Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 51

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Weekly Quiz Competition

Question 1
1 / -0

The lower end of a glass capillary tube is dipped in water. Water rises to a height of 8 cm. The angle of contact is $${ 0 }^{ \circ }$$. the tube is then broken at a height of 6 cm. The height of water column and angle of contact will be.

A
$$6m,{ sin }^{ -1 }\left( \dfrac { 3 }{ 4 } \right) $$

B
$$6m,{ sin }^{ -1 }\left( \dfrac { \sqrt { 7 } }{ 4 } \right) $$

C
$$4m,{ sin }^{ -1 }\left( \dfrac { \sqrt { 7 } }{ 8 } \right) $$

D
$$6m,{ 0 }^{ \circ }$$

Solution

we know that $$h = {(2T cos θ) / (ρrg)} $$
As radius and content of tube is same.

$$T, r, ρ, g $$ = constant

$$∴ {h / (cosθ)} $$= constant

$$ ∴ (h1 / h2) = {(cosθ_1) / (cosθ_2)} $$

Initial height = $$h1 = 8 cm $$

As capillary is broken at height 6 cm, meaning that water will rise to height 6 cm hence $$h2 = 6 cm$$

$$(h1 / h2) = {(cos 0) / (cos θ_2)} $$ --------- initial angle = 0°

$$ ∴ (8/6) = {1 / (cos θ_2)} $$

$$cos θ_2 = 3/4 $$
$$θ_2 = cos^{–1} (3/4) $$

∴ height = $$6 \,cm$$ & angle = $$cos^{–1} (3/4) $$

or $$sin^{–1} (\sqrt7/4) $$
Question 2
1 / -0

A steel ball of density 7800 kg/$${ m }^{ 3 }$$ falls through water of density 1000 kg/$${ m }^{ 3 }$$ and coefficient of viscosity 0.01 poise. Find its terminal velocity (approx) if the radius is 1 cm.

A
1500 m/s

B
500 m/s

C
1000 m/s

D
None of these
Question 3
1 / -0

Vessel shown in the figure has two sections of areas of cross section $$A{}_{1}$$ and $${A}_{2}$$. A liquid of density $$\rho$$ fills both sections up to a height $$h$$ in each. Neglect atmospheirc pressure

A
The pressure at the base of the vessel is $$2h\rho g$$

B
The force exerted by the liquid on the base is $$2h\rho {A}_{2}g$$

C
The weight of the liquid is less thatn $$2hg\rho {A}_{2}$$

D
Walls of the vessel at the level $$x$$ exert a downward force $$hg\rho({A}_{2}-{A}_{1})$$ on the liquid.

Solution

$$\begin{array}{l} pressure\, at\, the\, bottom\, =\rho g\Delta H=2\rho gh \\ Force\, at\, bottom=2\rho gh\times { A_{ 2 } } \\ weight\, of\, liquid\, is\, less\, than\, 2\rho gh\times { A_{ 2 } } \\ \Delta F=\rho gh{ A_{ 2 } }-\rho gh{ A_{ 1 } }=\rho gh\left( { { A_{ 0 } }-{ A_{ 1 } } } \right) . \end{array}$$
Hence, the option $$D$$ is the correct answer.
Question 4
1 / -0

Write true or false. Correct the false statements.

A
The perpendicular distance from the pivot of a rotating body to the line of known as the moment arm.

B
pressure is inversely proportional to the force applied

C
Wooden sleepers are placed under railway tracks in order to increase pres.

D
pressure is independent of the roughness of the surfaces in correct.

Solution
Question 5
1 / -0

If pressure at half the depth of a lake is equal to $$\cfrac {2}{3}$$ pressure at the bottom of the lake then what is the depth of the lake $$[RPET=2000]$$

A
$$10$$ m

B
$$20$$ m

C
$$60$$ m

D
$$30$$ m

Solution

The Pressure at bottom of the lake = $$P_0+h\rho g$$
The Pressure at half the depth of a lake = $$P_0+\dfrac{h}{2}\rho g$$
According to given condition
$$P_0+\dfrac{1}{2}h\rho g=\dfrac{2}{3}(P_0+h\rho g)\Rightarrow \dfrac{1}{3}P_0 = \dfrac{1}{6}h\rho g$$
$$h=\dfrac{2P_0}{\rho g}=\dfrac{2\times10^5}{10^3\times10}=20m$$
Question 6
1 / -0

The angle of contact for the pair of pure water with clean glass is

A
acute

B
obtuse

C
$$90^0$$

D
$$0^0$$

Solution
Question 7
1 / -0

The angle of contact between glass and mercury is :

A
$$
0^{0}
$$

B
$$
30^{0}
$$

C
$$
90^{0}
$$

D
$$
135^{0}
$$

Solution

The angle of contact between glass and mercury is $$135^{0}$$
Question 8
1 / -0

Radius of a soap bubble is $$r$$, surface tension of soap solution is $$T$$. Then without increasing the temperature, how much energy will be needed to double its radius.

