Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 57

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Question 1
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Directions For Questions

A cylindrical tank having cross-sectional area $$A=0.5 m$$ is filled with two liquids of density $$\rho_1=900 kg m^{-3}$$ and $$\rho_2=600 kg m^{-3}$$, to a height $$h=60 cm$$ each as shown in the figure. A small hole having area $$a=5 cm^2$$ is made in right vertical wall at a height $$y=20cm$$ from the bottom. A horizontal force $$F$$ is applied on the tank to keep it in static equilibrium. The tank is lying on a horizontal surface. Neglect mass of the cylindrical tank in comparison to the mass of the liquids (take $$g=10 ms^{-2}$$).

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The velocity of efflux is:

A
$$10 ms^{-1}$$

B
$$20 ms^{-1}$$

C
$$4 ms^{-1}$$

D
$$35 ms^{-1}$$

Solution

Since area of a hole is very small in comparison to base area A of the cylinder, velocity of liquid inside the cylinder is negligible. Let velocity of efflux be v and atmospheric pressure be $$P_0$$.

Consider two points A (inside the cylinder) and B (must outside the hole) in the same horizontal line as shown in the figure in question.
Pressure at A: $$P_A= P_0 + h\rho_2g + (h-y)\rho_1g$$
Pressure at A: $$P_B= P_0$$
Applying Bernoulli's theorem at points A and B
$$ P_A + \dfrac{1}{2}\rho_2v_2^2 + (\rho_2 gh + \rho_1 g(h-y))= P_B + \dfrac{1}{2}\rho_1v_1^2 + (\rho_2 gh + \rho_1 g(h-y) )$$ .....(1)
Substituting $$ P_A $$ and $$ P_B$$ in eqn(1) we get,
$$P_0 + h\rho_2g + (h-y)\rho_1g + \dfrac{1}{2}\rho_2v_2^2 + (\rho_2 gh + \rho_1 g(h-y))= P_0 + \dfrac{1}{2}\rho_1v_1^2 + (\rho_2 gh + \rho_1 g(h-y) )$$ ....(2)

Given,
$$A=0.5\: m^2$$
$$a =5 cm^2 = 25 \times 10^{-4}\: m^2$$
From continuity equation,
$$Av_2 = av_1$$
$$\therefore \dfrac{v_2}{v_1} = \dfrac{a}{A}= \dfrac{25 \times 10^{-4}}{0.5}=50 \times 10^{-4}$$
Given $$h= 60\: cm=0.6\: m$$
$$y=20 \: cm = 0.2 \:m$$
$$\rho_1= 900\: kg\:m^{-3}$$
$$\rho_2= 600\: kg\:m^{-3}$$
$$P_0 + h\rho_2g + (h-y)\rho_1g + \dfrac{1}{2}\rho_2v_2^2 = P_0 + \dfrac{1}{2}\rho_1v_1^2 $$
As given in the problem, we ignore $$v_2$$ since a is very small.
$$ \therefore P_0 + h\rho_2g + (h-y)\rho_1g = P_0 + \dfrac{1}{2}\rho_1v_1^2 $$
$$ \therefore h\rho_2g + (h-y)\rho_1g = \dfrac{1}{2}\rho_1v_1^2$$
Substituting the values, $$0.6 \times 600 \times 10 + (0.6-0.2) \times 900 \times 10 = \dfrac{1}{2} \times 900 \times v_1^2$$
$$ \Rightarrow v_1 = 4\: ms^{-1}$$
Question 2
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A cylindrical vessel of cross-sectional area $$1000 cm^2$$, is fitted with a frictionless piston of mass $$10 kg$$, and filled with water completely. A small hole of cross-sectional area $$10 mm^2$$ is opened at a point $$50 cm$$ deep from the lower surface of the piston. The velocity of efflux from the hole will be

A
$$10.5 m/s$$

B
$$3.4 m/s$$

C
$$0.8 m/s$$

D
$$0.2 m/s$$

Solution

Given that
$$A = 1000 cm^3$$
$$m = 10 kg =10000 g$$
$$A_h = 10 mm^2 = 0.1 m^2$$
$$(H-h) = 50 cm$$
Here, the force on piston is
$$F = mg$$

