Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 9

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Weekly Quiz Competition

Question 1
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A tank with a small hole at the bottom has been filled with water and kerosene (specific gravity $$0.8$$). The height of water is $$3 m$$ and that of kerosene $$2 m$$. When the hole is opened the velocity of fluid coming out from it is nearly : (take $$g= 10 ms^{-2}$$ and density of water $$ =10^3 kg m^{-3})$$

A
$$8.5 ms^{-1}$$

B
$$9.6 ms^{-1}$$

C
$$10.7 ms^{-1}$$

D
$$7.6 ms^{-1}$$

Solution

From Torricelli's equation,
$$\rho _{k}\times g\times h_{k} + \rho _{w}\times g\times h_{w} = \dfrac{1}{2}\times \rho\times v^{2}$$
$$\therefore (0.8\times 2 + 1\times 3 )\times 9.81 = \dfrac{1}{2}\times 1\times v^{2}$$
$$\therefore v = 9.6 m/s $$
Question 2
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Two liquids of densities $${ \rho }_{ 1 }$$ and $${ \rho }_{ 2 }\left( { \rho }_{ 2 }=2{ \rho }_{ 1 } \right) $$ are filled up behind a square wall of side $$10m$$ as shown in figure. Each liquid has a height of $$5m$$. The ratio of the forces due to these liquids exerted on upper part $$MN$$ to that at the lower part $$NO$$ is (Assume that the liquids are not mixing)

A
$$1/4$$

B
$$2/3$$

C
$$1/2$$

D
$$1/3$$

Solution

Let $$\rho_1 = \rho$$ and $$\rho_2 = 2\rho$$

Let $$P_A$$ be the pressure at the top $$P_B$$ be the pressure at the interface between the liquids and $$P_C$$ be the pressure at the bottom of the wall.

Let $$F_1$$ and $$F_2$$ denote the forces acting on the upper part of the wall and the lower part of the wall respectively.

$$P_A = P_0 = 0$$

$$P_B = \rho_1 g h=\rho g h$$

$$P_B = \rho g \times 5 = 5 \rho g$$

$$P_C = \rho_1 g\ 5 + \rho_2 g\ 5=\rho g\ 5 + 2\rho g\ 5$$

$$P_C=15Pg$$

$$F_1 = \left(\dfrac{P_A + P_B}{2}\right) = \left( \dfrac{0+5\rho g}{2}\right).A = \dfrac{5\rho gA}{2}$$

$$F_2 = \left( \dfrac{P_B + P_C}{2}\right) \times A = \left( \dfrac{5\rho g+15\rho g}{2}\right).A = \dfrac{20\rho gA}{2}$$

$$\dfrac{F_1}{F_2} = \dfrac{\tfrac{5\rho gA}{2}}{ \tfrac{20\rho g A}{2}} = \dfrac{1}{4}$$

$$\dfrac{F_1}{F_2} = \dfrac{1}{4}$$
Question 3
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lf the terminal speed of a sphere of gold (liquid density $$=1kg/{m^3,density\ of\ solid=5kg/{m^3}}$$ ,gold density=$$15kg/{m^3})$$of the same size in the same liquid.Take $$V_s=4m/s$$

A
0.2 $$\mathrm{m}/\mathrm{s}$$

B
0.4

C
0.133 $$\mathrm{m}/\mathrm{s}$$

D
$$1.14m/s$$

Solution

$${V_{g}}=\dfrac{(\rho_{s}-\rho_{l})}{(\rho_{g}-\rho_{l})}V_{s}=1.14m/s$$
Question 4
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A capillary tube (A) is dipped in water. Another identical tube (B) is dipped in a soap-water solution. Which of the following shows the relative nature of the liquid columns in the two tubes?

A

B

C

D

Solution

The weight of the water risen in capillary tube is balanced by the force exerted due to decrease in pressure inside the water column due to the curvature at the water surface.
The decrease in pressure=$$\dfrac{2\sigma}{r}$$
Hence from balancing the pressure, we get
$$\dfrac{2\sigma}{r}=h\rho g$$
Hence $$h\propto \sigma$$
Since surface tension($$\sigma$$) of water decreases as soap is mixed, the height upto which water rises decreases.
Question 5
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Directions For Questions

A spray gun is shown in the figure where a piston pushes air out of a nozzle. A thin tube of uniform cross section is connected to the nozzle. The other end of the tube is in a small liquid container. As the piston pushes air through the nozzle, the liquid from the container rises into the nozzle and is sprayed out. For the spray gun shown, the radii of the piston and the nozzle are $$20 mm$$ and $$1 mm$$ respectively. The upper end of the container is open to the atmosphere.

