Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 55

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Weekly Quiz Competition

Question 1
1 / -0

A boat travelling at $$6 m/s$$ in sea water has a $$250 mm$$ diameter propeller that discharges the water at a velocity of $$12 m/s$$. Given that the density of seawater is $$1030 \mathrm{k}\mathrm{g}/\mathrm{m}^{3}$$. The effect of propeller hub is negligible. The magnitude of thrust produced $$F$$ is (in $$N$$)

A
$$1030$$

B
$$2730$$

C
$$4660$$

D
$$980$$

Solution

Pressure difference across the propeller hubs
$$=p_{1}-p_{2}=\dfrac{1}{2}\left ( 1030 \right )\left ( 12^{2}-6^{2} \right )$$
$$=55620p_{a}$$
So,
the thrust produce=area x pressure difference
$$=\pi \times \left ( 125 \times 10^{-3} \right )^{2}\times 55620N$$
$$=2730.240N$$
$$\approx 2730N$$
Question 2
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A wide vessel with a small hole at the bottom is filled with two liquids. The density and height of one liquid are $$ {\rho}_{1}\ and\ {h}_{1} $$ and that of the other are $$ {\rho}_{2}$$ and $${h}_{2}$$. (Given $$ {\rho}_{1} > {\rho}_{2} ) $$. The velocity of liquid coming out of the hole is :

A
$$ v = \sqrt{2g({h}_{1} + {h}_{2})} $$

B
$$ v = \sqrt{2g({h}_{1}{\rho}_{1} + {h}_{2}{\rho}_{2})/ ({\rho}_{1} + {\rho}_{2})} $$

C
$$ v = \sqrt{2g[{h}_{1} + \dfrac{{h}_{2}{\rho}_{2}}{{\rho}_{1}}]}$$

D
$$ v = \sqrt{2g[\dfrac{{h}_{1}{\rho}_{1}}{{\rho}_{2}} + {h}_{2}]}$$

Solution

Applying Bernoulli's equation to points A and B
$$P_0 + \rho_1gh_1 + \rho_2gh_2 +0 = P_0 + \dfrac{1}{2} \rho_1v^2$$
where v is the velocity at point B
$$ \therefore \dfrac{1}{2} \rho_1v^2 =g( \rho_1h_1 + \rho_2h_2)$$

$$ \therefore v_1^2 = 2g\dfrac{( \rho_1h_1 + \rho_2h_2)}{\rho_1}= 2g(h_1 + \frac{\rho_2}{\rho_1} \times h_2)$$

$$ \therefore v_1 = \sqrt{2g(h_1 + \dfrac{\rho_2}{\rho_1} h_2)}$$
Question 3
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The hydraulic press shown in the figure is used to raise the mass $$m$$ through a height of $$0.05 cm$$ by performing $$500 J$$ of work at the small piston. The diameter of the large piston is 10 cm while that of the smaller one is $$2 cm.$$ The mass $$M$$ is

A
$$10^{4} kg$$

B
$$10^{3} kg$$

C
$$100 kg$$

D
$$10^{5}kg$$

Solution
Question 4
1 / -0

Bantu slips into a large lake and he doesn't know swimming. But he is a great fan of Superman. He shouted for help remembering Superman. As usual, Superman arrives on the top of a cliff and, due to some reason, Superman lost his flying power immediately after arrival on cliff. Due to shortage of time, somehow, Superman manages a strong and long straw and decided to drink whole water of lake to save Bantu.(Data:Atmospheric pressure $$=1.2 \times 10^{5} Pa, g=10m/s^{2}$$, density of water$$= 1000kg/m^{3})$$ Assume Superman has infinite power and ability to drink whole water. Which of the following statements is/are true?
(1) Superman cannot save Bantu by this way.
(2) Superman can drink some water but not whole water.
(3) Superman will save Bantu by drinking whole water.

