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Bernoulli's theorem is based on :

Conceptual

Two spherical soap bubbles of radii r_{1} and r_{2} in vaccum collapse under isothermal condition. The resulting bubble has radius equal to :

$\left(4\right)\phantom{\rule{0ex}{0ex}}\mathrm{for}\mathrm{isothermal}\mathrm{condition}\phantom{\rule{0ex}{0ex}}\mathrm{PV}={\mathrm{P}}_{1}{\mathrm{V}}_{1}+{\mathrm{P}}_{2}{\mathrm{V}}_{2}\phantom{\rule{0ex}{0ex}}\frac{4\mathrm{T}}{\mathrm{r}}\frac{4}{3}{\mathrm{\pi r}}^{3}=\frac{4\mathrm{T}}{{\mathrm{r}}_{1}}\frac{4}{3}{\mathrm{\pi r}}_{1}^{3}+\frac{4\mathrm{T}}{{\mathrm{r}}_{2}}.\frac{4}{3}{\mathrm{\pi r}}_{2}^{3}\phantom{\rule{0ex}{0ex}}{\mathrm{r}}^{2}={\mathrm{r}}_{1}^{2}+{\mathrm{r}}_{2}^{2}$

The energy needed in breaking of a drop of liquid of radius R into n drops of radius r is given by (T is surface tension and P is atmospheric pressure) :

$Surfaceareaoflargesphere=4{\mathrm{\pi R}}^{2}\phantom{\rule{0ex}{0ex}}Surfaceareaofnsmallspheres=n4\pi {r}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{Work}\mathrm{done}=\mathrm{T}\times \u2206\mathrm{A}=\mathrm{T}\times \left(\mathrm{n}4{\mathrm{\pi r}}^{2}-4{\mathrm{\pi R}}^{2}\right)$

A rectangular film of liquid is extended from (4 cm$\times 2cm$) to $\left(5cm\times 4\times cm\right)$. If the work done is $3\times {10}^{-4}$ J,the value of the surface tension of the liquid is

(b) Increases in surface energy= work done in area$\times $Surface tension

$\because $Increase in surface area,

$\u2206A=\left(5\times 4-4\times 2\right)\times 2\left(\because flimhastwosurfaces\right)$

=$\left(20-8\right)\times 2c{m}^{2}=24c{m}^{2}\phantom{\rule{0ex}{0ex}}=24\times {10}^{-4}{m}^{2}$

So, work done, W=T.$\u2206A$

$3\times {10}^{-4}=T\times 24\times {10}^{-4}\phantom{\rule{0ex}{0ex}}\therefore T=\frac{1}{8}=0.125N/m$

Three liquids of densities ${\rho}_{1},{\rho}_{2}and{\rho}_{3}$ (with ${p}_{1}>{p}_{2}>{p}_{3}$), having the same value of surface tension T, rise to the same height in three identical capillaries. The angles of contact ${\theta}_{1},{\theta}_{2}and{\theta}_{3}$ obey

(b) According to ascent formula for capillary tube,

$h=\frac{2T\mathrm{cos}\theta}{pgr}\phantom{\rule{0ex}{0ex}}\therefore \frac{\mathrm{cos}{\theta}_{1}}{{p}_{1}}=\frac{\mathrm{cos}{\theta}_{2}}{{p}_{2}}=\frac{\mathrm{cos}{\theta}_{3}}{{p}_{3}}\phantom{\rule{0ex}{0ex}}Thus,\mathrm{cos}\theta \propto \rho \phantom{\rule{0ex}{0ex}}\therefore {p}_{1}{p}_{2}{p}_{3}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{cos}{\theta}_{1}\mathrm{cos}{\theta}_{2}\mathrm{cos}{\theta}_{3}\phantom{\rule{0ex}{0ex}}0\le {\theta}_{1}{\theta}_{2}{\theta}_{3}\frac{\mathrm{\pi}}{2}$

Two non-mixing liquids of densities $\rho $ and n$\rho $(n>1) are put in a container. The height of each liquid is h. A solid cylinder floats with its axis vertical and length pL $\left(p<1\right)$ in the denser liquid. The density of the cylinder is d. The cylinder floats with its axis vertical and length pL (p<1) in the denser liquid. The density d is equal to

According to Archimedes' principle, the buoyant force on the cylinder is equal to the weight of the liquid displaced by the cylinder. The volume of the liquid displaced is pLπr^2, where r is the radius of the cylinder. The weight of this liquid is pLπr^2ρn. This weight must equal the weight of the cylinder, which is pLπr^2d. Therefore, d = ρn.

The approximate depth of an ocean is 2700 m. The compressibility of water is 45.4 x 10^{-11} Pa^{-1} and density of water is 10^{3}kg/m^{3}. What fractional compression of water will be obtained at the bottom of the ocean?

Given d=2700 m

ρ=10^{3}kg/m^{3}

Compressibility=45.4 x 10^{-11} Pa^{-1} per pascal

The pressure at the bottom of ocean is given by

P=ρgd=10^{3}x10x2700=27x10^{6}Pa

So, fractional compression=compressibility x pressure

=45.4 x 10^{-11}x 27 x10^{6}

=1.2x10^{-2}

A wind with speed 40 m/s blows parallel to the roof of a house. The area of the roof is 250 m^{2}. Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be (P_{air}=1.2kg/m^{3})

From Bernoulli's theorem

p_{1}+$\frac{1}{2}$ρv_{1}^{2}=p_{2}+$\frac{1}{2}$ρv_{2}^{2}

where, p_{1} and p_{2} are pressure inside and outside the roof and v_{1} v_{2} are velocities of wind inside and outside the roof, Neglect the width of the roof.

