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NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids

by aglasem
August 26, 2021
in Extras
Reading Time: 1 min read
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NCERT Solutions

NCERT Solutions Class 11 Physics Chapter 10 Mechanical Properties of Fluids – Here are all the NCERT solutions for Class 11 Physics Chapter 10. This solution contains questions, answers, images, explanations of the complete chapter 10 titled Of Mechanical Properties of Fluids taught in Class 11. If you are a student of Class 11 who is using NCERT Textbook to study Physics, then you must come across chapter 10 Mechanical Properties of Fluids After you have studied the lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids in one place.

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NCERT Solutions Class 11 Physics Chapter 10 Mechanical Properties of Fluids

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Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Physics for Class 11 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.

For a better understanding of this chapter, you should also see summary of Chapter 10 Mechanical Properties of Fluids , Physics, Class 11.

Class 11
Subject Physics
Book Physics Part I
Chapter Number 10
Chapter Name

Mechanical Properties of Fluids

NCERT Solutions Class 11 Physics chapter 10 Mechanical Properties of Fluids

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Class 11, Physics chapter 10, Mechanical Properties of Fluids solutions are given below in PDF format. You can view them online or download PDF file for future use.

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Question & Answer

Q.1: Explain why 
(a) The blood pressure in humans is greater at the feet than at the brain 
(b) Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at the sea level, though the height of the atmosphere is more than 100 km 
(c) Hydrostatic pressure is a scalar quantity even though pressure is force divided by area.

Ans : (a) The pressure Of a liquid is given by the relation: p = hpg where, p = Pressure h = Height of the liquid column p = Density of the liquid g = Acceleration due to the gravity It can be inferred that pressure is directly proportional to height. Hence, the blood pressure in human vessels depends on the height Of the blood column in the body. The height Of the blood column is more at the feet than it is at the brain. Hence, the blood pressure at the feet is more than it is at the brain. (b) Density of air is the maximum near the sea level. Density of air decreases with increase in height from the surface. At a height of about 6 km, density decreases to nearly half of its value at the sea level. Atmospheric pressure is proportional to density. Hence, at a height of 6 km from the surface, it decreases to nearly half of its value at the sea level. (c) When force is applied on a liquid, the pressure In the liquid is transmitted in all directions. Hence, hydrostatic pressure does not have a fixed direction and it is a scalar physical quantity.

Q.2: Explain why 
(a) The angle of contact of mercury with glass is obtuse, while that of water with glass is acute. 
(b) Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops. (Put differently, water wets glass while mercury does not.)
(c) Surface tension of a liquid is independent of the area of the surface 
(d) Water with detergent dissolved in it should have small angles of contact. 
(e) A drop of liquid under no external forces is always spherical in shape

Ans : (a) The angle between the tangent to the liquid surface at the point of contact and the surface inside the liquid is called the angle of contact (B), as shown in the given figure. \(\begin{array}{l}{S_{\text { lat }} S_{\text { sat }} \text { and } S_{\text { sl }} \text { are the respective interfacial tensions between the liquid-air, solid-air, }} \\ {\text { and solid-liquid interfaces. At the line of contact, the surface forces between the three }} \\ {\text { media must be in equilibrium, i.e., }}\end{array}\) \(\cos \theta=\frac{S_{\mathrm{a}}-S_{\mathrm{sl}}}{S_{\mathrm{la}}}\) \(\begin{array}{l}{\text { The angle of contact } \theta, \text { is obtuse if } S_{\mathrm{sa}} < S_{1 a}(\text { as in the case of mercury on glass). This }} \\ {\text { angle is acute if } S_{\mathrm{sl}} < S_{\mathrm{la}}(\text { as in the case of water on glass). }}\end{array}\) (b) Mercury molecules (which make an obtuse angle with glass) have a strong force Of attraction between themselves and a weak force of attraction toward solids. Hence, they tend to form drops. On the other hand, water molecules make acute angles with glass. They have a weak force of attraction between themselves and a strong force of attraction toward solids. Hence, they tend to spread out. (c) Surface tension is the force acting per unit length at the interface between the plane Of a liquid and any other surface. This force is independent Of the area Of the liquid surface. Hence, surface tension is also independent Of the area Of the liquid surface. (d) Water with detergent dissolved in it has small angles of contact (9). This is because for a small 9, there is a fast capillary rise of the detergent in the cloth. The capillary rise of a liquid is directly proportional to the cosine of the angle of contact (B). If \( \theta\) is small, then \(\cos \theta\) will be large and the rise of the detergent water in the cloth will be fast. (e) A liquid tends to acquire the minimum surface area because of the presence of surface tension. The surface area Of a sphere is the minimum for a given volume. Hence, under no external forces, liquid drops always take spherical shape.

