Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 26

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Weekly Quiz Competition

Question 1
1 / -0

The diagram shows a venturimeter through which water is flowing. The speed of water $$X$$ is $$2 \ \text{cms}^{-1}$$. The speed of water at $$Y$$ $$(\ taking\ g = 10 \text{ms}^{-2})$$ is

A
$$23\ \text{cms}^{-1}$$

B
$$32\ \text{cms}^{-1}$$

C
$$101\ \text{cms}^{-1}$$

D
$$1024\ \text{cms}^{-1}$$

Solution

$$p_x - p_y = \dfrac{\rho}{2}(v_y^2 - v_x^2) \\\Rightarrow \Delta h \rho g=\dfrac{\rho}{2}(v_y^2 - v_x^2) \\\Rightarrow \Delta h g=\dfrac{1}{2}(v_y^2 - v_x^2) \\\Rightarrow 0.51 \times 1000=\dfrac{1}{2}(v_y^2 - 2^2) \\\Rightarrow 1020= v_y^2 - 4 \\\Rightarrow 1024= v_y^2 \\\Rightarrow v_y=32 cm/s$$
Question 2
1 / -0

An ideal fluid is:

A
similar to a perfect gas.

B
non-viscous and incompressible.

C
one which satisfies the continuity equation.

D
one which obeys Newton's formula for viscous drag.

Solution

An ideal fluid is one which is incompressible and non viscous.
Question 3
1 / -0

Water stored in a tank flows out through a hole of radius 1mm at a depth 10 m below the surface of water.The rate of flow of water in $$m^3 /s$$ will be

A
$$4.4 \times 10^{-5}$$

B
$$4.4 \times 10^{-4}$$

C
$$4.4 \times 10^{-3}$$

D
$$4.4 \times 10^{-2}$$

Solution

According to Torricelli's Theorem velocity of efflux i.e. the velocity with which the liquid flows out of a hole is equal to $$\sqrt{2gh}$$ where $$h$$ is the depth of the hole below the liquid surface. here $$h=10m$$
$$\Rightarrow v=\sqrt{2 \times 9.8 \times 10}=\sqrt{196}=14m/s$$
and rate of flow $$=av=\pi (0.001)^2 \times 14=4.4 \times 10^{-5} m^3/s$$
Question 4
1 / -0

Bernoulli's equation includes as a special case:

A
Archimede's principle

B
Pascal's law

C
Toricelli's law

D
Hooke's law

Solution

Torricelli's law states that the speed of efflux, $$v$$ of a fluid through a sharp-edged hole at the bottom of a tank filled to a depth $$h$$ is the same as the speed that a body (in this case a drop of water) would acquire in falling freely from a height $$h$$.
This problem is solved using Bernoulli's equation,
$$gh+\dfrac{P_{atm}}{\rho}=\dfrac{v^2}{2}+\dfrac{P_{atm}}{\rho}$$
$$\implies v=\sqrt{2gh}$$
Question 5
1 / -0

Water flows steadily through a horizontal tube of variable cross-section. If the pressure of water is p at a point where the velocity of flow is v, what is the pressure at another point where the velocity of flow is 2v; $$\rho$$ being the density of water?

A
$$p - \displaystyle \frac {3}{2} \rho v^2$$

B
$$p + \displaystyle \frac {3}{2} \rho v^2$$

C
$$p - 2 \rho v^2$$

D
$$p + 2 \rho v^2$$

Solution

Applying Bernoullis's Equation at two points:
$${ p }_{ 1 }+\displaystyle\frac { \rho { { v }_{ 1 } }^{ 2 } }{ 2 } ={ p }_{ 2 }+\displaystyle\frac { \rho { { v }_{ 2 } }^{ 2 } }{ 2 }$$

Given, $${ p }_{ 1 }=p,{ \quad v }_{ 1 }=v\quad { v }_{ 2 }={ 2v }\\ { p }_{ 2 }={ p }+\displaystyle\frac { \rho { { v } }^{ 2 } }{ 2 } -\displaystyle\frac { 4\rho { { v } }^{ 2 } }{ 2 } \\ { p }_{ 2 }={ p }-\displaystyle\frac { 3\rho { { v } }^{ 2 } }{ 2 } $$
Question 6
1 / -0

