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Q75PE

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College Physics (Urone)
Found in: Page 398

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Short Answer

Pressure in the spinal fluid is measured as shown in Figure \[{\rm{11}}{\rm{.43}}\]. If the pressure in the spinal fluid is \[{\rm{10}}{\rm{.0 mm Hg}}\]: (a) What is the reading of the water manometer in cm water? (b) What is the reading if the person sits up, placing the top of the fluid \[{\rm{60 cm}}\] above the tap? The fluid density is \[{\rm{1}}{\rm{.05 g/mL}}\].

(a) The height of the water is \[{\rm{13}}{\rm{.6 cm}}\] of water.

(b) The new pressure will be \[{\rm{76}}{\rm{.6 cm}}\] of water.

See the step by step solution

Step by Step Solution

Step 1: Conceptual Introduction

Fluid statics, often known as hydrostatics, is a branch of fluid mechanics that investigates the state of balance of a floating and submerged body, as well as the pressure in a fluid, or imposed by a fluid, on an immersed body.

Step 2: Given data

The density of the stomach fluid is \[{\rm{1}}{\rm{.05 g/mL}}\].

The density in kg/m3 = \[{\rm{1}}{\rm{.05 \times 1}}{{\rm{0}}^{\rm{3}}}{\rm{ kg/}}{{\rm{m}}^{\rm{3}}}\].

Step 3: Reading of water manometer (Height of the water):

(a) By putting all the value into the equation we get:

\[\begin{array}{l}{P_{Hg}} = {P_w}\\{h_{Hg}}{\rho _{Hg}}g = {h_w}{\rho _w}g\\{h_w} = \frac{{{h_{Hg}}{\rho _{Hg}}}}{{{\rho _w}}}\\{h_w} = \frac{{1 \times 13600}}{{1000}}\end{array}\]

\[{h_w} = 13.6\,cm\,of\,{H_2}O\]

Hence, the height of the water is \[{\rm{13}}{\rm{.6 cm}}\] of water.

Step 4: Pressure when person sits up:

(b) The additional pressure of the height of the fluid when the new pressure will be the sum of the pressure when they are lying down and the pressure when they are standing up.

\(\begin{array}{l}{P_n} = {P_l} + {P_s}\\{P_n} = {P_l} + {h_s}{\rho _s}g\end{array}\)

Here again we need to convert the pressure into cm of water so:

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