...view full instructions

If the density of air is $$\rho_a$$ and that of the liquid $$\rho_l$$, then for a given piston speed the rate (volume per unit time) at which the liquid is sprayed will be proportional to

A
$$\displaystyle \sqrt{\dfrac{\rho_a}{\rho_l}}$$

B
$$\sqrt{\rho_a \rho_l}$$

C
$$\displaystyle \sqrt{\dfrac{\rho_l}{\rho_a}}$$

D
$$\rho_l$$

Solution

From the principle of continuity,
$$ \dfrac{1}{2} \rho_{1} v_{1} ^{2} = \dfrac{1}{2} \rho_{2} v_{2} ^{2} $$
$$ \therefore \dfrac{v_{1}}{v_{2}} = \sqrt{\dfrac{\rho_{2}}{\rho_{1}} } $$
Question 6
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Ratio of area of hole to beaker is $$ 0.1$$ Height of liquid in beaker is $$3m$$, and hole is at the height of $$52.5 cm$$ from the bottom of beaker, find the square of the velocity of liquid coming out from the hole:

A
$$50 \ (m/s)^2$$

B
$$50.5 \ (m/s)^2$$

C
$$51 \ (m/s)^2$$

D
$$42 \ (m/s)^2$$

Solution

Given : Ratio of area of hole to that of beaker $$\dfrac{A_h}{A_b} = 0.1$$
From the figure, $$h = 3 - 0.525 =2.475 m$$
$$\therefore$$ Velocity of liquid coming out of hole $$v = \sqrt{\dfrac{2gh}{1 - \frac{A_h^2}{A_b^2}}}$$
$$\therefore$$ $$v = \sqrt{\dfrac{2\times 10 \times 2.475}{1 - 0.01}} = \sqrt{50}$$
$$\implies$$ $$v^2 = 50$$ $$(m/s)^2$$
Question 7
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A glass capillary tube is of the shape of truncated cone with an apex angle $$\alpha$$ so that its two ends have cross sections of different radii. When dipped in water vertically, water rises in it to a height $$h$$, where the radius of its cross section is $$b$$. If the surface tension of water is $$S$$, its density is $$\rho$$, and its contact angle with glass is $$\theta$$, the value of $$h$$ will be ($$g$$ is the acceleration due to gravity)

A
$$\dfrac{2S}{b \rho g} \cos (\theta - \alpha )$$

B
$$\dfrac{2S}{b \rho g} \cos (\theta+ \alpha )$$

C
$$\dfrac{2S}{b \rho g} \cos (\theta - \alpha /2)$$

D
$$\dfrac{2S}{b \rho g} \cos (\theta + \alpha /2)$$

Solution

Let r = radius of curvature of meniscus
$$\displaystyle b = r\, cos \left ( \theta +\frac{\alpha}{2} \right )$$
Excess pressure on concave side of meniscus $$\displaystyle \frac {2S}{r}$$
$$\displaystyle \Rightarrow (P_0 + h\rho g) - P_0 = \frac {2S}{r}$$

$$\displaystyle h\rho g =\frac {2S cos \left (\theta +\frac{\alpha}{2} \right ) }{b}$$

or $$\displaystyle h =\frac {2S}{b\rho g} cos \left (\theta +\frac{\alpha}{2} \right )$$

Question 8

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A person in a lift is holding a water jar, which has a small hole at the lower end of its side. When the lift is at rest, the water jet coming out of the hole hits the floor of the lift at a distance $$d$$ of $$1.2 m$$ from the person. In the following, state of the lift's motion is given in List I and the distance where the water jet hits the floor of the lift is given in List II. Match the statements from List I with those in List II and select the correct answer using the code given below the lists.

	List I		List II
P.	Lift is accelerating vertically up.	1.	$$d= 1.2 m$$
Q.	Lift is accelerating vertically down with an acceleration less than the gravitational acceleration.	2.	$$d > 1.2 m$$
R.	Lift is moving vertically up with constant speed.	3.	$$d < 1.2 m$$
S.	Lift is falling freely.	4.	No water leaks out of the jar

P-2, Q-3, R-2, S-4

P-2, Q-3, R-1, S-4

P-1, Q-1, R-1, S-4

P-2, Q-3, R-1, S-1

Solution

Velocity of efflux $$V=\sqrt{2g_{eff}l}$$
where $$l$$ is length of jar from top of point.
time taken in covering horizontal distance is $$T= \sqrt{\dfrac{2h}{g_{eff}}}$$, where $$h$$ is distance of floor from jar
Horizontal range = $$VT==\sqrt{2g_{eff}l}\sqrt{\dfrac{2h}{g_{eff}}}= \sqrt{4lh}$$
which is independent of $$g_{eff}$$.
In free fall velocity of efflux will be zero.

