Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 34

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Question 1
1 / -0

A 20 cm long capillary tube is dipped vertically in water and the liquid rises upto 10 cm. If the entire system is kept in a freely falling platform, the length of the water column in the tube will be

A
5 cm

B
10 cm

C
15 cm

D
20 cm

Solution

The height raised by liquid in capillary tube
$$H=\dfrac{2l\,cos\theta}{\rho gh}$$
Where H is rise in the capillary tube.
As in freely falling platform, a body experience weightlessness.
So, the liquid will rise upto to length of the capillary.
i.e height raised by the liquid will be 20 cm.
Question 2
1 / -0

The work done in splitting a drop of water of $$1 mm$$ radius into $$\displaystyle { 10 }^{ 6 }$$ droplets is (surface tension of water $$\displaystyle 72\times { 10 }^{ -3 }{ N }/{ M }$$ ) ;

A
$$\displaystyle 5.98\times { 10 }^{ -5 }J$$

B
$$\displaystyle 10.98\times { 10 }^{ -5 }J$$

C
$$\displaystyle 16.95\times { 10 }^{ -5 }J$$

D
$$\displaystyle 8.95\times { 10 }^{ -5 }J$$

Solution

Radius of new droplet if be r then, $$\displaystyle { 10 }^{ 6 }\times \frac { 4 }{ 3 } \pi { r }^{ 3 }=\frac { 4 }{ 3 } \pi \times { \left( 0.001 \right) }^{ 3 }$$
$$\displaystyle { r }^{ 3 }={ 10 }^{ -15 }\Rightarrow r={ 10 }^{ -5 }$$
Increase in surface area
$$\displaystyle =\left[ 4\pi \times { \left( { 10 }^{ -5 } \right) }^{ 2 }\times { 10 }^{ 6 } \right] -\left[ 4\pi \times { \left( { 10 }^{ -3 } \right) }^{ 2 } \right] $$
$$\displaystyle =\left[ 4\pi \times { 10 }^{ -4 } \right] -\left[ 4\pi \times { 10 }^{ -6 } \right] =4\pi { 10 }^{ -6 }\left[ 100-1 \right] $$
$$\displaystyle =4\pi \times { 10 }^{ -6 }\times 99=4\pi \times { 10 }^{ -6 }\times 99$$
Work done = surface tension x increase in surface area
$$\displaystyle =72\times 4\pi \times 99\times { 10 }^{ -6 }\times { 10 }^{ -3 }=8.95\times { 10 }^{ -5 }J$$
Question 3
1 / -0

A spherical solid ball of volume V is made of a material of density $$\displaystyle { \rho }_{ 1 }$$ It is falling through a liquid of density $$\displaystyle { \rho }_{ 1 }\left( { \rho }_{ 2 }<{ \rho }_{ 1 } \right) $$. Assume that the liquid applies a viscous force on the ball that is Proportional to the square of its speed v, i.e., $$\displaystyle { F }_{ viscous }=-{ kv }^{ 2 }\left( k>0 \right) $$. The terminal speed of the ball is :

A
$$\displaystyle \sqrt { \frac { Vg\left( { \rho }_{ 1 }-{ \rho }_{ 2 } \right) }{ k } } $$

B
$$\displaystyle \frac { Vg{ \rho }_{ 1 } }{ k } $$

C
$$\displaystyle \sqrt { \frac { Vg{ \rho }_{ 1 } }{ k } } $$

D
$$\displaystyle \frac { Vg\left( { \rho }_{ 2 }<{ \rho }_{ 1 } \right) }{ k } $$

Solution

The condition for terminal speed $$(v_t)$$ is Weight = Buoyant force + Viscous force
$$W = V\rho_1g$$
$$V\rho_1g = V\rho_2 g + k v_t^2$$
$$v_t =\sqrt{ \dfrac{V g(\rho_1 -\rho_2)}{k}}$$
Question 4
1 / -0

