Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 56

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Question 1
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A capillary tube with inner cross-section in the form of a square of side a is dipped vertically in a liquid of density $$ \rho $$ and surface tension $$ \sigma $$ which wet the surface of capillary tube with angle of contact $$ \theta $$. The approximate height to which liquid will be raised in the tube is : (Neglect the effect of surface tension at the corners of capillary tube)

A
$$ \dfrac{2\sigma cos\theta }{a\rho g} $$

B
$$ \dfrac{4\sigma cos\theta }{a\rho g} $$

C
$$ \dfrac{8\sigma cos\theta }{a\rho g} $$

D
None of these

Solution

Upward force by capillary tube on top surface of liquid is equal to the product of the perimeter and surface tension.

$$F_{up}= 4a\sigma \cos\theta$$ ... ..(1)

In equilibrium, surface tension force is equal to the weight of the liquid column of height h, h being the height to to which the liquid rises

$$W= a^2\rho gh$$ .....(2)

Equating (1) and (2)

$$4a\sigma \cos\theta = a^2\rho gh$$

$$ \therefore h= \dfrac{4\sigma \cos\theta}{a\rho g}$$
Question 2
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The figure shows a crude type of atomizer. When bulb A is compressed, air flows swiftly through tube BC causing reduced pressure in the particles of the vertical tube. The 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 atmospheric pressure, $$v$$ is the speed of air in BC, find how large would $$v$$ need to be to cause the liquid to rise to BC.
(Density of air $$=1.3 kg/m^3$$.)

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

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

Absolute pressure in bulb $$=$$ Gauge pressure + Atmospheric pressure $$=P+P_a$$
Using Bernoulli's equation,
$$P_a+P=P_{BC}+\displaystyle\frac {\rho_a v^2}{2}$$
where $$\rho_a$$ is the density of air.
$$\therefore P_{BC}=P_a+P=\left (\displaystyle\frac {1.3}{2}\right )v^2$$
$$P_{BC}=P_a-\rho gh$$

So equating these two values for $$P_{BC}$$, we get:
$$v=\sqrt{(\displaystyle\frac {P+\rho gh}{{0.65}})}$$
Question 3
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In a cylindrical water tank here are two small holes Q and P on the wall at a depth of $$h_1$$ from the upper level of water and at a height of $$h_2$$ from the lower end of the tank, respectively, as shown in the figure. Water coming out from both the holes strike the ground at the same point. The ratio of $$h_1$$ and $$h_2$$ is

A
$$1$$

B
$$2$$

C
$$> 1$$

D
$$< 1$$

Solution

The two streams strike at the same point on the ground.
$$R_1=R_2=R$$
$$u_1t_1=u_2t_2$$
where $$u_1$$ velocity of efflux at $$Q=\sqrt {(2gh_1)}$$ and $$u_2=$$ velocity of efflux at $$P=\sqrt {[2g(H-h_2)]}$$
$$t_1=$$ time of fall of water stream through Q is
$$\sqrt {\displaystyle\frac {2(H-h_1)}{g}}$$
$$t_2=$$ time of fall of the water stream through $$P=\sqrt {\displaystyle\frac {2h_2}{g}}$$
Putting theses values is Eq. (i), we get
$$\sqrt{2gh_1}(\sqrt{\displaystyle\frac{2(H-h_1)}{g}})=\sqrt{2g(H-h_2)}\sqrt{\displaystyle\frac{2h_2}{g}}$$
$$(H-h_1)h_1=(H-h_2)h_2$$
or $$[H-(h_1+h_2)][h_1-h_2]=0$$
$$H=h_1+h_2$$ is irrelevant because the holes are at two different heights. Therefore, $$h_1=h_2$$ or $$h_1/h_2=1$$
Question 4
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A hole is made at the bottom of a tank filled with water (density $$=10^3 kg/m^3)$$. If the total pressure at the bottom of the tank is $$3 atm$$ ($$1 atm=10^5 N/m^2$$), then the velocity of efflux is

