Question

Figure below shows U-Tube manometer used to measure absolute pressure of the gas in the bulb centre "O". Mercury and water are immiscible liquids. At this unknown pressure of gas at O, the interface between the liquids is at the bottom most point of the U-Tube. Ignore the size of tube and radius of curvature of liquids (meniscus effects & capillarity effects). The height of mercury column is given as 75 cm. The pressure of gas at O is : (Neglect gas column since gas is of lower density).

• A. 89.5 kPa vaccum
• B. 10.5 kPa absolute
• C. both of these
• D. none of these

Answer 1

The correct option is C both of these

Height of water column

$=75+50=125cm$

Using principle of manometry,

Equating the pressure at the bottom most point,

$$P_{O,Abs}+{\rho }_{Hg}.g.h_{Hg}+{\rho }_{gas}.g.h_{gas}$$
$={\rho }_w.g.h_w+P_{Atm}$

$\Rightarrow P_{O,Abs}+13600\times 10\times 0.75+Neglected$

$=1000\times 10\times 1.25+100000$

$\therefore P_{O,Abs}=10500N/m^2=10.5kpa\left(abs\right)$

(or)

$P_{O,gauge}=P_{O,Abs}-P_{Atm}$

$=10500-100000$

$=-89500N/m^2$

$=-89.5kpagauge$

$\left(vacuumorsuction\right)$