Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 19

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Question 1
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A tank with vertical walls is mounted so that its base is at a height of $$1.2 m$$ above the horizontal ground. The tank is filled with water to a depth $$2.8m.$$ A hole is punched in the side wall of thetank at a depth $$x \ m$$ below the surface of water to have maximum range of the emerging stream.Then the value of $$x$$ in meter is

A
$$4.0$$

B
$$1.6$$

C
$$2.0$$

D
$$2.3$$

Solution

Velocity of ejection at a depth x is $$ \sqrt{2gx}$$
time of flight $$ = \sqrt{\dfrac{2H}{9}} = \sqrt{\dfrac{2(1.2+2.8-x)}{9}}$$
$$ = \sqrt{\dfrac{2(4-x)}{9}}$$
$$\therefore$$ Horizontal Range $$R = Vt = \sqrt{4 x (4-x)} $$
$$ = 2\sqrt{x(4-x)}$$
$$ \dfrac{dR}{dx} = 0 \Rightarrow x= \dfrac{4}{2} = 2m$$
Question 2
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A wind - powered generator converts wind energy into electrical energy. Assume that the generator converts a fixed fraction of the wind energy intercepted by its blades into electrical energy. For wind speed $$V$$, the electrical power output will be proportional to

A
$$V$$

B
$$V^{2}$$

C
$$V^{3}$$

D
$$V^{4}$$

Solution

Applying bernoullis eqn at point 1 and 2,
$$p_{0}+\dfrac{1}{2}f v^{2}=p_{r},\left [ p_{0}=0\right ]$$
So,
force on the wing$$=p_{1}\times A=\dfrac{1}{2}f v^{2}$$
So,
work done on wing$$=\dfrac{1}{2}f v^{2}.\dfrac{v}{k}=Kv^{3}$$
So,
power output in wing $$\propto v^{3}$$
Question 3
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Water stands at a height of $$100 cm$$ in a vessel whose side walls are vertical. A , B and C are holes at height $$80cm, 50 cm,$$ and $$20 cm$$ respectively from the bottom of the vessel. The correct system of water flowing out is :

A

B

C

D

Solution

At any height h, velocity of water ejection $$ = \sqrt{2gh}$$
time of flight $$ t = \sqrt{\dfrac{2(H-h)}{g}}$$
$$\therefore$$ range $$R = vT = \sqrt{2gh} \times \sqrt{\dfrac{2(H-h)}{g}} = 2\sqrt{h(H-h)}$$
$$\therefore$$ for $$ h = 20cm , 80cm $$ $$R$$ is same.
Therefore
$$R$$ is maximum for $$h = \dfrac{H}{2}$$
Question 4
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The pressure at the top of a building of height $$200m$$ is $$500kPa$$. The minimum pressure required at the base of the building to pump water to a closed tank on the top of the building is($$g=10m/s^{2}$$)

A
$$2.5 \times 10^{4}Pa$$

B
$$2.5 \times 10^{5}Pa$$

C
$$2.5 \times 10^{6} Pa$$

D
$$2.5 \times 10^{3} Pa$$

Solution

So,
Minimum pressure required at the base of building
$$=500\times 10^{3}+1000\times 10\times 200$$
$$=500\times 10^{5}+20\times 10^{5}=25\times 10^{5}Pa$$
$$=2.5\times 10^{6}Pa$$
Question 5
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There are two holes $$O_{1}$$ and $$O_{2}$$ in a tank of height H. The water emerging from $$O_{1}$$ and $$O_{2}$$ strikes the ground at the same points,as shown in fig. Then:

A
$$H=h_{1}+h_{2}$$

B
$$H=h_{2}-h_{1}$$

C
$$H=h_{1}h_{2}$$

D
$$H=h_{2}/h_{1}$$

Solution

At any height h below the surface , velocity of ejection $$ V = \sqrt{2gh}$$
time of flight $$t = \sqrt{2g (H-h)}$$
$$\therefore$$ Range $$R = vt = 2\sqrt{h (H-h)}$$
If $$ R_{1} = R_{2} \Rightarrow h_{1}(H-h_{1}) = h_{2} (H-h_{2})$$
$$ \Rightarrow h_{1} = H-h_{2}$$
$$h_{1}+h_{2} =H$$
Question 6
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A horizontal pipe of non-uniform cross-section allows water to flow through it with a velocity 1ms$$^{-1}$$ when pressure is 50kPa at a point. If the velocity of flow has to be 2 ms$$^{-1}$$ at some other piont,the pressure at that point should be:

A
50 kPa

B
100kPa

C
48.5 kPa

D
24.25 kPa

Solution

Applying Bernoulli's equation at point 1 and point 2,

$$\Rightarrow P_{1}+\dfrac{1}{2}\rho V_{1}^{2}=p_{2}+\dfrac{1}{2}\rho V_{2}^{2}$$
$$\Rightarrow 50\times 10^{3}+\dfrac{1}{2}\times 10^{3}\left ( 1 \right )^{2}=p_{2}+\dfrac{1}{2}\times 10^{3}\times 4$$
$$\Rightarrow 5\times 10^{4}+500-2000=p_{2}$$
$$\Rightarrow p_{2}=48.5kp{a}$$
Question 7
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Air of density $$1.3 kg m^{-3}$$ blows horizontally with a speed of $$108 km h^{-1}$$. A house has a plane roof of area $$40 m^{2}$$. The magnitude of aerodynamic lift on the roof is.

A
$$2.34 \times 10^{4}$$ Dynes

B
$$0.234 \times 10^{4} N$$

C
$$2.34 \times 10^{4} N$$

D
$$23.4 \times 10^{4} N$$

Solution

Applying bernoullis theorem above and below the roof,
$$p_{1}+\dfrac{1}{2}+v_{1}^{2}=p_{2}$$
$$\Rightarrow p_{2}-p_{1}=\dfrac{1}{2}+v_{1}^{2}=\dfrac{1}{2}\times 1.3\times \left ( 30 \right )^{2}$$
$$\Rightarrow$$ difference in pressure $$=585 N/m^{2}...........2$$
So,
Magnitude of aerodynamic life$$=1\times Area$$
$$=585\dfrac N{m^{2}}\times 40m^{2}$$
$$2.34\times 10^{4}N$$
Question 8
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In a plant, sucrose solution of coefficient of viscosity $$0.0015 N.m^{-2}$$ is driven at a velocity of $$10^{-3} m s^{-1}$$ through xylem vessels of radius $$2\mu m$$ and length $$5 \mu m$$ . The hydrostatic pressure difference across the length of xylem vessels in $$Nm^{-2}$$ is :

A
$$5$$

B
$$8$$

C
$$10$$

D
$$15$$

Solution

Fow rate $$q=v\times \pi r^2$$
also $$q=\dfrac{\pi Pr^4}{8\mu l}=v\pi r^2$$
now $$v=10^{-3}m/s\ ,\mu=0.0015N/m^2\ ,r=2\times 10^{-6}m$$
on solving above equation and putting value in SI unit we get
$$P=15 N/m^2$$
Question 9
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Two identical tall jars are filled with water to the brim. The first jar has a small hole on the side wall at a depth $$h/3$$ and the second jar has a small hole on the side wall at a depth of $$2h/3$$, where h is the height of the jar. The water issuing out from the first jar falls at a distance $$R_{1}$$ from the base and the water issuing out from the second jar falls at a distance $$R_{2}$$ from the base. The correct relation between $$R_{1} \ and \ R_{2}$$ is

A
$$R_{1} > R_{2}$$

B
$$R_{1} < R_{2}$$

C
$$R_{2} = 2 \times R_{1}$$

D
$$R_{1} = R_{2}$$

Solution

At any height h from surface, velocity $$V = \sqrt{2gh}$$
Time of flight $$ t = \sqrt{\dfrac{2(H-h)}{g}}$$
where $$H$$ is the total height of jar.
$$ \therefore$$ Range $$R = V\times t$$
$$R = 2\sqrt{h(H-h)}$$

Given, $$ h_{1} = \dfrac{h}{3} \ and \ H=h$$
$$ \Rightarrow $$ $$R_1 = 2\sqrt{\dfrac{h}{3}(h-\dfrac{h}{3})}$$ $$ = \dfrac{2\sqrt{2}h}{{3}}$$
For $$ \ h_{2} = \dfrac{2h}{3}\ and \ H=h$$
$$ \Rightarrow $$ $$R_2 = 2\sqrt{\dfrac{2h}{3}(h-\dfrac{2h}{3})}$$ $$ = \dfrac{2\sqrt{2}h}{{3}}$$
$$\therefore R_{1} = R_{2}$$
Question 10
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There is a hole at the bottom of a large open vessel.If water is filled upto a height $$h$$, it flows out in time $$t$$ .If water is filled to a height $$4h$$, it will flow out in time is

A
$$4t$$

B
$$t/4$$

C
$$t/2$$

D
$$2t$$

Solution

$$ v-u = -gt$$

$$ \Rightarrow 0 - \sqrt{2gh} = -gt$$

$$ t = \sqrt{\dfrac{2h}{g}}$$

For height $$ h^{1} = 4h t^{1} = \sqrt{\dfrac{2 (4h)}{g}} = 2t$$