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**NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids**: If you're a Class 11 student in search of NCERT solutions, you're in the right spot. Here, on the Careers360 page, you can find extensive class 11 physics chapter 10 exercise solutions available for free, complete with detailed step-by-step explanations. This collection covers a total of thirty questions, including those from 10.1 to 10.20 for exercise questions and 10.20 to 10.30 for additional exercises.

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This Story also Contains

- NCERT Solutions Class 11 Physics Chapter 10 – Download Free PDF
- NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids
- Mechanical properties of fluids class 11 NCERT solutions - Exercise
- NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids - Additional Exercise
- NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties Of Fluids
- Benefits of NCERT solutions for class 11 physics chapter 10 mechanical properties of fluids:
- NCERT Solutions for Class 11 Subject wise

We all know that fluids do not have a particular shape. The shape of the fluid is decided by the shape of the container. In the solutions of NCERT Class 11 Physics Chapter 10 Mechanical Properties of Fluids, questions related to the properties of fluids (liquid and gas) and certain laws associated with the properties are discussed. Also, questions on all the topics of the chapter Mechanical Properties of Fluids Class 11 are covered in the CBSE NCERT solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids and are very important from the exam point of view.

We all have heard the term pressure; Pascal's law is related to pressure and this law states that pressure in a fluid at rest is the same at all points which are at the same height. A change in pressure applied to an enclosed fluid is transmitted to every point of the fluid and the walls of the containing vessel container. One of the applications of this law is a Hydraulic Lift. Let's try to find a few more examples in NCERT by yourself. The main aim of the mechanical properties of fluids NCERT solutions is to check the depth of understanding of the concepts and formulas studied in the chapter. Careers360 has made the mechanical properties of fluids class 11 NCERT solutions available in PDF format, which can be accessed by students anytime and anywhere, to aid in their understanding of the chapter and its associated questions.

**NCERT Class 11 Physics Chapter 10 Exercise Solutions PDF download**** **for free.

****According to the CBSE Syllabus for the academic year 2023-24, the chapter you previously referred to as Chapter 10, " ****Mechanical Properties of Fluids** **," has been renumbered as Chapter 9.**

** Q 10.1 (a) ** Explain why

The blood pressure in humans is greater at the fe et than at the brain

** Answer: **

The pressure in a fluid column increases with the height of the column, as the height of the blood column is more than that for the brain the blood pressure in feet is more than the blood pressure in the brain.

** Q 10.1 (b) ** Explain why

** Answer: **

This is because the air density does not remain the same in the atmosphere. It decreases exponentially as height increases.

** Q 10.1 (c) ** Explain why

Hydrostatic pressure is a scalar quantity even though pressure is force divided by area.

** Answer: **

When a force is applied on fluid the pressure which gets generated gets uniformly transmitted to all directions and therefore has no particular direction and is a scalar quantity. We talk of division of force with area only while considering the magnitudes. The actual vector form of the relation is

** Q 10.2 (a) ** Explain why

(a) The angle of contact of mercury with glass is obtuse, while that of water with glass is acute.

** Answer: **

T _{ SL } = Surface tension corresponding to the solid-liquid layer

T _{ LA } = Surface tension corresponding to liquid-air layer

T _{ SA } = Surface tension corresponding to solid-air layer

The angle of contact is

Since the liquid is not flowing over the solid surface the components of T _{ SL } , T _{ LA } and T _{ SA } along the solid surface must cancel out each other.

In case of mercury T _{ SA } < T _{ SL } and therefore and therefore i.e the angle of contact of mercury with glass is obtuse.

** Q 10.2 (b) ** Explain why

** Answer: **

Cohesive forces between water molecules is much lesser than adhesive forces between water and glass molecules and that's why water tends to spread out on glass whereas cohesive forces within mercury is comparable to adhesive forces between mercury and glass and that's why mercury tends to form drops.