Pressure difference is

p_{1}-p_{2}=$\frac{1}{2}$ρ(v_{2}^{2}-v_{1}^{2})

=$\frac{1}{2}$x1.2(40^{2}-0)=960N/m^{2}

Force acting on the roof is given by

F=(p_{1}-p_{2})A=960 x 250=24x10^{4}N=24X10^{5}N

As the pressure inside the roof is more than outside to it. So the force will act m the upward direction, i.e. F=24x10^{5}N,upward

The cylindrical tube of a spray pump has radius R, one end of which has n fine holes, each of radius r. If the speed of the liquid in the tube is v, the speed of the ejection of the liquid through the holes is

The liquid flow is incompressible and steady. By the principle of conservation of mass, the volume flow rate must be constant throughout the tube. Therefore, the speed v in the tube is related to the speed v' through the holes by the equation πR^2v = nπr^2v', which gives v' = vR^2/nr^2.

The heart of a man pumps 5L of blood through the arteries per minute at a pressure of 150 mm of mercury. If the density of mercury be $13.6x{10}^{3}$ kg/m^{3} and g =10 m/s^{2}, then the power of heart in watt is

Given pressure=150mm of Hg

Pumping rate of heart of a man=$\frac{dV}{dt}$=$\frac{{\displaystyle 5x{10}^{3}}}{{\displaystyle 60}}$ m^{3}/s

Power of heart=P.$\frac{dV}{dt}$=ρgh.$\frac{dV}{dt}$ [P=ρgh]

=>(13.6x10^{3}kg/m^{3}) (10x0.15x5x10^{-3})/60

=1.70W

Water rises to a height h in capillary tube . If the length of capillary tube is above the surface of water is made less than h, then

In case of insufficient length, angle of contact adjusts itself so that water does not spill

A certain number of spherical drops of a liquid of radius r coalesce to form a single drop of radius R and volume V. If T is the surface tension of the liquid, then

$Ifthesurfaceareachanges,itwillchangethesurfaceenergyaswell.Asthesurfaceareaisdecrea\mathrm{sin}g,energywillbereleased.\phantom{\rule{0ex}{0ex}}Changeinsurfaceenergy=T\times \u2206A\phantom{\rule{0ex}{0ex}}Ifthereisnno.ifsmalldrops;\phantom{\rule{0ex}{0ex}}Volumeofalargedrop=volumeofnsmalldrops\phantom{\rule{0ex}{0ex}}\frac{4}{3}{\mathrm{\pi R}}^{3}=\mathrm{n}\times \frac{4}{3}{\mathrm{\pi r}}^{3}\phantom{\rule{0ex}{0ex}}\mathrm{Change}\mathrm{in}\mathrm{area};\phantom{\rule{0ex}{0ex}}\u2206\mathrm{A}=4{\mathrm{\pi R}}^{2}-\mathrm{n}\times 4{\mathrm{\pi r}}^{2}=3\left(\frac{4{\mathrm{\pi R}}^{3}}{3\mathrm{R}}-\mathrm{n}\times \frac{4{\mathrm{\pi r}}^{3}}{3\mathrm{r}}\right)=3\left(\frac{4{\mathrm{\pi R}}^{3}}{3\mathrm{R}}-\frac{4{\mathrm{\pi R}}^{3}}{3\mathrm{r}}\right)=3\left(\frac{\mathrm{V}}{\mathrm{R}}-\frac{\mathrm{V}}{\mathrm{r}}\right)=3\mathrm{V}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{r}}\right)\phantom{\rule{0ex}{0ex}}\mathrm{E}=\mathrm{T}\times \u2206\mathrm{A}=3\mathrm{VT}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{r}}\right),\mathrm{released}.$

The wettability of a surface by a liquid depends primarily on

(d) The value of angle of contact determines whether a liquid will spread on the surface

An engine pumps water continuously through a hose. Water leaves the hose with a velocity v and m is the mass per unit length of the water jet. What is the rate at which kinetic energy is imparted to water?

Let m is the mass per unit length then rate of mass per sec=$\frac{mx}{t}=mv$

Rate of KE=$\frac{1}{2}\left(mv\right){v}^{2}=\frac{1}{2}m{v}^{3}$

Two bodies are in equilibrium when suspended in water from the arms of a balance. The mass of one body is 36 g and its density is 9 g / cm^{3}. If the mass of the other is 48 g, its density in g / cm^{3} is

(c) Apparent weight = $\mathrm{V}(\mathrm{\rho}-\mathrm{\sigma})\mathrm{g}=\frac{\mathrm{m}}{\mathrm{\rho}}(\mathrm{\rho}-\mathrm{\sigma})\mathrm{g}$

where m = mass of the body,

$\mathrm{\rho}$ = density of the body

$\mathrm{\sigma}$ = density of water

If two bodies are in equilibrium then their apparent weight must be equal.

$\therefore \frac{{\mathrm{m}}_{1}}{{\mathrm{\rho}}_{1}}({\mathrm{\rho}}_{1}-\mathrm{\sigma})=\frac{{\mathrm{m}}_{2}}{{\mathrm{\rho}}_{2}}({\mathrm{\rho}}_{2}-\mathrm{\sigma})\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{36}{9}(9-1)=\frac{48}{{\mathrm{\rho}}_{2}}({\mathrm{\rho}}_{2}-1)$

By solving we get ${\mathrm{\rho}}_{2}=3.$