Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 25

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Question 1
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Directions For Questions

The charge of an electron was first measured by the American physicist Robert Millikan during 1909-1913. In his experiment, oil was sprayed in very fine drops (around $$10^{-4}$$ mm in diameter) into the space between two parallel horizontal plates separated by a distance $$d$$. A potential difference $$V_{AB}$$ is maintained between the parallel plates, causing a downward electric field between them. Some of the oil drops acquire a negative charge because of frictional effects or because of ionisation of the surrounding air by X-rays or radioactivity. The drops are observed through microscope. $$\rho$$ is density of oil and $$r$$ is radius of the drop under investigation.

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To measure the radius of the drop Millikan used _____ law of freely falling drops.

A
Poiseuille's

B
Ostwald's

C
Brewester's

D
Stoke's

Solution

Poise uille's law is applicable when a liquid flow a capillary tube.
Ostwald's law is a relationship between the dissociation constant and the degree of dissociation of a weak electrolyte.
Brewester's law is related to the light waves. It is used for calculating angle of polarization.
Stoke's law - when any object rises or falls through a fluid it will experience a viscous drag, whether it is a parachutist or spacecraft falling through air, a stone falling through water or a bubble rising through fizzy lemonade.
Question 2
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A spherical ball is dropped in a long column of a viscous liquid. The speed $$v$$ of ball as a function of time may be best represented by graph

A

B

C

D
Solution
$$\textbf{Explanation:}$$
- As soon as a spherical ball is started to fall in a viscous liquid, its speed will increase due to gravity, and at that time dragging force will be low as it is not that deep in liquid.
- When the ball falls in a long column, its speed will be zero at t = 0. So, options C and D are incorrect. Now, with acceleration, velocity will increase, but after enough depth is reached, dragging force will start to work and when its value becomes equal to gravity force mg, net acceleration will be zero. And velocity will remain constant. So, in the graph, A velocity keeps on increasing with time so it is wrong.
Answer:

Hence, option B is the correct answer.
Question 3
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There is a small hole of diameter 2 mm in the wall of a water tank at a depth of 10 m below the free water surface. The velocity of efflux of water from the hole will be:

A
0.14 m/s

B
1.4 m/s

C
0.014 m/s

D
14 m/s

Solution

According to Torricelli's theorem velocity of efflux i.e., the velocity with which the liquid flows out of a hole is equal to $$\sqrt{2gh}$$, where $$h$$ is the depth of the hole below the liquid surface.
Here $$h=10m$$
$$\Rightarrow v=\sqrt{2 \times 9.8 \times 10}=\sqrt{196}=14m/s$$
Question 4
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The difference of two liquid levels in a manometer is $$10 cm$$ and its density is $$0.8 gm/cm^3$$. If the density of air is $$1.3 \times 10^3 gm/cm^3$$ then the velocity of air will be (in $$cm/s$$)

A
$$347$$

B
$$34.7$$

C
$$3470$$

D
$$0.347$$

Solution

The manometer here gives the stagnation pressure i.e. the velocity of liquid is zero at point B.$$ \\ $$Hence the pressure drop is given by $$ \\ \Delta P=\rho gh={ P

}_{ B }-{ P }_{ A }\\ \rho =0.8{ gm }/{ { cm }^{ 3 } }=800kg/{ m }^{ 3 } ,

h=0.1m , { \rho }_{ air }=1.3\times { 10 }^{ 3 }{ gm }/{ { cm }^{ 3 }

}=1.3{ kg }/{ { m }^{ 3 } }\\ \Delta P=800\times 9.81\times 0.1\\ \Delta

P=784.8 N/{ m }^{ 2 }\\ $$
From Bernoulli's equation for stagnation pressure
$$\\ { P

}_{ A }+\dfrac { { \rho }_{ air }{ { V } }^{ 2 } }{ 2 } ={ P }_{ B }\\

V=\sqrt { \dfrac { 2\Delta P }{ { \rho }_{ air } } } \\ V=\sqrt {

\dfrac { 2\times 784.8 }{ 1.3 } } \\ V=34.74{ m }/{ s }\\ V\cong 3470{ cm

}/{ s } $$
Question 5
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One end of a horizontal pipe is closed with the help of a valve and the reading of a barometer attached to the pipe is $$3 \times 10^5$$ pascal. When the valve in the pipe is opened then the reading of the barometer falls to $$10^5$$ pascal. The velocity of water flowing through the pipe will be in m/s.

