Have you ever wondered why water comes out quicker when you push a garden pipe nozzle, or how much heavier aeroplanes can fly in the air? All these common illustrations can be described in the most admirable manner in NCERT Solutions for Class 11 Physics Chapter 9 - Mechanical Properties of Fluids, where you can discover the exciting nature of gases and liquids.
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The chapter addresses important concepts like fluid pressure, the law of Pascal, viscosity, force of buoyancy, the principle of Archimedes, and surface tension, and hence will develop a solid theoretical understanding. The NCERT solutions are created by subject matter experts and adhere to the most up-to-date topics and CBSE syllabus of the current academic year of the syllabus. In addition to that, the student is also provided with significant formulae, solved problems and further tips that can be used in preparing and practising JEE, NEET and Olympiads. Students can also download NCERT Solutions of Class 11 Physics Chapter 9 in a very simple format of PDF and revise them anytime, anywhere. Those solutions not only reinforce exam preparation but they also relate physics concepts to a real-world context, which results in deeper comprehension and achievement.
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NCERT Solutions of Class 11 Physics Chapter 9 Mechanical Properties of Fluids present detailed knowledge behind each and every textbook problem, thus helping in understanding concepts such as pressure, buoyancy, viscosity, and surface tension. Each solution is prepared based on the current syllabus of CBSE and can be downloaded easily because they are available in PDF format only.
The NCERT Solutions given for Class 11 Physics Chapter 9: Mechanical Properties of Fluids are a clear and definite answer to the exercises in the textbook. These will assist students in reinforcing their knowledge of principles such as Pascal law, Archimedes' principle and Bernoulli's principle, facilitating sufficient preparation to write exams.
Q 9.1 (a) Explain why
The blood pressure in humans is greater at the feet 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 9.1 (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 9.1 (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 $\vec{F}=p\vec{A}$
Q 9.2 (a) Explain why
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 $\theta$
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.
$T_{SL}+T_{LA}cos\theta =T_{SA}$
$ cos\theta =\frac{T_{SA}-T_{SL}}{T_{LA}}$
In case of mercury $T_{SA}$ < $T_{SL}$ and therefore $cos\theta <0$ and therefore $\theta >\frac{\pi }{2}$ i.e the angle of contact of mercury with glass is obtuse.
Q 9.2 (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 9.2 (c) Explain why
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 9.2 (d) Explain why
Water with detergent dissolved in it should have small angles of contact.
Answer:
As we know, detergent with water rises very fast in the capillaries of clothes, which is only possible when the cosine of the angle of contact is large, i.e. angle of contact must be small.
Q 9.2 (e) Explain why
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 9.3 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 temperature.
(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 in 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 9.4 (a) Explain why
T 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 paper remains the same and that's why it remains horizontal.
Q 9.4 (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 9.4 (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 9.4 (d) Explain why
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 9.4 (e) Explain why
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 a 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
$\\Pressure=\frac{Force}{Area}\\ =\frac{490}{\pi \left ( \frac{1\times 10^{-2}}{2} \right )^{2}}\\ =6.24\times 10^{6}Pa$
Answer:
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 h w would be
$h_{w}=\frac{P}{\rho _{w}g}$
$h_{w}=\frac{1.01\times 10^{5}}{984\times 9.8}$
$h_{w}=10.474m$
Answer:
The density of water is $\rho _{w}=1000\ kg\ m^{-3}$
The 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.
Answer:
The Maximum Pressure that the piston would have to bear is
$\\P_{max}=\frac{Maximum\ weight\ of\ a\ car}{Area\ of\ the\ piston}\\ =\frac{3000\times 9.8}{425\times 10^{-4}}\\ =6.917\times 10^{5} Pa$
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.
$\\h_{w}\rho _{w}g=h_{s}\rho _{s}g$
$ \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.
Answer:
Let the difference in the levels of mercury in the two arms be hHg
$\begin{aligned}
& \left(\rho_w-\rho_s\right) g h=\rho_{H g} g h_{H g} \\
& \left(1-\frac{\rho_s}{\rho_w}\right) h=\frac{\rho_{H g}}{\rho_w} h_{H g} \\
& (1-0.8) \times 15=13.68 h_{H g} \\
& h_{H g}=\frac{0.2 \times 15}{13.6} \\
& h_{H g}=0.22 \mathrm{~cm}
\end{aligned}$
Q 9.11 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 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$
Reynold's 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.
