Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 60

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Question 1
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Three capillaries of internal radii $$2r$$, $$3r$$ and $$4r$$, all of the same length, are joined end to end. A liquid passes through the combination and the pressure difference across this combination is $$20.2 cm$$ of mercury. The pressure difference across the capillary of internal radius $$2r$$ is

A
$$2 cm$$ of $$Hg$$

B
$$4 cm$$ of $$Hg$$

C
$$8 cm$$ of $$Hg$$

D
$$16 cm$$ of $$Hg$$

Solution
Question 2
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A small hole is made at a height of $$h = \dfrac{1}{2} m$$ from the bottom of a cylindrical water tank and at a depth of $$h = 2 m$$ from the upper level of water in the tank. The distance, where the water emerging from the hole strikes the ground, is:

A
$$22m$$

B
$$1 m$$

C
$$2 m$$

D
None of these.

Solution

Range of water coming out = 2$$\sqrt { h(H-h) } $$
given h = $$\dfrac { 1 }{ 2 } $$m & H-h = 2m.
So Range = 2$$\sqrt { \frac { 1 }{ 2 } \times 2 } $$ = 2m
Question 3
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If the terminal speed of a sphere of gold (density $$\displaystyle =19.5\quad { kg }/{ { m }^{ 3 } }$$) is $$\displaystyle 0.2{ m }/{ s }$$ in a viscous liquid (density $$\displaystyle =1.5\quad { kg }/{ { m }^{ 3 } }$$), find the terminal speed of a sphere of silver (density $$\displaystyle =10.5\quad { kg }/{ { m }^{ 3 } }$$) of the same size in the same liquid.

A
$$\displaystyle 0.4{ m }/{ s }$$

B
$$\displaystyle 0.133{ m }/{ s }$$

C
$$\displaystyle 0.1{ m }/{ s }$$

D
$$\displaystyle 0.2{ m }/{ s }$$

Solution

Terminal speed of spherical body in a viscous liquid is given by
$$\displaystyle { v }_{ T }=\frac { 2{ r }^{ 2 }\left( \rho -\sigma \right) g }{ 9\eta } $$
where $$\displaystyle \rho $$ density of substance of body and $$\displaystyle \sigma $$ $$\displaystyle $$ density of liquid.
From given data
$$\displaystyle \frac { { v }_{ T }\left( Ag \right) }{ { v }_{ T }\left( Gold \right) } =\frac { { \rho }_{ Ag }-{ \sigma }_{ 1 } }{ { \rho }_{ gold }-{ \sigma }_{ 1 } } $$
$$\displaystyle \Rightarrow \quad { v }_{ T }\left( Ag \right) =\frac { 10.5-1.5 }{ 19.5-1.5 } \times 0.2=\frac { 9 }{ 18 } \times 0.2$$ $$\displaystyle =0.1{ m }/{ s }$$
Question 4
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A tank is filled upto a height $$h$$ with a liquid and is placed on a platform of height $$h$$ from the ground. To get maximum range, $${x}_{m}$$ a small hole is punched at a distance of $$y$$ from the free surface of the liquid. Then

A
$${ x }_{ m }=2h$$

B
$${ x }_{ m }=1.5h$$

C
$$v=h$$

D
$$y=0.75h$$

Solution
Question 5
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A wooden cylinder of diameter $$4r$$, height $$h$$ and density $$\rho /3$$ is kept on a hole of diameter $$2r$$ of a tank filled with water of density $$\rho$$ as shown in the figure. The height of the base of cylinder from the base of tank is $$H$$. Now level of the liquid starts decreasing slowly. When the level of the liquid is at a height $$h_1$$, above the cylinder the block just starts to lift. At what value of $${h}_{1}$$, will the block rise.

