Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 24

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Question 1
1 / -0

The area of cross-section of the wider tube show in figure is $$800 { cm^2 }$$. If a mass of $$12 kg$$ is placed on the massless piston, the difference in heights $$h$$ in the level of water in the two tubes is :

A
$$10cm$$

B
$$6cm$$

C
$$15cm$$

D
$$2cm$$

Solution

Given that:
$$A =

800 cm^2$$
$$m =

12 kg =12000 g$$

Here, the force on piston is $$F=

mg$$
Hence, increase in pressure on the liquid in the wider tube is
$$P

= \dfrac{F}{A} = \dfrac{mg}{A}$$
Since, h is the difference in level of water in the two tubes, thus
$$P

= h\rho g$$
$$h =

\dfrac{P}{\rho g}$$
$$h =

\dfrac{mg}{A\rho g} = \dfrac{m}{A\rho}$$
$$h =

\dfrac{12000}{800}$$.......($$\because \rho = 1$$)
$$\therefore

h = 15 cm$$
Question 2
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The diagram (fig.) shows a venturimeter, through which water is flowing. The speed of water at X is 2 cm/sec. The speed of water at Y (taking $$g=1000 cm/sec^2)$$ is:

A
23 cm/sec

B
32 cm/sec

C
101 cm/sec

D
1024 cm/sec

Solution

$$p_x - p_y = \dfrac{\rho}{2}(v_y^2 - v_x^2) \\\Rightarrow \Delta h \rho
g=\dfrac{\rho}{2}(v_y^2 - v_x^2) \\\Rightarrow \Delta h
g=\dfrac{1}{2}(v_y^2 - v_x^2) \\\Rightarrow 0.51 \times
1000=\dfrac{1}{2}(v_y^2 - 2^2) \\\Rightarrow 1020= v_y^2 - 4
\\\Rightarrow 1024= v_y^2 \\\Rightarrow v_y=32 cm/s$$
Question 3
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Which of the following is the incorrect graph for a sphere falling in a viscous liquid? (Given at $$t=0$$, velocity $$v=0$$ and displacement $$x=0$$)

A

B

C

D

Solution

A sphere falling in a viscous liquid increases its speed with time and finally achieves a terminal velocity at infinite time.
Velocity varies with time as shown.
Net acceleration of the sphere is ($$mg-$$Buoyant force-Viscous Force) downwards
$$a=d\dfrac{4}{3}\pi R^3 g-\rho\dfrac{4}{3}\pi R^3 g-6\pi\eta Rv$$
Hence a varies linearly with velocity. Hence option C is the correct answer.
Question 4
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A long capillary tube of radius '$$r$$' initially just vertically completely imerged inside a liquid of angle of contact $${ 0 }^{ \circ }$$. If the tube is slowly raised then relation between radius of curvature of miniscus inside the capillary tube and displacement $$(h)$$ of tube can be represented by

A

B

C

D

Solution

The curvature equation relating the radius of meniscus inside the capillary tube and the displacement of the tube is given by,

$$ H = \frac {1}{2} \times (\frac {1}{{R}_{1}} + \frac {1}{{R}_{2}}) $$

For a hemispherical, meniscus (which is the case for $$ {0}^{o} $$ contact angle liquid,
$$ {R}_{1} = {R}_{2} = r $$

Substituting the values in the equation,

H = $$ \frac {1}{r} $$

Thus, the graph should resemble a hyperbolic parabola with a value r at a displacement h.

Hence, option B
Question 5
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The displacement of a ball falling from rest in a viscous medium is plotted against time. Choose a possible option:

A

B

C

D

Solution

A falling ball will eventually attain a constant velocity. Thus the st graph would eventually follow a straight line as shown. Thus D is correct.
Question 6
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A large tank is filled with water (density$$= { 10 }^{ 3 } kg/{ m }^{ 3 }$$). A small hole is made at a depth $$10 m$$ below water surface. the range of water issuing out of the hole is R on ground. What extra pressure must be applied on the water surface so that the range becomes $$2R$$ ( take $$1atm={ 10 }^{ 5 } Pa$$ and $$g=10 m/{ s }^{ 2 }$$)