Q.3: Fill in the blanks using the word(s) from the list appended with each statement: 
(a) Surface tension of liquids generally ... with temperatures (increases / decreases) 
(b) Viscosity of gases ... with temperature, whereas viscosity of liquids ... with temperature (increases / decreases) 
(c) For solids with elastic modulus of rigidity, the shearing force is proportional to ... , while for fluids it is proportional to ... (shear strain / rate of shear strain) 
(d) For a fluid in a steady flow, the increase in flow speed at a constriction follows (conservation of mass / Bernoulli’s principle) 
(e) For the model of a plane in a wind tunnel, turbulence occurs at a ... speed for turbulence for an actual plane (greater / smaller)

Ans : (a) decreases The surface tension of a liquid is inversely proportional to temperature. (b) increases; decreases Most fluids Offer resistance to their motion. This is like internal mechanical friction, known as viscosity. Viscosity Of gases increases with temperature, while viscosity Of liquids decreases with temperature. (c) Shear strain; Rate of shear strain With reference to the elastic modulus of rigidity for solids, the shearing force is proportional to the shear strain. With reference to the elastic modulus of rigidity for fluids, the shearing force is proportional to the rate of shear strain.

Q.4: Explain why 
(a) To keep a piece of paper horizontal, you should blow over, not under, it 
(b) When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers 
(c) The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection 
(d) A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel 
(e) A spinning cricket ball in air does not follow a parabolic trajectory

Ans : (a) When air is blown under a paper, the veloc ty Of air is greater under the paper than it is above it. As per Bernoulli's principle, atmospheric pressure reduces under the paper. This makes the paper fall. To keep a piece of paper horizontal, one should blow over it, This increases the velocity of air above the paper. As per Bernoulli's principle, atmospheric pressure reduces above the paper and the paper remains horizontal. (b) According to the equation of continuity: Area x Velocity = Constant For a smaller opening, the velocity Of flow Of a fluid is greater than it is when the opening is bigger. When we try to close a tap Of water with our fingers, fast jets Of water gush through the openings between our fingers, This is because very small openings are left for the water to flow out of the pipe. Hence, area and velocity are inversely proportional to each other. (c) The small opening Of a syringe needle controls the velocity Of the blood flowing Out. This is because Of the equation Of continuity. At the constriction point Of the syringe system, the flow rate suddenly increases to a high value for a constant thumb pressure applied. (d) When a fluid flows out from a small hole in a vessel, the vessel receives a backward thrust. A fluid flowing out from a small hole has a large velocity according to the equation of continuity : Area x Velocity Constant According to the law of conservation of momentum, the vessel attains a backward velocity because there are no external forces acting on the system. (e) A spinning cricket ball has two simultaneous motions — rotary and linear. These two types of motion oppose the effect of each other. This decreases the velocity of air flowing below the ball. Hence, the pressure on the upper side of the ball becomes lesser than that on the lower side. An upward force acts upon the ball. Therefore, the ball takes a curved path. It does not follow a parabolic path.

Q.5: A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor ?

Ans : \(\begin{array}{l}{\text { Mass of the girl, } m=50 \mathrm{kg}} \\ {\text { Diameter of the heel, } d=1 \mathrm{cm}=0.01 \mathrm{m}} \\ {\text { Radius of the heel, } r=\frac{d}{2}=0.005 \mathrm{m}}\end{array}\) \(\begin{array}{l}{\text { Area of the heel }=\pi r^{2}} \\ {=\pi(0.005)^{2}} \\ {=7.85 \times 10^{-5} \mathrm{m}^{2}} \\ {\text { Force exerted by the heel on the floor: }} \\ {F=m g}\end{array}\) \(\begin{array}{l}{=50 \times 9.8} \\ {=490 \mathrm{N}} \\ {\text { Pressure exerted by the heel on the floor: }} \\ {P=\frac{\text { Force }}{\text { Area }}}\end{array}\) \(\begin{array}{l}{=\frac{490}{7.85 \times 10^{-5}}} \\ {=6.24 \times 10^{6} \mathrm{Nm}^{-2}} \\ {\text { Therefore, the pressure exerted by the heel on the horizontal floor is }} \\ {6.24 \times 10^{6} \mathrm{Nm}^{-2}}\end{array}\)

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