Question

At a location where the acceleration due to gravity is 9.807 m/s2, the atmospheric pressure is 9.891 × 104 Pa. A barometer at the same location is filled with an unknown liquid. What is the density of the unknown liquid if its height in the barometer is 1.163 m?

Answer 1

Answer:

$8616.7468 \ kg/m^3$

Explanation:

Pressure is measured is $p=\rho gh$ here p is pressure $\rho$ is density and h is height

We have given pressure $p=9.891\times 10^4\ Pa$ acceleration due to gravity $g=9.9870\ m/sec^2$ height =1.163 m

$\rho =\frac{p}{gh}=\frac{9.891\times 10^4}{9.870\times 1.163}=8616.7468 \ kg/m^3$

Answer 2

Answer:

Density is 8669.44kg/m^3

Explanation:

At a location where the acceleration due to gravity is 9.807 m/s2, the atmospheric pressure is 9.891 × 104 Pa. A barometer at the same location is filled with an unknown liquid. What is the density of the unknown liquid if its height in the barometer is 1.163 m?

Solution

Pressure is the ratio of force per unit area. A barometer can used to derive the atmospheric pressure P in terms of the height h of the unknown liquid column.

Density is the ratio of mass to volume

Pressure is defined as

P=force/unit area

P=m*g/(A).......... 1

Recall that density=mass/volume.............2

Volume =A*h. Area *height

From equation 2

D=m/Ah

m=DAh...............3

Substituting equ 3 into 1

P=DAh *g/A

P=Dgh

9.891 × 10^4=D*1.163*9.81

D=9.891*10^(4)/(1.163*9.81)

D=8669.44kg/m^3

Density is 8669.44kg/m^3