** Q 10.2 (c) ** Explain why

(c) Surface tension of a liquid is independent of the area of the surface

** Answer: **

Surface tension is the force acting per unit length at the interface of a liquid and another surface. Since this force itself is independent of area, the surface tension is also independent of area.

** Q 10.2 (d) ** Explain why

(d) Water with detergent dissolved in it should have small angles of contact.

** Answer: **

As we know detergent with water rises very fast in capillaries of clothes which is only possible when cosine of the angle of contact is the large i.e. angle of contact must be small.

** Q 10.2 (e) ** Explain why

(e) A drop of liquid under no external forces is always spherical in shape

** Answer: **

While in a spherical shape the surface area of the drop of liquid will be minimum and thus the surface energy would be minimum. A system always tends to be in a state of minimum energy and that's why in the absence of external forces a drop of liquid is always spherical in shape.

** Q 10.3 ** 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)

** Answer: **

(a) The surface tension of liquids generally __ decreases __ with temperatures.

(b) The viscosity of gases __ increases __ with temperature, whereas the viscosity of liquids __ decreases __ with temperature.

(c) For solids with elastic modulus of rigidity, the shearing force is proportional to __ shear strain __ , while for fluids it is proportional to the __ rate of shear strain __ .

(d) For a fluid in a steady flow, the increase in flow speed at a constriction follows from __ conservation of mass __ while the decrease of pressure there follows from __ Bernoulli’s principle __ .

(e) For the model of a plane in a wind tunnel, turbulence occurs at a __ greater __ speed for turbulence for an actual plane.

** Q 10.4 (a) ** Explain why

(a) To keep a piece of paper horizontal, you should blow over, not under, it

** Answer: **

As per Bernoulli's principle when we blow over a piece of paper the pressure there decreases while the pressure under the piece of the paper remains the same and that's why it remains horizontal.

** Q 10.4 (b) ** Explain why

** Answer: **

This is because when we cover the tap there are very small gaps remaining for the water to escape and it comes out at very high velocity in accordance with the equation of continuity.

** Q 10.4 (c) ** Explain why

** Answer: **

Because of the extremely small size of the opening of a needle, its size can control the flow with more precision than the thumbs of a doctor.

According to the equation of continuity area * velocity= constant. if the area is very small the velocity must be large. Thus if the area is small flow becomes smooth

** Q 10.4 (d) ** Explain why

(d) A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel

** Answer: **

Through a small area, velocity will be large. A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel in accordance with the law of conservation of linear momentum.

** Q 10.4 (e) ** Explain why

(e) A spinning cricket ball in air does not follow a parabolic trajectory

** Answer: **

The ball while travelling rotates about its axis as well causing a difference in air velocities at different points around it thus creating pressure difference which results in external forces. In the absence of air, a ball would have travelled along the expected parabolic path.

** Answer: **

Mass of the girl m = 50 kg.

Gravitational acceleration g = 9.8 m s ^{ -2 }

Weight of the girl (W) , mg = 490 N

** Answer: **

Atmospheric pressure is

The density of French wine

Height of the wine column h _{ w } would be

** Answer: **

The density of water is

Depth of the ocean is 3 km

The pressure at the bottom of the ocean would be

The above value is much lesser than the maximum stress the structure can withstand and therefore it is suitable for putting up on top of an oil well in the ocean.

** Answer: **

Maximum Pressure which the piston would have to bear is

** Answer: **

Since the mercury columns in the two arms are equal the pressure exerted by the water and the spirit column must be the same.

Therefore the specific gravity of spirit is 0.8.

** Answer: **

Let the difference in the levels of mercury in the two arms be h _{ Hg }

** Q 10.11 ** Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river? Explain.

** Answer: **

No. Bernoulli's equation can be used only to describe streamline flow and the flow of water in a river is turbulent.

** Answer: **

No, unless the atmospheric pressures at the two points where Bernoulli’s equation is applied are significantly different.

