Calculate the absolute pressure at the bottom of a lake at a depth of 32.8 m. Assume the density of the water is 1 x 10^3 kgm-3 and the air above is at a pressure of 101.3 kPa.
[Take g = 9.8 ms-2]
[Take g = 9.8 ms-2]
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Correct Answer: Option A
Explanation:
h=32·8m, p= \(1×10^3kgm^-3\)
g=9·8 \(ms^-2\) , \(P_a\)=101·3KPa
\(P_T\)= ?
P=hpg
⇒\(32·8×1×10^3 × 9·8\)=321440 \(P_a\)
⇒321·44KPa
⇒ \(P_T\) =321.44+101.3
;\(P_T\)=422·7KPa
h=32·8m, p= \(1×10^3kgm^-3\)
g=9·8 \(ms^-2\) , \(P_a\)=101·3KPa
\(P_T\)= ?
P=hpg
⇒\(32·8×1×10^3 × 9·8\)=321440 \(P_a\)
⇒321·44KPa
⇒ \(P_T\) =321.44+101.3
;\(P_T\)=422·7KPa