Question

The heights of mercury in a manometer are $2cm$ and $8cm$. Find the pressure of the gas in the cylinder? $P_{atm}={10}^{5}N{m}^{-2}$

The density of mercury, ${\rho }_m$ = $13600kg{m}^{-3}$.
Acceleration due to gravity, $g=10m{s}^{-2}$

• A. $1.08\times {10}^{5}N{m}^{-2}$
• B. $2.08\times {10}^{5}N{m}^{-2}$
• C. $0.56\times {10}^{5}N{m}^{-2}$
• D. $3.48\times {10}^{6}N{m}^{-2}$

Answer 1

The correct option is A

$1.08\times {10}^{5}N{m}^{-2}$

Let the pressure of gas contained in the chamber be $P_g$.

Consider the lowest point of U-tube.
Pressure from the right column of U-tube is:
$P_1=P_{atm}+{\rho }_{m}g{h}_1$
Pressure from the left column of U-tube is:
$P_2=P_g+{\rho }_{m}g{h}_2$

According to Pascal's Law,
$P_1=P_2$
$P_{atm}+{\rho }_{m}g{h}_1=P_g+{\rho }_{m}g{h}_2$
$P_g=P_{atm}+{\rho }_{m}g(h_1-h_2)$
$P_g={10}^5+$$[13600\times 10\times (\frac{8}{100}-\frac{2}{100})]$
$P_g\approx 1.08\times {10}^{5}Pa$