physics-class-11-unit-10-chapter-02-mechanical-properties-of-fluids-2-lecture-2-5-xqox54rispm

### <Success style="font-size:25px"> Mechanical Properties Of Fluids 2 L-2 </Success>
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- The surface of water in a Storage tank is 20 $\{m}$ above the water tap in the kitchen of a house. Calculate the pressure at the tap.

- Given: density of water = $1 \times 10^3 \{kg} / \{m}^3=\rho_{\text {waler }}$

- $\Delta P = \rho g h = (1 \times 10^{3} kg/m^3)(98 m s^{-2}) (20 m)$

- = $1.96 \times 10^{5} N/m^2$

- $ = 1.96 \times 10^5 Pa$

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<footer style = "font-size: 18px">Mechanical Properties of Fluids $\rightarrow$ Pressure due to Liquid Column $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Problem $\rightarrow$ Problem </footer>

### <Success style="font-size:25px"> Mechanical Properties Of Fluids 2 L-2 </Success>
### <primary style="font-size:24px"> Problem </primary>
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- What is the difference in blood pressure between top of the head and the bottom of the feet of 1.60 m tall person standing vertically?

- Density of blood is 1060 $kg/m^3$.

- Pressure difference in given by

- $\Delta P = \rho g h $

- = $(1060 kg/m^3)(98 m s^{-2}) (1.60 m)$

- = 166208 $N/m^2$

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<footer style = "font-size: 18px">Pressure due to Liquid Column $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Problem $\rightarrow$ Problem </footer>

### <Success style="font-size:25px"> Mechanical Properties Of Fluids 2 L-2 </Success>
### <primary style="font-size:24px"> Problem </primary>
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- When you run up a  tall hill or run down the hii quickly the ears ' pop'. Suppose this did not happen, What would be the force in the eardrum of area $0.5 \{cm}^2$ if a change in attitude (height) of $1000 \{m}$ takes place? (Given : $P_{air} =1.29 \{kg} / \{m}^3$.)

- P = h $\rho$ g  = (1000 m )($1.29 kg / m^3$) ($9.8 m s^{-2}$)

- = 12642 $N/m^2$
  
- F =P A = (12642 $N/m^2$) $\times (0.5 \times 10^{-4} m^2)$

- F = 6.32 N
   
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<footer style = "font-size: 18px">Problem $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Problem $\rightarrow$ Problem </footer>

### <Success style="font-size:25px"> Mechanical Properties Of Fluids 2 L-2 </Success>
### <primary style="font-size:24px"> Problem </primary>
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- Determine the variation in pressure in the earth's atmosphere as a function of the height $y$ above sea level,
- assuming ' $g$ ' be constant and that the density of air is proportional to pressure. Also at what elevates the air pressure equal to half the pressure at sea level ? $(P_{atm}  = 1.013 \times 10^{5} N/m^3)$

- $\frac{\rho}{\rho_0} = \frac{P}{P_0} \quad \cdot$ eqn(1)

- $\frac{dP}{dy} = - \rho g \quad \cdot $ eqn(2)

- $\rho = \frac{P}{P_0} \times{\rho_0} \Rightarrow \frac{dp}{dy} = -\frac{P}{P_0} \rho_0 g \quad \cdot $ eqn(3)

- $\int_{P_0}^P = - \int_0^y \frac{\rho_0 g}{P_0} dy$ $\Rightarrow  ln \frac{P}{P_0} = \frac{\rho_0 g}{P_0} y $

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<footer style = "font-size: 18px">Problem $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Problem $\rightarrow$ Atmospheric Pressure and Gauge Pressure </footer>

### <Success style="font-size:25px"> Mechanical Properties Of Fluids 2 L-2 </Success>
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- $P=P_0 e^{-(\frac{\rho_0 g}{P_0}) y}$

- $\frac{\rho_0 g}{P_0}$ = $\frac{(1.29 {kg} / \{m}^3) \times(98 \{ms}^{-2})}{(1.013 \times 10^5 {N} / \{m}^2} $ = $1.25 \times 10^{-4} m^{-1}$

- $P = \frac{P_0}{2}$

- $P = P_0 e^{(1.25 \times 10^{-4} m^{-1})} y$

- $y = \frac{ln 2}{1.25 \times 10^{-4} m^{-1}}$,  (ln 2 = 0.693)

- = $\frac{0.693}{(1.25 \times 10^{-4} m^{-1})} m $

- = 5550 m = 18,000 ft

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<footer style = "font-size: 18px">Problem $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Atmospheric Pressure and Gauge Pressure $\rightarrow$ Measurement of Pressure </footer>

### <Success style="font-size:25px"> Mechanical Properties Of Fluids 2 L-2 </Success>
### <primary style="font-size:24px"> Atmospheric Pressure and Gauge Pressure </primary>
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- Atmospheric Pressure: $ P_0=1.013 \times 10^5 \{N} / \{m}^2 $

- $ 1 bar = 1 \times 10^5 \{N} / \{m}^2 $

- **Tire gauge**

- $P=P_{atm}+P_{\text{gauge pressure}}$

- If a tire gauge measure a pressure of 200 kPa then the partial pressure is 200 kPa + 100 kPa.

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<footer style = "font-size: 18px">Problem $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Atmospheric Pressure and Gauge Pressure </span> $\rightarrow$ Measurement of Pressure $\rightarrow$ Different Units of Pressure </footer>

### <Success style="font-size:25px"> Mechanical Properties Of Fluids 2 L-2 </Success>
### <primary style="font-size:24px"> Measurement of Pressure </primary>
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- $P = \rho_0 + \rho g h $

- mm Hg =$(13.6 \times 10^3 \{g} / \{m}^3) (9.8 \{m} / \{s}^2)$ $(10^{-3} \{m}) $

- =$133 {N} / {m}^2 $= 1 torr

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![alt text](https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/98082d3c-1aac-4f66-bed6-03697b1bd900/public)

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<footer style = "font-size: 18px">Problem $\rightarrow$ Atmospheric Pressure and Gauge Pressure $\rightarrow$ <span style = "color:blue"> Measurement of Pressure </span> $\rightarrow$ Different Units of Pressure $\rightarrow$ Mercury Barometer </footer>

### <Success style="font-size:25px"> Mechanical Properties Of Fluids 2 L-2 </Success>
### <primary style="font-size:24px"> Different Units of Pressure </primary>
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- $1 atm = 1.013 \times 10^5 \{N} / \{m}^2 $ = $1.013 \times 10^5 \{Pa}=101.3 \{kPa} $
 
- $1 bar = 1 \times 10^5 \{N} / \{m}^2 $

- =$1.013 \text { bar }=76 \{cm} \text { of } \{Hg}$
 
 - = 760 mm of Hg = 760 torr
 
 - = 1.03 $\times 10^4$ mm of$ {H_2 O}$ at $4^0 C$
 
- **Blood pressure** 120/80, systolic pressure is 120 mm Hg and diastolic pressure is 80 mm Hg.
 
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<footer style = "font-size: 18px">Atmospheric Pressure and Gauge Pressure $\rightarrow$ Measurement of Pressure $\rightarrow$ <span style = "color:blue"> Different Units of Pressure </span> $\rightarrow$ Mercury Barometer $\rightarrow$ Thank You </footer>