Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

A $$50kg$$ skydiver falls through the air and reaches terminal velocity after some time. The drag force is a function of velocity given by
$${F}_{drag}=-b{v}^{2}$$
where the negative sign denotes that the drag force is opposite to the direction of the velocity.
What is the terminal velocity of the skydiver (assuming the drag constant $$b$$ is $$0.2 {kg}/{m}$$)?

A
$$5\dfrac{m}{s}$$

B
$$50\dfrac{m}{s}$$

C
$$100\dfrac{m}{s}$$

D
$$250\dfrac{m}{s}$$

E
$$2500\dfrac{m}{s}$$

Solution

When the falling body through the air will reach to the terminal velocity, the drag force will be equal to weight of the body.
Thus, $$bv^2=mg$$
The terminal velocity, $$v=v_t=\sqrt{\frac{mg}{b}}=\sqrt{\frac{50\times 10}{0.2}}=50 m/s$$
Thus , option B will be correct.
Question 2
1 / -0

A $$300,000 kg$$ commercial airlines is flying through the air with $$150,000 N$$ of thrust.
For airplanes traveling at high speeds, the drag force is related to the velocity by the following equation
$${F}_{drag}=-b{v}^{2}$$
where drag force is opposite to the direction of the velocity.
If the drag constant $$b$$ is $$3 {kg}/{m}$$ and the airplane is moving at $$200 {m}/{s}$$, how would you describe its motion?

A
In equilibrium (cruising at constant speed)

B
Near equilibrium, positive acceleration

C
Near equilibrium, negative acceleration

D
Greatly accelerating during takeoff

E
Greatly slowing down during landing

Solution

Drag force acting at the given moment=$$F_{drag}=-bv^2=-(3)\times (200)^2=-120000N$$
Hence net force acting on the plane=$$150000N-120000N$$
$$=30000N$$
Hence the option is accelerating at the moment.
Question 3
1 / -0

A spherical body of diameter D is falling in viscous medium. Its terminal velocity is proportional to:

A
$${ V }_{ 1 }\propto { D }^{ 1/2 }$$

B
$${ V }_{ 1 }\propto { D }^{ 3/2 }$$

C
$${ V }_{ 1 }\propto { D }^{ 2 }$$

D
$${ V }_{ 1 }\propto { D }^{ 5/2 }$$

Solution

The formula for terminal velocity is $$v=\dfrac { 2g{ r }^{ 2 }({ \rho }-{ \sigma }) }{ 9{ \eta } } $$, where $${\eta}$$ is the viscosity.
So this clearly shows that terminal velocity is directly proportional to square of diameter.
Question 4
1 / -0

Water is flowing in streamline motion through a horizontal tube. The pressure at a point in the tube is $$P$$ where the velocity of flow is $$v$$. At another point, where the pressure is $$P/2$$, the velocity of flow is:
[Density of water $$=\rho$$]

A
$$\sqrt { { v }^{ 2 }+\cfrac { P }{ \rho } } $$

B
$$\sqrt { { v }^{ 2 }-\cfrac { P }{ \rho } } $$

C
$$\sqrt { { v }^{ 2 }+\cfrac { 2P }{ \rho } } $$

D
$$\sqrt { { v }^{ 2 }-\cfrac { 2P }{ \rho } } $$

Solution

As the water is flowing through the horizontal tube, so in the streamline flow of water, the sum of static pressure and dynamic pressure is constant
$$P+\cfrac { 1 }{ 2 } \rho { v }^{ 2 }=\cfrac { P }{ 2 } +\cfrac { 1 }{ 2 } \rho { v }_{ 1 }^{ 2 }$$
$$\Rightarrow { v }_{ 1 }=\sqrt { \cfrac { P }{ \rho } +{ v }^{ 2 } } $$
Question 5
1 / -0

A block of mass of a $$2kg$$ with dimensions $$5cm\times 20cm\times 10cm$$ respectively. The ratio of minimum to maximum pressure it exerts on the change in orientation is:

A
$$1:1$$

B
$$1:2$$

C
$$1:4$$

D
$$4:1$$

Solution

Pressure is the force per unit area. So, $$P=F/A$$
Thus, minimum surface area $$(5 cm \times 10 cm) $$ will give the maximum pressure and maximum area $$(20 cm\times 10 cm)$$ will give the minimum pressure.
So, $$P_{max}=\dfrac{2g}{50\times 10^{-4}} N/m^2$$ and $$P_{min}=\dfrac{2g}{200\times 10^{-4}} N/m^2$$
Thus, $$\dfrac{P_{min}}{P_{max}}=50/200=1/4$$
The option C will be right.
Question 6
1 / -0

Due to air a falling body faces a resistive force proportional to square of velocity $$v$$, consequently its effective downward acceleration is reduced and is given by $$a = g - kv^{2}$$ where $$k = 0.002m^{-1}$$. The terminal velocity of the falling body is (in m/s)

A
$$60$$

B
$$70$$

C
$$80$$

D
$$90$$

Solution

As velocity increases $$a$$ will decrease and at some point, it will become zero and velocity become constant that is terminal velocity.
At terminal velocity- $$g-kv_{ter}^2 =0 \implies v_{ter}^2= \dfrac{g}{k}=\dfrac{9.8}{0.002}=4900 \implies v_{ter}=70\dfrac{m}{s}$$
Question 7
1 / -0

The pressure exerted by a column of liquid of height $$h$$ and density $$p$$ is given by the hydrostatic pressure equation equal to

