What is the absolute pressure at a point 3 { m} below the free surface of a liquid having a density of 1.53 \\times10^{3} {kg} \/ {m}^{3} if the atmospheric pressure is equivalent to 750 {mm} of mercury? The specific gravity of mercury is 13.6 and density of water is 1000 {kg} \/ {m}^{3}.

Correct Answer is : 145090N∕m2

SSC JE Mechanical Engineering 27 Oct 2020 Shift 2 Papers

Question : 138 of 200

Marks: +1, -0

What is the absolute pressure at a point

3 \ { m}

below the free surface of a liquid having a density of

1.53 \×10^{3} {kg} \/ {m}^{3}

if the atmospheric pressure is equivalent to

750 {mm}

of mercury? The specific gravity of mercury is

13.6

and density of water is

1000 {kg} \/ {m}^{3}

Solution:

{h}=3 {m}, \ρ_{\text "liquid "}=1.53 \×10^{3} {kg} \/ {m}^{3}

{h}_{{atm}}=750 {mm}

of mercury

({Hg})=0.75 {m}

specific gravity

, {S}_{{Hg}}=13.6, \ρ=13.6 \×1000 {kg} \/ {m}^{3}

{P}_{{atm}}=\ρg {h}_{{atm}}=13.6 \×1000 \×9.81 \×0.75

=100,062 {N} \/ {m}^{2}

{P}_{{abs}}={P}_{{atm}}+{P}_{\text "gauge "}

{P}_{\text "gauge "}=(\ρg {h})_{\text "liquid "}

=1.53 \×10^{3} \×9.81 \×3=45027.9 {N} \/ {m}^{2}

{P}_{{abs}}={P}_{{atm}}+{P}_{\text "gauge "}

=100062+45027.9=145090 {N} \/ {m}^{2}

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