A
Only (1) & (2)

B
Only (3)

C
Only (1)

D
All (1), (2) & (3) are wrong

Solution

The water can rise in the straw only upto a level which would make the pressure at the bottom of the straw equal to the atmospheric pressure. Thus the maximum height of the liquid in the straw is given as $$\rho g h=1000\times 10\times h=P_{atm}=1.2\times 10^5$$
or
h=12 m.
As the straw is more than 12.5 m, Superman cannot save Bantu.
Question 5
1 / -0

Directions For Questions

Consider an airplane moving through the air at velocity $$v=200m/s$$. The streamlines which move just over the top of the airplane are compressed to eight-tenths their normal area and given that those under the wing are not compressed at all (Density of air 1.3kg/ms$$^{2}$$)

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The difference in the pressure between the air just over the wing $$\mathrm{P}_{1}$$ and that under the wing $$\mathrm{P}_{2}$$ is:

A
$$2.8\times 10^{3}$$ Pa

B
$$0.723\times 10^{2}$$ Pa

C
$$1.46\times 10^{4}$$ Pa

D
Zero

Solution

Compression of the streamlines means that the stream tube above the wing
has a smaller cross-sectional area than that in front of the plane and from the continuity equation , the velocity v' of the air must therefore be greater above the wing.
Area $${A}'= \dfrac{8}{10}A$$ $$\Rightarrow \dfrac{A}{{A}'}=\dfrac{10}{8}$$
From continuity equation,
$$Av={A}'{v}'$$
$$\therefore {v}' =\dfrac{A}{{A}'}v= \dfrac{10}{8} \times 200= 250 \: ms^{-1}$$
The greater velocity (v',as obtained above) implies lower pressure than the normal pressure of the air in front of the plane. Given that the flow-lines under the wing are not compressed at all. The pressure under the wing is just the normal pressure of the air in front of the wing.
From Bernoulli's equation , with both points 1 (for p1) and 2 (for p2) at effectively same elevation.
$$P_a + \dfrac{1}{2}\rho_av_1^2 = P_b + \dfrac{1}{2}\rho_bv_2^2$$
Given:
$$v_2=200\: ms^{-1}$$
$$\therefore P_2-P_1 = \dfrac{1}{2}\rho_a( v_1^2-v_2^2)= \dfrac{1}{2}\times 1.3( 250^2-200^2)= 1.46 \times 10^4\: Pa$$
Question 6
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Equal volumes of two immiscible liquids of densities $$\delta $$ and 3$$\delta $$ are filled in a vessel. Two small holes are punched at depth $$h/3$$ and $$4h/3$$ from upper surface of lighter liquid. If $$V_{1} $$ and $$ V_{2}$$ are velocities of efflux at these two holes, then $$V_{1}/V_{2}$$ is :

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

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

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

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

Solution

Applying Bernoulli's theorem to pints A( just at the surface of the liquid) and B( just outside the first hole in the vessel) taking the position of the hole as reference
$$P_0 + \rho_1g\dfrac{h}{3} + \dfrac{1}{2}\rho_1 v^2 = P_0 + 0 + \dfrac{1}{2}\rho_1v_1^2$$ .....(1)
$$P_0$$ - atmospheric pressure
$$v$$ - velocity of the liquid on the surface
$$v_1$$ - velocity of the liquid at the first hole
Applying Bernoulli's theorem to pints A( just at the surface of the liquid) and C( just outside the second hole in the vessel) taking the position of the hole as reference
$$P_0 +\rho_1 gh + \rho_2g\dfrac{h}{3} + \dfrac{1}{2}\rho_1 v^2 = P_0 + 0 + \dfrac{1}{2}\rho_2v_2^2$$
Putting $$\rho_2= 3 \rho_1$$,
the above eqn becomes
$$P_0 +\rho_1 gh + 3 \rho_1g\dfrac{h}{3} + \dfrac{1}{2}\rho_1 v^2 = P_0 + \dfrac{1}{2} 3\rho_1v_2^2$$ ......(2)
$$v_2$$ - velocity of the liquid at the second hole
$$v=0$$ since the liquid is at rest on on the surface
Putting $$v=0$$ in eqns (1) we get
$$v_1= \sqrt{\dfrac{2gh}{3}}$$
Putting $$v=0$$ in eqns (2) we get
$$v_2= \sqrt{\dfrac{4gh}{3}}$$
$$\therefore \dfrac{v_1}{v_2} = \sqrt{\dfrac{2}{4}}= \sqrt{\dfrac{1}{2}}$$
Question 7
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Water is filled upto height H units in a tank placed on the ground and whose side walls are vertical. A hole is made in one of the vertical wall such that the emerging stream of water strikes the ground at the maximum range. If the level in the tank is changing at the rate of R units per second, at that instant the rate at which range will be changing will be:

A
R/2 units per second

B
R units per second

C
2R units per second

D
Zero

Solution

According to to Torricelli's theorem, velocity of efflux at point A ( shown in the figure ) is
$$v_A=\sqrt{(2gh)}$$
After emerging from the orifice the water adopts parabolic path. if it takes time t secs in falling a vertical height (H=h) and covers a horizontal range $$R_1$$, then
$$H-h = \dfrac{1}{2}gt^2$$
$$ \Rightarrow t= \sqrt{\dfrac{2(H-h)}{g}}$$
$$R_1= v_At= \sqrt{(2gh)} * \sqrt{\dfrac{2(H-h)}{g}}= 2\sqrt{h(H-h)}$$ ....(1) where $$R_1$$ is the range.
The above results show that Range is same whether the hole is made at distance h from the top or at (H-h) from the top.
$$ \therefore h= H-h$$
$$ \Rightarrow h=\dfrac{1}{2}H$$ ....(2)
If h is changing at the rate $$\dfrac{dh}{dt}=R$$
differentiating eqn (1) wrt t, we get
$$ \dfrac{dR_1}{dt}= (h(H-h))^{-\dfrac{1}{2}}(\dfrac{dh}{dt}(H-h) + h( -\dfrac{dh}{dt}))=(h(H-h))^{-\dfrac{1}{2}} (R(H-h) -Rh)$$ ....(3)
Substituting $$ h=\dfrac{H}{2}$$ from eqn(2) in eqn(3) we get
$$ \dfrac {dR_1}{dt} = (\dfrac{H}{2})^{-1}R(H-2h)= (\dfrac{H}{2})^{-1}R(H-H)=0$$
Question 8
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A tank is filled up to a height 2H with a liquid and is placed on a platform of height H from the ground. The distance x from the ground where a small hole is punched to get the maximum range R is

A
H

B
1.25 H

C
1.5 H

D
2 H

Solution

We get the distance travelled in the y direction as
$$x=\dfrac {1}{2}gt^2\Rightarrow t=\sqrt {\dfrac {2x}{g}}$$
Velocity of efflux $$v=\sqrt{2gh}=\sqrt {2g(3H-x)}$$
Hence we get the range as
$$R=vt=2\sqrt {x(3H-x)}$$
For range to be maximum $$\dfrac {dR}{dx}=0$$
Thus we get
$$\dfrac{dR}{dx}=2(x(3H-x))^{-1/2}(3H-2x)$$
Which gives $$x=\dfrac {3}{2}H$$
Question 9
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In a container water is filled upto certain height with a hole at its bottom. A bird is coming towards free surface of the liquid with velocity $$\overrightarrow V_b=(12\hat i-10\hat j)m/s$$ and fish is rising upwards with velocity $$\overrightarrow V_{fish}(3\hat j)m/s$$.
The speed of bird as seen by fish under normal incidence condition at the moment when water level in container is $$20 m$$ from its bottom is: (given hole area $$=1 cm^2$$; surface area of liquid $$=20 cm^2)$$

A
$$20 m/s$$

B
$$40 m/s$$

C
$$10 m/s$$

D
$$12 m/s$$

Solution

$$\upsilon = \sqrt {2gh}=\sqrt {2\times 10\times 20}=20 m/s$$
$$A\times u=a\upsilon$$
$$u=\dfrac {1}{20}\times 20=1m/sec$$
$$V_{APP}=9\times \dfrac {4}{3}=12m/sec$$
So Velocity of bird w.r. t. fish
$$=16\hat j+12\hat i = 20m/sec$$
Question 10
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A long capillary glass tube of uniform diameter of $$1mm$$ is filled completely with water and then held vertically in air. It is now opened at both ends. Find the length of the water column remaining in the glass tube. Surface tension of water is $$0.075 N/m.$$ $$(g=10m/s^2$$ and density of water is $$10^3 kg/m^3)$$:

A
Zero

B
1.5 cm

C
3 cm

D
6 cm

Solution

Refer to the figure.
Pressure at a $$P_a= P_0-\dfrac{2T}{r}$$ .....(1)
Pressure at b $$P_b=P_0 + \dfrac{2T}{r}$$ .....(2)
Also, $$P_b -P_a = \rho gh$$
Substituting $$P_a$$ and $$P_b$$ in
$$ \therefore 2 \times \dfrac{2T}{r} = \rho gh$$
$$ \Rightarrow h= \dfrac{4T}{r\rho g}= \dfrac{4 \times 0.075}{0.5 \times 10^{-3} \times 10^3 \times 10}=0.6 \times 10^{-1} \:m=6 \:cm$$