A
$$pgh$$

B
$$pg$$

C
$$gh$$

D
Cannot be calculated

Solution

Let area of column = $$A$$
Total volume of mass = $$V\times \rho = (Ah)\rho $$
Pressure exerted =$$\dfrac{F}{A} = \dfrac{Ah\rho \times g}{A} = \rho gh$$
Question 8
1 / -0

Water is flowing in streamline motion through a horizontal tube. The pressure at a point in the tube is $$p$$ where the velocity of flow is $$v$$. At another point, where the pressure is $$p/2$$, the velocity of flow is [density of water = $$\rho$$]

A
$$\sqrt{v^2 + \dfrac{p}{\rho}}$$

B
$$\sqrt{v^2 - \dfrac{p}{\rho}}$$

C
$$\sqrt{v^2 + \dfrac{2p}{\rho}}$$

D
$$\sqrt{v^2 - \dfrac{2p}{\rho}}$$

Solution

Given : $$P_1 = p$$ $$v_1 = v$$ $$P_2 = p/2$$
Using Bernoulli's equation, $$P_1 + \rho gh_1 + \dfrac{1}{2}\rho v_1^2 = P_2 +\rho g h_2 + \dfrac{1}{2}\rho v_2^2$$
As water flows in a horizontal tube i.e. $$h_1 = h_2$$
We get $$P_1 + \dfrac{1}{2}\rho v_1^2 = P_2 + \dfrac{1}{2}\rho v_2^2$$
OR $$p+ \dfrac{1}{2}\rho v^2 = \dfrac{p}{2} + \dfrac{1}{2}\rho v_2^2$$
OR $$\dfrac{1}{2} \rho (v_2^2 - v^2 ) = \dfrac{p}{2}$$ $$\implies v_2 = \sqrt{v^2 +\dfrac{p}{\rho}}$$
Question 9
1 / -0

In the diagram the area of cross section of the pistons $$A$$ and $$B$$ are $$8{cm}^{2}$$ and $$320{cm}^{2}$$ respectively then the thrust on the piston at $$B$$ is

A
$$160kgf$$

B
$$320kgf$$

C
$$420kgf$$

D
$$80kgf$$

Solution

$${ A }_{ 1 }=8{ cm }^{ 2 }\quad \quad \quad { A }_{ 2 }=320{ cm }^{ 2 }$$
$${ F }_{ 1 }=4kgf\quad \quad \quad { F }_{ 2 }=xN$$
Since $$\dfrac { { F }_{ 1 } }{ { A }_{ 1 } } =\dfrac { { F }_{ 2 } }{ { A }_{ 2 } } \Rightarrow \dfrac { x }{ 320 } =\dfrac { 4 }{ 8 } $$
$$\boxed { x=160kgf } $$
Question 10
1 / -0

A cylinder is filled with non viscous liquid of density $$d$$ to a height $$h_o$$ and a hole is made at a height $$h_1$$ from the bottom of the cylinder. The velocity of liquid issuing out of the hole is

A
$$\sqrt(2gh_o)$$

B
$$\sqrt(2g(h_o-h_1))$$

C
$$\sqrt(dgh_1)$$

D
$$\sqrt(dgh_0)$$

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Re-Attempt Weekly Quiz Competition

Mechanical Properties of Fluids Test - 35

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Mechanical Properties of Fluids Test - 35

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SHARING IS CARING

Question 1

$$5\dfrac{m}{s}$$

$$50\dfrac{m}{s}$$

$$100\dfrac{m}{s}$$

$$250\dfrac{m}{s}$$

$$2500\dfrac{m}{s}$$

Solution

Question 2

In equilibrium (cruising at constant speed)

Near equilibrium, positive acceleration

Near equilibrium, negative acceleration

Greatly accelerating during takeoff

Greatly slowing down during landing

Solution

Question 3

$${ V }_{ 1 }\propto { D }^{ 1/2 }$$

$${ V }_{ 1 }\propto { D }^{ 3/2 }$$

$${ V }_{ 1 }\propto { D }^{ 2 }$$

$${ V }_{ 1 }\propto { D }^{ 5/2 }$$

Solution

Question 4

$$\sqrt { { v }^{ 2 }+\cfrac { P }{ \rho } } $$

$$\sqrt { { v }^{ 2 }-\cfrac { P }{ \rho } } $$

$$\sqrt { { v }^{ 2 }+\cfrac { 2P }{ \rho } } $$

$$\sqrt { { v }^{ 2 }-\cfrac { 2P }{ \rho } } $$

Solution

Question 5

$$1:1$$

$$1:2$$

$$1:4$$

$$4:1$$

Solution

Question 6

$$60$$

$$70$$

$$80$$

$$90$$

Solution

Question 7

$$pgh$$

$$pg$$

$$gh$$

Cannot be calculated

Solution

Question 8

$$\sqrt{v^2 + \dfrac{p}{\rho}}$$

$$\sqrt{v^2 - \dfrac{p}{\rho}}$$

$$\sqrt{v^2 + \dfrac{2p}{\rho}}$$

$$\sqrt{v^2 - \dfrac{2p}{\rho}}$$

Solution

Question 9

$$160kgf$$

$$320kgf$$

$$420kgf$$

$$80kgf$$

Solution

Question 10

$$\sqrt(2gh_o)$$

$$\sqrt(2g(h_o-h_1))$$

$$\sqrt(dgh_1)$$

$$\sqrt(dgh_0)$$

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