Mechanical Properties of Fluids

Physics Part II

Physics

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Do intermolecular or inter-atomic forces follow inverse square law?

No. Intermolecular and inter-atomic forces do not obey the inverse square law.

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Due to capillary action, the liquid rises into the tube. The angle of contact of liquid with tube is _________

Acute

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What is hydraulic lift? What is its principle? Discuss its working.

Hydraulic lift is a machine which is based on Pascal's law. In these devices fluids are used for transmitting presure.

Hydraulic lift consists of two cylinders of different cross-sectional areas connected with a pipe. The cylinders are filled with incompressible liquid and frictionless pistons are fitted in both the cylinders as shown in figure below.

Let 'a' and 'A' be the area of cross-section of smaller piston and bigger piston respectively. The load to be lifted is placed on bigger cross-section and effort is applied on smaller piston.

Let a force 'f' be applied on the smaller piston.

The pressure exerted by f on piston is,

$\mathrm{P}=\frac{\mathrm{f}}{\mathrm{a}}$ ... (1)

Now, according to Pascal's law, the pressure transmitted to bigger piston is also P.

Thus, force on bigger piston is, given by,

$\mathrm{F}=\mathrm{PA}=\frac{\mathrm{A}}{\mathrm{a}}\mathrm{f}\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{have},\mathrm{A}\mathrm{a}.\phantom{\rule{0ex}{0ex}}\mathrm{Let},\mathrm{A}=\mathrm{\eta a}\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{F}=\mathrm{\eta f}$

That is, the transmitted force gets multiplied by a factor $\mathrm{\eta}$ of applied force.

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State and prove Pascal's law.

PAscal's law states that, if some pressure is applied at any point of incompressible liquid then the same pressure is transmitted to all the points of liquid and on the walls of the container.

Let us imagine an arbitrary right angled prismatic triangle in the liquid of density ρ. This prismatic element is very small so, every part is considered at the same depth from the liquid surface. Therefore, effect of gravity is the same at all these points. That, the small element is in equilibrium.

The area of faces ABFE, ABDC and CDFE are ad, bd and cd respectively. Let the pressure of liquid on faces ABFE, ABDC and CDFE be P_{1}, P_{2} and P_{3}respectively.

The pressure of liquid exerts the force normal to the surface. Let us assume pressure P_{1}exerts the force F_{1} on the surface ABFE, pressure P_{2} exerts force F_{2} on the surface ABDC and pressure P_{3} exerts force on the surface CDFE.

So, Force F_{1} is given by,

${\mathrm{F}}_{1}={\mathrm{P}}_{1}\times \mathrm{Area}\mathrm{ABFE}={\mathrm{P}}_{1}\mathrm{ad}\phantom{\rule{0ex}{0ex}}{\mathrm{F}}_{2}={\mathrm{P}}_{2}\times \mathrm{Area}\mathrm{ABDC}={\mathrm{P}}_{2}\mathrm{bd}\phantom{\rule{0ex}{0ex}}{\mathrm{F}}_{3}={\mathrm{P}}_{3}\times \mathrm{Area}\mathrm{CDFE}={\mathrm{P}}_{3}\mathrm{cd}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Also},\mathrm{sin}\mathrm{\theta}=\raisebox{1ex}{$\mathrm{b}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{a}$}\right.\mathrm{and}\mathrm{sin}\mathrm{\theta}=\raisebox{1ex}{$\mathrm{c}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{a}$}\right.$

Since the prism is in equilibrium, so net force on the prism is zero.

Thus,

${\mathrm{F}}_{1}\mathrm{sin}\mathrm{\theta}={\mathrm{F}}_{2},\mathrm{and}\phantom{\rule{0ex}{0ex}}{\mathrm{F}}_{1}\mathrm{cos}\mathrm{\theta}={\mathrm{F}}_{3}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\phantom{\rule{0ex}{0ex}}{\mathrm{P}}_{1}\mathrm{ad}(\mathrm{b}/\mathrm{a})={\mathrm{P}}_{2}\mathrm{bd},\mathrm{and}...\left(1\right)\phantom{\rule{0ex}{0ex}}{\mathrm{P}}_{1}\mathrm{ad}(\mathrm{c}/\mathrm{a})\hspace{0.17em}={\mathrm{P}}_{3}\mathrm{cd}...\left(2\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{from}\mathrm{equations}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{have}\phantom{\rule{0ex}{0ex}}{\mathrm{P}}_{1}={\mathrm{P}}_{2},\mathrm{and}\phantom{\rule{0ex}{0ex}}{\mathrm{P}}_{1}={\mathrm{P}}_{3}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore {\mathrm{P}}_{1}={\mathrm{P}}_{2}={\mathrm{P}}_{3}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{Pascal}\text{'}\mathrm{s}\mathrm{law}\mathrm{is}\mathrm{proved}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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Obtain an expression for the pressure exerted by liquid column.

Consider a liquid of density ρ in a vessel as shown in figure.

To find: Pressure difference between two points A and B separated by vertical height h.

Consider an imaginary cuboid of area of cross-section a of liquid with upper and lower cap passing through A and B respectively in order to evaluate the pressure difference between points A and B.

Volume of the imaginary cylinder is, V = ah

Mass of liquid of imaginary cylinder, m = ρah

Let, P_{1} and P_{2} be the pressure on the upper and lower face of cylinder.

Forces acting on the imaginary cylinder are:

(i) Weight, mg = ρahg in vertically downward direction.

(ii) Downward thrust of F_{1} =P_{1}a on upper cap.

(iii) Upward thrust of F_{2} = P_{2}a on lower face.

As the imaginary cylinder in the liquid is in equilibrium, therefore the net force on the cylinder is zero.

$\mathrm{i}.\mathrm{e}.,{\mathrm{F}}_{1}+\mathrm{mg}={\mathrm{F}}_{2}\phantom{\rule{0ex}{0ex}}{\mathrm{P}}_{1}\mathrm{a}+\mathrm{\rho ahg}={\mathrm{P}}_{2}\mathrm{a}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{P}}_{2}-{\mathrm{P}}_{1}=\mathrm{\rho gh}$

Thus, the pressure difference between two points separated vertically by height h in the presence of gravity is ρgh.

Note: In the absence of gravity this pressure difference becomes zero.

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