Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 31

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Question 1
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The small piston of a hydraulic lift has an area of $$0.20m^2$$. A car weighing $$1.20\times 10^4N$$ sits on a rack mounted on the large piston. The large piston has an area of $$0.90m^2$$. How large a force must be applied to the small piston to support the car?

A
$$2.67\times 10^3N$$

B
$$2.67\times 10^4N$$

C
$$2.67\times 10^5N$$

D
$$2.67\times 10^6N$$

Solution

Hydraulic lift works on the principle of pascal's law
$$P=\frac { { F }_{ 1 } }{ a_{ 1 } } =\frac { { F }_{ 2 } }{ { a }_{ 2 } }$$
applying the given formula to the question.
$$\frac { 1.2\times { 10 }^{ 4 } }{ 0.90 } =\frac { F }{ 0.2 } \\ F=\frac { 0.24\times { 10 }^{ 4 } }{ 0.90 } =2.67\times { 10 }^{ 3 }\quad N$$
Question 2
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Eight spherical rain drops of the same mass and radius are falling down with a terminal speed of $$6cms^{-1}$$. If they coalesce to form one big drop, what will be its terminal speed? Neglect the buoyancy due to air:

A
$$1.5cms^{-1}$$

B
$$6cms^{-1}$$

C
$$24cms^{-1}$$

D
$$32cms^{-1}$$

Solution

When eight equal drops coalesces, volume of water remains same and a bigger drop is formed. Lets say $$r$$ is radius of drop before these combine and $$R$$ is radius after these combine.
We have
$$8 \times \frac{4}{3} \pi r^3 =\frac{4}{3} \pi R^3 \\\Rightarrow R=2r $$
and we know that terminal velocity $$V_T \propto$$ radius$$^2$$, so we have
$$\frac{V_T'}{V_T}=\frac{R^2}{r^2}

\\\Rightarrow V_T' =\left (\frac{R}{r} \right)^2 V_T = \left (2

\right)^2 \times 6 cms^{-1}=24 cms^{-1}$$
Question 3
1 / -0

Water rises in a capillary upto a height h. If now this capillary is tilted by an angle of $$45^{\circ}$$, then the length of the water column in the capillary becomes

A
2h

B
$$\displaystyle \frac{h}{2}$$

C
$$\displaystyle \frac{h}{\sqrt{2}}$$

D
$$h\sqrt{2}$$

Solution

The expression for the capillary rise in a tube is given by,

H = $$ \dfrac {2T cos \theta}{\rho g r} $$

for all other parameters kept constant, if I change the angle of inclination to $$ 45^o $$

$$ cos \theta $$ will go from 1 to $$ \dfrac {1}{\sqrt 2} $$

Therefore, the height of capillary tube rise will also change from $$ H to \dfrac{H}{\sqrt2} $$
Question 4
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A $$1.5m$$ wide by $$2.5m$$ long water bed weighs $$1025N$$. Find the pressure that the water bed exerts on the floor. Assume that the entire lower surface of the bed makes contact with the floor.

A
$$273.3Pa$$

B
$$373.5 Pa$$

C
$$173.3Pa$$

D
$$473.3Pa$$

Solution

Pressure exerted by water bed will be force applied per unit area.
$$P=\dfrac {F}{A}$$
Given:
Force= $$1025 \ N$$
Area= $$1.5 m \times 2.5m = 3.75 m^2 $$

Pressure
$$=\dfrac { 1025 \ N }{ 3.75 \ m^2} =273.3 \ N/m^2 = 273.3 \ Pa$$
Question 5
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A horizontal pipeline carries water in a streamline flow. At a point along the tube where the cross sectional area is $$10^{-2}m^{2}$$ , the water velocity is 2m/s and the pressure is 8000 Pa. The pressure of water at another point wher cross sectional area is $$0.5 \times 10^{-2} m^{2}$$ is

A
4000Pa

B
1000Pa

C
2000Pa

D
3000Pa

Solution

Area at other point is half. So speed will be double.
Now,
$$\displaystyle p_{1}+\frac{1}{2}\rho v_{1}^{2}=p_{2}+\frac{1}{2}\rho v_{2}^{2}$$
$$\therefore $$   $$\displaystyle p_{2}=p_{1}+\frac{1}{2}\rho \left ( v_{1}^{2}-v_{2}^{2} \right )$$
   $$\displaystyle =\left ( 8000 \right )+\frac{1}{2}\times 1000\left ( 4-16 \right )$$
   $$=2000$$ Pa
Question 6
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The manometer shown below is used to measure the difference between water level of the two tanks. Calculate this difference for the conditions indicated.

A
4 cm

B
40 cm

C
100 cm

D
12 cm

Solution

Answer is A.

$$P_a\, +\, h_1 pg\, -\, 40p_1g\, +\, 40pg\, =\, P_a\, +\, h_2 pg$$
$$h_2pg\, -\, h_1 pg\, =\, 40 pg\, -\, 40 p_1g$$
As p1 = 0.9p, the eation becomes
$$h_2pg\, -\, h_1 pg\, =\, 40 pg\, -\, 36 pg$$
Therefore, $$h_2\, -\, h_1 \, =$$ 4 cm.
Hence, the difference for the conditions indicated is 4 cm.
Question 7
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A 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 range, $$h$$ is equal to :-

A
$$\displaystyle \dfrac{H}{4}$$

B
$$\displaystyle \dfrac{H}{2}$$

C
$$\displaystyle \dfrac{3H}{4}$$

D
$$H$$

Solution

The cylinder is filled upto a height of $$H$$ and a hole is drilled at a depth h from the surface of water.

