NCERT Solutions for Class 11 Physics Chapter 10
NCERT Solutions for Class 11 Physics Chapter 10
Physics is the study of matter mainly dealing with what matter constitutes, its motion and behaviour, and energy and force. It also deals with the various properties of matter and nature. The NCERT solutions Class 11 Physics Chapter 10 curated by expert teachers helps students to understand the topics well. We have laid out the chapters in such a way that they can be downloaded directly to your computer. You have the option of downloading either the entire syllabus or each chapter separately.
Physics Chapter 10 NCERT Solutions Class 11 contains all the chapters laid out and answered by our expert teachers. With many years of expertise on board-issued textbooks, they mark out important questions and give you tips on how to solve them.
NCERT solution Class 11 Physics Chapter 10 discusses fluids and their mechanical properties. The topic of fluids in class 11 is a broad and comprehensive chapter. Class 11 Physics Chapter 10 NCERT Solutions mechanical properties of fluids aim to solve this common problem among students by making studies simpler and learning too. To score good marks, it is more important to understand the concept of mechanical properties of fluids in Class 11. NCERT Solution Class 11 Physics Chapter 10 is your guide and tool toward a better learning experience.
NCERT Solutions for Class 11 Physics Chapter 10 – Mechanical Properties Of Fluids
The Physics NCERT Solutions Class 11 aims at making the subject easier for each student as all students do not have the same distinction. The main motive of the solutions is to help students understand the subject and make it easier for the students. Class 11 Physics introduces students to topics like Thermodynamics, Laws of Motion, etc. These topics are extremely important for the students to prepare. With NCERT Solutions for Physics Class 11, it becomes easier for the students to learn the subject. Having a solutions book also makes it easier for the students to revise before the exam as all the syllabus is integrated at one place for the students to quickly go through.
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Access NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties Of Fluids
NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids is a resource study where you can expect to prepare for the final examination. In chapter 10, we will be studying the physical properties of gases and liquids. Both liquids and gases can flow and hence are called fluids.
This property differentiates gases and liquids from solids. The mechanical properties of these fluids are explained in brief, the NCERT Solutions for Class 11 Physics in a format which can be accessed by the students without restrictions of time and place. Students can easily access these solutions for Class 11 Physics Chapter 10 and practice various questions based on this topic. Those students who find difficulties in solving the problems can refer to this free Class 11 Physics Chapter 10 NCERT Solutions online. Liquids and gases can flow and are therefore called fluids. It is this property that differentiates liquids and gases from solids in a basic way. NCERT Solutions are one of the best tools to prepare for Class 11 Worksheets. Mechanical Properties of Fluids is a significant chapter in the second term of CBSE Class 11.
Students must prepare this chapter well to understand the topic in-depth and score high marks in their examination. Extramarks provides NCERT solutions Class 11 Physics Chapter 10 written in a comprehensive and easy-to-understand manner so that students can quickly grasp the topic and score well in the exams. Our core focus is to help students keep their concepts clear and to the point without any misunderstandings or misguidances.
NCERT Solutions For Class 11 Physics Chapter 10 Mechanical Properties Of Fluids – Free Download
NCERT Solutions Class 11 Physics Chapter 10 Mechanical Properties of Fluids is a prime study resource on which you can rely to prepare for the final examination as well as for the entrance examinations. NCERT provides free Solutions for all the subjects from Class 11. Students will be able to grasp the important topics in a shorter duration with the help of the solutions. All the concepts are explained in a simple and easy-to-understand language to help students irrespective of their intelligence level. In Chapter 10, we will be studying the physical properties of gases and liquids. Liquids and gases both can flow and hence are called fluids.
This property differentiates gases and liquids from solids. The mechanical properties of these fluids are explained in brief as per the latest update on the syllabus. For a good idea about this chapter, the NCERT Solutions for Class 11 Physics can be accessed by the students with utmost ease and convenience. They can highlight the important topics, formulas, and definitions to revise before exams.
This makes the preparation part smoother. These concepts are not only helpful with the board exams but they are also very important in terms of competitive exams. One can find these concepts being repeated in further competitive exams but with more detailed definitions and learnings. If students can keep their basics clear, they can surely crack competitive exams.
Class 11 Physics Mechanical Properties Of Fluids – Topics
Class 11 Mechanical Properties of Fluids NCERT solutions discuss the common physical properties of different fluids. Moreover, it clarifies that fluids refer to both liquids and gases, and not just liquids. It is these properties of fluids that distinguish them from solids.