Question 9
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A wind with speed $$40$$ m/s blows parallel to the roof of a house. The area of the roof is $$250\ m^2$$. Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be $$(\rho_{air}=1.2 \ kg/m^3)$$

A
$$2.4\times 10^5N$$, upward

B
$$2.4\times 10^5N$$, downward

C
$$4.8\times 10^5N$$, downward

D
$$4.8\times 10^5N$$, upward

Solution

Applying Bernoulli's theorem to two points just inside the house and outside the house,

$$P + \dfrac{1}{2}\rho v^2 = constant$$

Inside the pressure is $$P_{atm}$$

The pressure outside is $$P$$

v is zero inside.

$$ \therefore P - P_{atm} = \frac{1}{2}\rho v^2= \frac{1}{2} \times 1.2 \times 40^2=960$$

Force is : $$F = (P - P_{atm}) \times Area= 960 \times 250 = 2.4 \times 10^5\:N$$ in upward direction as pressure goes from higher to lower pressure area.
Question 10
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A small hole of an area of cross-section $$2 mm^2$$ is present near the bottom of a fully filled open tank of height 2 m. Taking $$g = 10 m/s^2$$, the rate of flow of water through the open hole would be nearly

A
$$12.6 \times 10^{-6} m^3/s$$

B
$$8.9 \times 10^{-6} m^3/s$$

C
$$2.23 \times 10^{-6} m^3/s$$

D
$$6.4 \times 10^{-6} m^3/s$$

Solution

Rate of flow liquid
$$Q = au = a \sqrt{2 gh}$$
$$= 2 \times 10^{-6} m^2 \times \sqrt{2 \times 10 \times 2} m/s$$
$$= 2 \times 2 \times 3.14 \times 10^{-6} m^3/s$$
$$= 12.56 \times 10^{-6} m^3/s$$
$$=12.6 \times 10^{-6} m^3/s$$

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Mechanical Properties of Fluids Test - 9

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Mechanical Properties of Fluids Test - 9

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Question 1

$$8.5 ms^{-1}$$

$$9.6 ms^{-1}$$

$$10.7 ms^{-1}$$

$$7.6 ms^{-1}$$

Solution

Question 2

$$1/4$$

$$2/3$$

$$1/2$$

$$1/3$$

Solution

Question 3

0.2 $$\mathrm{m}/\mathrm{s}$$

0.4

0.133 $$\mathrm{m}/\mathrm{s}$$

$$1.14m/s$$

Solution

Question 4

Solution

Question 5

$$\displaystyle \sqrt{\dfrac{\rho_a}{\rho_l}}$$

$$\sqrt{\rho_a \rho_l}$$

$$\displaystyle \sqrt{\dfrac{\rho_l}{\rho_a}}$$

$$\rho_l$$

Solution

Question 6

$$50 \ (m/s)^2$$

$$50.5 \ (m/s)^2$$

$$51 \ (m/s)^2$$

$$42 \ (m/s)^2$$

Solution

Question 7

$$\dfrac{2S}{b \rho g} \cos (\theta - \alpha )$$

$$\dfrac{2S}{b \rho g} \cos (\theta+ \alpha )$$

$$\dfrac{2S}{b \rho g} \cos (\theta - \alpha /2)$$

$$\dfrac{2S}{b \rho g} \cos (\theta + \alpha /2)$$

Solution

Question 8

P-2, Q-3, R-2, S-4

P-2, Q-3, R-1, S-4

P-1, Q-1, R-1, S-4

P-2, Q-3, R-1, S-1

Solution

Question 9

$$2.4\times 10^5N$$, upward

$$2.4\times 10^5N$$, downward

$$4.8\times 10^5N$$, downward

$$4.8\times 10^5N$$, upward

Solution

Question 10

$$12.6 \times 10^{-6} m^3/s$$

$$8.9 \times 10^{-6} m^3/s$$

$$2.23 \times 10^{-6} m^3/s$$

$$6.4 \times 10^{-6} m^3/s$$

Solution

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