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
$\begin{aligned}
& P_1+\rho g h+\frac{1}{2} \rho v_1^2=P_2+\rho g h+\frac{1}{2} \rho v_2^2 \\
& P_2-P_1=\frac{1}{2} \rho\left(v_1^2-v_2^2\right) \\
& \Delta P=\frac{1}{2} \times 1.3 \times\left(70^2-63^2\right) \\
& \Delta P=605.15 \mathrm{~Pa}
\end{aligned}$
The pressure difference between the regions below and above the wing is 605.15 Pa
The lift on the wing is F
$\begin{aligned}
& F=\Delta P \times \text { Area of wing } \\
& F=605.15 \times 2.5 \\
& F=1512.875 \mathrm{~N}
\end{aligned}$
Answer:
By the continuity equation, the velocity of the non-viscous liquid will be larger at the kink than at the rest of the tube, and therefore, pressure would be lower 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 $\mathrm{a}_1=8.0 \mathrm{~cm}^2$
The total area of the 40 fine holes is $a_2$
$\begin{aligned}
& a_2=40 \pi\left(\frac{d^2}{4}\right) \\
& a_2=40 \times \pi \times \frac{\left(10^{-3}\right)^2}{4} \\
& a_2=3.143 \times 10^{-5} \mathrm{~m}^2
\end{aligned}
$
Speed of liquid inside the tube is $\mathrm{v}_1=1.5 \mathrm{~m} \mathrm{~min}^{-1}$
Let the speed of ejection of fluid through the holes be $\mathrm{v}_2$
Using the continuity equation
$\begin{aligned}
& a_1 v_1=a_2 v_2 \\
& v_2=\frac{a_1 v_1}{a_2} \\
& v_2=\frac{8 \times 10^{-4} \times 1.5}{60 \times 3.143 \times 10^{-5}} \\
& v_2=0.636 \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}$
Answer:
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
$\begin{aligned}
T & =\frac{W}{l} \\
T & =\frac{1.5 \times 10^{-2}}{60 \times 10^{-2}} \\
T & =2.5 \times 10^{-2} \mathrm{Nm}^{-1}
\end{aligned}$
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 $4.5\times 10^{-2}\ N$.
Answer:
Surface Tension of Mercury is $T=4.65 \times 10^{-1} \mathrm{~N} \mathrm{~m}^{-1}$
The radius of the drop of Mercury is $r=3.00 \mathrm{~mm}$
Excess pressure inside the Mercury drop is given by
$\begin{aligned}
& \Delta P=\frac{2 T}{r} \\
& \Delta P=\frac{2 \times 4.65 \times 10^{-1}}{3 \times 10^{-3}} \\
& \Delta P=310 \mathrm{~Pa}
\end{aligned}$
Atmospheric Pressure is $P_0=1.01 \times 10^5 \mathrm{~Pa}$
Total Pressure inside the Mercury drop is given by
$\begin{aligned}
& P_T=\Delta P+P_0 \\
& P_T=310+1.01 \times 10^5 \\
& P_T=1.0131 \times 10^5
\end{aligned}$
Answer:
Excess Pressure inside a bubble is given by
$\Delta P=\frac{4 T}{r} \quad \text { (It's double the usual value because of the presence of } 2 \text { layers in case of soap bubble) }$
where $T$ is surface tension and $r$ is the radius of the bubble
$\begin{aligned}
& \Delta P=\frac{4 \times 2.5 \times 10^{-2}}{5 \times 10^{-3}} \\
& \Delta P=20 P a
\end{aligned}$
Atmospheric Pressure is $P_a=1.01 \times 10^{-5} \mathrm{~Pa}$
The density of soap solution is $\rho_s=1.2 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}$
The pressure at a depth of 40 cm (h) in the soap solution is
$\begin{aligned}
& P_s=P_a+\rho_s g h \\
& P_s=1.01 \times 10^5+1.2 \times 10^3 \times 9.8 \times 40 \times 10^{-2} \\
& P_s=1.01 \times 10^5+4704 \\
& P_s=1.057 \times 10^5 \mathrm{~Pa}
\end{aligned}$
Total Pressure inside an air bubble at that depth
$\begin{aligned}
P_t & =P_s+\frac{2 T}{r} \\
P_t & =1.057 \times 10^5+\frac{2 \times 2.5 \times 10^{-2}}{5 \times 10^{-3}} \\
P_t & =1.057 \times 10^5+10 \\
P_t & =1.0571 \times 10^5 \mathrm{~Pa}
\end{aligned}$
Class 11 Physics Chapter 9: Mechanical Properties of Fluids – HOTS Questions are designed to test deep conceptual understanding and application of principles like pressure, buoyancy, and fluid flow. These challenging questions enhance problem-solving skills and prepare students for competitive exams like JEE and NEET.
Q1: A particle of mass m is dropped inside a liquid medium which applies a bouncy force v and a drag force k.v². If acceleration due to gravity g is constant. What is the terminal velocity attained by the particle?
Answer:
From the figure,
$\begin{aligned} & m g=B+k v_t^2 \\ & m g-B=k v_t^2 \\ & v_t^2=\frac{m g-B}{k} \\ & v_t=\sqrt{\frac{m g-B}{k}}\end{aligned}$
Q2: For the arrangement shown in the figure, what is the density of oil?
