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Step-by-Step Solution
Step 1: Understand the Physical Situation
An air bubble rises from the bottom of a lake to the surface. As it rises, the pressure around it decreases and the bubble’s volume increases from radius $r$ to $5r/4$. We are to determine the approximate depth of the lake, given that atmospheric pressure is equivalent to a 10 m water column.
Step 2: Note the Key Assumptions
• Atmospheric pressure, $P_0$, corresponds to a 10 m column of water, i.e., $P_0 = 10\,\rho g$.
• Surface tension and temperature effects are ignored (so any excess pressure due to surface tension is taken as negligible).
• The relationship for a bubble expanding in an ideal manner is given by the ideal gas relation $P V = \text{constant}$ (assuming the amount of gas in the bubble remains the same).
Step 3: Write Down the Pressure-Volume Relation
At the bottom of the lake (depth $h$), pressure is:
$$
P_\text{bottom} = P_0 + \rho g\,h,
$$
and the radius of the bubble is $r$. Hence, its volume is
$$
V_\text{bottom} = \frac{4}{3}\pi r^3.
$$
At the surface of the lake, the pressure is just atmospheric:
$$
P_\text{top} = P_0,
$$
and the radius of the bubble is $5r/4$. So its volume is
$$
V_\text{top} = \frac{4}{3}\pi\left(\frac{5r}{4}\right)^3 = \frac{4}{3}\pi \frac{125\,r^3}{64}.
$$
Given that the bubble expansion follows $P_1 V_1 = P_2 V_2$, we have:
$$
\bigl(P_0 + \rho g\,h\bigr)\left(\frac{4}{3}\pi r^3\right)
=
\bigl(P_0\bigr)\left(\frac{4}{3}\pi \frac{125\,r^3}{64}\right).
$$
Step 4: Simplify and Solve for h
Cancel the common factor $\frac{4}{3}\pi r^3$ on both sides, yielding:
$$
P_0 + \rho g\,h = P_0 \times \frac{125}{64}.
$$
But $P_0 = 10\,\rho g,$ thus
$$
10\,\rho g + \rho g\,h = 10\,\rho g \times \frac{125}{64}.
$$
Divide through by $\rho g$ to eliminate common factors:
$$
10 + h = 10 \times \frac{125}{64}.
$$
So
$$
10 + h = \frac{1250}{64}.
$$
Subtract 10 from both sides:
$$
h = \frac{1250}{64} - 10.
$$
Perform the calculation:
$$
\frac{1250}{64} \approx 19.53 \quad\text{(this is slightly incorrect in isolation; see correct step below)},
$$
but more precisely:
$$
\frac{1250}{64} = 19.53125,\quad
19.53125 - 10 = 9.53125 \approx 9.5\,\text{m}.
$$
Hence,
$$
h \approx 9.5\,\text{m}.
$$
Step 5: Conclude the Depth
Therefore, the depth of the lake is approximately 9.5 m.
Reference Illustration from the Provided Solution
