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(D) At depth $h$,the total pressure $P_1$ is given by $P_1 = P + \rho gh$,where $P$ is the atmospheric pressure,$\rho$ is the density of water,and $g$

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$V_0(1 + \frac{\rho gh}{P})$

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(D) At depth $h$,the total pressure $P_1$ is given by $P_1 = P + \rho gh$,where $P$ is the atmospheric pressure,$\rho$ is the density of water,and $g$ is the acceleration due to gravity.At the surface,the pressure $P_2$ is equal to the atmospheric pressure $P$.Since the temperature is constant,we apply Boyle's Law: $P_1 V_1 = P_2 V_2$.Here,$V_1 = V_0$ and $V_2 = V$ (the volume at the surface).Substituting the values: $(P + \rho gh) V_0 = P V$.Solving for $V$: $V = V_0 \frac{(P + \rho gh)}{P} = V_0 (1 + \frac{\rho gh}{P})$.

$A$ bubble of volume $V_0$ is released from a depth $h$ in a lake. If the atmospheric pressure is $P$ and the temperature is assumed to be constant,what is the new volume of the bubble when it reaches the surface? (Density of water is $\rho$)

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