A
$$4\pi {r^2}T$$

B
$$2\pi {r^2}T$$

C
$$12\pi {r^2}T$$

D
$$24\pi {r^2}T$$

Solution

As we know ,
without increasing temperature increasing radius will be done by performing some work on the bubble,

$$W=8\pi T(R_{2}^{2}-R_{1}^{2})$$

$$=8\pi T[(2r)^{2}-(r)^{2}]=24\pi r^{2}T$$
Question 9
1 / -0

By sucking through a straw, a student can reduce the pressure in his lungs to 750 mm of Hg (density $$13.6 gcm^{-3}$$) using the straw, he can drink water from a glass up to a maximum depth of

A
$$10.2 cm$$

B
$$75.3 cm$$

C
$$13.6 cm$$

D
$$1.96 cm$$

Solution

Normal pressure is $$760 mm\,Hg$$ so the boy can create a depression of $$10 mm\,Hg$$ in his lungs.

According to the law of flotation

Weight of floating body $$=$$ Weight of displaced liquid

Wt. Of water displaced from glass $$=$$ Wt of air of lung reduced (interm of mmHg)

Wt of water displaced from glass = density of water $$(d_1) \times$$ volume of water displaced $$(v_1)$$

Wt of air reduce from lung (interm of mmHg) $$=$$ density of Hg $$(d_2)\times$$ volume of air displaced $$(v_2)$$

Now $$d_1 \times v_1 = d_2 \times v_2$$

And $$v_1 = d_2 \times \dfrac{v_2}{d_1} = 13.6 g/cm –3 \times \dfrac{10mm}{ 1g/cm} –3 = 136mm = 13.6cm$$
Question 10
1 / -0

Anomalous expansion of water is demonstrated by using

A
Jolly's apparatus

B
Hopes apparatus

C
Regnaults apparatus

D
Specific apparatus

Solution

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Mechanical Properties of Fluids Test - 51

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Mechanical Properties of Fluids Test - 51

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Question 1

$$6m,{ sin }^{ -1 }\left( \dfrac { 3 }{ 4 } \right) $$

$$6m,{ sin }^{ -1 }\left( \dfrac { \sqrt { 7 } }{ 4 } \right) $$

$$4m,{ sin }^{ -1 }\left( \dfrac { \sqrt { 7 } }{ 8 } \right) $$

$$6m,{ 0 }^{ \circ }$$

Solution

Question 2

1500 m/s

500 m/s

1000 m/s

None of these

Question 3

The pressure at the base of the vessel is $$2h\rho g$$

The force exerted by the liquid on the base is $$2h\rho {A}_{2}g$$

The weight of the liquid is less thatn $$2hg\rho {A}_{2}$$

Walls of the vessel at the level $$x$$ exert a downward force $$hg\rho({A}_{2}-{A}_{1})$$ on the liquid.

Solution

Question 4

The perpendicular distance from the pivot of a rotating body to the line of known as the moment arm.

pressure is inversely proportional to the force applied

Wooden sleepers are placed under railway tracks in order to increase pres.

pressure is independent of the roughness of the surfaces in correct.

Solution

Question 5

$$10$$ m

$$20$$ m

$$60$$ m

$$30$$ m

Solution

Question 6

acute

obtuse

$$90^0$$

$$0^0$$

Solution

Question 7

$$ 0^{0} $$

$$ 30^{0} $$

$$ 90^{0} $$

$$ 135^{0} $$

Solution

Question 8

$$4\pi {r^2}T$$

$$2\pi {r^2}T$$

$$12\pi {r^2}T$$

$$24\pi {r^2}T$$

Solution

Question 9

$$10.2 cm$$

$$75.3 cm$$

$$13.6 cm$$

$$1.96 cm$$

Solution

Question 10

Jolly's apparatus

Hopes apparatus

Regnaults apparatus

Specific apparatus

Solution

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$$
0^{0}
$$

$$
30^{0}
$$

$$
90^{0}
$$

$$
135^{0}
$$