A
$$0.2$$

B
$$2$$

C
$$20$$

D
$$200$$

Solution

Applying Bernoulli's principle on a horizontal pipe, we have
$$P_1+\frac{1}{2} \rho {v_1}^2=P_2+\frac{1}{2} \rho {v_2}^2$$
here $$P_1= 3 \times 10^5$$ pascal , $$v_1=0$$, $$P_2=10^5$$ Pascal, $$\rho=1000 kg/m^3$$
$$P_1+\frac{1}{2} \rho {v_1}^2=P_2+\frac{1}{2} \rho {v_2}^2 \\\Rightarrow 3 \times 10^5=10^5+\frac{1}{2} \times 1000 \times {v_2}^2 $$
Solving the above equation we have, $$v_2=20m/s$$
Question 6
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Water rises in a vertical capillary tube upto a length of $$10cm.$$ If the tube is inclined at $$45^o$$, the length of water risen in the tube will be,

A
$$10 cm$$

B
$$10 \sqrt2 cm$$

C
$$\displaystyle \dfrac {10}{\sqrt2}$$

D
none of these

Solution

The vertical rise in the level of liquid is constant.
Hence for an inclined tube, the effective length is $$ \displaystyle\dfrac {h}{\cos\theta} $$
So, the length of water risen in the tube will be $$ 10 \sqrt 2 cm $$
Question 7
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In a streamline flow.

A
The speed of the particle always remains same

B
The velocity of the particle always remains same

C
The kinetic energies of all the particles arriving at a given point are same

D
The momenta of all the particles arriving at the given point are same

Solution

Streamline flow are a family of curves that are instantaneously tangent to the velocity vector of the flow.These show the direction a massless fluid element will travel in at any point in time.
so by the defination we can say that at any point kinetic energy will be same and if kinetic energy is same then moments(momenta) is also same at that point.
Question 8
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The height of water level in a tank is $$H$$. The range of water stream coming out of a hole at depth $$\dfrac{H}{4}$$ from upper water level will be.

A
$$\displaystyle \frac {\sqrt 3 H}{2}$$

B
$$\displaystyle \frac {2H}{\sqrt 3}$$

C
$$\displaystyle \frac {H}{\sqrt 3}$$

D
$$\sqrt 3 H$$

Solution

The horizontal range of the liquid coming out a hole is equal to $$2\sqrt{hh'}$$, where
$$h=$$ depth of the hole below the free surface of the liquid
$$h'=$$ height of the hole above the bottom of the tank

Here, $$h=\dfrac{H}{4}$$ and $$h'=H-\dfrac{H}{4} =\dfrac{3H}{4}$$

So range of the water stream $$=2\sqrt{\dfrac{H}{4} \times \dfrac{3H}{4}}=\dfrac{\sqrt{3}H}{2}$$
Question 9
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The profile of advancing liquid is the same at a given point at all times, then the flow is:

A
turbulent

B
rapid

C
viscous

D
streamlined

Solution

$$ Steamline\quad flow\quad is\quad defined\quad as\quad that\quad type\quad of\quad flow\quad in\quad which\quad the\quad fluid\quad $$
$$ characteristics\quad like\quad velocity,\quad pressure,\quad density,\quad etc.\quad at\\ a\quad point\quad do\quad not\quad change\quad with\quad time.\\ The\quad profile\quad being\quad same\quad states\quad that\\ \quad { \left( \displaystyle \frac { \partial V }{ \partial t } \right) }_{ { x }_{ 0 },{ y }_{ 0 },{ z }_{ 0 } }=0\quad ,\quad which\quad is\quad same\quad as\quad streamline\quad flow.$$
Question 10
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Two equal drops are falling through air with a steady velocity of $$5 cms^{-1}$$. If the drop coalesces, the new terminal velocity will become:

A
$$5 \times 2 cms^{-1}$$

B
$$5 \times \sqrt 2 cms^{-1}$$

C
$$5 \times (4)^{\frac {1}{3}} cms^{-1}$$

D
$$\displaystyle \frac {5}{\sqrt 2} cms^{-1}$$

Solution

when the two equal drops coalesces, volume of water remains same and a bigger drop is formed. Lets say $$r$$ is radius of drop before these combine and $$R$$ is radius after these combine. we have
$$2 \times \frac{4}{3} \pi r^3 =\frac{4}{3} \pi R^3 \\\Rightarrow R=2^{1/3}r $$
and we know that terminal velocity $$V_T \propto$$ radius$$^2$$, so we have
$$\dfrac{V_T'}{V_T}=\dfrac{R^2}{r^2} \\\Rightarrow V_T' =\left (\frac{R}{r} \right)^2 V_T = \left (2^{1/3} \right)^2 \times 5 cms^{-1}=5 \times 4^{\frac{1}{3}} cms^{-1}$$