Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

The velocity of the wind over the surface of the wing of an aeroplane is 80 ms$$^{-1}$$ and under the wing 60 ms$$^{-1}$$. If the area of the wing is 4m$$^{2}$$, the dynamic lift experienced by the wing is [ density of air $$=$$ 1.3 kg. m$$^{-3}$$]:

A
3640 N

B
7280 N

C
14560 N

D
72800 N

Solution

Dynamic lift $$ = \dfrac{1}{2}A\rho \left ( v^{2}_{2}- v^{2}_{1}\right )$$

$$ = \dfrac{1}{2}\times 1.3\times \left ( 2800 \right )\times 4$$
$$ = 7280$$ N
Question 2
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There is a hole at the side-bottom of a big water tank. The area of the hole is $$2\ mm^{2}$$. Through it, a pipe is connected. The upper surface of the water is $$5\ m$$ above the hole. The rate of flow of water through the pipe is ( in $$m^{3}s^{-1}$$) ( $$g= 10 \ ms^{-2}$$)

A
$$4 \times 10^{-5}$$

B
$$4 \times 10^{5}$$

C
$$4 \times 10^{-6}$$

D
$$28 \times 10^{-6}$$

Solution

Velocity of water at the side bottom hole $$=\sqrt{2gh}$$$$=\sqrt{2 \times 10 \times 5}=10 \ m/s$$
Area of hole = $$4 mm^{2}= 4 \times 10^{-6}\ m^{2}$$
So, rate of flow of water through pipe = $$4 \times 10^{-6} \times 10 \ m^{3}/s$$$$=4 \times 10^{-5} \ m^{-3}/s$$
Question 3
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Directions For Questions

A cylindrical tank of $$1$$ meter radius rests on a platform $$5m$$ high. Initially, the tank is filled with water to a height of $$5m. $$ A plug whose area is $$10^{-4}m^{2}$$ is removed from an orifice on the side of the tank at the bottom.

...view full instructions

The initial speed with which water strikes the ground in $$ms^{-1}$$ is:

A
$$10$$

B
$$5$$

C
$$5\sqrt{2}$$

D
$$10\sqrt{2}$$

Solution

Velocity of efflux for a given body = $$ ( \dfrac{ 2gh }{ 1 - \dfrac {a ^{2}} {A^{2}} } )^{0.5}$$ where "h" is the height from the ground , 'A' is the area of the tank and "a" is the area of the orifice .
now here "h" = 10m , a = $$ 10^{-4} m^{2}$$ , A = $$ \pi r^{2} $$ = $$\pi m^{2}$$
so $$ ( 1 - \dfrac{ a^{2} }{ A ^{2} } ) $$ is approximately = 1 in this case
so velocity of efflux in this case = $$ ( 2(10)(10) )^ {\dfrac{1}{2} }$$ = $$ (200)^{0.5}$$
so option (D) is correct
Question 4
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Calculate the velocity of efflux of kerosene oil from an orifice of a tank in which pressure is $$4 atm$$. The density of kerosene oil $$= 720 kg m^{-3}$$ and $$1$$ atmospheric pressure $$= 1.013 \times 10^{5}N m^{-2}$$.

A
$$3.355 m/s$$

B
$$33.55 m/s$$

C
$$335.5 m/s$$

D
$$33.55 cm/s$$

Solution

From Bernoulli's Theorem,
$$P_{1}=\dfrac{1}{2}\rho v^{2}$$
$$\Rightarrow 4 atm = \dfrac{1}{2}\times \rho V^{2}$$
$$\Rightarrow \dfrac{4 \times 1.013 \times 10^{5} \times 2}{720}=V^{2}$$
$$\Rightarrow V=33.549 m/s$$
$$\Rightarrow V \approx 33.55 m/s$$
Question 5
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In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are $$70m s^{-1}$$ and $$63 ms^{-1}$$ respectively. What is the lift on the wing, if its area is $$2.5 m^{2}$$ ? Take the density of air to be $$1.3kg m^{-3}$$.