** Answer: **

The volumetric flow rate of glycerine flow would be given by

The viscosity of glycerine is

Assuming Laminar flow for a tube of radius r, length l, having pressure difference P across its ends a fluid with viscosity would flow through it with a volumetric rate of

Reynolds number is given by

Since Reynolds Number is coming out to be 0.3 our Assumption of laminar flow was correct.

** Answer: **

The speed of air above and below the wings are given to be v _{ 1 } = 70 m s ^{ -1 } and v _{ 2 } = 63 m s ^{ -1 } respectively.

Let the pressure above and below the wings be p _{ 1 } and p _{ 2 } and let the model aeroplane be flying at a height h from the ground.

Applying Bernoulli's Principle on two points above and below the wings we get

The pressure difference between the regions below and above the wing is 605.15 Pa

The lift on the wing is F

** Answer: **

By the continuity equation, the velocity of the non-viscous liquid will be large at the kink than at the rest of the tube and therefore pressure would be lesser here by Bernoulli's principle and the air column above it, therefore, should be of lesser height. Figure (a) is therefore incorrect.

** Answer: **

Cross-sectional area of cylindrical tube is a _{ 1 } = 8.0 cm ^{ 2 }

The total area of the 40 fine holes is a _{ 2 }

Speed of liquid inside the tube is v _{ 1 } = 1.5 m min ^{ -1 }

Let the speed of ejection of fluid through the holes be v _{ 2 }

Using the continuity equation

** Answer: **

Total weight supported by the film

Since a soap film has two surfaces, the total length of the liquid film is 60 cm.

Surface Tension is T

** Answer: **

As the liquid and the temperature is the same in all three the surface tension will also be the same. Since the length is also given to be equal (40 cm) in all three cases the weight being supported is also the same and equal to .

** Answer: **

Surface Tension of Mercury is

The radius of the drop of Mercury is r = 3.00 mm

Excess pressure inside the Mercury drop is given by

Atmospheric Pressure is

Total Pressure inside the Mercury drop is given by

** Answer: **

Excess pressure inside a bubble is given by

(It's double the usual value because of the presence of 2 layers in case of soap bubble)

where T is surface tension and r is the radius of the bubble

Atmospheric Pressure is

The density of soap solution is

The pressure at a depth of 40 cm (h) in the soap solution is

Total Pressure inside an air bubble at that depth

** Answer: **

Pressure in the waterside at the bottom is

Pressure in the acid side at the bottom is

The pressure difference across the door is

Area of the door, a = 20 cm ^{ 2 }

The force necessary to keep the door closed is

Note: The dimensions of the door are small enough to neglect pressure variations near it.

(a) Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and (b), in units of cm of mercury.

** Answer: **

In figure (a)

Gauge Pressure = 20 cm of Mercury

Absolute Pressure = Atmospheric Pressure + Gauge Pressure

Absolute Pressure = 76 + 20= 96 cm of Mercury

In figure (b)

Gauge Pressure = -18 cm of Mercury

Absolute Pressure = Atmospheric Pressure + Gauge Pressure

Absolute Pressure = 76 + (-18)= 58 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?

** Answer: **

As we know Specific Gravity of Mercury is 13.6 therefore 13.6 cm of water column would be equal to 1 cm of Mercury column.

The pressure at the Mercury Water interface in the right column = Atmospheric Pressure + 1 cm of Mercury = 77 cm of Mercury

The difference in Pressure due to the level of the Mercury column = Pressure at the Mercury Water interface - Absolute Pressure of the Glass enclosure

= 77 - 58 = 19 cm of Mercury.

The difference in the two limbs would, therefore, become 19 cm.

** Answer: **

Since the height of the water level in the vessels is the same the Pressure at the bottom would be equal. As the area of the bottom is also the same the Force exerted by the water on the bottom would be the same.

The difference in the reading arises due to the fact that the weight depends on the volume of the water inside the container which is more in the first vessel. The vertical component of the force exerted by the fluid on the sidewalls would be more in the first vessel and the difference in this vertical component is equal to the difference in the readings on a weighing scale.