A sphere of mass M and radius R is falling in a viscous fluid. The terminal velocity attained by the falling object will be proportional to :

A
$$\displaystyle { R }^{ 2 }$$

B
$$\displaystyle R$$

C
$$\displaystyle 1/R$$

D
$$\displaystyle { 1 }/{ { R }^{ 2 } }$$

Solution
Question 5
1 / -0

$$64$$ spherical rain drops of equal size are falling vertically through air with a terminal velocity $$1.5 {ms}^{-1}$$. If these drops coalesce to form a big spherical drop, then terminal velocity of big drop is:

A
$$8 \ {ms}^{-1}$$

B
$$16 \ {ms}^{-1}$$

C
$$24 \ {ms}^{-1}$$

D
$$32 \ {ms}^{-1}$$

Solution

Terminal velocity is directly proportional to the square of the radius of the spherical body.
Since mass is conserved and density remains same, upon coalesceing, the volumes will add up.
Thus, volume of bigger drop $$=$$ 64 times the volume of each small drop. Since volume is directly proportional to cube of radius, this implies that the radius of the big drop is 4 times the radius of each small drop.
Therefore, terminal velocity of the big drop is 4 square $$=$$ 16 times the terminal velocity of each small drop $$=1.5\times16=24ms^{-1}$$
Question 6
1 / -0

If the terminal speed of a sphere of gold (density =$$\displaystyle 19.5{ kg }/{ { m }^{ 3 } }$$) is $$0.2 m/s$$ in a viscous liquid(density = $$\displaystyle 1.5{ kg }/{ { m }^{ 3 } }$$), find the terminal speed of a sphere of silver (density = $$\displaystyle 10.5{ kg }/{ { m }^{ 3 } }$$) of the same size in the same liquid

A
$$0.4 m/s$$

B
$$0.133 m/s$$

C
$$0.1 m/s$$

D
$$0.2m/s$$

Solution

$$V_T = \dfrac{2}{9\eta}r^2(d_1 - d_2)g$$
$$\dfrac{V_{T_2}}{0.2} = \dfrac{10.5 - 1.5}{19.5 - 1.5}$$
$$V_{T_2} = 0.1 m/s$$
Question 7
1 / -0

A force of $$100N$$ is applied on an area of $$4m^2$$. the pressure being applied on the area is

A
$$50Pa$$

B
$$25 Pa$$

C
$$12.5 Pa$$

D
$$400Pa$$

Solution

Given, $$froce=100N$$
and $$area=4m^2$$
$$\therefore$$$$pressure=\dfrac{force}{area}=\dfrac{100N}{4m^{2}} =25 Pa$$
Question 8
1 / -0

A hole is made at the bottom of the tank filled with water (density $$\displaystyle 1000{ kg }/{ { m }^{ 3 } }$$), If the total pressure at the bottom ofthe tankis 3 atmosphere
( 1 atmosphere = $$\displaystyle { 10 }^{ 5 }{ N }/{ { m }^{ 2 } }$$ ), then the velocity of efflux is :

A
$$\displaystyle \sqrt { 200 } { m }/{ s }$$

B
$$\displaystyle \sqrt { 400 } { m }/{ s }$$

C
$$\displaystyle \sqrt { 500 } { m }/{ s }$$

D
$$\displaystyle \sqrt { 800 } { m }/{ s }$$

Solution

We know that velocity of efflux,
$$v = \sqrt{2gh}$$
At the bottom of tank pressure is 3 atmosphere. So, total pressure due to water column
$$h\rho g = 2 \times 10^5$$
$$gh = 2 \times 10^2$$
$$v = \sqrt{2gh} = \sqrt{2 \times 2\times 10^2} = \sqrt{400} m/s$$
Question 9
1 / -0