A
$$1513 N$$

B
$$1513\ dynes$$

C
$$151.3 N$$

D
$$151.3\ dynes$$

Solution

Applying Bernoulli's principle to above and below the using,
$$P_{1}+\dfrac{1}{2}PV_{1}^{2}=P_{2}+\dfrac{1}{2}PV_{2}^{2}$$
$$\Rightarrow P_{1}-P_{2}=\dfrac{1}{2}\times 1.3 \times (70^{2}-63^{2})$$
$$\Rightarrow$$ Pressure differnece across sing = $$605.15N/m^{2}$$
Lift on the using = pressure difference across surface of using $$ \times $$ Area.
$$= 605.15 N/m^{2} \times 2.5 m^{2}$$
$$=1512.875 N$$
$$\approx 1513 N$$
Question 6
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Section-A Section-B
a) Incompressible liquid e) Density constant
b) Turbulent flow f) Stream lines
c) Tube of flow g) Constant
d) Fluid flux rate in laminar flow h) Reynold's no $$> 2000$$

A
a-f; b-e; c-g; d-h

B
a-e; b-h; c-f; d-g

C
a-g; b-f; c-e; d-h

D
a-h; b-g; c-e; d-f

Solution

Incompressibility means that the density of fluid is same at all the points and remains same as time passes.
The motion of water in high fall of at high speed is called turbulent flow. Reynold's no > $$2000$$ for turbulent flow.
Streamline flow is flow of liquid in a line of flow.
In laminar flow, fluid flux rate is constant.
Question 7
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Directions For Questions

A cylindrical tank of $$1$$ meter radius rests on a platform $$5m$$ high. Initially, the tank is filled with water to a height of $$5m. $$ A plug whose area is $$10^{-4}m^{2}$$ is removed from an orifice on the side of the tank at the bottom.

...view full instructions

The initial speed with which water flows out from the orifice in ms$$^{-1}$$ is (g $$=$$ 10ms$$^{-2}$$):

A
10

B
5

C
5.$$\sqrt{2}$$

D
10.$$\sqrt{2}$$

Solution

Speed at which water flows out from the orifice= $$\sqrt{2 \times g \times h}=\sqrt{5 \times 2 \times 10}$$ =10 m/s
where h is the height of water in the tank.
Question 8
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In a horizontal pipe line of uniform cross-section,pressure falls by $$5\ Pa$$ between two points separated by $$1\ km$$. The change in the kinetic energy per kg of the oil flowing at these points is :(Density of oil $$= 800\ kgm^{-3}$$)

A
$$6.25 \times 10^{-3}\ Jkg^{-1}$$

B
$$5.25 \times 10^{-4}\ Jkg^{-1}$$

C
$$3.25 \times 10^{-5}\ Jkg^{-1}$$

D
$$4.25 \times 10^{-2}\ Jkg^{-1}$$

Solution

from Bernoulli's therom,
$$P_{1}+\dfrac{1}{2}\rho V_{1}^{2}=P_{2}+\dfrac{1}{2}\rho V_{2}^{2}$$
$$\Rightarrow 5 Pa- \dfrac{1}{2} \times 800 \times (V_{2}^{2}-V_{1}^{2})$$
$$\Rightarrow \dfrac{100}{800}= V_{2}^{2}-V_{1}^{2}\Rightarrow 0.0125=V_{2}^{2}-V_{1}^{2}$$
$$\Rightarrow \dfrac{1}{2}m(V_{2}^{2}-V_{1}^{2})=\dfrac{1}{2} \times 1 \times0.0125$$[where m=1kgs]
$$\Rightarrow$$ Change in K.E per kg $$= 6.24 \times 10^{-3}/kh^{-1}$$
Question 9
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A horizontal pipe of non uniform cross section has water flow through it such that the velocity is $$2ms^{-1}$$ at a point where the pressure is $$40 kPa$$. The pressure at a point where the velocity of water flow is $$3 ms^{-1}$$ is : ( in $$kPa$$)

