure equals pressure due to a $10$ m water column, $P_{\text{atm}} = \rho g \cdot 10$

Hence, $P_1 = \rho g(10+h), \qquad P_2 = \rho g \cdot 10$

3. Apply Boyle’s law in terms of water-column heights

$\rho g(10+h) = \rho g \cdot 10 \cdot \frac{125}{64}$

Cancel $\rho g$: $10+h = 10\cdot \frac{125}{64}$

$10+h = \frac{1250}{64} = 19.53125$

$h = 19.53125 - 10 = 9.53125 \text{ m}$

4. Approximate value

$h \approx 9.5 \text{ m}$

5. Option check

• A: $11.2$ m ✗
• B: $8.7$ m ✗
• C: $9.5$ m ✓
• D: $10.5$ m ✗

Therefore, the correct answer is Option C.