Question

In the manometer as shown, $2$ immiscible fluids - mercury $(\rho =13600{\text{kg/m}}^3)$ and water $(\rho =1000{\text{kg/m}}^3)$ are used as manometric fluids. The water end is exposed to the atmosphere $\left(100\text{kPa}\right)$ and the mercury end is exposed to a gas. At this position, the interface between the fluids is at the bottom most point of the manometer. The value of $h$ is found to be $9.45\text{m}$. The height of the mercury column is given to be $75\text{cm}$. Find the gauge pressure of the gas. $(g=9.8{\text{m/s}}^2)$

• A. $100\text{kPa}$
• B. $50\text{kPa}$
• C. $200\text{kPa}$
• D. $0\text{kPa}$

Answer 1

The correct option is D $0\text{kPa}$

Height of water column $=0.75+9.45$
$=10.2\text{m}$
As we know,
$P_g$ is the absolute pressure of the gas
Gauge pressure of the gas $=P_g-P_a$

Pressures at bottommost point will be the same. Hence, equating pressure at the bottommost point
$P_a+{\rho }_w\times g\times \left(10.2\right)=P_g+{\rho }_m\times g\times \left(0.75\right)$
$\Rightarrow P_a+{10}^3\times g\times \left(10.2\right)=P_g+13.6\times {10}^3\times g\times 0.75$
$\Rightarrow P_g-P_a=0\text{kPa}$

$\therefore$ Gauge pressure of the gas $=P_g-P_a=0\text{kPa}$