Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 58

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Question 1
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Directions For Questions

A container of large uniform cross-sectional area A resting on a horizontal surface, holds, two immissicible, non-viscous and incompressible liquids of densities $$d$$ and $$2d$$ each of height $$H/2 $$ as shown in the figure. The lower density liquid is open to the atmosphere having pressure $$P_{0}. A$$ homogeneous solid cylinder of length $$ L(L< H /2)$$, cross-sectional area $$A/5 $$ is immersed such that it floats with its axis vertical at the liquid-liquid interface with length $$L/4$$ in the denser liquid.
The cylinder is then removed and the original arrangement is restored. A tiny hole of area $$s(s < < A)$$ is punched on the vertical side of the container at a height $$h(h< H/2)$$. As a result of this, liquid starts flowing out of the hole with a range $$x$$ on the horizontal surface.

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The density $$D$$ of the material of the floating cylinder is:

A
$$5d/4$$

B
$$3d/4$$

C
$$4d/5$$

D
$$4d/3$$

Solution

Weight=upthrust
$$\therefore $$ $$\displaystyle \left (

L\frac{A}{5}Dg \right )=\frac{L}{4}\times \frac{A}{5}\times 2d\times

g+\frac{3L}{4}\times \frac{A}{5}\times d\times g$$
$$\therefore $$ $$\displaystyle D=\frac{5d}{4}$$
Question 2
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A container of large uniform cross-sectional area $$A$$ resting on a horizontal surface, holds, two immiscible, non-viscous and incompressible liquids of densities $$d$$ and $$2d$$ each of height $$H/2 $$ as shown in the figure. The lower density liquid is open to the atmosphere having pressure $$P_{0}. A$$ homogeneous solid cylinder of length $$ L(L< H /2)$$, cross-sectional area $$A/5 $$ is immersed such that it floats with its axis vertical at the liquid-liquid interface with length $$L/4$$ in the denser liquid.
The cylinder is then removed and the original arrangement is restored. A tiny hole of area $$s(s < < A)$$ is punched on the vertical side of the container at a height $$h(h< H/2)$$. As a result of this, liquid starts flowing out of the hole with a range $$x$$ on the horizontal surface.

The horizontal distance traveled by the liquid, initially, is :

A
$$\sqrt{(3H+4h) h}$$

B
$$\sqrt{(3h+4H) h}$$

C
$$\sqrt{(3H-4h) h}$$

D
$$\sqrt{(3H-3h) h}$$

Solution

Applying Bernoulli's equation just inside and just outside the hole,$$\displaystyle P_{0}+\left ( \frac{H}{2} \right )\left ( d \right )g+\left ( \frac{H}{2}-h \right )\left ( 2d \right )\left ( g \right )=\frac{1}{2}\left ( 2d \right )v^{2}+P_{0}$$

$$\therefore $$$$\displaystyle v=\sqrt{\frac{g}{2}\left ( 3H-4h \right )}$$

$$\displaystyle t=\sqrt{\frac{2h}{g}}$$

$$\therefore $$ $$x=vt=\sqrt{\left ( 3H-4h \right )h}$$
Question 3
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An open vessel full of water is falling freely under gravity. There is a small hole in one filee of the vessel as shown in the figure. The water which comes out from the hole at the instant when hole is at height H above the ground, strikes the ground at a distance of $$x$$ from $$P$$. Which of the following is correct for the situation described?

A
The value of x is $$\displaystyle 2\sqrt{\frac{2hH}{3}}$$

B
The value of x is $$\displaystyle \sqrt{\frac{4hH}{3}}$$

C
The value of $$x$$ can't be computed from information provided

D
No water comes out from the hole

Solution

Since the whole vessel is falling under gravity, apparent pressure at the hole inside the vessel=$$P_0+\rho(g-a)h_1=P_0+\rho(g-g)h_1=P_0$$.
The pressure at hole outside the vessel is $$P_0$$.
Since there is no pressure difference, no water comes out of the hole.
Question 4
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A liquid flows along a horizontal pipe AB of uniform cross-section. The difference between the level of the liquid in tubes P and Q is $$10\ cm$$. The diameter of the tubes P and Q are the same. Then:
($$\displaystyle \\ g=9.8{ ms }^{ -2 }$$)

A
level in P is greater than that of Q and velocity of flow is $$1.4 \ m/s$$

B
level in Q is greater than that of P and velocity of flow is $$1.4\ m/s$$

C
level in P is greater than that of Q and velocity of flow is $$0.7\ m/s$$

D
level in Q is greater than that of P and velocity of flow is $$0.7\ m/s$$

Solution

Assuming laminar flow, velocity of liquid inside P is zero as it is perpendicular to the direction of flow of liquid. Hence, liquid level in P is same as the height of the pipe AB.
For pipe Q, it is inclined against the direction of water and there is extra pressure on the bottom of Q. This causes rise in level in Q.
Applying Bernoulli theorem for Q,
$$P_o+\rho g h=P_o+\dfrac{1}{2}\rho v^2$$
$$v=\sqrt {2gh}=\sqrt {2 \times 9.8 \times 10}=1.4 m/s$$
Question 5
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A cylindrical vessel open at the top is 20 cm high and 10 cm in diameter. A circular hole of cross sectional area $$1 cm^{2}$$ is cut at the centre of the bottom of the vessel. Water flows from a tube above it in to the vessel at the rate of $$10^{2} cm^{3}/s$$. The height of water in the vessel under steady state is
(Take $$g= 10m/s^{2}$$)

