# NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties Of Fluids

NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids: We all know that the 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 properties of fluids (liquid and gas) and certain laws associated with the properties are discussed. Also, questions on all the topics of the chapter are covered in the CBSE NCERT solutions for class 11 physics chapter 10 mechanical properties of fluids and are very important for 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. The main aim of the NCERT solutions is to check the depth of understanding of the concepts and formulas studied in the chapter. 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

## NCERT solutions for class 11 physics chapter 10 mechanical properties of fluids exercise

Q 10.1 (a) Explain why

The blood pressure in humans is greater at the feet than at the brain

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

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

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.

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$\vec{F}=p\vec{A}$

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.

TSL = Surface tension corresponding to the solid-liquid layer

TLA = Surface tension corresponding to liquid-air layer

TSA = Surface tension corresponding to solid-air layer

The angle of contact is $\Theta$

Since the liquid is not flowing over the solid surface the components of TSL, TLA and TSA along the solid surface must cancel out each other.

$\\T_{SL}+T_{LA}cos\Theta =T_{SA}\\ cos\Theta =\frac{T_{SA}-T_{SL}}{T_{LA}}$

In case of mercury TSA < TSL and therefore $cos\Theta <0$ and therefore $\Theta >\frac{\pi }{2}$ i.e the angle of contact of mercury with glass is obtuse.

Q 10.2 (b) Explain why

(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.)

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

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.

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

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)

(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

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

(b) When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers

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

(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

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

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

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.

Mass of the girl m = 50 kg.

Gravitational acceleration g = 9.8 m s-2

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

$\\Pressure=\frac{Force}{Area}\\ =\frac{490}{\pi \left ( \frac{1\times 10^{-2}}{2} \right )^{2}}\\ =6.24\times 10^{6}Pa$

Atmospheric pressure is  $P=1.01\times 10^{5}Pa$

The density of French wine $\rho_{w}= 984\ kg\ m^{-3}$

Height of the wine column hw would be

$\\h_{w}=\frac{P}{\rho _{w}g}\\h_{w}=\frac{1.01\times 10^{5}}{984\times 9.8}\\ h_{w}=10.474m$

The density of water is $\rho _{w}=1000\ kg\ m^{-3}$

Depth of the ocean is 3 km

The pressure at the bottom of the ocean would be

$\\P=\rho _{w}gh\\ P=1000\times 9.8\times 3\times 1000 \\P=2.94\times 10^{7}Pa$

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.

Maximum Pressure which the piston would have to bear is

$\\P_{max}=\frac{Maximum\ weight\ of\ a\ car}{Area\ of\ the\ piston}\\ =\frac{3000\times 9.8}{425\times 10^{-4}}\\ =6.917\times 10^{5} Pa$

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

$\\h_{w}\rho _{w}g=h_{s}\rho _{s}g\\ \frac{\rho _{s}}{\rho _{w}}=\frac{h_{w}}{h_{s}}\\ \frac{\rho _{s}}{\rho _{w}}=\frac{10}{12.5}\\ \frac{\rho _{s}}{\rho _{w}}=0.8$

Therefore the specific gravity of spirit is 0.8.

Let the difference in the levels of mercury in the two arms be hHg

$\\(\rho _{w}-\rho _{s})gh=\rho _{Hg}gh_{Hg}\\ \left ( 1-\frac{\rho _{s}}{\rho _{w}} \right )h=\frac{\rho _{Hg}}{\rho _{w}}h_{Hg}\\ (1-0.8)\times 15=13.68h_{Hg}\\ h_{Hg}=\frac{0.2\times 15}{13.6}\\ h_{Hg}=0.22cm$

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

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

The volumetric flow rate of glycerine flow would be given by

$\\V=\frac{Amount\ of\ glycerine\ flow\ per\ second\ }{Density\ of\ glycerine}\\ V=\frac{M}{\rho }\\ V=\frac{4\times 10^{-3}}{1.3\times 10^{3}}\\ V=3.08\times 10^{-6}ms^{-1}$

