Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 32

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Question 1
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The area of cross-section of the two arms of a hydraulic press are 1 $$cm^2$$ and 10 $$cm^2$$ respectively (figure). A force of 50 N is applied on the water in the thicker arm. What force should be applied on the water in the thinner arm so that the water may remain in equilibrium?

A
5 N

B
10 N

C
25 N

D
50 N

Solution

Answer is A.

In equilibrium, the pressures at the two surfaces should be equal as they lie in the same horizontal level. If the atmospheric pressure is P and a force F is applied to maintain the equilibrium, the pressures are
$$P_0\, +\, \displaystyle \frac{50N}{10cm^2}\, and\, P_0\, +\, \displaystyle \frac{F}{1 cm^2}$$
This givens F = 5 N.
Hence, force that should be applied on the water in the thinner arm so that the water may remain in equilibrium is 5 N.
Question 2
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Two parallel glass plates are dipped partly in a liquid of density $$'d'$$ keeping them vertical. If the distance between the plates is $$'x'$$ Surface tension for liquid is $$T$$ & angle of contact is $$\displaystyle \theta $$ then rise of liquid between the plates due to capillary will be

A
$$\displaystyle \dfrac{T\cos \theta }{xd}$$

B
$$\displaystyle \dfrac{2T\cos \theta }{xdg}$$

C
$$\displaystyle \dfrac{2T}{xdg\cos \theta}$$

D
$$\displaystyle \dfrac{T\cos \theta }{xdg}$$

Solution

weight of liquid of height 'h' = (area of tube x h) x g x d= (3.14/4)hdg$${x}^2$$
vertical component of surface tension force=(1/2)x(Txcircumference)x cosθ=3.14Txcosθ
therefore, (3.14/4)hgd$${x}^2$$=(1/2)Tx3.14xcosθ
h=(2Tcosθ)/(gdx)
Question 3
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A force of $$500 N$$ acts on a square piece of plywood, each of whose sides is $$5 m$$ long. Calculate the pressure acting on the piece of plywood.

A
$$500 Nm^{-2}$$

B
$$100 Nm^{-2}$$

C
$$2500 Nm^{-2}$$

D
$$20 Nm^{-2}$$

Solution

The pressure is given as Force/Area.
In this case, the force is 500 N and area =$$5\times 5\quad =\quad 25{ m }^{ 2 }$$ as the side of the square plywood is given as 5 m.
Therefore, $$Pressure=\dfrac { Force }{ Area } =\dfrac { 500N }{ 25{ m }^{ 2 } } =20N/{ m }^{ 2 }$$.
Hence, the pressure acting on the piece of plywood is $$20N/{ m }^{ 2 }$$.
Question 4
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A tank is filled with water up to height $$H$$. Water is allowed to come out of a hole P in one of the walls at a depth $$D$$ below the surface of water. The horizontal distance $$x$$ in terms of $$H$$ and $$D$$ will be expressed as :

A
$$x\, =\, \sqrt {D(H - D)}$$

B
$$x\, =\, \sqrt {\displaystyle \dfrac {D(H - D)}{2}}$$

C
$$x\, =\, 2\sqrt {D(H - D)}$$

D
$$x\, =\, 4\sqrt {D(H - D)}$$

Solution

Answer is C.

The velocity of water stream coming out of P = $$\sqrt { 2gD } $$.
The time taken by the water stream to fall through the height H-D = $$\sqrt { \dfrac { 2(H-D) }{ g } } $$.
Therefore, $$x=\sqrt { 2gD } \times \sqrt { \dfrac { 2(H-D) }{ g } } \quad =2\sqrt { D(H-D) } $$.
Hence, the horizontal distance x in terms of H and D is $$2\sqrt { D(H-D) } $$.
Question 5
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A force of $$400 N$$ exerts a pressure of $$20 N/cm^{2}$$. What is the area on which the force acts?

