Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 23

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Question 1
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The area of cross-section of the wider tube shown in figure is $$800{ cm }^{ 2 }$$. if mass of $$12 kg$$ is placed on the massless piston, the difference in heights h in the level of water in the two tubes is:

A
$$10 cm$$

B
$$6 cm$$

C
$$15 cm$$

D
$$2 cm$$

Solution

Given that
$$A = 800 cm^3$$
$$m = 12 Kg =12000 g$$
Here, the force on piston is
$$F = mg$$
Hence, Increase in pressure on the liquid in the wider tube is
$$P = \dfrac{F}{A} = \dfrac{mg}{A}$$
since, h is the difference in level of water in the two tubes, thus
$$P = h\rho g$$
$$h = \dfrac{P}{\rho g}$$
$$h = \dfrac{mg}{A\rho g} = \dfrac{m}{A\rho}$$
$$h = \dfrac{12000}{800}$$.......($$\because \rho = 1$$)
$$\therefore h = 15 cm$$
Question 2
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A large cylindrical tank has a hole of area $$A$$ at its bottom and water is poured into the tank through a tube of cross-sectional area $$A$$ ejecting water at the speed $$v$$. Which of the following is true?

A
Water level in tank keeps on rising.

B
No water can be stored in the tank.

C
Water level will rise to a height $$\dfrac{v^2}{2g}$$ and then stop.

D
The water level will be oscillating.

Solution

Initially, the water flowing out will be less than that flowing into it. Hence, the water level will go on rising. When the water level is h, the velocity of efflux $$=\sqrt {2gh}$$. When this becomes equal to the velocity of inflow, the level will become steady as the area of cross section of the filling tube and area of cross section of the hole are equal. This height is given by Torricelli's theorem,
$$v=\sqrt {2gh}$$ or $$\frac {v^2}{2g}=h$$
Question 3
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A solid metallic sphere of radius $$r$$ is allowed to fall freely through air. If the frictional resistance due to air is proportional to the cross-sectional area and to the square of the velocity, then the terminal velocity of the sphere is proportional to which of the following?

A
$${ r }^{ 2 }$$

B
$$r$$

C
$${ r }^{ { 3 }/{ 2 } }$$

D
$${ r }^{ { 1 }/{ 2 } }$$

Solution

The frictional resistance $$f_r$$ is given as:
$$f_r\propto A$$ and $$f_r\propto v^2$$

$$f_r=kAv^2$$(here k is a constant)

The downwards force is given as: $$mg=\displaystyle\frac{4}{3}\pi r^3\rho g$$

Balancing the above we get the terminal velocity as:
$$\displaystyle\frac{4}{3}\pi r^3\rho g=k(\pi r^2)v^2$$

$$r\propto v^2$$

$$v\propto r^{1/2}$$
Question 4
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Equal volumes of two immiscible liquids of densities $$\rho$$ and $$2\rho$$ are filled in a vessel as shown in figure. Two small holes are punched at depth h/2 and 3h/2 from the surface of lighter liquid. If $$V_{1}$$ and $$V_{2}$$ are the velocities of a flux at these two holes, then $$V_{1}/V_{2}$$ is :

A
$$\displaystyle \frac{1}{2\sqrt2}$$

B
$$\displaystyle \frac{1}{2}$$

C
$$\displaystyle \frac{1}{4}$$

D
$$\displaystyle \frac{1}{\sqrt2}$$

Solution

Here, two small holes are punches at depth $$\dfrac{h}{2}$$ and $$\dfrac{3h}{2}$$ from the surface of lighter liquid. Hence, the height of first hole from the surface of lighter liquid is
$$h_1 = \dfrac{h}{2}$$
and the height of second hole from the first hole is(i.e. from surface of remaining liquid)
$$h_2 = \dfrac{3h}{2} - \frac{h}{2} = h$$
Now, we have the velocity of flux from first hole is
$$V_1 = \sqrt 2h_1g$$
$$V_1 = \sqrt 2(\dfrac{h}{2})g$$
$$V_1 = \sqrt hg$$
Also, the velocity of flux from second hole is
$$V_2 = \sqrt 2h_2g$$
$$V_2 = \sqrt 2hg$$
Therefore,
$$\dfrac{V_1}{V_2} = \dfrac{\sqrt hg}{\sqrt 2hg}$$
$$\dfrac{V_1}{V_2} = \dfrac{1}{\sqrt 2}$$
Question 5
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The area of cross-section of the wider tube shown in figure is $$800 cm^2$$. If a mass of 12 kg is placed on the massless piston, the difference in heights h in the level of water in the two tubes is :

A
10 cm

B
6 cm

C
15 cm

D
2 cm

Solution

Given,

$$m=12kg=12000g$$

$$g=980g/cm^2$$

$$A=800cm^2$$

$$\rho =1g/cm^3$$

Pressure, $$P=\rho g h=\dfrac{mg}{A}$$

$$1\times 980\times h=\dfrac{12\times 1000\times 980}{800}$$

$$h=\dfrac{12\times 10}{8}=15cm$$

The correct option is C.
Question 6
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A tube is attached as shown in closed vessel containing water. The velocity of water coming out from a small hole is

A
$$\sqrt { 2 } m/s$$

B
$$2 m/s$$

C
Depends on pressure of air inside vessel

D
None of these

Solution

Hint: Use Torricelli's equation for velocity

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

Torricelli's theorem states that speed v by which water comes out from the tank is proportional to the square toot of twice the acceleration created by gravity and vertical height between surface and center of the opening.

