Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

A piston of cross-sectional area $$100 cm^2$$ is used in a hydraulic press to exert a force of $$10^7$$ dyne on the water. The cross-sectional area of the other piston which supports an object having a mass of 2000 kg is :

A
$$100 cm^2$$

B
$$10^9 cm^2$$

C
$$2\times 10^4 cm^2$$

D
$$2\times 10^{10} cm^2$$

Solution

As we know , 1 dyne = $$10 ^{-5}$$ N so $$ 10 ^{7} $$ dyne = 100 N
hence , for Hydraulic press we have
$$\dfrac{100N}{100 cm^{2}}$$ = $$\dfrac{2000g}{A}$$ $$\Rightarrow$$ Area = $$2\times 10 ^{4} cm^{2}$$
So option C is correct.
Question 2
1 / -0

A solid sphere falls with a terminal velocity of 10 m/s in air. If it is allowed to fall in vacuum:

A
terminal velocity will be more than 10 m/s

B
terminal velocity will be less than 10 m/s

C
terminal velocity will be 10 m/s

D
there will be no terminal velocity

Solution

Terminal velocity is reached in air since there is viscous force while moving through air( viscous force is balanced by the weight )
In vacuum, terminal velocity is not achieved since velocity of the sphere will keep on increasing with time.
Question 3
1 / -0

A uniformly tapering vessel shown in figure, is filled with liquid of density $$900 kg/m^3$$. The force that acts on the base of the vessel due to liquid is :

A
3.6 N

B
7.2 N

C
9.0 N

D
12.6 N

Solution

This cylinder so formed has volume exactly half the total volume,
$$\therefore$$ force due to liquid
$$= 2\times A \times 8gh$$
$$= 2\times 10^{-3} \times 9.8\times 900\times 0.4$$
$$= 7.056\ N = 7.2\ N$$ (Taking $$g = 10\ m/s^{2})$$.
Question 4
1 / -0

A girl stands on a box having 60 cm length, 40 cm breadth and 20 cm width in three ways. The pressure exerted by the box will be:

A
maximum when length and breadth form the base

B
maximum when breadth and width form the base

C
maximum when width and length form the base

D
the same in all the above three cases

Solution

The force exerted by the box on base=Weight of object=$$W$$
The pressure that it exerts on the base$$=\dfrac{W}{A}$$
where $$A$$ is the area of the base.
Hence the pressure is maximum for smallest area of the base, which is when formed by breadth and width.
Question 5
1 / -0

Equal volumes of two immiscible liquids of densities $$\rho$$ and $$2\rho$$ are filled in a vessel as shown in figure. Two small holes are punched at depth $$\frac{h}{2}$$ and $$\dfrac{3h}{2}$$ from the surface of lighter liquid. If $$v_1$$ and $$v_2$$ are the velocities of a flux at these two holes, then $$v_1/v_2$$ is:

A
$$\dfrac { 1 }{ 2\sqrt { 2 } } $$

B
$$\dfrac { 1 }{ 2 } $$

C
$$\dfrac { 1 }{ 4 } $$

D
$$\dfrac { 1 }{ \sqrt { 2 } } $$

Solution

Here, two small holes are punchesd at depth $$\frac{h}{2}$$ and $$\frac{3h}{2}$$ from the surface of lighter liquid. Hence, the height of first hole from the surface of lighter liquid is
$$h_1 = \dfrac{h}{2}$$
and the height of second hole from the first hole is(i.e. from surface of remaining liquid)
$$h_2 = \dfrac{3h}{2} - \dfrac{h}{2} = h$$
Now, we have the velocity of flux from first hole is
$$v_1 = \sqrt 2h_1g$$
$$v_1 = \sqrt 2(\frac{h}{2})g$$
$$v_1 = \sqrt hg$$
Also, the velocity of flux from second hole is
$$v_2 = \sqrt 2h_2g$$
$$v_2 = \sqrt 2hg$$
Therefore,
$$\dfrac{v_1}{v_2} = \dfrac{\sqrt hg}{\sqrt 2hg}$$
$$\dfrac{v_1}{v_2} = \dfrac{1}{\sqrt 2}$$
Question 6
1 / -0

With the increase in the area of contact of an object, the pressure:
(Note : Thrust remains same)

A
Increases

B
Decreases

C
Is not affected

D
None of these

Solution

Since,
$$P = \frac{F}{A}$$
Thus pressure is inversely proportional to area,
$$i.e. \,\,P \propto \frac{1}{A}$$
With the increase in the area of contact of an object the pressure decreases because pressure is inversely proportional to area of contact.
Question 7
1 / -0

Figure shows water filled in a symmetrical container. Four pistons of equal area $$A$$ are used at the four openings to keep the water in equilibrium. Now an additional force $$F$$ is applied at each piston. The increase in the pressure at the centre of the container due to this addition is

A
$$\dfrac {F}{A}$$

B
$$\dfrac {2F}{A}$$

C
$$\dfrac {4F}{A}$$

D
0

Solution

The four pistons are initially in equilibrium. As additional force $$F$$ is applied to each piston, the pressure in the fluid at each point must be increased by $$F.A$$ so that each piston retains state of equilibrium.
Thus, the increment in the pressure at each point is
$$\Delta P=\dfrac {F}{A}$$ (by Pascal's law)
So, the pressure increment at the centre is $$ 4\dfrac {F}{A}$$
Question 8
1 / -0

A Cylindrical vessel of cross-sectional area $$1000cm^{2}$$, is fitted with a frictionless piston of mass 10 kg, and filled with water completely. A small hole of cross-sectional area $$10mm^{2}$$ is opened at a point 50 cm deep from the lower surface of the piston. The velocity of effluent from the hole will be