** Answer: **

The density of whole blood

Gauge Pressure

Height at which the blood container must be placed so that blood may just enter the vein

(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?

** Answer: **

The diameter of the artery is

The viscosity of blood is

The density of blood is

The average velocity is given by

Taking the Maximum value of Reynold's Number ( N _{ Re } = 2000) at which Laminar Flow takes place we have

(b) Do the dissipative forces become more important as the fluid velocity increases? Discuss qualitatively.

** Answer: **

As the fluid velocity increases the dissipative forces become important as turbulence rises due to which drag due to friction forces increases.

** Answer: **

The diameter of the artery is d

The viscosity of blood is

The density of blood is

The average velocity is given by

Taking the Maximum value of Reynold's Number ( N _{ Re } = 2000) at which Laminar Flow takes place we have

** Q 10.26 (b) ** What is the corresponding flow rate? (Take viscosity of blood to be 2.084 × 10 ^{ -3 } Pa s).

** Answer: **

Volumetric flow rate is given as

** Answer: **

Speed of the wind above the upper wing surface is v _{ 1 } = 234 km h ^{ -1 }

Speed of the wind over the lower wing is v _{ 2 } = 180 km h ^{ -1 }

Let the pressure over the upper and lower wing be P _{ 1 } and P _{ 2 }

Let the plane be flying at a height of h

The density of air is

Applying Bernoulli's Principle at two points over the upper and lower wing we get

Area of each wing is a = 25 m ^{ 2 }

The net upward force on the plane is F

This upward force is equal to the weight m of the plane.

The mass of the plane is about 4400 kg

** Answer: **

Neglecting buoyancy due to air the terminal velocity is

Viscous Force F _{ v } at this speed is

** Answer: **

Since the angle of contact is obtuse the Pressure will be more on the Mercury side.

This pressure difference is given as

The dip of mercury inside the narrow tube would be equal to this pressure difference

The mercury dips down in the tube relative to the liquid surface outside by an amount of 5.34 mm.

** Answer: **

For the angle of contact , radius of the tube r, surface tension t, the density of fluid the rise in the column is given by

The radii of the two limbs r _{ 1 } and r _{ 2 } are 3.0 mm and 1.5 mm respectively

The level in the limb of diameter 6.0 mm is

The level in the limb of diameter 3.0 mm is

The difference in the heights is h _{ 2 } - h _{ 1 } = 4.96 mm

Providing class 11 physics chapter 10 exercise solutions is invaluable for Class 11 students, as it aids in their preparation. Additionally, understanding 'Mechanical Properties of Fluids' is crucial for competitive exams like JEE and NEET, where physics plays a significant role. This chapter of mechanical properties of fluids exercise equips students with essential knowledge about fluid behaviour under mechanical forces, helping them tackle complex questions in these exams with confidence.

**NCERT Solutions for Class 11 Physics Chapter Wise **

Chapter 1 | |

Chapter 2 | |

Chapter 3 | |

Chapter 4 | |

Chapter 5 | |

Chapter 6 | |

Chapter 7 | |

Chapter 8 | |

Chapter 9 | |

Chapter 10 | Mechanical Properties of Fluids |

Chapter 11 | |

Chapter 12 | |

Chapter 13 | |

Chapter 14 | |

Chapter 15 |

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Download EBook#### Modulus of rigidity

#### Poisson’s ratio

#### Elastic energy

U = 1/2*strain*stress*volume

#### Heat Capacity

Where:

ΔQ is the amount of heat supplied to the substance and ΔT change in its temperature

#### Specific heat capacity

These formulas are essential because they play a crucial role in solving problems and understanding the concepts covered in Chapter ch 10 physics class 11 ncert solutions.

**Mechanical Properties Of Fluids Class 11 NCERT Physics Topics**

The problems discussed in the NCERT solutions for Class 11 Physics chapter 10 Mechanical Properties of Fluids are based on the following topics.