Two liquid jets coming out of the small holes at P and Q intersect at the point R. Find the position of R if we maintain the liquid level constant

A
$$\sqrt 2h$$

B
$$2\sqrt 2h$$

C
$$3\sqrt 2h$$

D
$$5\sqrt 2h$$

Solution

Velocity of liquid coming out of point P $$V_p = \sqrt{2gh}$$
Time taken by liquid stream to reach R $$T_1 = \sqrt{\dfrac{2 (h+y)}{g}}$$
$$\therefore$$ $$x = V_p T_1 = \sqrt{4h (h+y)}$$ .........(1)

Velocity of liquid coming out of point Q $$V_Q = \sqrt{2g(2h)}$$
Time taken by liquid stream to reach R $$T_2 = \sqrt{\dfrac{2 y}{g}}$$
$$\therefore$$ $$x = V_Q T_2 = \sqrt{4h (2h)}$$ .........(2)
From (1) and (2) we get $$4h (h+y) = 4h (2h)$$ $$\implies y = h$$
$$\therefore$$ Position of R $$x = \sqrt{4h (h+y)} = \sqrt{4h (h+ h)} =2\sqrt{2} h$$
Question 7
1 / -0

A spherical ball falls through viscous medium with terminal velocity $$v$$. If this ball is replaced by another ball of the same mass but half the radius, then the terminal velocity will be (neglect the effect of buoyancy)

A
$$v$$

B
$$2v$$

C
$$4v$$

D
$$8v$$

Solution

In case of spherical ball
$$m=\dfrac{4}{3}\pi r^3 \rho$$
keeping $$m$$ constant, if $$r$$ is halved, $$\rho$$ will increased by a factor of $$8$$;
If buoyancy is neglected,
$${ V }_{ T }\propto { r }^{ 2 }\rho; \text{ otherwise } { V }_{ T }\propto { r }^{ 2 }\left ( \rho-\rho_o \right ) \\ \Rightarrow \dfrac { { V }_{ T }' }{ { V }_{ T } } =\dfrac { { r' }^{ 2 }\rho ' }{ { r }^{ 2 }\rho } \\ \Rightarrow \dfrac { { V }_{ T }' }{ { V }_{ T } } =\dfrac { { \left( \dfrac { r }{ 2 } \right) }^{ 2 }8\rho }{ { r }^{ 2 }\rho } \\ \Rightarrow \dfrac { { V }_{ T }' }{ { V }_{ T } } =\dfrac { 2{ r }^{ 2 }\rho }{ { r }^{ 2 }\rho } \\ \Rightarrow \dfrac { { V }_{ T }' }{ { V }_{ T } } =2\\ \Rightarrow { V }_{ T }'=2{ V }_{ T }\\ \Rightarrow \boxed{{ V }_{ T }'=2v}$$
Question 8
1 / -0

Water is filled in a container up to height $$3\ \text{m}.$$ A small hole of area '$$a$$' is punched in the wall of the container at a height $$52.5\ \text{cm}$$ from the bottom. The cross sectional area of the container is $$A.$$ If $$\dfrac{a}{A}=0$$; then $$v^2$$ is $$($$where $$v$$ is the velocity of water coming out of the hole$$)$$

A
50

B
51

C
48

D
51.5

Solution

$$\text{According to Torricelli's Theorem velocity of efflux i.e. the velocity with which the liquid}$$
$$\text{flows out of a hole is equal to}\ \sqrt{2gh}\ \text{where}\ h\ \text{is the depth of the hole below the liquid surface.}$$
Here $$h=300-52.5=247.5cm=2.475m$$
$$\Rightarrow v^2=2gh=2 \times 10 \times 2.475 = 49.5 \approx 50$$
Hence correct choice is 'A'.
Question 9
1 / -0

In a horizontal pipeline of uniform cross section the pressure falls by $$8N/m^2$$ between two points separated by 1 km. If oil of density $$800 kg/m^3$$ flows through the pipe, find the change in KE per kg of oil at these points.