A
10.5 m/s

B
3.4 m/s

C
0.8 m/s

D
0.2 m/s

Solution

Given that
$$A = 1000 cm^3$$
$$m = 10 Kg =10000 g$$
$$A_h = 10 mm^2 = 0.1 m^2$$
$$(H-h) = 50 cm$$
Here, the force on piston is
$$F = mg$$
Hence, Increase in pressure on the liquid in the wider tube is
$$P = \frac{F}{A} = \frac{mg}{A}$$
Let, H is the level of water to which piston will move, thus
$$P = H\rho g$$
$$H = \frac{P}{\rho g}$$
$$H = \frac{mg}{A\rho g} = \frac{m}{A\rho}$$
$$H = \frac{10000}{1000}$$.......($$\because \rho = 1$$)
$$\therefore H = 10 cm$$ from surface of tube.
Since, (H-h) = 50cm, h = 60cm = 0.6 m
The velocity of effluent from the hole is
$$v = \sqrt h\rho g$$
$$v = (\sqrt 2gh) $$
$$v = \sqrt (2)(9.8)(0.6) = 3.4 ms^{-1}$$
Question 9
1 / -0

Match column I with column II :
List I List II
Magnus energy Pascal's Law
Loss of Energy Archimedes' principle
Pressure is same at same level in a liquid Viscous force
Hydraulic Machines Lifting of asbestos roofs

A
B$$\rightarrow$$4, C$$\rightarrow$$2, D$$\rightarrow$$1, A$$\rightarrow$$1

B
A$$\rightarrow$$4, B$$\rightarrow$$3, C$$\rightarrow$$1, D$$\rightarrow$$1

C
C$$\rightarrow$$4, A$$\rightarrow$$2, B$$\rightarrow$$1, D$$\rightarrow$$1

D
D$$\rightarrow$$4, A$$\rightarrow$$2, B$$\rightarrow$$1, C$$\rightarrow$$1

Solution

Magnus effect talks of difference in pressure caused due to difference in velocity to lift a structure. Lifting of asbestos roofs during cyclonic storms in based on this phenomenon. So, $$A\rightarrow 4$$
Viscous force is wet friction and brings loss in internal energy. So, $$B\rightarrow 3$$
At the same level in the same liquid in a container, the pressure is the same according to Pascal's law. So, $$C\rightarrow 1$$
Hydraulic machines use the idea of Pascal's law to transfer equal pressure in all directions from the free surface where it is applied. So, $$D\rightarrow 1$$
Question 10
1 / -0

The velocity of the liquid coming out of a small hole of a vessel containing two different liquids of densities $$2\rho$$ and $$\rho$$ as shown in the figure is

A
$$\sqrt {6gh}$$

B
$$2\sqrt {gh}$$

C
$$2\sqrt {2gh}$$

D
$$\sqrt {gh}$$

Solution

Pressure at (1), $$P_1=P_{atm}+\rho g(2h)+2\rho gh$$
Applying Bernoulli's theorem between points (1) and (2),
$$(P_{atm}+2\rho gh)+(2\rho) gh=P_{atm}+\dfrac {1}{2}(2\rho )v^2$$
$$\Rightarrow v=2\sqrt {gh}$$

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Re-Attempt Weekly Quiz Competition

List I	List II
Magnus energy	Pascal's Law
Loss of Energy	Archimedes' principle
Pressure is same at same level in a liquid	Viscous force
Hydraulic Machines	Lifting of asbestos roofs

Mechanical Properties of Fluids Test - 22

Result

Mechanical Properties of Fluids Test - 22

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Question 1

$$100 cm^2$$

$$10^9 cm^2$$

$$2\times 10^4 cm^2$$

$$2\times 10^{10} cm^2$$

Solution

Question 2

terminal velocity will be more than 10 m/s

terminal velocity will be less than 10 m/s

terminal velocity will be 10 m/s

there will be no terminal velocity

Solution

Question 3

3.6 N

7.2 N

9.0 N

12.6 N

Solution

Question 4

maximum when length and breadth form the base

maximum when breadth and width form the base

maximum when width and length form the base

the same in all the above three cases

Solution

Question 5

$$\dfrac { 1 }{ 2\sqrt { 2 } } $$

$$\dfrac { 1 }{ 2 } $$

$$\dfrac { 1 }{ 4 } $$

$$\dfrac { 1 }{ \sqrt { 2 } } $$

Solution

Question 6

Increases

Decreases

Is not affected

None of these

Solution

Question 7

$$\dfrac {F}{A}$$

$$\dfrac {2F}{A}$$

$$\dfrac {4F}{A}$$

0

Solution

Question 8

10.5 m/s

3.4 m/s

0.8 m/s

0.2 m/s

Solution

Question 9

B$$\rightarrow$$4, C$$\rightarrow$$2, D$$\rightarrow$$1, A$$\rightarrow$$1

A$$\rightarrow$$4, B$$\rightarrow$$3, C$$\rightarrow$$1, D$$\rightarrow$$1

C$$\rightarrow$$4, A$$\rightarrow$$2, B$$\rightarrow$$1, D$$\rightarrow$$1

D$$\rightarrow$$4, A$$\rightarrow$$2, B$$\rightarrow$$1, C$$\rightarrow$$1

Solution

Question 10

$$\sqrt {6gh}$$

$$2\sqrt {gh}$$

$$2\sqrt {2gh}$$

$$\sqrt {gh}$$

Solution

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