Hence, Increase in pressure on the liquid in the wider tube is
$$P = \dfrac{F}{A} = \dfrac{mg}{A}$$
Let, $$H$$ is the level of water to which piston will move, thus
$$P = H\rho g$$
$$H = \dfrac{P}{\rho g}$$
$$H = \dfrac{mg}{A\rho g} = \dfrac{m}{A\rho}$$
$$H = \dfrac{10000}{1000}$$.......($$\because \rho = 1$$)
$$\therefore H = 10 cm$$ from surface of tube.

Since, $$(H-h) = 50cm, H = 60cm = 0.6 m$$
The velocity of effluent from the hole is
$$v = \sqrt {h\rho g}$$
$$v = (\sqrt {2gh}) $$
$$v = {\sqrt {(2)(9.8)(0.6) }= 3.4 ms^{-1}}$$
Question 3
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Directions For Questions

Figure shows a large closed cylindrical tank containing water. Initially, the air trapped above the water surface has a height $$h_0$$ and pressure $$2p_0$$ wherre $$p_0$$ is the atmospheric pressure. There is a hole in the wall of the tank at a depth $$h_1$$ below the top from which water comes out. A long vertical tube is connected as shown.

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Find the speed with which water comes out of the hole.

A
$$\dfrac {1}{\rho}[p_0-\rho g(h_1-2h_0)]^{\dfrac {1}{2}}$$

B
$$\left [\dfrac {2}{\rho}[p_0+\rho g(h_1-h_0)]\right ]^{\dfrac {1}{2}}$$

C
$$\left [\dfrac {3}{\rho}[p_0+\rho g(h_1-h_0)]\right ]^{\dfrac {1}{2}}$$

D
$$\left [\dfrac {4}{\rho}[p_0+\rho g(h_1-h_0)]\right ]^{\dfrac {1}{2}}$$

Solution

Equating the pressure due air and due to water we get
$$2p_0=(h_2+h_0)\rho g+p_0$$ (since liquids at the same level have the same pressure)
$$p_0=h_2\rho g+h_0\rho g$$
$$h_2\rho g=p_0-h_0\rho g$$
$$h_2=\displaystyle\frac {p_0}{\rho g}-\frac {h_0\rho g}{\rho g}=\frac {p_0}{\rho g}-h_0$$
KE of the water $$=$$ Pressure energy of the water at that layer
$$\displaystyle\frac {1}{2}mv^2=m\times \frac {P}{\rho}$$
$$v^2=\displaystyle\frac {2P}{\rho}=\frac {2}{\rho}[p_0+\rho g(h_1-h_0)]$$
$$v=\left [\displaystyle\frac {2}{\rho}\left \{p_0+\rho g(h_1-h_0)\right \}\right ]^{\dfrac {1}{2}}$$
Question 4
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Directions For Questions

Figure shows a large closed cylindrical tank containing water. Initially, the air trapped above the water surface has a height $$h_0$$ and pressure $$2p_0$$ wherre $$p_0$$ is the atmospheric pressure. There is a hole in the wall of the tank at a depth $$h_1$$ below the top from which water comes out. A long vertical tube is connected as shown.

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Find the height of the water in the long tube above the top when the water stops coming out of the hole.

A
$$-2h_0$$

B
$$h_0$$

C
$$h_2$$

D
$$-h_1$$

Solution

Equating the pressure due air and due to water we get
$$2p_0=(h_2+h_0)\rho g+p_0$$ (since liquids at the same level have the same pressure)
$$p_0=h_2\rho g+h_0\rho g$$
$$h_2\rho g=p_0-h_0\rho g$$
$$h_2=\displaystyle\dfrac {p_0}{\rho g}-\dfrac {h_0\rho g}{\rho g}=\dfrac {p_0}{\rho g}-h_0$$
KE of the water $$=$$ Pressure energy of the water at that layer
$$\displaystyle\frac {1}{2}mv^2=m\times \dfrac {P}{\rho}$$
$$v^2=\displaystyle\dfrac {2P}{\rho}=\dfrac {2}{\rho}[p_0+\rho g(h_1-h_0)]$$
$$v=\left [\displaystyle\dfrac {2}{\rho}\left \{p_0+\rho g(h_1-h_0)\right \}\right ]^{\dfrac {1}{2}}$$

