A cylindrical container with a cross-sectional area of 66.2 cm2 holds a fluid of density 856 kg/m3 . At the bottom of the container the pressure is 119 kPa . Assume Pat = 101 kPa A) What is the depth of the fuild?

B) Find the pressure at the bottom of the container after an additional 2.35×10−3 m3 of this fluid is added to the container. Assume that no fluid spills out of the container.

(A)  The depth of the fluid is 2.14 m.

(B)  The new pressure at the bottom of container is 121972 Pa.

Given data:

The cross-sectional area of the container is, .

The density of fluid is, .

The container pressure at bottom is, .

The atmospheric pressure is, .

(A)

The given problem is based on the net pressure on the container, which is equal to the difference between the pressure at the bottom and the atmospheric pressure. Then the expression is,

Here, h is the depth of fluid.

Solving as,

Thus, the depth of the fluid is 2.14 m.

(B)

For an additional volume of to the liquid, the new depth is,

Now, calculate the new pressure at the bottom of the container as,

Thus, we can conclude that the new pressure at the bottom of container is 121972 Pa.

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A. h = 2.15 m

B. Pb' = 122 KPa

Explanation:

The computation is shown below:

a)  Let us assume the depth be h

As we know that

After solving this,  

h = 2.15 m

Therefore the depth of the fluid is 2.15 m

b)

Given that  

height of the extra fluid is

h' = 0.355 m

Now let us assume the pressure at the bottom is Pb'

so, the equation would be

Pb' = 122 KPa

and 1and 1