Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

Applications of Bernoulli's theorem can be seen in the

A
dynamic lift of Aeroplane

B
hydraulic press

C
helicopter

D
none of these

Solution

The shape of the aeroplane wings is such that when it moves forward, the air molecules at the top of the wings have a greater velocity (relative to the wings) compared to the air molecules at the bottom. Therefore in accordance with Bernoulli's principle, the pressure at the top of the wings is less than that at the bottom. This results in a dynamic lift of the wings which balances the weight of the plane.
Question 2
1 / -0

A 50 kg girl wearing heel shoes balances on a single heel. The heel is circular with a diameter 1 cm. The pressure exerted by the heel on the horizontal floor is then (Take g = 10 m s$$^{-2}$$)

A
6.4 x 10$$^4$$ Pa.

B
6.4 x 10$$^5$$ pa

C
6.4 x 10$$^6$$ Pa

D
6.4 x 10$$^7$$ Pa

Solution

Here,$$m =50 kg, $$

$$D = 1 cm = 10^{-2} m$$,

$$g= 10 m s^{-2}$$

$$\therefore$$ Pressure exerted by the heel on the horizontal floor is
$$Pressure=\dfrac{Force}{Area}$$ Force here is WEIGHT and area is area of circular portion of heel.

$$\displaystyle P=\frac{F}{A} = \frac{mg}{\pi \left(\dfrac{D}{2}\right)^{2}}=\frac{4 mg}{\pi D^2}$$

$$\displaystyle = \frac{4\times 50 kg \times 10\,m\,s^{-2}}{3.14\times(10^{-2}\,m)^2}=6.4 \times 10^6\,pa$$
Question 3
1 / -0

In a wind tunnel experiment, the pressures on the upper and lower surfaces of the wings are 0.90 x$$10^5$$ Pa and 0.91 x 10$$^5$$ Pa respectively. If the area of the wing is 40 m$$^2$$ the net lifting force on the wing is

A
2 x 10$$^4$$ N

B
4 x 10$$^4$$ N

C
6 x 10$$^4$$ N

D
8 x 10$$^4$$ N

Solution

Given data,
Pressure of the upper surface $$P_{2}=0.91\times10^{5}Pa$$
Pressure of the lower surface $$P_{1}=0.90\times10^{5}Pa$$
Area$$=40m^{2}$$
Force$$=?$$
Pressure difference which provides the lift
= pressure difference xx Area of the wing
$$F=(P_2−P_1)\times A$$

$$F=(0.91\times 10^{5}−0.90\times 10^{5})Pa\times ms^{−2}$$

$$F=0.01\times10^{5}Pa\times 40m^{2}$$

$$F=4\times 10^{4}N.$$
Question 4
1 / -0

Hydraulic brakes are based on

A
Pascal's law

B
Torricelli's law

C
Newton's law

D
Boyle's law

Solution

Hydraulic brakes work on the principle of Pascal’s law. According to this law whenever pressure is applied on fluid it travels uniformly in all the directions. Therefore when we apply force on a small piston, the pressure gets created which is transmitted through the fluid to a larger piston. As a result of this larger force, uniform braking is applied on all four wheels. As braking force is generated due to hydraulic pressure, they are known as hydraulic brakes. Liquids are used instead of gas as liquids are incompressible.
Question 5
1 / -0

A steam of water flowing horizontally with a speed of $$25 ms^{-1}$$ gushes out of a tube of cross-sectional area $$10^{-3}m^2$$, and hits at a vertical wall nearby. What is the force exerted on the wall by the impact of water?

A
$$125\ N$$

B
$$625\ N$$

C
$$-650\ N$$

D
$$-1125\ N$$

Solution

Here, $$u=25 ms^{-1}, v= 0, t= 1s, A=10^{-3}m^2$$
Density of water, $$rho=1000 kg m^{-3}$$

Mass of water gushed out per second, m =$$\dfrac{volume\times density}{time}=\dfrac{area\times distance\times density}{time}$$

= $$Area \times velocity \times density$$

$$=A v\rho = 10^{-3}\times 25\times1000=25 kg$$
Force exerted on the wall by the impact of water,

$$F=ma= m(\dfrac{v-u}{t})= 25\times (\dfrac{-2}{1})= 625 N$$

F'=-F= 625 N
Question 6
1 / -0

Spherical balls of radius R are falling in a viscous fluid of velocity v. The retarding viscous force acting on the spherical ball is

A
directly proportional to R but inversely proportional to v.

B
directly proportional to both radius R and velocity v.

C
inversely proportional to both radius R and velocity v

D
inversely proportional to R but directly proportional to velocity v

Solution

Retarding force acting on a ball falling into a viscous fluid $$F=6\pi\eta Rv$$
Where $$R=$$Radius of ball
$$v=$$velocity of ball
$$\eta=$$coefficient of viscosity
$$\therefore F\propto R$$ and $$F\propto v$$ or in words,
Retarding force is directly proportional to both $$R$$ and $$v$$
Question 7
1 / -0

At what velocity does water emerge from an orifice in a tank in which gauge pressure is
$$3\times 10^5N m^{-2}$$ before the flow starts? (Take the density of water= 1000 kg m$$^{-3}$$.)

A
$$24.5 m s^{-1}$$

B
$$14.5 m s^{-1}$$

C
$$34.5 m s^{-1}$$

D
$$44.5 m s^{-1}$$

Solution

Here,$$P = 3 \times 10^5 Nm^{-2},$$
$$ \rho = 1000 kgm^{-3},$$
$$g = 9.8 ms^{-2}$$
As $$P = h\rho g$$

$$\therefore \displaystyle h = \frac{P}{\rho g}= \frac {3 \times 10^5}{1000 \times 9.8} m$$

Velocity of efflux,

$$\displaystyle v = \sqrt {2gh} = \sqrt \frac {2 \times 9.8 \times 3 \times 10^5}{1000 \times 9.8}$$

$$= \sqrt {600} = 24.495 \,ms^{-1} = 24.5 \,ms^{-1}$$
Question 8
1 / -0

Which of the following device is used to measure the rate of flow of liquid through a pipe?

