Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 15

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Question 1
1 / -0

A capillary tube is dipped in water vertically.Water rises to a height of 10mm. The tube is now tilted and makes an angle 60$$^{o}$$ with vertical.Now water rises to a height of:

A
10 mm

B
5 mm

C
20 mm

D
40 mm

Solution

If capillary tube is tilted the vertical height of water in tube remains same but volume of the water increases in the tube. So, height of water column will be 10mm.
Question 2
1 / -0

The liquid meniscus in a capillary tube will be convex, if the angle of contact is :

A
greater than $$90^{o}$$

B
less than $$90^{o}$$

C
equal to $$90^{o}$$

D
equal to zero

Solution

A convex meniscus occurs when the particles in the liquid have a stronger attraction to each other(cohesion) than to the material of the container.
It occurs when mercury is placed in a glass capillary.
The types of meniscus are shown in the figure.
In a convex meniscus, the angle of contact is greater than $$90^{\circ}$$, which is angle between tangent to the liquid surface and the capillary surface, inside the liquid.
Question 3
1 / -0

A capillary tube when immersed vertically in a liquid rises to 3 cm. If the tube is held immersed in the liquid at an angle of 60$$^{o}$$ with the vertical,the length of the liquid column along the tube will be:

A
2 cm

B
4.5 cm

C
6 cm

D
7.5 cm

Solution

$$h = \dfrac {2T cos \theta}{r \rho g}$$
$$h \alpha cos \theta$$
for $$\theta = 60^0 cos \theta = \dfrac {1}{2}$$
$$\therefore$$ h is double $$\Rightarrow h = 6 cm$$.
Question 4
1 / -0

A vessel whose bottom has round holes with diameter of $$1mm $$ is filled with water. Assuming that surface tension acts only at holes, then the maximum height to which the water can be filled in vessel without leakage is
(Given surface tension of water is $$75 \times 10 ^{-3} N/m$$ and $$g =10m / s^{2}$$)

A
$$3 cm$$

B
$$0.3 cm$$

C
$$3 mm$$

D
$$3 m$$

Solution

Using formula
$$ h = \dfrac{2T \cos\theta}{\rho rg}$$
= $$\dfrac{2 \times 75 \times 10^{-3} \times 1}{1000 \times \dfrac{1}{2} \times 10^{-3} \times 10} \quad(\cos 0^o = 1)$$
$$= 0.03m = 3 cm$$
Question 5
1 / -0

Four identical capillary tubes $$a, b, c$$ and $$d$$ are dipped in four beakers containing water with tube ‘$$a$$’ vertically, tube ‘$$b$$’ at $$30^{o}$$, tube ‘$$c$$’ at $$45^{o}$$ and tube ‘$$d$$’ at $$60^{o}$$ inclination with the vertical. Arrange the lengths of water column in the tubes in descending order.

A
$$d, c, b, a$$

B
$$d, a, b, c$$

C
$$a, c, d, b$$

D
$$a, b, c, d$$

Solution

In capillary tube fluid always rise to the same vertical height as when the tube is perfectly vertical. So, the tube which is making greater angle with vertical will get more water in it.
So, order of lengths of water column will be $$d > c > b > a$$.

Question 6

1 / -0

Match List I with List - II

List - I	List - II
a) Meniscus of water in a glass Capillary tube	e) convex
b) Meniscus of water in silver capillary tube	f) do not over flow
c) Meniscus of mercury in glass capillary tube	g) flat
d) Water in glass capillary tube of insufficient length	h) concave

The correct match is :

a -- h; b-- g; c-- e; d-- f

a -- e; b-- f;c-- g; d -- h

a -- f; b-- e; c--g; d-- h

a-- h; b-- g; c-- f; d-- e

Solution

In a mercury meniscus, the cohesive forces within the drops are stronger than the adhesive forces between the drops and glass. Hence the meniscus is convex.

Opposite is the case in water meniscus, and hence concave meniscus is formed.

However in water in silver capillary tube, the two forces are equal and hence a flat meniscus is formed.

When length of capillary tube is insufficient of water, the water increases the angle of contact to increase the pressure inside the tube, balancing the excess pressure, and hence prevents the overflow.

