Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 44

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Question 1
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Directions For Questions

Fluids at rest exert a normal force to the walls of the container or to the surface of the body immersed in the fluid. The pressure exerted by this force at a point inside the liquid is the sum of atmospheric pressure and a factor which depends on the density of the liquid, the acceleration due to gravity and the height of the liquid above that point. The upthrust acting on a body immersed in a stationary liquid is the net force acting on the body in the upward direction. A number of phenomenon of liquids in motion can be explain by Bernoullis theorem which relates the pressure, flow speed and height for flow of an ideal incompressible fluid.
.A container of large uniform cross-sectional area $$A$$ resting on a horizontal surface holds two immiscible, non-viscous and incompressible liquids of densities $$d$$ and $$2d$$, each of height $$H/2$$ as shown in the figure. The lower density liquid is open to the atmosphere having pressure $${P}_{0}$$.
Situation II:
The cylinder is removed and the original arrangement is restored. A tiny hole of area $$s (s\ll A)$$ is punched on the vertical side of the container at a height $$h (h<H/2)$$.

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The height $${h}_{m}$$ at which the hole should be punched so that the liquid travels the maximum distance is

A
$$\dfrac { 2H }{ 3 }$$

B
$$\dfrac { 3H }{8 }$$

C
$$\dfrac { 4H }{ 3 }$$

D
$$\dfrac { 5H }{ 3 }$$

Solution

$$x = v\sqrt{\dfrac{2h}{g}}$$

$$= \sqrt{\dfrac{(3H-4h)g}{2} \times \dfrac{2h}{g}}$$

$$= \sqrt{(3H-4h)h}$$

$$x^2 = 3Hh - 4h^2$$

$$\dfrac{dx^2}{dh} = 3H-8h= 0$$

$$\Rightarrow h = \dfrac{3H}{8}$$
Question 2
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Calculate the pressure inside a small air bubble of radius $$r$$ situated at a depth $$h$$ below the free surface of liquids of densities $${\rho}_{1}$$ and $${\rho}_{2}$$ and surface tension $${T}_{1}$$ and $${T}_{2}$$. The thickness of the first and second liquids are $${h}_{1}$$ and $${h}_{2}$$, respectively. Take atmosphere pressure $$={P}_{0}$$.

A
$${ P }_{ 0 }+{ \rho }_{ 1 }{ gh }_{ 1 }+{ \rho }_{ 2 }g\left( h-{ h }_{ 1 } \right) -\dfrac { 2{ T }_{ 2 } }{ r }$$

B
$${ P }_{ 0 }+{ \rho }_{ 1 }{ gh }_{ 1 }+{ \rho }_{ 2 }g\left( h-{ h }_{ 1 } \right) +\dfrac { 2{ T }_{ 2 } }{ r }$$

C
$${ P }_{ 0 }-{ \rho }_{ 1 }{ gh }_{ 1 }+{ \rho }_{ 2 }g\left( h-{ h }_{ 1 } \right) +\dfrac { 2{ T }_{ 2 } }{ r }$$

D
None of these

Solution

$$P_c=P_B+\dfrac{2T}{r}$$

$$P_B=P_A+\rho_1gh_1+\rho_2g(h-h_1)$$

where $$P_A=P_0$$

$$\boxed{P_c=P_0+\rho_1gh_1+\rho_2g(h-h_1)+\dfrac{2T_2}{r}}$$
Question 3
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Figure 4.191 shown water filled in a symmetrical container. Four pistons of equal area $$A$$ are used at the four openings to keep the water in equilibrium. Now an additional force $$F$$ is applied at each piston. The increase in the pressure at the center of the container due to this addition is

A
$$\dfrac { F }{ A }$$

B
$$\dfrac { 2F }{ A }$$

C
$$\dfrac { 4F }{ A }$$

D
$$0$$

Solution

As at each piston additional force $$F$$ is introduced so change in pressure will be the net additional force upon total area upon which it acts:
$$\Delta P= \dfrac{\Delta F}{Total\, area }$$

$$=\dfrac{4F}{4A}=\dfrac{F}{A}$$

so, net increase in pressure at cenftre $$ = \dfrac{F}{A}$$
Question 4
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Water is filled in a container up to height $$3\ m$$. A small hole of area $$a$$ is punched in the wall of the container at a height $$52.5\ cm$$ from the bottom. The cross sectional area of the container is $$A$$. If $$a/A =0$$; then $${v}^{2}$$ is (where $$v$$ is the velocity of water coming out of the hole)

A
$$50$$

B
$$51$$

C
$$48$$

D
$$51.5$$

Solution

Given $$\cfrac{a}{A}=0\\ \Rightarrow V_x=\sqrt{2gH}\\ V^2=2gH=2\times10\times2.5=50$$
Question 5
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A cylindrical vessel of cross-section $$A$$ contains water to a height $$h$$. There is a hole in the bottom of radius $$a$$. The time in which it will be emptied is

