physics-class-11-unit-10-chapter-01-mechanical-properties-of-fluids-1-lecture-1-5-nnnugkgeixu

### <Success style="font-size:25px"> Mechanical Properties Of Fluids 1 L-1 </Success>
### <primary style="font-size:24px"> State of Matter </primary>
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- 1.Solids

- 2.Liquids

- 3.Gases

- 4.Plasma

- Solid have definite shape and size.
 
- Fluids do not have definite shape and size.

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<footer style = "font-size: 18px"> $\rightarrow$ Mechanical Properties of Fluids $\rightarrow$ <span style = "color:blue"> State of Matter </span> $\rightarrow$ Density $\rightarrow$ Solid </footer>

### <Success style="font-size:25px"> Mechanical Properties Of Fluids 1 L-1 </Success>
### <primary style="font-size:24px"> Solid </primary>
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| Material | $\rho(kg/m^3)$   |
| -------- | --------       |
| iron     | $7.8\times10^3$|
| Aluminium| $.7\times10^3$ |
| Nood(pine)| $0.5\times10^3$|
| Glass    | $2.5\times10^3$|

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<footer style = "font-size: 18px">State of Matter $\rightarrow$ Density $\rightarrow$ <span style = "color:blue"> Solid </span> $\rightarrow$ Liquid $\rightarrow$ Gases </footer>

### <Success style="font-size:25px"> Mechanical Properties Of Fluids 1 L-1 </Success>
### <primary style="font-size:24px"> Liquid </primary>
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| Material    | $\rho (kg/m^3)$ |
| --------    | --------        |
|Water        |$1\times10^3$    |
|Sea walve    |$1.025\times10^3$|
|Meracury     |$13.6\times10^3$ |
|Ethyl Alcohole|$0.79\times10^3$|

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<footer style = "font-size: 18px">Density $\rightarrow$ Solid $\rightarrow$ <span style = "color:blue"> Liquid </span> $\rightarrow$ Gases $\rightarrow$ Problem </footer>

### <Success style="font-size:25px"> Mechanical Properties Of Fluids 1 L-1 </Success>
### <primary style="font-size:24px"> Gases </primary>
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| Material    | $\rho (kg/m^3)$|
| --------    | ----    --- |
|Air          |1.29         |
|Helium       |0.179        |
|Carbondioxide|1.98         |

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<footer style = "font-size: 18px">Solid $\rightarrow$ Liquid $\rightarrow$ <span style = "color:blue"> Gases </span> $\rightarrow$ Problem $\rightarrow$ Specific Gravity </footer>

### <Success style="font-size:25px"> Mechanical Properties Of Fluids 1 L-1 </Success>
### <primary style="font-size:24px"> Problem </primary>
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- Consider that a person whose weight is 60 kgs and his weight is equally distributed by his two legs and each foot has an area of a 600 $cm^2$.

- $W - m g = 60 \times 10 \mathrm{N}=600 \mathrm{N}=\mathrm{F}$

- $P=\frac{F}{2 \times 600 cm^2}=\frac{600 \mathrm{N}}{2 \times 600 \times 10^{-4}}$

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<footer style = "font-size: 18px">Specific Gravity $\rightarrow$ Concept of Pressure $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Pressure due to Static Fluids $\rightarrow$ Pressure due to Static Fluids </footer>

### <Success style="font-size:25px"> Mechanical Properties Of Fluids 1 L-1 </Success>
### <primary style="font-size:24px"> Pressure due to Static Fluids </primary>
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- Fluid exerts a force from all sides and this force acts normally on the surface.

- We have assumed that the fluid is static, so there can't be any component of force which is acting at a given angle to the surface.

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<footer style = "font-size: 18px">Concept of Pressure $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Pressure due to Static Fluids </span> $\rightarrow$ Pressure due to Static Fluids $\rightarrow$ Pressure Due to Liquid Columns </footer>

### <Success style="font-size:25px"> Mechanical Properties Of Fluids 1 L-1 </Success>
### <primary style="font-size:24px"> Pressure due to Static Fluids </primary>
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- Density of the liquid $=\rho$

- $F=m g=V \rho g= hApg$, A is the area of cross section.