Now we know:
$$2P_0+\rho g(h_1-h_0)=p_0+\rho gX$$
$$\Rightarrow X=\displaystyle\frac {p_0}{\rho g}+(h_1-h_0)=h_2+h_1$$
i.e, X is $$h_1$$ metre below the top or X is $$-h_1$$ above the top.
Question 5
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Directions For Questions

Figure shows a large closed cylindrical tank containing water. Initially, the air trapped above the water surface has a height $$h_0$$ and pressure $$2p_0$$ wherre $$p_0$$ is the atmospheric pressure. There is a hole in the wall of the tank at a depth $$h_1$$ below the top from which water comes out. A long vertical tube is connected as shown.

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Find the height $$h_2$$ of the water in the long tube above the top initially.

A
$$\dfrac {3p_0}{\rho g}-\dfrac {h_0}{3}$$

B
$$\dfrac {2p_0}{\rho g}-\dfrac {h_0}{2}$$

C
$$\dfrac {p_0}{\rho g}-h_0$$

D
$$\dfrac {p_0}{2\rho g}-2h_0$$

Solution

Equating the pressure due air and due to water we get
$$2p_0=(h_2+h_0)\rho g+p_0$$ (since liquids at the same level have the same pressure)
$$p_0=h_2\rho g+h_0\rho g$$
$$h_2\rho g=p_0-h_0\rho g$$
$$h_2=\displaystyle\dfrac {p_0}{\rho g}-\dfrac {h_0\rho g}{\rho g}=\dfrac {p_0}{\rho g}-h_0$$
Question 6
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A cubical box of wine has a small spout located in one of the bottom corners. When the box is full and placed on a level surface, opening the spout results in a flow of wine with an initial speed of $${ v }_{ 0 }$$ (see figure). When the box is half empty, someone tilts it at $${ 45 }^{ \circ }$$ so that the spout is at the lowest point (see figure). When the spout is opened the wine will flow out with a speed of

A
$${ v }_{ 0 }$$

B
$${ v }_{ 0 }/2$$

C
$${ v }_{ 0 }/\sqrt { 2 } $$

D
$${ v }_{ 0 }/\sqrt [ 4 ]{ 2 } $$

Solution

Hint: Use Torricelli’s equation for velocity
$$\textbf{Step1: Torricelli's equation for velocity}$$

Torricelli's theorem stats that speed v by which water come out from tank is proportional to square toot of twice the acceleration created by gravity and vertical height h between surface and center of opening.
$$V = \sqrt{2\ \ast\ g\ \ast\ h}$$
$$\textbf{Step2: Calculation of speed with which wine will flow}$$
So, if box is having height h, then $$V_{0} = \sqrt{2\ \ast\ g\ \ast\ h}$$

When box is half empty, then half of diagonal of the cube’s face will have same height as height of wine above spout.
So, it will be $$\frac{\sqrt2}{2} h$$ = $$\frac{h}{\sqrt2}$$
Now, on putting the values in the original formula of velocity,
$$V' = \sqrt{2g\ (\frac{h}{\sqrt2})} = \frac{V_{0}}{\sqrt[4]{2}}$$
Answer:

Hence, option D is the correct answer.
Question 7
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Figure shows a capillary tube of radius $$r$$ dipped into water. If the atmospheric pressure is $$P_0$$, the pressure at point $$A$$ ( just below the meniscus ) is

