Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 27

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Question 1
1 / -0

The figure shows capillary tube of radius $$r$$ dipped into water. The atmospheric pressure is $$P_{0}$$ and the capillary rise of water is $$h. S$$ is the surface tension for water-glass.
Initially, $$h = 10\ cm$$. If the capillary tube is now inclined at $$45^{\circ}$$, the length of water rising in the tube will be

A
$$10 cm$$

B
$$10\sqrt {2} cm$$

C
$$\dfrac {10}{\sqrt {2}}cm$$

D
None of these

Solution

When the capillary tube is tilted by an angle $$\theta$$ with the vertical, the capillary rise to distance $$l$$ in the tube will be such that the meniscus will remain at the same vertical height above the level of water in the container.
Hence, $$h = l\cos \theta$$
$$l = \dfrac {h}{\cos \theta} = \dfrac {10}{\cos 45^{\circ}} = 10\sqrt {2} cm$$.
Question 2
1 / -0

The figure shows capillary tube of radius $$r$$ dipped into water. The atmospheric pressure is $$P_{0}$$ and the capillary rise of water is $$h. s$$ is the surface tension for water-glass.
The pressure inside water at the point $$A$$ (lowest point of the meniscus) is

A
$$P_{0}$$

B
$$P_{0} + \dfrac {2s}{r}$$

C
$$P_{0} - \dfrac {2s}{r}$$

D
$$P_{0} - \dfrac {4s}{r}$$

Solution

Since the pressure above the meniscus is $$ P_0$$.
Difference between pressures outside and inside of the meniscus is 2s/r
Since Meniscus is concave upwards, this means pressure inside meniscus is less as compared to outside
So Pressure below the meniscus of radius $$ r$$ is $$ P_0-2s/r $$
( assuming meniscus to be spherical)
Question 3
1 / -0

The incident intensity on a horizontal surface at sea level from the sun is about $$1 kW m^{-2}$$. Find the ratio of this pressure to atmospheric pressure $$p_0$$ (about $$1\times 10^5 Pa)$$ at sea level.
[Assuming that $$50 %$$ % of this intensity is reflected and $$50 % $$ % is absorbed]

A
$$5\times 10^{-11}$$

B
$$4\times 10^{-8}$$

C
$$6\times 10^{-12}$$

D
$$8\times 10^{-11}$$

Solution

Given,
$$I=1k W/m^2$$
$$c=3\times 10^8 m/s$$
$$P_0=1\times 10^5 Pa$$
Pressure exerted due to absorbed light, $$P_a= \dfrac{I}{2c}$$

Pressure exerted due to reflected light, $$P_r=\dfrac{2I}{2c}$$
Hence, the total radiation pressure on the surface can be given as:
$$P_t=P_a+P_r=\dfrac{3I}{2c}$$

$$P_t=\dfrac{3\times 10^3}{3\times2\times 10^8}=\dfrac{ 10^3}{2\times 10^8}$$

$$\Rightarrow P_t=5\times 10^{-6}Pa$$

So, the ratio of the radiation and atmospheric pressure will be:
$$\dfrac{P}{P_0}=\dfrac{5\times 10^{-6}}{1\times 10^5}$$

$$\Rightarrow \dfrac{P}{P_0}=5\times 10^{-11}$$
Hence, the correct option is $$(A)$$
Question 4
1 / -0

Mercury in a test tube forms ______ meniscus

A
concave

B
convex

C
no

D
cant say

Solution

Answer: $$convex $$

Mercury in a test tube forms $$ convex $$ meniscus.

The $$cohesive \space force$$ between the molecules of the material i.e $$mercury$$ is higher than the $$ adhesive \space force.$$
The molecules tend to bind each other tightly hence the force in the test tube is inwards due to the surface tension.
Question 5
1 / -0

Directions For Questions

Molecular forces exist between the molecules of a liquid in container. The molecules on the surface have unequal force leading to a tension on the surface. If this is not compensated by a force, the equilibrium of teh liquid will be a difficult task. This leads to an excess pressure on the surface. The nature of the meniscus can inform us of the direction of the excess pressure. The angle of contact of the liquid decided by the forces between the molecules, air and container can make the angle of contact.

...view full instructions

The direction of the excess pressure in the meniscus of a liquid of angle of contact $$2\pi/3$$ is:

A
Upward

B
Downward

C
Horizontal

D
Cannot determined

Solution

Excess pressure is always on the concave side of the surface. For angle of contact of $$2\pi/3$$, the liquid should have a convex surface. So, the excess pressure should be in the upward direction.
Question 6
1 / -0

A small ball (menu) falling under gravity in a viscous medium experiences a drag force proportional to the instantaneous speed u such that $$F_{drag}$$ = Ku. Then the terminal speed of the ball within the viscous medium is:

A
$$ \dfrac {K}{mg}$$

B
$$ \dfrac {mg}{K}$$

C
$$ \sqrt{\dfrac {mg}{K}}$$

D
$$ \left ( \dfrac{mg}{K} \right)^2$$

Solution

For terminal velocity , weight should be balanced by upward force.
Here we have excluded buoyancy force.
Now, $$Ku=mg$$
$$u=\dfrac{mg}{K}$$
Question 7
1 / -0

There is a hole in the bottom of tank having water. If total pressure at bottom is 3 atm (1 atm = $$10^5 N/m^2$$) then the velocity of water flowing from hole is:

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

B
$$\sqrt{600}m/s$$

C
$$\sqrt{60}m/s$$

D
None of these

Solution

Pressure at the bottom of tank $$ P = h\rho g = 3 x 10^5 \dfrac{N}{m^2}$$
Pressure due to liquid column
$$ P_1 = 3 x 10^5 - 1 x 10^5 = 2 x 10^5 $$
and velocity of water $$ v = \sqrt{2gh} $$
$$ \therefore v = \sqrt{\dfrac{2P_1}{\rho}}=\sqrt{\dfrac{2x 2 x 10^5}{10^3}}= \sqrt{400}m/s$$
Question 8
1 / -0

A cylinder is filled with non-viscous liquid of density d to a height $$h_0$$ and a hole is made at a height $$h_1$$ from the bottom of the cylinder. The velocity of the liquid issuing out of the hole is:

A
$$\sqrt{2gh_0}$$

B
$$\sqrt{2g(h_0-h_1)}$$

C
$$\sqrt{dgh_1}$$

D
$$\sqrt{dgh_0}$$

Solution

$$\text{Velocity of liquid flowing out of hole} = \sqrt{2gh}$$;
Here $$h= (h_0 - h_1)$$
Question 9
1 / -0

A spherical ball of iron of radius $$2\ \text{mm}$$ is falling through a column of glycerine. If densities of glycerine and iron are respectively $$1.3\times 10^3\ \text{kg/m}^3$$ and $$8\times 10^3\ \text{kg/m}^3$$. $$\eta\ {for \ glycerine} = 0.83\ \text{Nm}^{-2}\ \text{sec},$$ then the terminal velocity is:

A
$$0.7\ \text{m/s}$$

B
$$0.07\ \text{m/s}$$

C
$$0.007\ \text{m/s}$$

D
$$0.0007\ \text{m/s}$$

Solution

Terminal velocity$$, v_0 = \dfrac{2r^2 (\rho -\rho_0)g}{9\eta }$$

$$v_0= \dfrac{2\times (2 \times 10^{-3})^2\times (8 -1.3) \times 10^3 \times 9.8} {9 \times 0.83} = 0.07\ \text{ms}^{-1}$$
Question 10
1 / -0

The rain drops falling from the sky neither injure us nor make holes on the ground because they move with:

A
constant acceleration

B
variable acceleration

C
variable speed

D
constant terminal velocity
Solution
Gravity does cause things to fall with increasing speed, but as they speed up air resistance increases. Eventually the force of the air resistance is enough to balance the force of gravity, so the acceleration stops and the raindrop reaches a constant terminal velocity. Apparently the terminal velocity of rain isn't high enough to cause damage. Since $$F=Ma$$ so the force required to stop the raindrop wouldn't be that much because
- a raindrop doesn't have a lot of mass
- it's speed (terminal velocity) isn't that high.

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Re-Attempt Weekly Quiz Competition

Mechanical Properties of Fluids Test - 27

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Mechanical Properties of Fluids Test - 27

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Question 1

$$10 cm$$

$$10\sqrt {2} cm$$

$$\dfrac {10}{\sqrt {2}}cm$$

None of these

Solution

Question 2

$$P_{0}$$

$$P_{0} + \dfrac {2s}{r}$$

$$P_{0} - \dfrac {2s}{r}$$

$$P_{0} - \dfrac {4s}{r}$$

Solution

Question 3

$$5\times 10^{-11}$$

$$4\times 10^{-8}$$

$$6\times 10^{-12}$$

$$8\times 10^{-11}$$

Solution

Question 4

concave

convex

no

cant say

Solution

Question 5

Upward

Downward

Horizontal

Cannot determined

Solution

Question 6

$$ \dfrac {K}{mg}$$

$$ \dfrac {mg}{K}$$

$$ \sqrt{\dfrac {mg}{K}}$$

$$ \left ( \dfrac{mg}{K} \right)^2$$

Solution

Question 7

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

$$\sqrt{600}m/s$$

$$\sqrt{60}m/s$$

None of these

Solution

Question 8

$$\sqrt{2gh_0}$$

$$\sqrt{2g(h_0-h_1)}$$

$$\sqrt{dgh_1}$$

$$\sqrt{dgh_0}$$

Solution

Question 9

$$0.7\ \text{m/s}$$

$$0.07\ \text{m/s}$$

$$0.007\ \text{m/s}$$

$$0.0007\ \text{m/s}$$

Solution

Question 10

constant acceleration

variable acceleration

variable speed

constant terminal velocity

Solution

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