A
$$\sqrt {400}m/s$$

B
$$\sqrt {200}m/s$$

C
$$\sqrt {600}m/s$$

D
$$\sqrt {500}m/s$$

Solution

Applying Bernoulli's equation to two points one on the surface and the other just outside the hole at the bottom,
$$P_{atm} + \rho gh + \dfrac{1}{2}\rho v^2 = P_{atm} + 0 + \dfrac{1}{2}\rho v'^2$$
Given,
$$ v=0$$
Pressure difference between top and bottom is $$2 atm$$ i.e equivalent of $$20m$$ of water column
$$ \therefore h= 20$$

$$\dfrac{1}{2}\rho v'^2 = \rho gh$$

$$ \therefore \dfrac{1}{2} \times 1 \times v'^2 = 10 \times 20$$

$$ \Rightarrow v' = \sqrt{400}$$
Question 5
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Equal volume of two immiscible liquids of densities $$\rho$$ and $$2\rho$$ are filled in a vessel as shown in the figure. Two small holes are punched at depths h/2 and 3h/2 from the surface of lighter liquid. If $$v_1$$ and $$v_2$$ are the velocities of efflux at these two holes, then $$v_1/v_2$$ is:

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

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

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

D
$$\displaystyle\frac {1}{\sqrt 2}$$

Solution

Using the velocity of efflux for the upper hole we get
$$v_1=\sqrt {2g\left (\displaystyle\frac {h}{2}\right )}=\sqrt {gh}$$
For the second hole we apply the Bernoulli's equation, thus we get
$$\rho gh+2\rho g\left (\displaystyle\frac {h}{2}\right )=\displaystyle\frac {1}{2}(2\rho )v_2^2$$
$$\Rightarrow v_2=\sqrt {2gh}$$
$$\therefore \displaystyle\frac {v_1}{v_2}=\frac {1}{\sqrt 2}$$
Question 6
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A cylindrical vessel contains a liquid of density $$\rho$$ up to height h. The liquid is closed by a piston of mass $$m$$ and area of cross section $$A$$. There is a small hole at the bottom of the vessel. The speed $$v$$ with which the liquid comes out of the hole is

A
$$\sqrt {2gh}$$

B
$$\sqrt {2\left (gh+\frac {mg}{\rho A}\right )}$$

C
$$\sqrt {2\left (gh+\frac {mg}{A}\right )}$$

D
$$\sqrt {2gh+\frac {mg}{A}}$$

Solution

Applying Bernoulli's theorem at points 1 and 2, difference in pressure energy between 1 and 2 $$=$$ difference in kinetic energy between 1 and 2.
Hence, $$h\rho g+\displaystyle\frac {mg}{A}=\frac {1}{2}\rho v^2$$
or
$$v=\sqrt {2gh+\displaystyle\frac {2mg}{\rho A}}=\sqrt {2\left (gh+\displaystyle\frac {mg}{\rho A}\right )}$$
Question 7
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A square box of water has a small hole located in one of the bottom corners. When the box is full and sitting on a level surface, complete opening of the hole results in a flow of water with a speed $$v_0$$, as shown in fig. When the box is still half empty, it is tilted by $$45^o$$ so that the hole is at the lowest point. Now the water will flow out with a speed of:

A
$$v_0$$

B
$$v_0\cdot 2$$

C
$$\dfrac {v_0}{\sqrt 2}$$

D
$$\dfrac {v_0}{\sqrt [4]{2}}$$

Solution

Using Bernoulli's principle we know that the velocity at the bottom is given as $$\sqrt {2gh}$$ where h is the height of the water level.
$$v_{o} = \sqrt{2gh}$$
Now when square is tilted, effective height = $$\displaystyle\frac{h}{\sqrt{2}}$$
Thus substituting this as as the height we get the answer as D.
$$v_1 = \displaystyle\sqrt{2g\frac{h}{\sqrt{2}}} = \displaystyle\frac{v_{o}}{\sqrt[4]{2}} $$
Question 8
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In the figure, the cross-sectional area of the smaller tube is $$a$$ and that of the larger tube is $$2a$$. A block of mass $$m$$ is kept in the smaller tube having the same base area $$a$$, as that of the tube. The difference between water levels of the two tubes is

