Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

The height of water in a capillary tube of radius $$2 cm$$ is $$4 cm$$. What should be the radius of capillary, if the water rises to $$8 cm$$ in tube?

A
$$1cm$$

B
$$2 cm$$

C
$$3 cm$$

D
$$4 cm$$

Solution

Since we know that height of capillary rise in inversely proportional to radius of capillary.
i.e. height $$\alpha $$ $$\dfrac { 1 }{ radius } $$
$$\dfrac { { h }_{ 1 } }{ { h }_{ 2 } } =\dfrac { { r }_{ 2 } }{ { r }_{ 1 } } $$
$$\dfrac { 4 }{ 8 } =\dfrac { { r }_{ 2 } }{ 2 } \Rightarrow \boxed { { r }_{ 2 }=1cm } $$
Question 2
1 / -0

For tap water and clean glass, the angle of contact is

A
$$0$$

B
$$90$$

C
$$140$$

D
$$8$$

Solution

Angle of contact between pure water and glass is $${ 0 }^{ 0 }$$ but in tap water there are some foreign particles, which reduces the cohesive force, leading to on acute angle contact between $${ 8 }^{ 0 }$$ to $${ 18 }^{ 0 }$$.
Question 3
1 / -0

If every particle of fluid has irregular flow, then flow is said to be

A
laminar flow

B
turbulent flow

C
fluid flow

D
both $$A$$ and $$B$$

Solution

Turbulent flow is the irregular flow of particles of a liquid due to high velocity movement.
Question 4
1 / -0

Venturi relation is one of the applications of

A
equation of continuity.

B
Bernoulli's equation.

C
light equation.

D
speed equation.

Solution

Venturi relation is one of the application of energy balance which is Bernoulli's equation.
Question 5
1 / -0

A liquid is kept in a glass vessel. If the liquid solid adhesive force between the liquid and the vessel is very weak as compared to the cohesive force in the liquid, then the shape of the liquid surface near the solid should be

A
Concave

B
Convex

C
Horizontal

D
Almost vertical

Solution

If cohesive forces are stronger than adhesive force then we can observe obtuse angle of contact leading to convex meniscus of liquid near to the solid.
Question 6
1 / -0

Water and mercury are filled in two cylindrical vessels upto same height. Both vessels have a hole in the wall near the bottom. The velocity of water and mercury coming out of the holes are $$v_1$$ and $$v_2$$ respectively. Then :

A
$$v_1 = v_2$$

B
$$v_1 = 13.6v_2$$

C
$$v_1 = \dfrac{v_2}{13.6}$$

D
$$v_1 = (13.6v_2)$$

Solution

The velocity of water coming out of the holes is independent of density or nature of substance, i.e.,
Velocity = $$\sqrt { 2gh } $$
Hence $${ { v }_{ 1 }={ v }_{ 2 } } $$
Question 7
1 / -0

When a capillary tube is immersed vertically in water the capillary rise is $$3 cm$$. if the same capillary tube is inclined at angle of $$60^o$$ to the vertical, the length of the water column in the capillary tube above that of the outside level is

A
$$6 cm$$

B
$$1 cm$$

C
$$8 cm$$

D
Zero

Solution

Vertical Capillary rise will be constant
$$lcos{ 60 }^{ 0 }={ h }_{ 1 }$$
$$l=2\times 3cm=6cm$$
Question 8
1 / -0

Two capillary tubes of the same material but of different radii are dipped in a liquid. The heights to which the liquid rises in the two tubes are $$2.2 cm$$ and $$6.6 cm$$. The ratio of radii of the tubes will be

A
$$1:9$$

B
$$1:3$$

C
$$9:1$$

D
$$3:1$$

Solution

Since we know that height of capillary rise is inversely proportional to radii of tube, i.e.,
height $$\propto \dfrac { 1 }{ radius } $$

$$\dfrac { { h }_{ 1 } }{ { h }_{ 2 } } =\dfrac { { r }_{ 2 } }{ { r }_{ 1 } } \Rightarrow \dfrac { 2.2cm }{ 6.6cm } =\dfrac { { r }_{ 2 } }{ { r }_{ 1 } } $$