10.1 Introduction

10.2 Pressure

10.3 Streamline flow

10.4 Bernoulli’s principle

10.5 Viscosity

10.6 Surface tension

The CBSE is widely recognized as the leading educational board in India, and it follows the NCERT Syllabus to administer examinations for both Class 10 and 12 students. The class 11 physics ch 10 ncert solutions, which focuses on Mechanical Properties of Fluids, have been provided to allow students to gain a comprehensive understanding of the concepts discussed in this chapter.

Fluids possess the fundamental property of being able to flow, and they lack resistance to alterations in their shape. Consequently, the shape of a fluid is determined by the shape of its container. The following are some important points about the Mechanical Properties of Fluids of Matter.

- Pressure is a scalar quantity, and although the definition of pressure as "force per unit area" might give the impression that it is a vector, the "force" in the numerator is the component of the force perpendicular to the area on which it is applied.
- As the temperature increases, the atoms in a liquid become more mobile, causing the coefficient of viscosity η to decrease. In the case of a gas, the temperature increase results in an increase in the random motion of atoms, which increases η.

**Comprehensive Coverage:** The ch 10 physics class 11 ncert solutions cover all the topics and concepts presented in the chapter, ensuring a thorough understanding of mechanical properties of fluids.

**Step-by-Step Solutions: **Detailed step-by-step chapter 10 physics class 11 solutions for each exercise question are provided, making it easy for students to follow and grasp the concepts.

**Clarity and Simplicity: **The solutions are explained in clear and simple language, making complex concepts easy to understand.

**Diagrams and Graphs:** Wherever necessary, diagrams and graphs are included to enhance visual understanding and clarify concepts related to fluid properties.

**Concise Summaries:** NCERT solutions often include concise summaries of the main points in each section, helping students revise and retain important information.

**Examination-Oriented:** The ch 10 physics class 11 are designed to help students prepare for exams, including CBSE board exams and various competitive exams that may include questions related to fluid mechanics.

- If we are familiar with the CBSE NCERT solutions for Class 11 Physics chapter 10 Mechanical Properties of Fluids then it is very easy to solve problems from other reference books too.
- The solutions of NCERT solutions for Class 11 will help to score well in the Class 11 final exam and in competitive exams like JEE Main and NEET also.

- NCERT Books Class 11 Physics
- NCERT Syllabus Class 11 Physics
- NCERT Books Class 11
- NCERT Syllabus Class 11

1. Whether the Class 11 chapter Mechanical Properties of Fluids is useful for higher studies

Yes, the topics studied are the basics of fluid dynamics which is mainly studied in mechanical and civil related branches.

2. How many questions are asked from Mechanical Properties of Fluids for NEET and JEE Main

On an average one question is asked from the chapter Mechanical Properties of Fluids for both JEE Main and NEET exam. It can be morethan one also. All the formulas given in the NCERT book for Mechanical Properties of Fluids chapter are important. To get more problems related to the formulas students can use NCERT exemplar and JEE main and NEET previous year papers.

3. Name the subtopic covered in mechanical properties of fluids class 11

The subtopic covered in mechanical properties of fluids are give below:

1. Introduction

2. Pressure

3. Streamline flow

4. Bernoulli’s principle

5. Viscosity

6. Surface tension

4. How can I easily solve the questions given in chapter 10 class 11 physics

To solve the questions given in physics class 11 chapter 10, you can follow these steps:

Read the question carefully and understand what is being asked.

Identify the concept or formula required to solve the question.

Make sure you understand the formula and its variables.

Substitute the given values in the formula and solve for the unknown variable.

Check your answer and make sure it is reasonable and accurate.

Practice similar problems to improve your understanding and speed.

5. How physics class 11 chapter 10 notes is important for JEE or NEET?

Chapter 10 notes of Physics Class 11 are important for JEE and NEET exams as they cover fundamental concepts related to fluid mechanics. Understanding these concepts can help students solve questions and excel in related fields of engineering.

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