A
$$9 atm$$

B
$$4 atm$$

C
$$5 atm$$

D
$$3 atm$$

Solution

Range will become twice, if the velocity of efflux becomes twice. Since velocity of efflux is $$\sqrt{2gh}$$, it will become twice, if $$h$$ becomes $$4$$ times, or $$40m.$$
Thus, an extra pressure equivalent to $$30m$$ of water should be applied.
$$1 atm =0.76 \times 13.6$$ m of water $$=10.336 m$$ of water
$$30m$$ of water $$\approx 3 atm$$
Question 7
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A barometer tube, containing mercury, is lowered in a vessel containing mercury until only $$50 cm$$ of the tube is above the level of mercury in the vessel. If the atmospheric pressure is $$75 cm$$ of mercury, what is the pressure at the top of the tube

A
$$33.3 kPa$$

B
$$66.7kPa$$

C
$$3.33 MPa$$

D
$$6.67 MPa$$

Solution

Since the mercury rises only $$50cm$$ it implies the rest of the excess atmospheric pressure is being countered by the pressure at the top of the tube.
So pressure at top=atmospheric pressure - $$50cm$$ of Hg=$$(75-50)cm=25cm$$ of Hg=$$1/3\times 100kPa= 33.33kPa$$
Question 8
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Equal volumes of two immiscible liquids of densities $$\rho$$ and $$2\rho$$ are filled in a vessel as shown figure. Two small holes are punches at depth $$h/2$$ and $$3h/2$$ 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 punches at depth $$\dfrac{h}{2}$$ and $$\dfrac{3h}{2}$$ 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
$$v_1 = \sqrt {2h_1g}$$
$$v_1 = \sqrt {2(\dfrac{h}{2})g}$$
$$v_1 = \sqrt {hg}$$
Applying Bernoulli's equation to point just inside and outside the second hole,
$$ P_{a} + \rho gh + 2\rho g \displaystyle \dfrac{h}{2} = P_{a} + \displaystyle \dfrac{1}{2} 2\rho v^2 $$
$$ v = \sqrt {2gh} $$
So, the velocity of flux from second hole is
$$v_2 = \sqrt {2hg}$$
Therefore,
$$\dfrac{v_1}{v_2} = \dfrac{\sqrt {hg}}{\sqrt {2hg}}$$
$$\dfrac{v_1}{v_2} = \dfrac{1}{\sqrt 2}$$
Question 9
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A cylidrical drum, open at the top, contains 30 litres of water. It drains out through a small opening at the bottom. 10 litres of water comes out in time $$t_1$$, the next 10 litres in further time $$t_2$$ and the last 10 litres in further time $$t_3$$. Then,

A
$$t_1 = t_2 = t_3$$

B
$$t_1 > t_2 > t_3$$

C
$$t_1 < t_2 < t_3$$

D
$$t_2 < t_1 = t_3$$

Solution

Using the torricelli's theorem we know that the velocity of efflux is given as $$\sqrt{2gh}$$. Thus we see that the velocity is dependent upon height. As the height decreases the velocity of efflux will also decrease, so time taken by water to drain out of the drum will increase.
Thus option C is correct.
Question 10
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A water tank is filled with water upto height H. A hole is made in a tank wall at a depth D from the surface of water. The distance X from the lower end of wall where the water stream from tank strikes the ground is:

A
$$2 \sqrt {gD}$$

B
$$2 \sqrt {D \left ( H + D \right)}$$

C
$$2 \sqrt {D \left ( H - D \right)}$$

D
$$\sqrt D$$

Solution

Velocity of efflux at point D = $$ \sqrt {2gD} $$
Time taken to reach the ground = $$ \sqrt{ \displaystyle\frac{2(H-D)}{g}} $$
So horizontal range = velocity $$\times$$ time = $$ \sqrt {2gD} \times \sqrt{ \displaystyle\frac{2(H-D)}{g}} $$
So range of the water stream $$X=2\sqrt{D(H-D)}$$