Chapter 10 Class 11 physics NCERT solutions effectively cover the following topics:
The NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids of Matter is given so that students can understand the concepts of this chapter in-depth. The basic property of a fluid is that it can flow. The fluid does not have any resistance to change in its shape. Thus, the shape of the fluid is governed by the shape of its container. Some key points on Mechanical Properties of Fluids of Matter are given here. The pressure is a scalar quantity. The definition of pressure as “force per unit area” may give one false impression that pressure is a vector.
The “force” in the numerator of the definition is the component of the force normal to the area upon which it is impressed. As the temperature rises, the atoms of the liquid become more mobile and the coefficient of viscosity η falls. In a gas, the temperature rise increases the random motion of atoms, and η increases. The chapter holds quite a good amount of weightage from an exam perspective. If the students can score well in this chapter, they can properly focus on other important topics as well to increase their overall scores.
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How Does Temperature Affect The Surface Tension Of A Liquid?
Class 11 Physics NCERT solutions chapter 1 explains all aspects of surface tension. Surface tension in liquids arises due to the force of attraction between molecules that are on the surface and underneath. The molecules on the surface of a liquid experience an inwards pull by the molecules on the interior. However, liquids have a tendency to assume a shape that covers the least surface area. NCERT solutions offer explanations for various physical properties of fluids.
Mechanical Properties of Fluids Class 11 NCERT describes fluids as any substance that continually deforms under external force. Fluids include liquids, gases, and plasma. Fluid Mechanics Class 11 NCERT solutions effectively describe the effect of temperature on the surface tension of liquids. With the rise in temperature, the kinetic energy of the molecules increases. As a result, the intermolecular attraction decreases, therefore lowering the surface tension of the liquid. Similarly, the surface tension of liquids rises with a fall in temperature. You can refer to our Class 11 Physics Chapter 10 NCERT solutions for a more elaborate explanation.
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.
(a) Pressure of a liquid is given as:
P = hρg
Here, ρ = density of liquid
We can conclude that pressure is directly proportional to height. Thus, the blood pressure at the feet is more than it is at the brain.
(b) The density of air is the maximum near the sea level and it decreases rapidly with increase in height. At a height of about 6 km, density decreases to approximately half of its value at the sea level..
(c) As force is applied on a liquid, the pressure in the liquid is transmitted equally in all directions. Therefore, hydrostatic pressure does not have a fixed direction. It is a scalar 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
(a) The angle formed between the tangent to the liquid surface at the point of contact and the surface inside the liquid is called as the angle of contact (θ). In the given figure, Sla, Ssa, and Ssl are the surface tensions between the liquid-air, solid-air, and solid-liquid interfaces respectively. The surface forces between the three media should be in equilibrium at the line of contact, i.e.,
(b) The mercury makes an obtuse angle of contact with glass. The molecules of mercury have a strong force of attraction between themselves called cohesion and a weak force of attraction toward solids called adhesion. Therefore, they have tendency to form drops. On the contrary, the molecules of water have a weak force of cohesion and a strong force of adhesion. Therefore, they make acute angles with glass and have tendency to spread out.
(c) Surface tension is defined as the force acting per unit length at the interface between the plane of a liquid and any other surface. Surface tension does not depend on the area of the surface of the liquid; therefore, it is also independent of the area of the liquid surface.
(d) The cloth has small spaces in the form of capillaries. The liquid with detergent dissolved in it has small angles of contact (θ), because for a small θ, there is a fast capillary rise of detergent in the cloth. The rise of a liquid in a capillary tube is directly proportional to the cosine of the angle of contact (θ). If angle of contact is small, then cosine of angle of contact will be large and the penetration of the detergent water in the cloth will be fast.
(e) The liquid drop has the tendency to acquire the minimum surface area due to the presence of surface tension. Since the surface area of a sphere is the minimum for a given volume, therefore, in the absence of 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)
(b) Increases; decreases
(c) Shear strain; Rate of shear strain
(d) Conservation of mass; Bernoulli’s principle
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
(a) When we blow under a paper, the velocity of air blow increases under the paper. According to Bernoulli’s principle, atmospheric pressure decreases under the paper and the paper falls. On blowing over the paper the velocity increases above the paper. According to Bernoulli’s principle, atmospheric pressure decreases above the paper and the paper remains horizontal.
(b) From the equation of continuity:
Area × Velocity = Constant
If we try to close a tap of water with our fingers, the area of the outlet of the water jet is reduced, so velocity of water increases. Therefore, area is inversely proportional to velocity.
(c) The size of the needle of the syringe controls the velocity of the blood flowing outside
From the above equation, it can be observed that the velocity have more influence on the flow of a liquid as it occurs with power two. Therefore, the needle has a better control over the flow.
