Answer
a) $2.72\;\rm m$
b) $1.11\;\rm MPa$
Work Step by Step
As we see in the figure below, initially, at 1, the pressure is the atmospheric pressure and the air volume is the volume of the whole bell.
And finally, the pressure increases, and the volume of the air decreases.
Also, we are told that the air temperature drops after a while to 10$^\circ$C.
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a) To find $h$, we need to find the final volume of the air. We assume that the air inside the bell is an ideal gas.
Thus,
$$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$$
The volume of the air is given by $V=Ah'$ where $A$ is the cross-sectional area of the bell which is constant and $h'$ is the length of the air column inside the bell. Thus, $h'_1=L$ and $h'_2=L-h$
$$\dfrac{P_1 \color{red}{\bf\not} AL}{T_1}=\dfrac{P_2 \color{red}{\bf\not} A(L-h)}{T_2}$$
$$\dfrac{P_1 L}{T_1}=\dfrac{P_2 (L-h)}{T_2}$$
Recall that $P_1=P_a$, and from the figure below we can see that
$P_2=P_a+(H-h)\rho g$
$$\dfrac{P_a L}{T_1}=\dfrac{[P_a+(H-h)\rho g](L-h)}{T_2}$$
$$\dfrac{P_a L}{T_1}=\dfrac{[P_a+H\rho g-h\rho g](L-h)}{T_2}$$
$$\dfrac{P_a LT_2}{T_1}= P_aL-P_ah+H\rho gL-H\rho gh-h\rho gL+h^2\rho g $$
$$\dfrac{P_a LT_2}{T_1}- P_aL-H\rho gL=-P_ah-H\rho gh-h\rho gL+h^2\rho g $$
$$(\rho g)h^2 -(P_a+H\rho g+L\rho g)h\\
-\left(\dfrac{P_a LT_2}{T_1}- P_aL-H\rho gL\right)=0$$
It is a quadratic equation now and we need to plug the known to find the two roots of $h$.
$$(1030\times 9.8)h^2 -([1.013\times 10^5]+[100\times 1030\times 9.8]+[3\times 1030\times 9.8])h-\left(\dfrac{(1.013\times 10^5)(3)(10+273)}{(20+273)}- [(1.013\times 10^5)(3)]-[100\times 1030\times 9.8\times 3]\right)=0$$
Using any software calculator:
$$10094h^2-1140982h+(3.03\times 10^6)=0$$
Hence,
$$h=2.72\;{\rm m}\;\;\;\;\;\;{\rm Or}, \;\;\;\;\;h=110.3$$
The second root is declined since it is meaningless here.
Therefore,
$$h=\color{red}{\bf 2.72}\;\rm m$$
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b) To expel the water from the bell we need to increase the air pressure until it equals the pressure of the water at 100 m below the surface.
Thus, the minimum air pressure needed is
$$P=P_a+H\rho g$$
Plugging the known;
$$P=(1.013\times 10^5)+(100)(1030)(9.8)$$
$$P=\color{red}{\bf1.11\times 10^6}\;\rm Pa$$
