The U-tube device connected to the tank in Fig. 13-6 is called a manometer. As you can see, the mercury in the tube stands higher in one side than the other. What is the pressure in the tank if atmospheric pressure is $76 \mathrm{~cm}$ of mercury? The density of mercury is $13.6 \mathrm{~g} / \mathrm{cm}^{3}$.
$$
\begin{aligned}
\text { Pressure at } A_{1} &=\text { Pressure at } A_{2} \\
(P \text { in } \operatorname{tank})+(P \text { due to } 5 \mathrm{~cm} \text { mercury }) &=(P \text { due to atmosphere }) \\
P+(0.05 \mathrm{~m})\left(13600 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right) &=(0.76 \mathrm{~m})\left(13600 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)
\end{aligned}
$$
from which $P=95 \mathrm{kPa}$.
Or, more simply perhaps, we could note that the pressure in the tank is $5.0 \mathrm{~cm}$ of mercury lower than atmospheric. So the pressure is $71 \mathrm{~cm}$ of mercury, which is $94.6 \mathrm{kPa}$.