Question 7
1 / -0

Rain drops fall with terminal velocity due to:

A
Buoyancy

B
Viscosity

C
Low weight

D
surface tension

Solution

Rain drops fall with terminal velocity due to the viscosity of air. Terminal velocity is speed of the object when net force on the object is zero. In case of rain drops due to mass of drop weight acts in downward direction and viscous force of air acts in upward direction.
Question 8
1 / -0

Three tubes A, B, C are connected to a horizontal pipe in which liquid is flowing. The radii of the pipes at the joints of A, B and C are 2 cm, 1 cm and 2 cm respectively. The height of the liquid:

A
in A is maximum

B
in A and C is equal

C
is same in all the three

D
in A and B is same

Solution

According to the Bernoulli's theorem, for the flow of liquid through a horizontal pipe
$$P+\frac{1}{2}\rho V^2$$ = constant
where variables have their usual meanings.
Hence, when the tubes A, B and C are connected to the pipe with different radii, the height of liquid in tubes is different and depends on radii of tubes in inverse proportion. Hence, height of the liquid in the tubes A and C is the same.
Question 9
1 / -0

A water barrel having water up to depth '$$d$$' is placed on a table of height '$$h$$'. A small hole is made on the wall of the barrel at its bottom. If the stream of water coming out of the hole falls on the ground at a horizontal distance '$$R$$' from the barrel, then the value of '$$d$$' is:

A
$$\dfrac{4h}{R^{2}}$$

B
$$4hR^{2}$$

C
$$\dfrac{R^{2}}{4h}$$

D
$$\dfrac{h}{4R^{2}}$$

Solution

By Bernoulli's equation:
mgd = $$\dfrac{1}{2}mv^2$$
v = $$ \sqrt{2gd}$$ ....... (1)

By Projectile motion equation:
$$\dfrac{1}{2}gt^2 = h$$
t = $$ \sqrt{\dfrac{2h}{g}}$$

Therefore, $$R = vt$$
= $$ \sqrt{2gd} \sqrt{\dfrac{2h}{g}}$$
= 2 $$ \sqrt{dh} $$

Hence d = $$\dfrac{R^2}{4h}$$
Question 10
1 / -0

A solid rubber ball of density 'd' and radius 'R' falls vertically through air. Assume that the air resistance acting on the ball is F $$=$$ KRV where K is constant and V is its velocity. Because of this air resistance the ball attains a constant velocity called terminal velocity $$V_{T}$$ after some time.Then $$V_{T}$$ is:

A
$$\dfrac{4\pi R^{2}dg}{3K}$$

B
$$\dfrac{3K}{4\pi R^{2}dg}$$

C
$$\dfrac{4}{3}\dfrac{\pi r^{3}dg}{K}$$

D
$$\pi rdgk$$

Solution

Let any time t, the velocity be V.
Acceleration is then $$\dfrac {mg - KRV}{m} = \dfrac {dV}{dt}$$
but once it attains terminal velocity $$V_T$$; $$\dfrac {dV}{dt} = 0$$
$$\Rightarrow mg = KRV_T$$
$$(\dfrac {4}{3} \pi R^3)dg = KRV_T$$
$$\Rightarrow V_T = \dfrac {4 \pi R^2dg}{3K}$$

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Mechanical Properties of Fluids Test - 15

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Mechanical Properties of Fluids Test - 15

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Question 1

10 mm

5 mm

20 mm

40 mm

Solution

Question 2

greater than $$90^{o}$$

less than $$90^{o}$$

equal to $$90^{o}$$

equal to zero

Solution

Question 3

2 cm

4.5 cm

6 cm

7.5 cm

Solution

Question 4

$$3 cm$$

$$0.3 cm$$

$$3 mm$$

$$3 m$$

Solution

Question 5

$$d, c, b, a$$

$$d, a, b, c$$

$$a, c, d, b$$

$$a, b, c, d$$

Solution

Question 6

a -- h; b-- g; c-- e; d-- f

a -- e; b-- f;c-- g; d -- h

a -- f; b-- e; c--g; d-- h

a-- h; b-- g; c-- f; d-- e

Solution

Question 7

Buoyancy

Viscosity

Low weight

surface tension

Solution

Question 8

in A is maximum

in A and C is equal

is same in all the three

in A and B is same

Solution

Question 9

$$\dfrac{4h}{R^{2}}$$

$$4hR^{2}$$

$$\dfrac{R^{2}}{4h}$$

$$\dfrac{h}{4R^{2}}$$

Solution

Question 10

$$\dfrac{4\pi R^{2}dg}{3K}$$

$$\dfrac{3K}{4\pi R^{2}dg}$$

$$\dfrac{4}{3}\dfrac{\pi r^{3}dg}{K}$$

$$\pi rdgk$$

Solution

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