Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 46

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Question 1
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In a car lift, compressed air exerts a force $$F$$ on a small piston having a radius of $$5\ cm$$. This pressure is transmitted to a second piston of radius $$15\ cm$$. If the mass of the car to be lifted is $$1350\ kg$$, what is $$F$$?

A
$$1.5\ \times 10^{3}N$$

B
$$2.5\ \times 10^{3}N$$

C
$$3.5\ \times 10^{3}N$$

D
$$4.5\ \times 10^{3}N$$

Solution

pascal's law states that
$${ F }_{ 1 }{ d }_{ 1 }={ F }_{ 2 }{ d }_{ 2 }$$
$${ P }_{ 1 }=\cfrac { { F }_{ 1 } }{ { A }_{ 1 } } ,{ P }_{ 2 }=\cfrac { { F }_{ 2 } }{ { A }_{ 2 } } $$
Let $${ F }_{ 1 }=F$$
$${ F }_{ 2 }=1350\times 10N$$
$${ d }_{ 1 }=10cm$$
$${ d }_{ 1 }=30cm$$
$$F\times 10=13500\times 30$$
$$F=\cfrac { 405000 }{ 10 } $$
$$F=4.5\times { 10 }^{ 3 }N$$
Question 2
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The lower end of glass capillary tube is dipped in water.Water rises to a height of $$8cm$$.The Tube is then broken at a height of $$6cm$$. The height of water column and angle of contact will be

A
$$6m\sin - 1\dfrac{3}{4}$$

B
$$6m\cos - 1\dfrac{3}{4}$$

C
$$4m\sin - 1\dfrac{1}{2}$$

D
$$6m\tan - 1\dfrac{3}{4}$$

Solution
Question 3
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The velocity of air over the upper surface of the wing of an aerplane is $$40m/s$$ and that on the lower surface is $$30m/s$$. If the area of the wing is $$5m^{2}$$ and the mass of the wing is $$300\ kg$$, the net force acting on the wing is (Density of air $$=1.3kg/m^{3}$$ and $$g=10m/s^{2}$$)

A
$$725N$$ upward

B
$$725N$$ downward

C
$$2275N$$ upward

D
$$2275\ N$$ downward

Solution

consider the wing be of negligible height
from Bernoulli's equation
Pressure developed across top and bottom of th ewing is
$$P=\cfrac{1}{2}P({V}_{1}^{2}-{V}_{2}^{2})$$
$$=\cfrac{1}{2}\times 1.3\times ({40}^{2}-{30}^{2})=455Pa$$
This pressure provides an upthrust $$F$$ of:
$$F=P\times A=455\times 5=2275N$$ (A: Area of wing)
Thus net force on the wing
$$N=mg-F$$
$$=300\times 10-2275$$
$$N=725N$$ (downward)
Question 4
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Consider a soap film on a rectangular frame of wire of area $$4 \times 4 cm^2$$. If the area of the soap film is increased to $$4 \times 5cm^2$$, the work done in the process will be (The surface tension of the soap film is $$3 \times 10^{-2} N/m$$)

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

B
$$2.4 \times 10^{-5} J$$

C
$$60 \times 10^{-6} J$$

D
$$96 \times 10^{-6} J$$

Solution

Consider the problem

Given,
initial area
$$\begin{array}{l} { A_{ 1 } }=4cm\times 4cm=16c{ m^{ 2 } } \\ =16\times { 10^{ -4 } }{ m^{ 2 } } \end{array}$$

Final area
$$\begin{array}{l} { A_{ 2 } }=4cm\times 5cm=20c{ m^{ 2 } } \\ =20\times { 10^{ -4 } }{ m^{ 2 } } \end{array}$$

Surface tension $$T = 3 \times {10^{ - 2}}N/m$$

Work done $$=?$$

In case of rectangular frame the water wets it from two sides,

Hence, the change in area $$=dA=2$$

$$\begin{array}{l} \left( { { A_{ 2 } }-{ A_{ 1 } } } \right) =2\times \left( { 20\times { { 10 }^{ -4 } }-16\times { { 10 }^{ -4 } } } \right) \\ \therefore dA=2\left( { { A_{ 2 } }-{ A_{ 1 } } } \right) =2\times 4\times { 10^{ -4 } } \\ =8\times { 10^{ -4 } }{ m^{ 2 } } \end{array}$$

Work done $$=$$change in surface energy

$$\begin{array}{l} =TdA=3\times { 10^{ -2 } }\times 8\times { 10^{ -4 } } \\ =2.4\times { 10^{ -5 } }J \end{array}$$

Hence, Option $$B$$ is the correct answer.
Question 5
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A man of mass 100 kg stands on a wood plank of area $$4\, m^2$$. What is the pressure exerted on the floor? Assume the area of a human foot to be $$200\, cm^2$$.

A
500 N

B
25 N

C
50000 N

D
250 N

Solution
Question 6
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A liquid is coming out from the orifice of tank and falls up to a maximum horizon distance of $$6m$$ . The height $$h$$ is equal to

A
$$1.5m$$

B
$$3.0m$$

C
$$4.5m$$

D
$$6.0m$$

Solution
Question 7
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The base area of a boat is $$2m^{2}$$. A man weighing $$76kg$$ weight steps in to the boat. Calculate the depth in to which the boat sinks further in to water.

A
$$1.2\ cm$$

B
$$2.5\ cm$$

C
$$3.8\ cm$$

D
$$4.2\ cm$$

Solution
Question 8
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If the surface tension of a liquid increases proportional to the n th power of its density then what is the value of n -

A
$$1$$

B
$$2$$

C
$$3$$

D
$$9$$

Solution

If the surface tension of a liquid increases proportional to the n th power of its density then what is the value of n is 9.
so the option D.
Question 9
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A tank full of water has a small hole at its bottom. Let $${t_1}$$ be the time taken to empty first one thbird of the tank and $${t_2}$$ be the time taken to empty second one third of the tank and $${t_3}$$ be the time taken to empty rest of the tank then

A
$${t_1} = {t_2} = {t_3}$$

B
$${t_1} > {t_2} > {t_3}$$

C
$${t_1} < {t_2} < {t_3}$$

D
$${t_1} > {t_2} < {t_3}$$

Solution

As the water level is decreasing the pressure at the bottom due to water (pgh) is decreasing. Because of this, inner pressure and o'cater pressure at the small hole are having decreasing difference. Due to this lowering pressure difference, the velocity of efflux will decrease. Thus it will take more time to sap out same amount of water at lower water level than that at higher.
So, $$ t_{3}> t_{2}> t_{1}$$ Option - C is correct.
Question 10
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A liquid enters a tube of variable cross section with a velocity $$3\operatorname { ms } ^ { - 1 }$$ through the wider end and leaves from the narrow and whose radius is half of the wider one. The velocity with which it leaves the tube is:

A
$$1.2 \operatorname { ms } ^ { - 1 }$$

B
$$12 \operatorname { ms } ^ { - 1 }$$

C
$$10 \operatorname { ms } ^ { - 1 }$$

D
$$20 \operatorname { ms } ^ { - 1 }$$

Solution

The equation of continuity,

$$A_1v_1=A_2v_2$$

$$\pi(2r)^2 v_1=\pi r^2 v_2$$

$$4\pi r^2v_1=\pi r^2 v_2$$

$$v_2=4\times 3 m/s$$

$$v_2=12m/s$$

The correct option is B.