Question

Pressure 3m below the free surface of a liquid is 15KN/m² in excess of atmosphere pressure. Determine its density and specific gravity. [g=10m/s²]​

Answer 1

$\huge{\underline{\sf{\red{Answer}}}}$

Given:

$P_\dash$Pressure = 15KN/m²

Height = 3 metre

$\rho_{water}=10^3$

$\rho =Absolute\:Pressure= 10^5$

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Formula

• $P = P_\not +\rho gh$
• Abosulte Gravity = $\rho_{s} = \dfrac{\rho}{\rho_{water}}$

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Solution:

KiloNewton $\implies$ Newton

1 KN = 1000N

$\therefore$ 15KN = 15 x 10³

Putting the Values given in the formula

$\implies 10^5=15\times 10^3 + \rho(10)(3)\\ \\ 30\rho= 15\times 10^3-10^5 \\ \\ \rho= \dfrac{\cancel{15}\times 10^2 }{\cancel{3}}\\ \\ \rho=5\times(10^2)\\ \\ \huge{\leadsto}{\boxed{500kg}}$

Now, By the Formula

$Specific \:Gravity = \rho_{s} = \dfrac{\cancel{500}}{\cancel{10^3}}=\dfrac{1}{2}\\ \\ \huge{\leadsto}{\boxed{0.5}}$

Answer 2

Explanation:

Given:

Pressure is in excess of the atmospheric pressure. Therefore,

P = 15 kN/m^2 = 15000 Pa

now,

since we know that:

P = hdg

15000 = 3 x d x 9.8

d = 510.204 kg/m^3