An air bubble of volume V0 is released by a fish at a depth h in a lake. The bubble rises to the surface. Assume constant temperature and standard atmospheric pressure P above the lake. The volume of the bubble just before touching the surface will be (density of water is ρ)

The correct option is V₀(1+ ρgh/P).

To find the volume of the bubble just before it reaches the surface, we can apply the principles of fluid pressure and gas laws.

Let's define the parameters at the initial depth and at the surface of the lake:
At depth $h$:
- The pressure acting on the bubble, $P_1$, is the sum of the atmospheric pressure $P$ above the lake and the hydrostatic pressure exerted by the water column of depth $h$:
$P_1=P+\rho gh$
- The initial volume of the bubble is $V_1=V_0$.

At the surface of the lake:
- The pressure acting on the bubble is only the standard atmospheric pressure:
$P_2=P$
- Let the final volume of the bubble just before touching the surface be $V_2$.

Since the temperature of the lake is assumed to remain constant, we can apply Boyle's Law, which states that for a fixed quantity of gas at constant temperature, the pressure of the gas is inversely proportional to its volume:
$P_1{V}_1=P_2{V}_2$

Substituting the values into the equation:
$(P+\rho gh)V_0=P{V}_2$

Now, we solve for the final volume $V_2$:
$V_2=\frac{(P+\rho gh)V_0}{P}$
By dividing each term in the numerator by the denominator $P$, we get:
$V_2=V_0(1+\frac{\rho gh}{P})$