Answer :
$\begin{aligned}
& \rho_0+\rho_w g l=\rho_0+\rho_{\text {oil }}(l+d) \cdot g \\
& \rho_{\text {oil }}=\frac{\rho_{\omega l}}{l+d}=\frac{1000 \times 135}{(135+12.3)} \\
& \rho_{\text {oil }}=916 \mathrm{~kg} / \mathrm{m}^3
\end{aligned}$
Q:3 In the given figure, the velocity $V_3$ will be
Answer:
From the continuity equation
$
\begin{aligned}
& A_1 V_1=A_2 V_2+A_3 V_3 \\
& \therefore \quad V_3=\frac{A_1 V_1-A_2 V_2}{A_3} \\
& V_3=\frac{(0.2 \times 4)-(0.2 \times 2)}{0.4} \\
& V_3=\frac{0.8-0.4}{0.4}=\frac{0.4}{0.4} \\
& V_3=1 \mathrm{~m} / \mathrm{s}
\end{aligned}
$
Q4: A tank filled with water (density $\rho_w$ = 1000 kg/m3) and oil (density $\rho_o$ = 900 kg/m3). The height of water is 1.0m and of the oil is 4.0m. Find the velocity of efflux through a hole at the bottom of the tank.
Answer :
The height of water which exerts the same pressure on the interface as whatever oil exerts:
$h \rho_w g=4 \rho_{o i l} g$
or
$h=\frac{4 * 900}{1000}=3.6 \mathrm{~m}$
Effective height of water over the hole:
$H = 1 + 3.6 = 4.6 m$
$\begin{aligned}
\therefore v_e & =\sqrt{2 g h}=\sqrt{2 * 10 * 4.6} \\
& =\sqrt{2 * 46}=\sqrt{96} \\
v_e & =9.5 \mathrm{~m} / \mathrm{s}
\end{aligned}$
Q5: The velocity of liquid coming out of a small hole of a vessel containing two different liquids of densities $2 \rho$ and $\rho$ as shown in the figure is-
Answer: With the help of Bernoulli's equation, we get-
$\begin{aligned}
P_0+\rho g(2 h)+(2 \rho) g h & =\frac{1}{2}(2 \rho) v^2+P_0 \\
P_0+4 \rho g h & =\rho v^2+P_0 \\
4 \rho g h & =\rho v^2 \\
v & =2 \sqrt{g h}
\end{aligned}$
NCERT Solutions for Class 11 Physics Chapter 9: Mechanical Properties of Fluids – The chapter is based on the concepts such as Pascal’s law, Archimedes’ principle, viscosity, surface tension, Bernoulli’s theorem, etc. These topics lay foundation to understand real world applications like fluid pressure in dam, flying of aircraft and blood circulation in living system.
9.1 Introduction
9.2 Pressure
9.2.1 Pascal’s law
9.2.2 Variation of pressure with depth
9.2.3 Atmospheric pressure and gauge pressure
9.2.4 Hydraulic machines
9.3 Streamline flow
9.4 Bernoulli’s principle
9.4.1 Speed of efflux: Torricelli’s law
9.4.2 Dynamic lift
9.5 Viscosity
9.5.1 Stokes’ law
9.6 Surface tension
9.6.1 Surface energy
9.6.2 Surface energy and surface tension
9.6.3 Angle of contact
9.6.4 Drops and bubbles
9.6.5 Capillary rise
To answer the NCERT Class 11 Physics Chapter 9: Mechanical Properties of Fluids questions, it is important to have a clear concept first about some of the basic laws, which include Pascal's law, Archimedes' principle, and Bernoulli's law. It is always best to begin by writing down what you are presented with (pressure, density, height, velocity, etc.) and what you need to find. Always sketch diagrams where possible in order to represent flows of fluids or forces that act upon bodies in fluids. Apply the right formula in a consistent manner, maintaining units in mind. In numerical problems, it is advisable to concentrate on the substitution of values step by step rather than going straight to the answer. To understand conceptual questions, compare the principle with real-life situations such as hydraulics, the buoyancy of a ship or the lift of a plane. By practising both theoretical and numerical questions regularly, it would be possible to solve any problem of fluid mechanics more easily and correctly.
NCERT textbook gives basic knowledge of the Mechanical Properties of Fluids in JEE/NEET preparation, but to perform better, students require advanced knowledge of the subject. The competitive exams require more knowledge on practical impact, higher order problem solving and mathematics on its application to concepts such as Bernoulli theorem, viscosity and surface tension.
NCERT Solutions for Class 11 Physics are designed to make learning simple and effective by providing clear, step-by-step answers to every chapter. These solutions not only help in preparing for school exams but also strengthen the base for competitive exams like JEE and NEET. Here you can access chapter-wise links for easy navigation and focused study.
Frequently Asked Questions (FAQs)
You can find Numericals on Mechanical Properties of Fluids Class 11 PDF from the above given link. These PDFs include step-by-step solutions to numerical problems based on pressure, viscosity, buoyancy, and other related concepts.
Yes, many websites offer the Mechanical Properties of Fluids Class 11 NCERT PDF for free. It usually includes theory, solved questions, and summary notes to help students understand the chapter better.
Pascal’s Law states that any change in pressure applied to an enclosed fluid is transmitted equally in all directions. It is used in hydraulic brakes and lifts.
Surface tension arises due to cohesive forces between liquid molecules at the surface, causing the liquid to behave like a stretched elastic sheet, as seen in water droplets forming spherical shapes.
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