A
20cm

B
15cm

C
10cm

D
5cm

Solution

In steady state, Volume flow rate entering the vessel=volume flow rate leaving the vessel
$$\therefore $$ $$Q=av=a\sqrt{2gh}$$ or $$\displaystyle h=\frac{Q^{2}}{2ga^{2}}$$
$$\displaystyle =\frac{\left ( 10^{3} \right )^{2}}{\left ( 2\times 1000 \right )\left ( 1 \right )^{2}}=5$$ cm
Question 6
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Equal volumes of a liquid is poured into containers A and B such that the area of cross-section of container A is double the area of cross-section of container B. If $$P_A$$ and $$P_B$$ are the pressures exerted at the bottom of the containers then $$P_A : P_B = $$__________

A
1 : 2

B
1 : 1

C
3 : 2

D
2 : 1

Solution
Question 7
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A cylinder of height h filled with water and is kept on lock of height h/2. The level of water in the cylinder is kept constant. Four holes numbered 1, 2, 3 and 4 are at the side of the cylinder and at height 0, h/4, h/2 and 3h/4, respectively. When all four holes are opened together, the hole from which water will reach farthest distance on the plane PQ is the hole number.

A
1

B
2

C
3

D
4

Solution

We know from Torricelli's theorem, that the range of the liquid falling from a certain height is given by:
$$ R = 2 \times \sqrt{h(H - h)} $$
where $$ H $$ is the total height of the container and
$$ h $$ is the height where the hole is.

For $$ R = R_{max} $$;
$$ \dfrac{dR}{dh} = 0 $$

$$ \dfrac{dR}{dh} = 2 \times \Big( \dfrac{1}{2 \sqrt{h}} \sqrt{H - h} + \sqrt{h} \dfrac{-1}{2 \sqrt{H - h}} \Big) $$

$$ \dfrac{dR}{dh} = \Big( \dfrac{ \sqrt{H - h}}{ \sqrt{h}} + \dfrac{- \sqrt{h}}{ \sqrt{H - h}} \Big) $$

$$ \dfrac{dR}{dh} = \dfrac{H - h -h}{\sqrt{h(H - h)}} = 0 $$

$$ \Rightarrow H - h -h = 0 $$
$$ \Rightarrow H = 2h $$

For $$ R = R_{max} $$
$$ h = \dfrac{H}{2} $$

Taking PQ as the reference,
$$ H = h + \dfrac{h}{2} = \dfrac{3h}{2} $$

So hole must be at height
$$ \dfrac{H}{2} = \dfrac{\Big( \dfrac{3h}{2} \Big)}{2} = \dfrac{3h}{4} $$

For hole 1,
$$ h_1 = \dfrac{h}{2} + 0 = \dfrac{h}{2} $$

For hole 2,
$$ h_2 = \dfrac{h}{2} + \dfrac{h}{4} = \dfrac{3h}{4} $$

For hole 3,
$$ h_3 = \dfrac{h}{2} + \dfrac{h}{2} = h $$

For hole 4,
$$ h_4 = \dfrac{h}{2} + \dfrac{3h}{4} = \dfrac{5h}{4} $$

Since, hole 2 is at the $$ \dfrac{H}{2} $$ height required for longest range.
$$ \therefore $$ Hole 2 is from which water will reach farthest distance on the plane PQ.

Hence, the correct answer is OPTION B.
Question 8
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When pressure is applied through a piston at the top of a closed tube containing water, the pressure is transmitted to:

A
Only the bottom of container

B
All directions

C
Only the side faces and the bottom of the container

D
None of these

Solution
Question 9
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A capillary tube of radius $$r$$ is immersed in water and water rises to a height of $$h$$. Mass of water in the capillary tube is $$5\times 10^{-3}kg$$. The same capillary tube is now immersed in a liquid whose surface tension is $$\sqrt{2}$$ times the surface tension of water. The angle of contact between the capillary tube and this liquid is $$45^o$$. The mass of liquid which rises into the capillary tube now is (in kg):

A
$$5\times 10^{-3}$$

B
$$2.5\times 10^{-3}$$

C
$$5\sqrt{2}\times 10^{-3}$$

D
$$3.5\times 10^{-3}$$

Solution

Since the force on liquid is proportional to $$Scos(\theta )$$, in the latter case, force will be $$\sqrt { 2 } cos(\dfrac { \pi }{ 4 } )=1$$ times the force in first case. Hence, since the same force is applied, mass of water rising in the tube will be same.
Question 10
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A given shaped glass tube having uniform cross section is filled with water and is mounted on a rotatable shaft as shown in figure. If the tube is rotated with a constant angular velocity $$\displaystyle \omega $$, then

A
Water levels in both sections A and B go up

B
Water level in section A goes up and that in B comes down

C
Water level in section A comes down and that in B it goes up

D
Water levels remain same in both sections

Solution

Water level in both A and B will go up. The pressure difference thus created will provide the necessary centripetal force for the water body to rotate around the vertical axis.