Water is filled in a container upto height of $$3 m$$. A small hole of area '$$\displaystyle { A }_{ 0 }$$' is punched in the wall of the container at a height $$52.5 cm$$ from the bottom. The cross sectional area of the container is A. If $$\displaystyle { { A }_{ 0 } }/{ A }=0.1$$ then $${ v }^{ 2 }$$ is (where $$v$$ is the velocity of water coming out of the hole)

A
$$\displaystyle 50{ { m }^{ 2 } }/{ { s }^{ 2 } }$$

B
$$\displaystyle 50.5{ { m }^{ 2 } }/{ { s }^{ 2 } }$$

C
$$\displaystyle 51{ { m }^{ 2 } }/{ { s }^{ 2 } }$$

D
$$\displaystyle 52{ { m }^{ 2 } }/{ { s }^{ 2 } }$$

Solution

The square of the velocity of flux
$$v^2 = \dfrac{2gh}{1 - (\dfrac{A_o}{A})^2}$$
$$ = \dfrac{2 \times 10 \times 2.475}{1 - (0.1)^2} = 50 m^2/s^2$$
Question 10
1 / -0

The angle of contact in case of liquid depends upon which of the following?

A
nature of liquid and solid

B
material which exists above free surface of liquid

C
both of them above

D
None

Solution

Angle of contact depends on both liquid liquid interaction & liquid solid interaction & the medium above them.

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Re-Attempt Weekly Quiz Competition

Mechanical Properties of Fluids Test - 34

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Mechanical Properties of Fluids Test - 34

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Question 1

5 cm

10 cm

15 cm

20 cm

Solution

Question 2

$$\displaystyle 5.98\times { 10 }^{ -5 }J$$

$$\displaystyle 10.98\times { 10 }^{ -5 }J$$

$$\displaystyle 16.95\times { 10 }^{ -5 }J$$

$$\displaystyle 8.95\times { 10 }^{ -5 }J$$

Solution

Question 3

$$\displaystyle \sqrt { \frac { Vg\left( { \rho }_{ 1 }-{ \rho }_{ 2 } \right) }{ k } } $$

$$\displaystyle \frac { Vg{ \rho }_{ 1 } }{ k } $$

$$\displaystyle \sqrt { \frac { Vg{ \rho }_{ 1 } }{ k } } $$

$$\displaystyle \frac { Vg\left( { \rho }_{ 2 }<{ \rho }_{ 1 } \right) }{ k } $$

Solution

Question 4

$$\displaystyle { R }^{ 2 }$$

$$\displaystyle R$$

$$\displaystyle 1/R$$

$$\displaystyle { 1 }/{ { R }^{ 2 } }$$

Solution

Question 5

$$8 \ {ms}^{-1}$$

$$16 \ {ms}^{-1}$$

$$24 \ {ms}^{-1}$$

$$32 \ {ms}^{-1}$$

Solution

Question 6

$$0.4 m/s$$

$$0.133 m/s$$

$$0.1 m/s$$

$$0.2m/s$$

Solution

Question 7

$$50Pa$$

$$25 Pa$$

$$12.5 Pa$$

$$400Pa$$

Solution

Question 8

$$\displaystyle \sqrt { 200 } { m }/{ s }$$

$$\displaystyle \sqrt { 400 } { m }/{ s }$$

$$\displaystyle \sqrt { 500 } { m }/{ s }$$

$$\displaystyle \sqrt { 800 } { m }/{ s }$$

Solution

Question 9

$$\displaystyle 50{ { m }^{ 2 } }/{ { s }^{ 2 } }$$

$$\displaystyle 50.5{ { m }^{ 2 } }/{ { s }^{ 2 } }$$

$$\displaystyle 51{ { m }^{ 2 } }/{ { s }^{ 2 } }$$

$$\displaystyle 52{ { m }^{ 2 } }/{ { s }^{ 2 } }$$

Solution

Question 10

nature of liquid and solid

material which exists above free surface of liquid

both of them above

None

Solution

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