A
$$\dfrac {P_0}{\rho g}+\dfrac {m}{a\rho}$$

B
$$\dfrac {P_0}{\rho g}+\dfrac {m}{2a\rho}$$

C
$$\dfrac {m}{a\rho}$$

D
$$\dfrac {m}{2\rho a}$$

Solution

By Pascal's Law, pressure is same at the same horizontal level in a liquid.
Let the heights of fluid above this level be $$ h $$ and $$ h_1 $$ as shown in the figure.
Pressure from part 1 is $$ P_0+h_1\rho g+\dfrac{mg}{a} $$
Pressure from part 3 is $$ P_0+h\rho g $$
Equating these two we have
$$ P_0+h_1\rho g+\dfrac{mg}{a}= P_0+h\rho g $$
$$ \therefore h-h_1=\dfrac{m}{a\rho} $$
Therefore difference in height of liquid in the two arms is $$ \dfrac{m}{a\rho} $$
Question 9
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A tank in filled with water of density $$10^3 kg/m^3$$ and oil of density $$0.9\times 10^3 kg/m^3$$. The height of water layer is $$1 m$$ and that of the oil layer is $$4 m$$. The velocity of efflux from an opening in the bottom of the tank is

A
$$\sqrt {85}m/s$$

B
$$\sqrt {88}m/s$$

C
$$\sqrt {92}m/s$$

D
$$\sqrt {98}m/s$$

Solution

Let $$d_w$$ and $$d_0$$ be the densities of water and oil, respectively. Then the pressure at the bottom of the tank is $$h_wd_wg+h_0d_0g$$.

Let this pressure be equivalent to pressure due to water of height h. Then,
$$hd_wg=h_wd_2g+h_0d_0g$$
$$\therefore h=h_w+\displaystyle\dfrac {h_0d_0}{d_w}=1+\dfrac {4\times 0.9}{1}$$
$$=1+3.6=4.6 m$$

According to Toricelli's theorem,
$$v=\sqrt {2gh}=\sqrt {2\times 10\times 4.6 m/s}=\sqrt {92}m/s$$
Question 10
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In a cylindrical vessel containing liquid of density $$\rho$$, there are two holes in the side walls at heights of $$h_1$$ and $$h_2$$ respectively such that the range of efflux at the bottom of the vessel is same. Find the height of a hole, for which the range of efflux would be maximum.

A
$$\dfrac {h_2+h_1}{2}$$

B
$$\dfrac {h_2-h_1}{2}$$

C
$$\dfrac {h_2h_1}{2}$$

D
$$\dfrac {h_2}{h_1}$$

Solution

Hint: The range of efflux is $$x=2\sqrt{(H-h)h}$$
Correct Answer: Option A
Explanation of correct option:
$$\textbf{Step1: Find height of cylinder}$$
Two holes are present in the side walls at heights $$h_1$$ and $$h_2$$ in such a way that the range of efflux at the bottom of the vessel is same,
$$\therefore x=2\sqrt{(H-h_1)h_1}=2\sqrt{(H-h_2)h_2}$$

$$\therefore (H-h_1)h_1=(H-h_2)h_2$$

$$H(h_1-h_2)=(h_1)^2-(h_2)^2$$

$$\therefore H=\dfrac{{h_1}^2-{h_2}^2}{h_1-h_2}$$

Multiplying and dividing by $$(h_1+h_2)$$

$$H=\dfrac{(h_1)^2-(h_2)^2}{h_1-h_2} \dfrac{(h_1+h_2)}{(h_1+h_2)}$$

Since, $$(h_1+h_2)(h_1-h_2)=(h_1)^2-(h_2)^2$$

$$\therefore H=h_1+h_2$$
$$\textbf{Step2: Find height of hole}$$
Also, maximum height for which the range of efflux is maximum is half of the heights of the holes,

$$h=\dfrac{H}{2}$$

$$\therefore h=\dfrac{h_1+h_2}{2}$$