Question

The heights of mercury surface in the two arms of the manometer shown in figure are $h_2$ and $h_1$, where $(h_1>h_2)$ and atmospheric pressure is equal to $P_a$. Pressure of the gas in the cylinder is equal to,

• A. $P_a+{\rho }_{Hg}\times g(h_1-h_2)$
• B. $P_a+{\rho }_{Hg}\times g\left(h_1\right)$
• C. $P_a+{\rho }_{Hg}\times g\left(h_2\right)$
• D. $P_a-{\rho }_{Hg}\times g(h_1-h_2)$

Answer 1

The correct option is A $P_a+{\rho }_{Hg}\times g(h_1-h_2)$

As $h_1>h_2$, the liquid levels in the two arms will be as shown in the diagram.

The liquid levels, in both of the arms of the manometer, are in equilibrium. So the pressure on both arms of the manometer must be equal.
$P_{gas}+{\rho }_{Hg}\times h_2\times g=P_a+{\rho }_{Hg}\times h_1\times g$

$\Rightarrow P_{gas}=P_a+{\rho }_{Hg}\times g(h_1-h_2)$

Hence, $\left(A\right)$ is the correct answer.

$\begin{array}{c}\text{Why this Question?}\\ \text{To solve manometer questions try equating the pressure at the bottommost point by considering each limb seperately.}\end{array}$