Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 61

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Question 1
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A narrow tube completely filled with a liquid is lying on a series of cylinders as shown in the figure. Assuming no sliding between any surfaces, the value of acceleration of the cylinders for which liquid will not come out of the tube from anywhere is given by

A
$$\frac{gH}{2L}$$

B
$$\frac{gH}{L}$$

C
$$\frac{2gH}{L}$$

D
$$\frac{gH}{\sqrt{2L}}$$

Solution
Question 2
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A liquid flows through a horizontal tube as shown in figure. The velocities of the liquid in the two sections, which have areas of cross-section $$A_1$$ and $$A_2$$, are $$v_1$$ and $$v_2$$, respectively The difference in the levels of the liquid in the two vertical tubes is h. then

A
$$v_2^2 - v_1^2 = 2gh$$

B
$$v_2^2 + v_1^2 = 2gh$$

C
$$v_2^2 - v_1^2 = gh$$

D
$$v_2^2 + v_1^2 = gh$$

Solution

Hint: Use Bernoulli’s equation to find the difference of pressure at both point.
Step – 1: From Bernoulli’s equation, find the difference of pressure at both point
Bernoulli’s equation is given by,
$$\dfrac{P_1}{\rho} + \dfrac{{V_1}^2}{2} = \dfrac{P_2}{\rho} + \dfrac{{V_2}^2}{2}$$
$$ \dfrac{P_1}{\rho} -\ \dfrac{P_2}{\rho} = \dfrac{{V_2}^2}{2} - \dfrac{{V_1}^2}{2}$$
$$ P_1 - P_2 = \dfrac{\rho}{2} ({V_2}^2 - {V_1}^2)$$
Step – 2: Find the difference between the pressure at two different pressure
There is difference h in height of 2 points. So, difference in pressure will be $$\rho gh$$.
Step – 3: Put the value in the equation
$$\rho gh = \dfrac{\rho}{2} ({V_2}^2 - {V_1}^2)$$
$$ ({V_2}^2 - {V_1}^2) = 2gh$$
$$Answer:$$
Hence, option A is the correct answer.
Question 3
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Directions For Questions

Figure 4.240 shows a large closed cylindrical tank containing water. Initially, the air trapped above the water surface has a height $${h}_{0}$$ and pressure $${2p}_{0}$$ where $${p}_{0}$$ is the atmospheric pressure. There is a hole in the wall of the tank at a depth $${h}_{1}$$ below the top from which water comes out. A long vertical tube is connected as shown.

...view full instructions

Find the speed with which water comes out of the hole.

A
$$\dfrac { 1 }{ \rho } { \left[ { p }_{ 0 }-\rho g({ h }_{ 1 }-2{ h }_{ 0 }) \right] }^{ 1/2 }$$

B
$${ \left[ \dfrac { 2 }{ \rho } { \left[ { p }_{ 0 }+\rho g({ h }_{ 1 }-{ h }_{ 0 }) \right] } \right] }^{ 1/2 }$$

C
$${ \left[ \dfrac { 3 }{ \rho } { \left[ { p }_{ 0 }+\rho g({ h }_{ 1 }+{ h }_{ 0 }) \right] } \right] }^{ 1/2 }$$

D
$${ \left[ \dfrac { 4 }{ \rho } { \left[ { p }_{ 0 }-\rho g({ h }_{ 1 }-{ h }_{ 0 }) \right] } \right] }^{ 1/2 }$$

Solution

KE of water=pressure energy of the water at the layer
$$\cfrac { 1 }{ 2 } m{ v }^{ 2 }=$$Pressure $$\times $$ volume
$$\cfrac { 1 }{ 2 } m{ v }^{ 2 }=\cfrac { m\times P }{ \rho } $$
$${ v }^{ 2 }=\cfrac { 2P }{ \rho } =\left[ \cfrac { 2 }{ \rho } \left[ { P }_{ 0 }+\rho g\left( { h }_{ 1 }-{ h }_{ 0 } \right) \right] \right] $$
$$v=\sqrt { \cfrac { 2 }{ \rho } \left[ { P }_{ 0 }+\rho g\left( { h }_{ 1 }-{ h }_{ 0 } \right) \right] } $$
Question 4
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A vertical cylinder is filled with liquid. A small hole is made in the wall of the cylinder at a depth $$H$$ below the free surface of the liquid. The force exerted on the cylinder by the liquid flowing out of the hole initially will be proportional to :

A
$$ \sqrt {H} $$

B
$$ H $$

C
$$ H^{3/2}$$

D
$$H^2$$

Solution

By using Bernoulli’s principle

The velocity of the liquid coming out is given as,

$$v = \sqrt {2gh} $$

The mass of the liquid will be

$$m = \rho AV$$

The change in momentum will be

$$P = mv$$

The force will be proportional of $${v^2}$$ which in turn is proportional to $$h$$
Question 5
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Directions For Questions

A cylindrical container of length $$L$$ is full of the brim with a liquid which has mass density $$\rho$$. It is placed on a weigh-scale; the scale reading is $$W$$. A light ball of volume $$V$$ and mass $$m$$ which would float on the liquid, if allowed to do so, is pushed gently down and held beneath the surface of the liquid with a rigid rod of negligible volume as shown on the left.