A
$$\dfrac { 2A }{ { \pi a }^{ 2 } } \sqrt { \dfrac { h }{ g } }$$

B
$$\dfrac { \sqrt { 2 } A }{ { \pi a }^{ 2 } } \sqrt { \dfrac { h }{ g } }$$

C
$$\dfrac { 2\sqrt { 2 } A }{ { \pi a }^{ 2 } } \sqrt { \dfrac { h }{ g } }$$

D
$$\dfrac { A }{ { \sqrt { 2 } \pi a }^{ 2 } } \sqrt { \dfrac { h }{ g } }$$

Solution
Question 6
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Directions For Questions

The vessel shown in the figure contains water, (take $${\rho}_{water}=1000\ kg/{m}^{3}, g=10\ m/{s}^{2}$$). For this arrangement, answer the following questions:

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Mark out the correct statement(s).

A
Net force acting on the base of the vessel $$>$$ weight of the liquid inside the vessel.

B
Net force acting on the base of the vessel $$=$$ weight of the liquid inside the vessel.

C
Net pressure force acting on the liquid $$=$$ weight of the vessel.

D
Both (a) and (c) are correct.

Solution
Question 7
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Air is streaming past a horizontal airplane wing such that its speed is $$90\ ms^{-1}$$ at the lower surface and $$120\ ms^{-1}$$ over the upper surface. If the wing is $$10\ m$$ long and has a average width of $$2\ m$$, the difference of pressure on the two sides and the gross lift on the wing respectively, are (density of $$air=\ 1.3\ kg\ { m }^{ -3 }$$)

A
$$5\ Pa$$

B
$$95\ Pa$$

C
$$4098\ Pa$$

D
$$4095\ Pa$$

Solution
Question 8
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A jar contains a gas and a few drops of water at TK. The pressure in the jar is $$830$$mm of Hg. The temperature of the jar is reduced by $$1\%$$. The saturated vapour pressure of water at the two temperatures are $$30$$ and $$25$$mm of Hg. Calculate the new pressure in the jar.

A
$$917$$ mm of Hg.

B
$$817$$ mm of Hg.

C
$$1017$$ mm of Hg.

D
$$777$$ mm of Hg.

Solution
Question 9
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A section of pipe is shown in the figure. Area of cross and height of A and B are $$0.1{m}^{2}, 1 m $$ and $$0.06 {m}^{2}$$, $$2m$$ respectively. find The pressure difference between A and B if velocity at B is $$5 m/s$$ and liquid flowing is water.

A
$$20 kpa$$

B
$$18 kpa$$

C
$$16 kpa$$

D
$$22 kpa$$

Solution

From Bernoulli's equation
$$p+\dfrac { 1 }{ 2 } \rho { v }^{ 2 }+\rho gh=$$ constant

$$Av=$$ constant by equation of continuity

$${ p }_{ A }-{ p }_{ B }=\rho g({ h }_{ B }-{ h }_{ A })+\dfrac { 1 }{ 2 } \rho({ v }_{ B }^{ 2 }-{ v }_{ A }^{ 2 })$$

$$=\rho g[2-1]+\dfrac { 1 }{ 2 } \rho [25-{ \left( 5\times \dfrac { 0.06 }{ 0.1 } \right) }^{ 2 }]$$

$$=\rho g+\dfrac { \rho }{ 2 } [25-9]=\rho [10+8]$$

$$\Delta p=18\times { 10 }^{ 3 }pa$$

$$\Delta p=18\quad kpa$$
Question 10
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The U tube having identical limbs contains mercury (density $${ \rho }_{ m }$$) to a level as shown in the figure. If the left limbs is filled to the top with water ( $${ \rho }_{ w }$$), then the rise of mercury level in the right limb will be

A
12ρm

B
$$\dfrac { l{ \rho }_{ w } }{ 2{ \rho }_{ m }+{ \rho }_{ w } } $$

C
$$\dfrac { { \rho }_{ m }l }{ { \rho }_{ w } } $$

D
$$\dfrac { l{ \rho }_{ w } }{ 2{ \rho }_{ m }-{ \rho }_{ w } } $$

Solution

The correct option is D.

Given,

U tube contains mercury density $$\rho_m$$

So,

The pressure at the bottom should be the same due to both the limbs.

$$\rho_A=\rho_wg(1+x)$$ And

$$\rho_B=\rho_mg(2x)$$

Since

$$\rho_A=\rho_B$$

Thus,

$$x=\dfrac{1\rho_w}{2\rho_m-\rho_w}$$