A
$$10^{-2}J/kg$$

B
$$10^{-3}J/kg$$

C
$$10^{-4}J/kg$$

D
$$10^{-1}J/kg$$

Solution

Applying Bernoulli's principle on a horizontal pipe, we have
$$P_{ 1 }+\dfrac { 1 }{ 2 } \rho { v_{ 1 } }^{ 2 }=P_{ 2 }+\dfrac { 1 }{ 2 } \rho { v_{ 2 } }^{ 2 }\\ \Rightarrow \dfrac { 1 }{ 2 } \rho { v_{ 2 } }^{ 2 }-\dfrac { 1 }{ 2 } \rho { v_{ 1 } }^{ 2 }=P_{ 1 }-P_{ 2 }\\ \Rightarrow \dfrac { 1 }{ 2 } { v_{ 2 } }^{ 2 }-\dfrac { 1 }{ 2 } { v_{ 1 } }^{ 2 }=\dfrac { P_{ 1 }-P_{ 2 } }{ \rho } \\ \Rightarrow \text{Change in kinetic energy per unit mass } (\Delta K.E.)=\dfrac { P_{ 1 }-P_{ 2 } }{ \rho } \\\Rightarrow \Delta K.E.=\dfrac{8}{800}=10^{-2}\mathrm{J/Kg}$$
Question 10
1 / -0

A pressure meter attached to a closed water tap reads $$1.5\times 10^5\ \text{Pa}$$. When the tap is opened, the velocity of flow of water is $$10\ \text{ms}^{-1}$$ and the reading of the pressure meter is

A
$$1.5\times 10^5\ \text{Pa}$$

B
$$3\times 10^5\ \text{Pa}$$

C
$$0.5\times 10^5\ \text{Pa}$$

D
$$10^5\ \text{Pa}$$

Solution

Applying Bernoulli's principle on a horizontal pipe, we have
$$P_1+\frac{1}{2} \rho {v_1}^2=P_2+\frac{1}{2} \rho {v_2}^2$$
here $$P_1= 1.5 \times 10^5\ \text{pascal},\ v_1=0,\ v_2=10 \text{m/s}$$, $$\rho=1000\ \text{kg/m}^3$$
$$P_1+\dfrac{1}{2} \rho {v_1}^2=P_2+\dfrac{1}{2} \rho {v_2}^2 \\\Rightarrow 1.5 \times 10^5=P_2+\dfrac{1}{2} \times 1000 \times {(10)}^2 $$
solving the above equation we have $$P_2= 10^5 \text{pascal}$$

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Re-Attempt Weekly Quiz Competition

Mechanical Properties of Fluids Test - 26

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Mechanical Properties of Fluids Test - 26

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Question 1

$$23\ \text{cms}^{-1}$$

$$32\ \text{cms}^{-1}$$

$$101\ \text{cms}^{-1}$$

$$1024\ \text{cms}^{-1}$$

Solution

Question 2

similar to a perfect gas.

non-viscous and incompressible.

one which satisfies the continuity equation.

one which obeys Newton's formula for viscous drag.

Solution

Question 3

$$4.4 \times 10^{-5}$$

$$4.4 \times 10^{-4}$$

$$4.4 \times 10^{-3}$$

$$4.4 \times 10^{-2}$$

Solution

Question 4

Archimede's principle

Pascal's law

Toricelli's law

Hooke's law

Solution

Question 5

$$p - \displaystyle \frac {3}{2} \rho v^2$$

$$p + \displaystyle \frac {3}{2} \rho v^2$$

$$p - 2 \rho v^2$$

$$p + 2 \rho v^2$$

Solution

Question 6

$$\sqrt 2h$$

$$2\sqrt 2h$$

$$3\sqrt 2h$$

$$5\sqrt 2h$$

Solution

Question 7

$$v$$

$$2v$$

$$4v$$

$$8v$$

Solution

Question 8

50

51

48

51.5

Solution

Question 9

$$10^{-2}J/kg$$

$$10^{-3}J/kg$$

$$10^{-4}J/kg$$

$$10^{-1}J/kg$$

Solution

Question 10

$$1.5\times 10^5\ \text{Pa}$$

$$3\times 10^5\ \text{Pa}$$

$$0.5\times 10^5\ \text{Pa}$$

$$10^5\ \text{Pa}$$

Solution

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