A
Thermometer

B
Barometer

C
Manometer

D
Venturimeter

Solution

A venturimeter is a device that is used for measuring the rate flow of a fluid flowing throughout the pipe.
Question 9
1 / -0

Two capillaries of same length and radii in the ratio 1: 2 are.connected in series. A liquid flows through them in streamlined condition. If the pressure across the two extreme ends of the combination is 1 m of water, the pressure difference across first capillary is

A
9.4 m

B
4.9 m

C
0.49 m

D
0.94 m

Solution

Here, $$l_1 = l_2 = 1m$$ and $$\displaystyle \frac{r_1}{r_2} = \frac{1}{2}$$

As $$V = \displaystyle \frac{ \pi P_1 r_1^4 }{8 \eta l} = \frac{ \pi P_2 r_2^4 }{8 \eta l}$$ or $$\displaystyle \frac{ P_1 }{P_2 } = \left( \frac{ r_2 }{ r_1} \right)^4 = 16$$

$$\therefore P_1 = 16 P_2$$

Since, both tubes are connected in series, hence pressure difference across the combination is
$$P = P_1 + P_2$$ $$\Rightarrow$$ $$\displaystyle 1 = P_1 + \frac{P_1}{16}$$
or $$\displaystyle P_1 = \frac{16}{17} = 0.94 m$$
Question 10
1 / -0

Directions For Questions

Fluids at rest exert a normal force to the walls of the container or to the surface of the body immersed in the fluid. The pressure exerted by this force at a point inside the liquid is the sum of atmospheric pressure and a factor which depends on the density of the liquid, the acceleration due to gravity and the height of the liquid above that point. The upthrust acting on a body immersed in a stationary liquid is the net force acting on the body in the upward direction. A number of phenomenon of liquids in motion can be explain by Bernoullis theorem which relates the pressure, flow speed and height for flow of an ideal incompressible fluid.
.A container of large uniform cross-sectional area $$A$$ resting on a horizontal surface holds two immiscible, non-viscous and incompressible liquids of densities $$d$$ and $$2d$$, each of height $$H/2$$ as shown in the figure. The lower density liquid is open to the atmosphere having pressure $${P}_{0}$$.
Situation II:
The cylinder is removed and the original arrangement is restored. A tiny hole of area $$s (s\ll A)$$ is punched on the vertical side of the container at a height $$h (h<H/2)$$.

...view full instructions

The horizontal distance $$x$$ travelled by the liquid initially is

A
$$\sqrt { \left( H-3h \right) h }$$

B
$$\sqrt { 3Hg }$$

C
$$\sqrt { \left( 3H-4h \right) h }$$

D
$$2h$$

Solution

Applying Bernoulli's equation at $$A$$ & $$B$$-
$${ P }_{ 0 }+dg\dfrac { H }{ 2 } +2d.g.\left( \dfrac { H }{ 2 } -h \right) ={ P }_{ 0 }+\dfrac { 1 }{ 2 } \left( 2d \right) { V }^{ 2 }$$

$$\dfrac { dgH }{ 2 } +dgH-2dgh=d{ V }^{ 2 }$$

$$\Rightarrow \dfrac { 3dHg }{ 2 } -2dgh={ V }^{ 2 }d$$

$$\Rightarrow \dfrac { 3Hg }{ 2 } -2gh={ V }^{ 2 }$$

$$\Rightarrow V=\sqrt { \left( 3H-4gh \right) \dfrac { g }{ 2 } } $$
$$x=V(t)$$
$$\Rightarrow x=V\sqrt { \dfrac { 2h }{ g } } $$
$$\Rightarrow x=\sqrt { \left( 3H-4h \right) \dfrac { g }{ 2 } } \sqrt { \dfrac { 2h }{ g } } $$
$$\Rightarrow x=\sqrt { \left( 3H-4h \right) .h } $$

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Mechanical Properties of Fluids Test - 43

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Mechanical Properties of Fluids Test - 43

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SHARING IS CARING

Question 1

dynamic lift of Aeroplane

hydraulic press

helicopter

none of these

Solution

Question 2

6.4 x 10$$^4$$ Pa.

6.4 x 10$$^5$$ pa

6.4 x 10$$^6$$ Pa

6.4 x 10$$^7$$ Pa

Solution

Question 3

2 x 10$$^4$$ N

4 x 10$$^4$$ N

6 x 10$$^4$$ N

8 x 10$$^4$$ N

Solution

Question 4

Pascal's law

Torricelli's law

Newton's law

Boyle's law

Solution

Question 5

$$125\ N$$

$$625\ N$$

$$-650\ N$$

$$-1125\ N$$

Solution

Question 6

directly proportional to R but inversely proportional to v.

directly proportional to both radius R and velocity v.

inversely proportional to both radius R and velocity v

inversely proportional to R but directly proportional to velocity v

Solution

Question 7

$$24.5 m s^{-1}$$

$$14.5 m s^{-1}$$

$$34.5 m s^{-1}$$

$$44.5 m s^{-1}$$

Solution

Question 8

Thermometer

Barometer

Manometer

Venturimeter

Solution

Question 9

9.4 m

4.9 m

0.49 m

0.94 m

Solution

Question 10

$$\sqrt { \left( H-3h \right) h }$$

$$\sqrt { 3Hg }$$

$$\sqrt { \left( 3H-4h \right) h }$$

$$2h$$

Solution

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