(d) From the Torricelli’s theorem, the velocity of a fluid flowing outside from a small hole is given as:
Here, H = Height of fluid above the hole
When a fluid flows out from a small hole in a vessel, it acquires a large velocity. Therefore, its momentum increases to a large amount in the outward direction.
Since no external forces are acting on the system, therefore, according to the law of conservation of momentum, the vessel experiences a recoil momentum in the backward direction. As a result, the vessel experiences a backward thrust.
(e) The spinning cricket ball has two types of simultaneous motions. These are rotatory motion and linear motion. These motions oppose the effect of each other. As a result, the velocity of air flowing below the ball decreases. Therefore, the pressure on the lower side of the ball becomes more than that on the upper side. Because of it, the spinning ball experiences an upward force. Therefore, the ball takes a curved path and it deviates from its 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?
Q.6 Torricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984 kg m–3. Determine the height of the wine column for normal atmospheric pressure.
Q.7 A vertical off-shore structure is built to withstand a maximum stress of 109 Pa. Is the structure suitable for putting up on top of an oil well in the ocean? Take the depth of the ocean to be roughly 3 km, and ignore ocean currents.
Here, Maximum stress for the off-shore structure, P = 109 Pa
Depth of ocean, d = 3 km = 3 × 103 m
Density of the sea water, ρ = 103 kgm-3
Acceleration due to gravity, g = 9.8 ms-2
Hydrostatic pressure exerted due to sea water at the bottom of sea,
d = ρdg = 3 × 103 m × 103 kgm-3 × 9.8ms-2
d = 2.94 × 107 Pa
The maximum stress which the off-shore structure can withstand is more than the pressure of the sea water (2.94 × 107 Pa). Hence, the pressure exerted by the ocean is less than the pressure that the off shore structure can withstand. Thus, the off shore structure will be suitable for putting up on top of the oil well in the ocean.
Q.8 A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm2. What maximum pressure would the smaller piston have to bear?
Q.9 A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. What is the specific gravity of spirit?
Q.10 In the previous problem, if 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms? (Specific gravity of mercury = 13.6)
Q.11 Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river? Explain.
No, Bernoulli’s equation can only be applied to streamline flow, Rapid in a river is a turbulent flow.
Q.12 Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation? Explain.
It does not matter if one uses gauge instead of absolute pressure while using Bernoulli’s equation. The two points where bernoulli’s equation will be applied will have different atmospheric pressures.
Q.13 Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0 × 10–3 kg s–1, what is the pressure difference between the two ends of the tube? (Density of glycerine = 1.3 × 103 kg m–3 and viscosity of glycerine = 0.83 Pa s). [You may also like to check if the assumption of lamina r flow in the tube is correct].
Q.14 In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s–1 and 63 m s-1 respectively. What is the lift on the wing if its area is 2.5 m2? Take the density of air to be 1.3 kg m–3.
Q.15 Figures 10.23 (a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect? Why?
Let us take the case given in figure (b).
From the equation of continuity, we have
av = constant
When the cross-sectional area in the middle portion of the venturimeter is small, the speed of the flow of liquid is more through this part.
From Bernoulli’s theorem, if speed is more, then pressure will be less. Since pressure is directly proportional to the height; therefore the water level in pipe 2 is less.
Hence, figure (a) is incorrect.
Q.16 The cylindrical tube of a spray pump has a cross-section of 8.0 cm2 one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m min–1, what is the speed of ejection of the liquid through the holes?
Q.17 A U-shaped wire is dipped in a soap solution, and removed. The thin soap film formed between the wire and the light slider supports a weight of 1.5 × 10–2 N (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film?
Q.18 Figure 10.24 (a) shows a thin liquid film supporting a small weight = 4.5 × 10–2 N. What is the weight supported by a film of the same liquid at the same temperature in Fig. (b) and (c)? Explain your answer physically.
Q.19 What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20°C) is 4.65 × 10–1 N m-1. The atmospheric pressure is 1.01 × 105 Pa. Also give the excess pressure inside the drop.
Q.20 What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20°C) is 2.50 ×10–2 N m–1? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble? (1 atmospheric pressure is 1.01 × 105 Pa).
Q.21 A tank with a square base of area 1.0 m2 is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 20 cm2. The tank is filled with water in one compartment, and an acid (of relative density 1.7) in the other, both to a height of 4.0 m. compute the force necessary to keep the door close.