So, $$\boxed { \dfrac { { r }_{ 1 } }{ { r }_{ 2 } } =3 } $$
Question 9
1 / -0

Water rises up to a height $$h_1$$ in a capillary tube of radius $$r$$. The mass of the water lifted in the capillary tube is $$M$$. If the radius of the capillary tube is doubled, the mass of water that will rise in the capillary tube will be

A
$$M$$

B
$$2M$$

C
$$\cfrac{M}{2}$$

D
$$4M$$

Solution

Since we know that mass of water rise is proportional to volume of water.
Mass $$\infty $$ volume
$$\dfrac { { M }_{ 1 } }{ { M }_{ 2 } } =\dfrac { { V }_{ 1 } }{ { V }_{ 2 } } =\dfrac { \pi { r }_{ 1 }^{ 2 }{ h }_{ 1 } }{ \pi { r }_{ 2 }^{ 2 }{ h }_{ 2 } } =\dfrac { { r }_{ 1 }^{ 2 }{ h }_{ 1 } }{ { r }_{ 2 }^{ 2 }{ h }_{ 2 } } \quad \rightarrow (1)$$
and for capillary tube, we know that height $$\alpha $$ $$\dfrac { 1 }{ radius } $$
So, $$\dfrac { { h }_{ 1 } }{ { h }_{ 2 } } =\dfrac { { r }_{ 2 } }{ { r }_{ 1 } } \quad \rightarrow (II)$$
hence from (1) & (II)
$$\dfrac { { M }_{ 1 } }{ { M }_{ 2 } } =\dfrac { { r }_{ 1 }^{ 2 } }{ { r }_{ 2 }^{ 2 } } \times \dfrac { { r }_{ 2 } }{ { r }_{ 1 } } =\dfrac { { r }_{ 1 } }{ { r }_{ 2 } } $$
So $${ M }_{ 2 }=\dfrac { { r }_{ 2 } }{ { r }_{ 1 } } \times { M }_{ 1 }=\dfrac { 2r }{ r } \times M=2M$$
$$\boxed { { M }_{ 2 }=2M } $$
Question 10
1 / -0

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

A
$$0.4 m/s$$

B
$$0.133 m/s$$

C
$$0.1 m/s$$

D
$$0.2 m/s$$

Solution

Terminal Velocity,
$$\displaystyle V_{T}=\frac{2r^{2}\left ( d_{1}-d_{2} \right )g}{9\eta }$$
$$\displaystyle \frac{V_{T_{2}}}{0.2}=\frac{\left ( 10.5-1.5 \right )}{\left ( 19.5-1.5 \right )}\Rightarrow V_{T_{2}}=0.2\times \frac{9}{18}$$
$$\displaystyle \therefore V_{T_{2}}=0.1m/s$$

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Re-Attempt Weekly Quiz Competition

Mechanical Properties of Fluids Test - 38

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Mechanical Properties of Fluids Test - 38

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SHARING IS CARING

Question 1

$$1cm$$

$$2 cm$$

$$3 cm$$

$$4 cm$$

Solution

Question 2

$$0$$

$$90$$

$$140$$

$$8$$

Solution

Question 3

laminar flow

turbulent flow

fluid flow

both $$A$$ and $$B$$

Solution

Question 4

equation of continuity.

Bernoulli's equation.

light equation.

speed equation.

Solution

Question 5

Concave

Convex

Horizontal

Almost vertical

Solution

Question 6

$$v_1 = v_2$$

$$v_1 = 13.6v_2$$

$$v_1 = \dfrac{v_2}{13.6}$$

$$v_1 = (13.6v_2)$$

Solution

Question 7

$$6 cm$$

$$1 cm$$

$$8 cm$$

Zero

Solution

Question 8

$$1:9$$

$$1:3$$

$$9:1$$

$$3:1$$

Solution

Question 9

$$M$$

$$2M$$

$$\cfrac{M}{2}$$

$$4M$$

Solution

Question 10

$$0.4 m/s$$

$$0.133 m/s$$

$$0.1 m/s$$

$$0.2 m/s$$

Solution

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