A
$$P_0$$

B
$$P_0 + \displaystyle \frac {2S}{r}$$

C
$$P_0 - \displaystyle \frac {2S}{r}$$

D
$$P_0 - \displaystyle \frac {4S}{r}$$

Solution

The pressure just above the meniscus is atmospheric pressure and is greater than the pressure just below the meniscus since the pressure on the concave side is greater than that on the convex side.
$$P_{concave} - P_{convex} = S(\dfrac{1}{R_1} + \dfrac{1}{R_2})$$
For a spherical surface $$R_1=R_2=r$$
Hence, $$P_{concave} - P_{convex} =\dfrac{2S}{r}$$
In the given scenario, $$P_{concave}= P_0$$ where $$P_0$$ is the atmospheric pressure.
$$P_{convex} = P_0 -\dfrac{2S}{r}$$
Question 8
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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 an average width of 2m, 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, 900 N

B
95 Pa, 900 N

C
4095 Pa, 900 N

D
4095 Pa, 81900 N

Solution

Applying Bernoulli's principle, we have
$$P_1+\frac{1}{2} \rho {v_1}^2=P_2+\frac{1}{2} \rho {v_2}^2$$
$$\Rightarrow P_2-P_1=\frac{1}{2} \rho \left ( v_1^2 - v_2^2 \right ) \\\Rightarrow \Delta P=\frac{1}{2} \times 1.3 \times \left ( 120^2 - 90^2 \right ) \\\Rightarrow \Delta P=0.65 \times (120+90) \times (120-90) \\\Rightarrow \Delta P=0.65 \times 210 \times 30=4095 Pa$$
The difference in pressure provides the lift to the aeroplane. So
lift on the aeroplane=pressure difference $$\times$$ area of wings
$$= \Delta P \times A = 4095 \times (10 \times 2)=81900N$$
Question 9
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Directions For Questions

A container of large uniform cross-sectional area A resting on a horizontal surface, holds, two immissicible, 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 $$P_{0}. A$$ homogeneous solid cylinder of length $$ L(L< H /2)$$, cross-sectional area $$A/5 $$ is immersed such that it floats with its axis vertical at the liquid-liquid interface with length $$L/4$$ in the denser liquid.
The cylinder is then removed and the original arrangement is restored. A tiny hole of area $$s(s < < A)$$ is punched on the vertical side of the container at a height $$h(h< H/2)$$. As a result of this, liquid starts flowing out of the hole with a range $$x$$ on the horizontal surface.

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The initial speed of efflux without cylinder is:

A
$$\displaystyle v=\sqrt{\frac{g}{3}[3H+4h]}$$

B
$$\displaystyle v=\sqrt{\frac{g}{2}[4H+3h]}$$

C
$$\displaystyle v=\sqrt{\frac{g}{2}[3H-4h]}$$

D
None of these

Solution

Applying Bernoulli's equation just inside and just outside the hole,
$$\displaystyle

P_{0}+\left ( \frac{H}{2} \right )\left ( d \right )g+\left (

\frac{H}{2}-h \right )\left ( 2d \right )\left ( g \right

)=\frac{1}{2}\left ( 2d \right )v^{2}+P_{0}$$
$$\therefore $$ $$\displaystyle v=\sqrt{\frac{g}{2}\left ( 3H-4h \right )}$$
Question 10
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What work should be done in order to squeeze all water from a horizontally located cylinder (figure shown above) during the time $$t$$ by means of a constant force acting on the piston? The volume of water in the cylinder is equal to $$V$$, the cross-sectional area of the orifice to $$s$$, with $$s$$ being considerably less than the piston area. The friction and viscosity are negligibly small.

A
$$\displaystyle A=\frac{1}{2}\rho\frac{V^3}{{(St)}^2}$$

B
$$\displaystyle A=\frac{3}{2}\rho\frac{V^3}{{(St)}^2}$$

C
$$\displaystyle A=\frac{5}{2}\rho\frac{3V^3}{{(St)}^2}$$

D
None of these

Solution

Volume is $$V$$, Orifice area is $$S$$.
Let the discharge velocity be $$v$$.
So, $$v\times St=V\Rightarrow v=V/St\ -(1)$$
Total mass is $$M=\rho\times V\ -(2)$$.
So, work done is equal to kinetic energy which is $$Mv^2/2$$.
From $$(1)\ and\ (2)$$,
kinetic energy is $$KE=\dfrac{1}{2}\rho\dfrac{V^3}{(St)^2}$$