Q.22 A manometer reads the pressure of a gas in an enclosure as shown in Fig. 10.25 (a) When a pump removes some of the gas, the manometer reads as in Fig. 10.25 (b)
The liquid used in the manometers is mercury and the atmospheric pressure is 76 cm of mercury.
(a) Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and (b), in units of cm of mercury.
(b) How would the levels change in case (b) if 13.6 cm of water (immiscible with mercury) are poured into the right limb of the manometer? (Ignore the small change in the volume of the gas).
Q.23 Two vessels have the same base area but different shapes. The first vessel takes twice the volume of water that the second vessel requires to fill up to a particular common height. Is the force exerted by the water on the base of the vessel the same in the two cases? If so, why do the vessels filled with water to that same height give different readings on a weighing scale?
Two vessels with same base area will have same force and equal pressure acting on their common base area. The force exerted on the sides of the vessels has non-zero vertical components as their shapes are different. On adding these vertical components the total force on one vessel comes out to be more than that on the other vessel. Thus, the two vessels filled with water to the same height, give different readings on a weighing scale.
Q.24 During blood transfusion the needle is inserted in a vein where the gauge pressure is 2000 Pa. At what height must the blood container be placed so that blood may just enter the vein? [Use the density of whole blood from Table 10.1].
Q.25 In deriving Bernoulli’s equation, we equated the work done on the fluid in the tube to its change in the potential and kinetic energy. (a) What is the largest average velocity of blood flow in an artery of diameter 2 × 10–3 m if the flow must remain laminar? (b) Do the dissipative forces become more important as the fluid velocity increases? Discuss qualitatively.
Q.26 (a) What is the largest average velocity of blood flow in an artery of radius 2 × 10–3 m if the flow must remain laminar? (b) What is the corresponding flow rate? (Take viscosity of blood to be 2.084 × 10–3 Pa s).
Q.27 A plane is in level flight at constant speed and each of its two wings has an area of 25 m2. If the speed of the air is 180 km h-1 over the lower wing and 234 kmh-1 over the upper wing surface, determine the plane’s mass. (Take air density to be 1 kg m–3).
Q.28 In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0 × 10–5 m and density 1.2 × 103 kg m–3? Take the viscosity of air at the temperature of the experiment to be 1.8 × 10–5 Pa s. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.
Q.29 Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside? Surface tension of mercury at the temperature of the experiment is 0.465 Nm–1. Density of mercury = 13.6 × 103 kgm–3.
Q.30 Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is 7.3 × 10–2 N m–1. Take the angle of contact to be zero and density of water to be 1.0 × 103 kg m–3 (g = 9.8 m s–2).
Q.31 (a) It is known that density ρ of air decreases with height y as
ρ = ρ0e-y/y0
Where ρ0 =1.25 kg m–3 is the density at sea level, and y0 is a constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of atmosphere remains a constant (isothermal conditions). Also assume that the value of g remains constant.
(b) A large He balloon of volume 1425 m3 is used to lift a payload of 400 kg. Assume that the balloon maintains constant radius as it rises. How high does it rise?
[Take y0 = 8000 m and ρHe= 0.18 kgm–3].
FAQs (Frequently Asked Questions)
1. Why should you prefer NCERT solutions Class 11 Physics Chapter 10?
NCERT solutions offer you well-explained solutions that are solved by highly experienced teachers. The solutions are narrated in a simple language to help students understand the concepts easily. The NCERT Solutions Physics Class 11 Chapter 10 is a perfect guide for students looking forward to scoring good marks in exams. If the concepts are clear, the students can score well even in competitive exams.
2. What is the main highlight of Chapter 10 Physics Class 11?
The key highlight of the NCERT solutions Chapter 10 Physics Class 11 is that the students will learn to differentiate between solids, liquids and gases. It will help students understand the chapter properly without any doubts. They will gain a wide knowledge about solids, liquids and gases.
3. What do you mean by fluids?
According to NCERT solutions Class 11, fluids are substances that can flow even with the smallest of pressure exerted by an external force. Since gases and liquids have the potential to flow easily and take a container’s shape, they are also called fluids. Fluids are shapeless and take the shape of the container they are kept in.
4. Is referring to NCERT solutions necessary for students?
We highly recommend students to refer to NCERT solutions Class 11 Physics Chapter 10 that are prepared by highly experienced teachers to make students understand the topic in a simple manner.
5. What do you mean by Streamline Flow?
Fluids have a natural tendency of flowing in separate layers. The layers flow independently without any collision or disruption. This is known as Streamline flow. We highly recommend you study the concept from our NCERT Solutions for Class 11 to make your concepts clear and more solid.
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