The viscosity of glycerine is $\eta =0.83\ Pa\ s$

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

$\\V=\frac{\pi Pr^{4}}{8\eta l}\\ P=\frac{8V\eta l}{\pi r^{4}}\\ P=\frac{8\times 3.08\times 10^{-6}\times 0.83}{\pi \times (10^{-2})^{4}}\\ P=9.75\times 10^{2}\ Pa$

Reynolds number is given by

$\\R=\frac{4\rho V}{\pi d\eta }\\ R=\frac{2\rho V}{\pi r\eta }\\ R=\frac{2\times 1.3\times 10^{3}\times 3.08\times 10^{-6}}{\pi \times 0.01\times 0.83}\\ R\approx 0.3$

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

The speed of air above and below the wings are given to be v1 = 70 m s-1 and v2 = 63 m s-1 respectively.

Let the pressure above and below the wings be p1 and p2 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

$\\P_{1}+\rho gh+\frac{1}{2}\rho v_{1}^{2}=P_{2}+\rho gh+\frac{1}{2}\rho v_{2}^{2}\\ P_{2}-P_{1}=\frac{1}{2}\rho (v_{1}^{2}-v_{2}^{2})\\ \Delta P=\frac{1}{2}\times 1.3\times \left ( 70^{2}-63^{2} \right )\\ \Delta P=605.15\ Pa$

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

The lift on the wing is F

$\\F=\Delta P\times Area\ of\ wing\\ F=605.15\times 2.5\\ F=1512.875\ N$

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.

Cross-sectional area of cylindrical tube is a1 = 8.0 cm2

The total area of the 40 fine holes is a2

$\\a_{2}=40\pi (\frac{d^{2}}{4})\\ a_{2}=40\times \pi \times \frac{(10^{-3})^{2}}{4}\\ a_{2}=3.143\times 10^{-5}m^{2}$

Speed of liquid inside the tube is v1 = 1.5 m min-1

Let the speed of ejection of fluid through the holes be v2

Using the continuity equation

$\\a_{1}v_{1}=a_{2}v_{2}\\ v_{2}=\frac{a_{1}v_{1}}{a_{2}}\\ v_{2}=\frac{8\times 10^{-4}\times 1.5}{60\times 3.143\times 10^{-5}}\\ v_{2}=0.636\ m\ s^{-1}$

Total weight supported by the film $W =1.5\times 10^{-2}\ N$

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

Surface Tension is T

$\\T=\frac{W}{l}\\ T=\frac{1.5\times 10^{-2}}{60\times 10^{-2}}\\ T=2.5\times 10^{-2}\ N\ m^{-1}$

## Q 10.18 Figure (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.

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 $4.5\times 10^{-2}\ N$.

Surface Tension of Mercury is $T=4.65\times 10^{-1}\ N\ m^{-1}$

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

Excess pressure inside the Mercury drop is given by

$\\\Delta P=\frac{2T}{r}\\ \Delta P=\frac{2\times 4.65\times 10^{-1}}{3\times 10^{-3}}\\ \Delta P=310\ Pa$

Atmospheric Pressure is $P_{0}=1.01\times 10^{5}\ Pa$

Total Pressure inside the Mercury drop is given by

$\\P_{T}=\Delta P + P_{0}\\ P_{T}=310+1.01\times 10^{5}\\ P_{T}=1.0131\times 10^{5}$

Excess pressure inside a bubble is given by

$\Delta P=\frac{4T}{r}$            (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

$\\\Delta P=\frac{4\times 2.5\times 10^{-2}}{5\times 10^{-3}}\\ \Delta P=20\ Pa$

Atmospheric Pressure is $P_{a}=1.01\times 10^{-5}\ Pa$

The density of soap solution is $\rho _{s}=1.2\times 10^{3}\ kg\ m^{-3}$

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

$\\P_{s}=P_{a}+\rho _{s}gh\\ P_{s}=1.01\times 10^{5}+1.2\times 10^{3}\times 9.8\times 40\times 10^{-2}\\ P_{s}=1.01\times 10^{5}+4704\\ P_{s}=1.057\times 10^{5}\ Pa$

Total Pressure inside an air bubble at that depth

$\\P_{t}=P_{s}+\frac{2T}{r}\\ P_{t}=1.057\times 10^{5}+\frac{2\times 2.5\times 10^{-2}}{5\times 10^{-3}}\\ P_{t}=1.057\times 10^{5}+10\\ P_{t}=1.0571\times 10^{5}\ Pa$