...view full instructions

What is the reading of the scale when the ball is fully immersed?

A
$$W-\rho Vg$$

B
$$W$$

C
$$W + mg -\rho Vg$$

D
None

Solution
Question 6
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A large open top container of negligible mass and uniform cross-sectional area-A has a small hole of cross-sectional area A/100 in its side wall near the bottom. The container is kept on a smooth horizontal floor and contains a liquid of density $$\rho$$ and mass $$m_0$$. Assuming that the liquid starts flowing out horizontally through the hole at t = 0, calculate the acceleration of the container

A
$$\dfrac{g}{50}$$

B
$$\dfrac{g}{60}$$

C
$$\dfrac{g}{70}$$

D
$$\dfrac{g}{80}$$

Solution
Question 7
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A fixed thermally conducting cylinder has a radius $$R$$ and height $$L_{0}$$. The cylinder is open at its bottom and has a small hole at its top. A piston of mass $$M$$ is held at a distance $$L$$ from the top surface as shown in Fig. The atmospheric pressure is $$p_{0}$$.
The piston is taken completely out of the cylinder. The hole at the top is sealed. A water tank is brought below the cylinder and put in a position so that the water surface in the tank is at the same level as the top of the cylinder as shown in Fig. The density of the water is $$\rho$$. In equilibrium, the height $$H$$ of the water column in the cylinder satisfies.

A
$$\rho g(L_{0} - H)^{2} + P_{0}(L_{0} - H) + L_{0}P_{0} = 0$$

B
$$\rho g(L_{0} - H)^{2} - P_{0} (L_{0} - H) - L_{0}P_{0} = 0$$

C
$$\rho g(L_{0} - H)^{2} + P_{0}(L_{0} - H) - L_{0}P_{0} = 0$$

D
$$\rho g(L_{0} - H)^{2} - P_{0}(L_{0} - H) + L_{0}P_{0} = 0$$

Solution

$$p_{1} = p_{2}$$
$$p_{0} + \rho g(L_{0} - H) = p ... (i)$$
Now, applying $$p_{1}V_{1} = p_{2}V_{2}$$ for the air inside the cylinder, we have
$$p_{0}(L_{0}) = p(L_{0} - H)$$
$$p = \dfrac {p_{0}L_{0}}{L_{0} - H}$$
Substituting in Eq. (i), we have
$$p_{0} + \rho g (L_{0} - H) = \dfrac {p_{0}L_{0}}{L_{0} - H}$$
$$\Rightarrow \rho g (L_{0} - H)^{2} + P_{0} (L_{0} - H) - L_{0}P_{0} = 0$$
Therefore, option (c) is correct.
Question 8
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The excess pressure inside on soap bubble is three times inside that of the second soap bubble.The ratio of their surface area is:

A
1:9

B
1:3

C
3:1

D
1:27

Solution
Question 9
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A capillary tube $$(A)$$ is dropped in water . Another identical tube $$(B)$$ is dipped in a soap water solution. Which of the following shows the relative nature of the liquid columns in the two tubes?

A

B

C

D

Solution
Question 10
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Directions For Questions

If the container filled with liquid gets accelerated horizontally or vertically, pressure in liquid gets changed. In liquid, for calculation of pressure, effective $$g$$ is used. A closed box with horizontal base $$6\ m$$ by $$6\ m$$ and a height $$2\ m$$ is half filled with liquid. It is given a constant horizontal acceleration $$g/2$$ and vertical downward acceleration $$g/2$$.

...view full instructions

Water pressure at the bottom of centre of the box is equal to (atmospheric pressure =$${10}^{5}\ N/{m}^{2}$$, density of water= $$1000\ kg/{m}^{3}, g= 10\ m/{s}^{2}$$)

A
$$ 1.1\ MPa$$

B
$$ 0.11\ MPa$$

C
$$0.101\ MPa$$

D
$$0.021\ MPa$$

Solution