A
$$\cfrac { 4h }{ 9 } $$

B
$$\cfrac { 5h }{ 9 } $$

C
$$\cfrac { 5h }{ 3 } $$

D
None of these

Solution
Question 6
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The block is maintained at the position by external means and the level of liquid is lowered. The height $${h}_{2}$$ when the external force reduces to zero is

A
$$\cfrac { 4h }{ 9 } $$

B
$$\cfrac { 5h }{ 9 } $$

C
$$\cfrac { 2h }{ 3 } $$

D
None of these

Solution
Question 7
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Figure shows a crude type of atomizer. When bulb $$A$$ is compressed, air flows swiftly through tube $$BC$$ causing a reduced pressure in the particle of the vertical tube. Liquid rises in the tube, enters $$BC$$ and is sprayed out. If the pressure in the bulb is $${ P }_{ a }+P,$$ where $$P$$ is the gauge pressure and $${ P }_{ a }$$ is the atmosphere pressure, $$v$$ is the speed of air in $$BC$$. Density of air$$=1.3\ kg/{ m }^{ 3 }$$.

A
$$\dfrac { P+\rho gh }{ \sqrt { 0.65 } } $$

B
$$\dfrac { Ph\rho g }{ \sqrt { 0.65 } } $$

C
$$\dfrac { P\sqrt { h\rho g } }{ 0.65 } $$

D
$$\dfrac { P }{ h\rho g\sqrt { 0.65 } } $$

Solution
Question 8
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Two fully blown balloons are suspended as shown in the diagram. A stream of air is passed in between the balloons. What will happen to the balloons?

A
They will remain as they are

B
They will go apart

C
They will come closer

D
They will go up

Solution
Question 9
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In the adjoining figure, the cross-sectional area of the smaller tube is a and the larger tube is 2a. A block of mass 2 m is kept in the smaller tube have the same base area an as that of the tube. The difference between water levels of the two tubes are (density of water is $$\rho$$ and in $${ \rho }_{ 0 }$$ is atmospheric pressure)

A
$$\dfrac { { \rho }_{ 0 } }{ pg } +\dfrac { m }{ ap }$$

B
$$\dfrac { { \rho }_{ 0 } }{ pg } +\dfrac { m }{ 2ap }$$

C
$$\dfrac { 2m }{ ap } $$

D
$$\dfrac { m }{ ap }$$

Solution
Question 10
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A plane is in level flight at constant speed and each of its two wings has an area of 25 m$$^2$$. If the speed of the air on the upper and lower surfaces of the wing are 270 km h$$^{-1}$$ and 234 km h$$^{-1}$$ respectively, then the mass of the plane is (Take the density of the air = 1 kg m$$^{-3}$$)

A
1550 kg

B
1750 kg

C
3500 kg

D
3200 kg

Solution

Let $$v_1$$, $$v_2$$ are the speed of air on the lower and upper surfaces of the wings of the plane $$P_1$$ and $$P_2$$ are the pressure there.
According to Bernoulli's theorem
$$\displaystyle P_1 + \frac {1}{2} \rho v_1^2 = P_2 + \frac {1}{2} \rho v_2^2$$

$$\displaystyle P_1 - P_2 = \frac{1}{2} \rho (v_2^2 - v_1^2)$$

Here, $$v_1$$ = 234 km h$$^{-1}$$ = 234 x 58 m s$$^{-1}$$ =65 m s$$^{-1}$$
$$v_2$$ = 270 km h$${-1}$$ = 270 x $$\displaystyle \frac{5}{18}$$ = 75 m s$$^{-1}$$

Area of wings = 2 x 25 m$$^2$$ = 50 m$$^2$$

$$\therefore \,P_1 - P_2=\frac {1}{2} \times 1 (75^2 -65^2)$$ Upward force on the plane = $$(P_1 - P_2) A$$

$$ =\dfrac {1}{2} \times 1 \times (75^2 -65^2) \times 50m$$
As the plane is in level flight, therefore upward force balances the weight of the plane.
$$\therefore mg = (P_1 - P_2) A$$

Mass of the plane,
$$\therefore \displaystyle m = \dfrac{(P_1 - P_2)}{g} \times = \dfrac{1}{2} \times \dfrac{1 \times (75^2 - 65^2)}{10} \times 50$$

$$\displaystyle = \dfrac{(75+65)(75-65) \times 50}{2 \times 10} = 3500 kg$$