- $P=\frac{F}{A}=h \rho g$

- Greater theheight of the object, the pressure will be more.

- $ P \propto h$, for gases, which are largely compressible, there could be significant variation of density with the height.

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

### <Success style="font-size:25px"> Mechanical Properties Of Fluids 1 L-1 </Success>
### <primary style="font-size:24px"> Pressure Due to Liquid Columns </primary>
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- Pressure at a height $y=p^{\prime}$ (distance)

- Pressure at a height $y+ dy = P+dP$ (distance)

- $dF_g = dmg$

- $= \rho g d v=\rho g A d y$

- Net force upgrade = PA – (P + dP) A
 
- $\frac{d P}{d y}=-\rho g$

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![alt text](https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/069a2a08-8b23-4e87-8c37-9de4cb7f9300/public)

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<footer style = "font-size: 18px">Pressure due to Static Fluids $\rightarrow$ Pressure due to Static Fluids $\rightarrow$ <span style = "color:blue"> Pressure Due to Liquid Columns </span> $\rightarrow$ Pressure Due to Liquid Columns $\rightarrow$ Question </footer>

### <Success style="font-size:25px"> Mechanical Properties Of Fluids 1 L-1 </Success>
### <primary style="font-size:24px"> Pressure Due to Liquid Columns </primary>
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- $\int_{P_1}^{P_2} dP=-\int_{y_1}^{y_2} \rho g d y$

- $P_2-P_1=-\rho g\left(y_2-y_1\right)$

- $\begin{aligned} & P_2=P_0 \\ & y_2=0\end{aligned}$

- $\begin{aligned} & P_1=P \\ & y_1=h\end{aligned}$

- <span style="color:blue">Atmospheric pressure</span>
 
- $P_2-P_1=P_0-P$
 
- $P=P_0+\rho g h$
 
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![alt text](https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/220c6ca2-4faa-4320-51f9-de1728563100/public)

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<footer style = "font-size: 18px">Pressure due to Static Fluids $\rightarrow$ Pressure Due to Liquid Columns $\rightarrow$ <span style = "color:blue"> Pressure Due to Liquid Columns </span> $\rightarrow$ Question $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Mechanical Properties Of Fluids 1 L-1 </Success>
### <primary style="font-size:24px"> Question </primary>
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- The surface of water in a storage tank is $20 \mathrm{~m}$ above the water tap in the kitchen of a house. Calculate the pressure at the tap.
 
- Given: Density of water $= 1 \times 10^3kg/m^3 = \rho_{water}$.

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![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/physics-class-11-unit-10-chapter-01-mechanical-properties-of-fluids-1-lecture-1_5-nnnugkgeixu-09.jpg)
		
![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/physics-class-11-unit-10-chapter-01-mechanical-properties-of-fluids-1-lecture-1_5-nnnugkgeixu-09-1.jpg)

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<footer style = "font-size: 18px">Pressure Due to Liquid Columns $\rightarrow$ Pressure Due to Liquid Columns $\rightarrow$ <span style = "color:blue"> Question </span> $\rightarrow$ Solution $\rightarrow$ Question </footer>

### <Success style="font-size:25px"> Mechanical Properties Of Fluids 1 L-1 </Success>
### <primary style="font-size:24px"> Question </primary>
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- What is the difference in blood pressure between lop of the bead and the bottom of the feet of $1.60 \mathrm{m}$   (all person standing vertically)?

- Density of blood is $1060 kg/m^3$
 
- Pressure difference is given by
 
- $\Delta P=\rho g h$
 
 - $=\left(1060kg /m^3\right)\left(9.8 ms^{-2}\right)(1.60 m)$
 
 - $=166208 N / m^2$
 
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<footer style = "font-size: 18px">Question $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Question </span> $\rightarrow$ Question $\rightarrow$ Solution </footer>