A
$$20m^{2}$$

B
$$20cm^{2}$$

C
$$1cm^{2}$$

D
$$10cm^{2}$$

Solution

The pressure is given as Force/Area.
In this case, the force is 400 N and the pressure exerted is $$20N/{ cm }^{ 2 }$$.
Therefore, $$Area=\dfrac { Force }{ Pressure } =\dfrac { 400N }{ 20N/{ cm }^{ 2 } } =20{ cm }^{ 2 }$$.
Hence, the area on which the force acts is $$20c{ m }^{ 2 }$$.
Question 6
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A fixed cylindrical vessel is filled with water up to height $$H$$. A hole is bored in the wall at a depth $$h$$ from the free surface of water. For maximum horizontal range $$h$$ is equal to :

A
$$H$$

B
$$3H/4$$

C
$$H/2$$

D
$$H/4$$

Solution

Let velocity at hole be $$v$$ , density be $$\rho$$. Applying Bernoulli theorem for top-most layer and hole,

           $$P_{atm}+ \rho gH+ 1/2\rho 0^{2} = P_{atm}+ \rho gh+1/2\rho V^{2}$$
           $$   V=\sqrt {2g(H-h)}$$

Time taken to reach ground $$t= \sqrt {2h/g}$$

Horizontal range
              $$( Horizontal_{range})$$ $$=V \times t$$

                                                        $$ = \sqrt {2gH-2gh}   \times \sqrt{2h/g}$$

Differentiating $$ Horizontal_{range}$$ and equating it to zero for maximum $$ Horizontal_{range}$$ we get maximum $$ Horizontal_{range}$$ when $$h=H/2.$$
Question 7
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A boy stands on the ground. The area below his feet is $$70 cm^{2}$$. The pressure he exerts on the ground is $$7 N/cm^{2}$$. Calculate the total force acting on the ground.

A
$$0.1 N$$

B
$$10 N$$

C
$$490 N$$

D
$$4.9 N$$

Solution

The pressure is given as Force/Area.
In this case, the area =$$70{ cm }^{ 2 }$$and the presure acting is $$7N/{ cm }^{ 2 }$$.
Therefore, $$Force=Pressure\times Area\quad =\quad 7N/{ cm }^{ 2 }\times 70{ cm }^{ 2 }\quad =\quad 490N$$.
Hence, the total force acting on the ground is 490 N.
Question 8
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Water flows in a horizontal tube as shown in figure. The pressure of water changes by 600 $$N/m^2$$ between x and y where the areas of cross - section are 3 $$cm^2$$ and 1.5$$cm^2$$ respectively. Find the rate of flow of water through the tube.

A
$$ 189\, cm^3/s$$

B
$$ 159\, cm^3/s$$

C
$$ 189\, cm^2/s$$

D
$$ 159\, cm^2/s$$

Solution

Let the velocity at x = $$v_x$$ and that at y = $$v_y$$.
By the equation of continuity, $$\displaystyle \frac{v_y}{v_x}\, =\, \displaystyle \frac{3 cm^2}{1.5 cm^2}\, =\, 2$$
By Bernoulli's equation,
$$P_x\, +\, \displaystyle \frac{1}{2}\, p\, v_x^2\, =\, P_y\, +\, \displaystyle \frac{1}{2}\, pv_y^2$$
or,$$P_x\, -\, O_y\, =\, \displaystyle \frac{1}{2}\, p(2v_y)^2\, -\, \displaystyle \frac{1}{2}\, pv_y^2\, =\, \displaystyle \frac{3}{2}\, pv_y^2$$
or, $$600\, \displaystyle \frac{N}{m^2}\, =\, \displaystyle

\frac{3}{2}\, \left (1000\, \displaystyle \frac{kg}{m^3} \right )\,

v_x^2$$
or, $$v_x\, =\, \sqrt{0.4\, m^2/ s^2}\, =\, 0.63\, m/s$$
The rate of flow $$=\, (3\, cm^2)\, (0.63\, m/s)\, =\, 189\, cm^3/s$$.
Question 9
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A force exerts a pressure of $$45 N/m^{2}$$ when it acts on an area of $$10 m^{2}$$. Calculate the total force

A
$$4.5 N$$

B
$$450 N$$

C
$$45 N$$

D
$$.45 N$$

Solution

The pressure is given as Force/Area.
In this case, the area =$$10{ m }^{ 2 }$$and the presure acting is $$45N/{ m }^{ 2 }$$.
Therefore, $$Force=Pressure\times Area\quad =\quad 45N/{ m }^{ 2 }\times 10{ m }^{ 2 }\quad =\quad 450N$$.
Hence, the total force is 450 N.
Question 10
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Water flows through a frictionless duct with a cross section varying as shown in fig. Pressure $$p$$ at points along the axis is represented by

A

B

C

D

Solution

Answer is A.

When cross section of duct decreases the velocity of water increases and in accordance with Bernoulli's theorem the pressure decreases at that place.
Therefore, in this case, the pressure remains constant initially and then decreases as the area of cross section decreases along the neck of the tube and then remains constant along the mouth of the tube.
Hence, graph in option A is correct.