## NCERT solutions for class 11 physics chapter 10 mechanical properties of fluids additional exercise

Pressure in the waterside at the bottom is

$\\P_{w}=\rho _{w}gh\\ P_{w}=10^{3}\times 9.8\times 4\\ P_{w}=3.92\times 10^{4}$

Pressure in the acid side at the bottom is

$\\P_{a}=\rho _{a}gh\\ P_{w}=1.7\times 10^{3}\times 9.8\times 4\\ P_{w}=6.664\times 10^{4}$

The pressure difference across the door is

$\\\Delta P=P_{a}-P_{w}\\ \Delta P=6.664\times 10^{4}-3.92\times 10^{4}\\ \Delta P=2.774\times 10^{4}\ Pa$

Area of the door, a = 20 cm2

The force necessary to keep the door closed is

$\\F=a\Delta P\\ F=20\times 10^{-4}\times 2.774\times 10^{4}\\ F=54.88\ N$

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

## Q 10.22 (a) A manometer reads the pressure of a gas in an enclosure as shown in Figure (a) When a pump removes some of the gas, the manometer reads as in Figure (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.

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?

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.

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.

The density of whole blood $\rho _{w}=1.06\times 10^{3}\ kg\ m^{-3}$

Gauge Pressure $\Delta P=2000\ Pa$

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

$\\h=\frac{\Delta P}{\rho _{w}g}\\ h=\frac{2000}{1.06\times 10^{3}\times 9.8}\\ h=0.1925\ m$

(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?

The diameter of the artery is $d=2\times 10^{-3}m$

The viscosity of blood is $\eta =2.08\times 10^{-3}\ Pa\ s$

The density of blood is $\rho =1.06\times 10^{3}\ kg\ m^{-3}$

The average velocity is given by $v_{avg}=\frac{N_{Re}\eta }{\rho d}$

Taking the Maximum value of Reynold's Number ( NRe = 2000) at which Laminar Flow takes place we have

$\\v_{avg,max}=\frac{2000\times 2.08\times 10^{-3}}{1.06\times 10^{3}\times 2\times 10^{-3}}\\ v_{avg,max}=1.97ms^{-1}$

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

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

The diameter of the artery is d

$\\d=2r\\ d=2\times 2\times 10^{-3}\\ d=4\times 10^{-3}$

The viscosity of blood is $\eta =2.08\times 10^{-3}\ Pa\ s$

The density of blood is $\rho =1.06\times 10^{3}\ kg\ m^{-3}$

The average velocity is given by $v_{avg}=\frac{N_{Re}\eta }{\rho d}$

Taking the Maximum value of Reynold's Number ( NRe = 2000) at which Laminar Flow takes place we have

$\\v_{avg,max}=\frac{2000\times 2.08\times 10^{-3}}{1.06\times 10^{3}\times 4\times 10^{-3}}\\ v_{avg,max}=0.98ms^{-1}$

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

Volumetric flow rate is given as

$\\\frac{dV}{dt}=v_{avg}\pi r^{2}\\ =0.98\times \pi \times (0.02)^{2}\\ =1.235\times 10^{-5}\ m^{3}\ s^{-1}$

Speed of the wind above the upper wing surface is v1 = 234 km h-1

$\\v_{1}=234\times \frac{1000}{3600}\\v_{1}=65ms^{-1}$

Speed of the wind over the lower wing is v2 = 180 km h-1

$\\v_{2}=180\times \frac{1000}{3600}\\ v_{2}=50ms^{-1}$

Let the pressure over the upper and lower wing be P1 and P2

Let the plane be flying at a height of h

The density of air is $\rho =1\ kg\ m^{-3}$

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

$\\P_{1}+\frac{1}{2}\rho v_{1}^{2}+\rho gh=P_{2}+\frac{1}{2}\rho v_{2}^{2}+\rho gh\\ P_{2}-P_{1}=\frac{1}{2}\rho (v_{1}^{2}-v_{2}^{2})\\ \Delta P=\frac{1}{2}(65^{2}-50^{2})\\ \Delta P=862.5\ Pa$