A
$$27$$

B
$$60$$

C
$$37.5$$

D
$$40$$

Solution

Using the bernoulli's therom,
$$P_{1}+\dfrac{1}{2}pV_{1}^{2}=P_{2}+\dfrac{1}{2}pV_{2}^{2}$$
$$\Rightarrow 40 \times 10^{3}+\dfrac{1}{2} \times 500 \times 4=P_{2}+\dfrac{1}{2}\times{1000}\times 9$$
$$\Rightarrow 40 \times 10^{3}+2000 - 45000 =P_{2}$$
$$\Rightarrow 37.5 \times 10^{3} Pa= P_{2}$$
$$\Rightarrow P_{2}=37.5 kPa$$
Question 10
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A room has a window of area $$A$$. Out side of the room wind is blowing parallel to the window with a velocity $$v$$. If the density of air is $$\rho $$ , then the force acting on the window is:

A
$$\dfrac{1}{2}\dfrac{\rho v^{2}}{A}$$

B
$$\dfrac{1}{2}\rho v^{2}A$$

C
$$\rho v^{2}A$$

D
$$2\rho v^{2}A$$

Solution

As the wind is blowing with velocity $$v$$ outside the window so, pressure will decrease from Bernoulli's equation as following :
Applying bernouillis theorem across the window ,
$$p_{1}+\cfrac{1}{2}\rho v^{2}=p_{2}$$
$$\Rightarrow p_{2}-p_{1}=\cfrac{1}{2}\rho v^{2}$$
So, force on the window =pressure difference x Area $$ = \Delta p\times A$$$$=\cfrac{1}{2}\rho Av^{2}$$

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Re-Attempt Weekly Quiz Competition

Section-A	Section-B
a) Incompressible liquid	e) Density constant
b) Turbulent flow	f) Stream lines
c) Tube of flow	g) Constant
d) Fluid flux rate in laminar flow	h) Reynold's no $$> 2000$$

Mechanical Properties of Fluids Test - 17

Result

Mechanical Properties of Fluids Test - 17

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Rank

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SHARING IS CARING

Question 1

3640 N

7280 N

14560 N

72800 N

Solution

Question 2

$$4 \times 10^{-5}$$

$$4 \times 10^{5}$$

$$4 \times 10^{-6}$$

$$28 \times 10^{-6}$$

Solution

Question 3

$$10$$

$$5$$

$$5\sqrt{2}$$

$$10\sqrt{2}$$

Solution

Question 4

$$3.355 m/s$$

$$33.55 m/s$$

$$335.5 m/s$$

$$33.55 cm/s$$

Solution

Question 5

$$1513 N$$

$$1513\ dynes$$

$$151.3 N$$

$$151.3\ dynes$$

Solution

Question 6

a-f; b-e; c-g; d-h

a-e; b-h; c-f; d-g

a-g; b-f; c-e; d-h

a-h; b-g; c-e; d-f

Solution

Question 7

10

5

5.$$\sqrt{2}$$

10.$$\sqrt{2}$$

Solution

Question 8

$$6.25 \times 10^{-3}\ Jkg^{-1}$$

$$5.25 \times 10^{-4}\ Jkg^{-1}$$

$$3.25 \times 10^{-5}\ Jkg^{-1}$$

$$4.25 \times 10^{-2}\ Jkg^{-1}$$

Solution

Question 9

$$27$$

$$60$$

$$37.5$$

$$40$$

Solution

Question 10

$$\dfrac{1}{2}\dfrac{\rho v^{2}}{A}$$

$$\dfrac{1}{2}\rho v^{2}A$$

$$\rho v^{2}A$$

$$2\rho v^{2}A$$

Solution

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