This means that the velocity of water flowing out of the hole is $$ \sqrt {2g(H-h)} $$

The time taken for the stream of water to reach ground,

$$ h = 0 + 0.5 g t^2 $$

This gives, $$t = \sqrt {h/5} $$

The range of this stream would be, $$x = ut$$

$$x = \sqrt {2g h(H-h)/5} $$

For maximum $$x$$, $$dx/dh = 0$$ , which gives $$h = H/2$$
Question 8
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In a hydraulic lift, used at a service station, the radius of the large and small piston is in the ratio of 20 : 1. What weight placed on the small piston will be sufficient to lift a car of mass 1500 kg?

A
3.75 kg

B
37.5 kg

C
7.5 kg

D
75 kg

Solution

Using pascal's law
$$\frac{F_1}{A_1}=\frac{F_2}{A_2}$$
$$\frac{1500}{\pi r_1^2}=\frac{W}{\pi r_2^2}=\frac{1500}{20^2}=\frac{W}{1^2}$$
$$W=3.75kg$$
Question 9
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The diagram (fig.) shows a venturimeter, through which water is flowing. The speed of water at a 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

The height difference between the points x and y is 5.1 mm

Therefore, the pressure difference between the 2 points will be,

$$ P = \rho g h $$

$$ P = 1 \times 1000 \times 0.51 $$

P = 510

Applying Bernoulli's theorem,

$$ \delta P + 0.5 \rho \delta v^2 = 0 $$

$$ \delta P = 510 $$

$$ 510 = 0.5 \times 1 \times (v_2^2 - 2^2) $$

Hence, $$ v_2^2 = 1024 \ or \ v_2 = 32 cm/sec $$
Question 10
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(a) A force of $$200\;N$$ acts on an area of $$0.02\;m^2$$. Find the pressure in Pascal.
(b) What force will exert a pressure of $$50,000\ Pa$$ on an area of $$0.5\;m^2$$?
(c) Find out the area of a body which experiences a pressure of $$500\ Pa$$ by a force of $$100\;N$$.

A
(a)$$\;10000\;Pa$$, (b)$$\;25000\;N$$, (c)$$\;0.2\;m^2$$

B
(a)$$\;1000\;Pa$$, (b)$$\;2500\;N$$, (c)$$\;2\;m^2$$

C
(a)$$\;100000\;Pa$$, (b)$$\;250000\;N$$, (c)$$\;0.2\;m^2$$

D
(a)$$\;100\;Pa$$, (b)$$\;250\;N$$, (c)$$\;0.2\;m^2$$

Solution

We know
$$Pressure(P)=\dfrac{Force(F)}{Area(A)}$$
(a) $$F=200N$$ $$A=0.02m^2$$
$$P=\dfrac{200}{0.02}=10000Pa$$
(b) $$P=50000Pa$$ $$A=0.5m^2$$
$$F=PA=50000\times 0.5=25000N$$
(c) $$P=500Pa$$ $$F=100N$$
$$A=\dfrac{F}{P}=\dfrac{100}{500}=0.2m^2$$

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Mechanical Properties of Fluids Test - 31

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Mechanical Properties of Fluids Test - 31

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Question 1

$$2.67\times 10^3N$$

$$2.67\times 10^4N$$

$$2.67\times 10^5N$$

$$2.67\times 10^6N$$

Solution

Question 2

$$1.5cms^{-1}$$

$$6cms^{-1}$$

$$24cms^{-1}$$

$$32cms^{-1}$$

Solution

Question 3

2h

$$\displaystyle \frac{h}{2}$$

$$\displaystyle \frac{h}{\sqrt{2}}$$

$$h\sqrt{2}$$

Solution

Question 4

$$273.3Pa$$

$$373.5 Pa$$

$$173.3Pa$$

$$473.3Pa$$

Solution

Question 5

4000Pa

1000Pa

2000Pa

3000Pa

Solution

Question 6

4 cm

40 cm

100 cm

12 cm

Solution

Question 7

$$\displaystyle \dfrac{H}{4}$$

$$\displaystyle \dfrac{H}{2}$$

$$\displaystyle \dfrac{3H}{4}$$

$$H$$

Solution

Question 8

3.75 kg

37.5 kg

7.5 kg

75 kg

Solution

Question 9

$$23\:cm/sec$$

$$32\:cm/sec$$

$$101\:cm/sec$$

$$1024\:cm/sec$$

Solution

Question 10

(a)$$\;10000\;Pa$$, (b)$$\;25000\;N$$, (c)$$\;0.2\;m^2$$

(a)$$\;1000\;Pa$$, (b)$$\;2500\;N$$, (c)$$\;2\;m^2$$

(a)$$\;100000\;Pa$$, (b)$$\;250000\;N$$, (c)$$\;0.2\;m^2$$

(a)$$\;100\;Pa$$, (b)$$\;250\;N$$, (c)$$\;0.2\;m^2$$

Solution

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