Area of each wing is a = 25 m2

The net upward force on the plane is F

$\\F=2a\Delta P\\ F=2\times 25\times 862.5\\ F=43125\ N$

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

$\\mg=F\\ m=\frac{F}{g}\\ m=\frac{43125}{9.8}\\ m=4400.51\ kg$

The mass of the plane is about 4400 kg

Neglecting buoyancy due to air the terminal velocity is

$\\v_{T}=\frac{2}{9}\frac{r^{2}\rho g}{\eta }\\ v_{T}=\frac{2\times (2\times 10^{-5})^{2}\times 1.2\times 10^{3}\times 9.8}{9\times 1.8\times 10^{-5}}\\ v_{T}=5.807\times 10^{-2}\ m\ s^{-1}$

Viscous Force Fv at this speed is

$\\F_{v}=6\pi \eta rv_{T}\\ F_{v}=6\pi \times 1.8\times 10^{-5}\times 2\times 10^{-5}\times 5.807\times 10^{-2}\\ F_{v}=3.94\times 10^{-10}N$

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

This pressure difference is given as

$\\\Delta P=\frac{2Tcos(180-\theta )}{r}\\ \Delta P=\frac{2\times 0.465cos40^{o}}{10^{-3}}\\\Delta P=\frac{2\times 0.465\times 0.766}{10^{-3}}\\ \Delta P=712.38\ Pa$

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

$\\\Delta P=\rho _{Hg}hg\\ h=\frac{\Delta P}{\rho _{Hg}g}\\ h=\frac{712.38}{13.6\times 10^{3}\times 9.8}\\ h=5.34\times 10^{-3}m$

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

## Q 10.30 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).

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

$h=\frac{2Tcos\theta }{r\rho g}$

The radii of the two limbs r1 and r2 are 3.0 mm and 1.5 mm respectively

The level in the limb of diameter 6.0 mm is

$\\h_{1}=\frac{2Tcos\theta }{r_{1}\rho g}\\ h_{1}=\frac{2\times 7.3\times 10^{-2}cos0^{o}}{3\times 10^{-3}\times 10^{3}\times 9.8}\\h_{1}=4.97\times 10^{-3}m$

The level in the limb of diameter 3.0 mm is

$\\h_{2}=\frac{2Tcos\theta }{r_{2}\rho g}\\ h_{2}=\frac{2\times 7.3\times 10^{-2}cos0^{o}}{1.5\times 10^{-3}\times 10^{3}\times 9.8}\\h_{2}=9.93\times 10^{-3}m$

The difference in the heights is h2 - h1 = 4.96 mm

## NCERT solutions for class 11 physics chapter wise

 Chapter 1 NCERT solutions for class 11 physics chapter 1 Physical world Chapter 2 Solutions of NCERT for class 11 physics chapter 2 Units and Measurement Chapter 3 CBSE NCERT solutions for class 11 physics chapter 3 physics Motion in a straight line Chapter 4 NCERT solutions for class 11 physics chapter 4 Motion in a Plane Chapter 5 Solutions of NCERT for class 11 physics chapter 5 Laws of Motion Chapter 6 CBSE NCERT solutions for class 11 physics chapter 6 Work, Energy and Power Chapter 7 NCERT solutions for class 11 physics chapter 7 System of Particles and Rotational motion Chapter 8 Solutions of NCERT for class 11 physics chapter 8 Gravitation Chapter 9 CBSE NCERT solutions for class 11 physics chapter 9 Mechanical Properties of Solids Chapter 10 NCERT solutions for class 11 physics chapter 10 Mechanical Properties of Fluids Chapter 11 Solutions of NCERT for class 11 physics chapter 11 Thermal Properties of Matter Chapter 12 CBSE NCERT solutions for class 11 physics chapter 12 Thermodynamics Chapter 13 NCERT solutions for class 11 physics chapter 13 Kinetic Theory Chapter 14 Solutions of NCERT for class 11 physics chapter 14 Oscillations Chapter 15 CBSE NCERT solutions for class 11 physics chapter 15 Waves

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## Benefits of NCERT solutions for class 11 physics chapter 10 mechanical properties of fluids:

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