$$V = \sqrt{2\ *\ g\ *\ h}$$

$$\textbf{Step2: Calculation of velocity of water coming out}$$

On putting the values, g = 10 $$m/s^{2}$$ and h = 20 cm = 0.2 m.

$$V = \sqrt{2\ *\ 10\ *\ 0.2}$$
= 2 m/s

Answer:

Hence, option B (2 m/s) is the correct answer.
Question 7
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A water barrel stands on a table of height $$h$$. If a small hole is punched in the side of the barrel at its base, it is found that the resultant stream of water strikes the ground at a horizontal distance $$R$$ from the barrel. The depth of water in the barrel is

A
$$\cfrac { R }{ 2 } $$

B
$$\cfrac { { R }^{ 2 } }{ 4h } $$

C
$$\cfrac { { R }^{ 2 } }{ h } $$

D
$$\cfrac { h }{ 2 } $$

Solution

Hint:

Use Torricelli’s equation for velocity and projectile motion’s horizontal range formula
$$\textbf{Step1: Torricelli's equation for velocity}$$
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 between surface and center of opening.
V = $$\sqrt{2\ \ast\ g\ \ast\ h}$$
$$\textbf{Step2:Expression of horizontal range}$$
The horizontal range where the water strikes the ground R is given by,
R = $$V * \sqrt{\frac{2\ h}{g}}$$
$$\textbf{Step3:Calculation of the depth of water in the barrel}$$
On putting the values and solving the equation,
R = $$\sqrt{2\ \ast\ g\ \ast\ h}\ * \sqrt{\frac{2\ h}{g}}$$
So, $$R^2 = 4hH$$
$$H = \frac{R^2}{4h}$$

Answer:

Hence, option B is the correct answer.
Question 8
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Some liquid is filled in a cylindrical vessel of radius $$R$$. Let $${ F }_{ 1 }$$ be the force applied by the liquid on the bottom of the cylinder. Now the same liquid is poured into a vessel of uniform square cross-section of side $$R$$. Let $${ F }_{ 2 }$$ be the force applied by the liquid on the bottom of this new vessel. (Neglect atmosphere pressure). Then

A
$${ F }_{ 1 }=\pi { F }_{ 2 }$$

B
$${ F }_{ 1 }=\cfrac { { F }_{ 2 } }{ \pi } $$

C
$${ F }_{ 1 }=\sqrt { \pi { F }_{ 2 }}$$

D
$${ F }_{ 1 }={ F }_{ 2 }$$

Solution

Force applied on the base is equal to the weight of the liquid, since in both cases same liquid has been poured in the container, it will exert same force on the base of the container.
$$F_1 = F_2$$
Question 9
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Water is flowing steadily through a horizontal tube of non-uniform cross-section.If the pressure of water is $$4\times10^4N/mm^2$$ at a point when cross section is $$0.02m^2$$ velocity of flow is 2m/s what is pressure at a point where cross section reduces to $$0.01m^2$$.

A
$$1.4\times { 10 }^{ 4 } N{ m }^{ 2 }$$

B
$$3.4\times { 10 }^{ 4 } N{ m }^{ 2 }$$

C
$$2.4\times { 10 }^{ -4 } N{ m }^{ 2 }$$

D
none of these

Solution

As we know
$$A_1V_1=A_2V_2$$
Where $$A_1=0.02\ m^2$$
$$V_1=2\ m/s$$
$$A_2=0.01\ m^2$$
$$\implies 0.02\times 2=0.01\times V_2$$
$$\implies V_2=4\ m/s$$
Now using Bernoulli's theorem
$$\cfrac{P_1}{\rho}+\cfrac{1}{2}V_1^2=\cfrac{P_2}{\rho}+\cfrac{1}{2}v_2^2$$
$$\implies P_2=P_1+\cfrac{\rho_1}{2}[V_1^2-V_1^2]$$
$$\implies P_2=4\times 10^{10}+\cfrac{1000}{2}[2^2-4^2]-4\times 10^{10}=3.4\times 10^{4}\ N/m^2$$
Question 10
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A rectangular tank is placed on a horizontal ground and is filled with water to a height $$H$$ above the base. A small hole is made on one vertical side at a depth $$D$$ below the level of water in the tank. The distance $$x$$ from the bottom of the tank at which the water jet from the tank will hit the ground is

A
$$2\sqrt { D(H-D) } $$

B
$$2\sqrt { DH } $$

C
$$2\sqrt { D(H+D) } $$

D
$$\cfrac { 1 }{ 2 } \sqrt { DH } $$

Solution

This setup can be visualised as a horizontal projectile from height h .
here $$h=$$ height of the hole above the bottom of the tank.
so $$h=H-D$$
let Horizontal velocity of water be $$u_x$$.
so velocity of liquid falling at depth d is given by $$\sqrt{2gd}$$
here $$d=D$$ so $$u_x=\sqrt{2gD}$$
Time taken to fall at surface can be found using
$$s=ut+\dfrac{1}{2}at^2$$
$$h=u_yt+\dfrac{1}{2}gt^2$$
here $$u_y=0$$, no component of velocity in vertical direction.
and $$h=H-D$$

$$H-D=\dfrac{1}{2}gt^2$$

$$t=\sqrt{\dfrac{2(H-D)}{g}}$$

since there is no acceleration in horizontal direction
Range in Horizontal direction= $$u_xt$$
putting values from above
$$r=\sqrt{2gD}\times $$$$\sqrt{\dfrac{2(H-D)}{g}}$$

So, range of the water stream $$x=2\sqrt{D(H-D)}$$