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NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties Of Fluids

NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties Of Fluids

Edited By Vishal kumar | Updated on Jun 24, 2025 11:40 AM IST

The NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids are provided here to help students prepare effectively for their Class 11 final exams and competitive exams like JEE and NEET. These solutions include detailed answers to all exercise questions, along with important problems, important topics, important formulas, and what you need to learn extra for the competitive exam. These NCERT solutions are prepared by subject experts as per the latest CBSE Syllabus.

Chapter 9 of Class 11 Physics explains the mechanical properties of fluids, including the behaviour of liquids and gases, which can flow and are collectively called fluids. This chapter helps students understand key concepts like pressure, viscosity, and surface tension. To make learning easier, NCERT Solutions for class 11 physics are prepared in easy language by our expert faculty. Students can also download the solution PDF from the given below link.

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This Story also Contains
  1. NCERT Solutions for Class 11 Physics Chapter 9: Download PDF
  2. Mechanical Properties of Fluids Class 11 NCERT Solutions: Exercise Questions
  3. Class 11 Physics NCERT Chapter 9: Higher Order Thinking Skills (HOTS) Questions
  4. NCERT Solutions for Class 11 Physics Chapter 9: Topics
  5. Mechanical Properties of Fluids Class 11 NCERT Solutions: Important Formulas
  6. Approach to Solve the NCERT Class 11 Physics Chapter 9 Questions
  7. What Extra Should Students Study Beyond NCERT for JEE/NEET?
  8. NCERT Solutions for Class 11 Physics Chapter-Wise
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties Of Fluids
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties Of Fluids

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NCERT Solutions for Class 11 Physics Chapter 9: Download PDF

NCERT Class 11 Physics Chapter 9 Exercise Solutions PDF is available here for free download. These solutions cover all textbook exercise questions with clear, step by step explanations

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Mechanical Properties of Fluids Class 11 NCERT Solutions: Exercise Questions

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

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

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 F=pA

Q 9.2 (a) Explain why

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

Answer:

1650445487018

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 θ

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.

TSL+TLAcosθ=TSA
cosθ=TSATSLTLA

In case of mercury TSA < TSL and therefore cosθ<0 and therefore θ>π2 i.e the angle of contact of mercury with glass is obtuse.

Q 9.2 (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.)

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

(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 9.2 (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 the 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 9.2 (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 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 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 9.4 (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 9.4 (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

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

(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

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

(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 9.4 (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.

Q 9.5 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?

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=ForceArea=490π(1×1022)2=6.24×106Pa

Q 9.6 Torricelli's barometer used mercury. Pascal duplicated it using French wine of density 984 kg m3 . Determine the height of the wine column for normal atmospheric pressure.

Answer:

Atmospheric pressure is P=1.01×105Pa

The density of French wine ρw=984 kg m3

Height of the wine column h w would be

hw=Pρwg
hw=1.01×105984×9.8
hw=10.474m

Q 9.7 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.

Answer:

The density of water is ρw=1000 kg m3

Depth of the ocean is 3 km

The pressure at the bottom of the ocean would be

P=ρwgh
P=1000×9.8×3×1000
P=2.94×107Pa

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.

Q 9.8 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?

Answer:

Maximum Pressure which the piston would have to bear is

Pmax=Maximum weight of a carArea of the piston=3000×9.8425×104=6.917×105Pa

Q 9.9 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?

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.

hwρwg=hsρsg
ρsρw=hwhs
ρsρw=1012.5
ρsρw=0.8

Therefore the specific gravity of spirit is 0.8.

Q 9.10 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)

Answer:

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

(ρwρs)gh=ρHgghHg(1ρsρw)h=ρHgρwhHg(10.8)×15=13.68hHghHg=0.2×1513.6hHg=0.22 cm

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.

Q 9.12 Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation? Explain

Answer:

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

Q 9.13 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 × 103 kg s1 , what is the pressure difference between the two ends of the tube ? (Density of glycerine = 1.3 × 103 kg m3 and viscosity of glycerine = 0.83 Pa s). [You may also like to check if the assumption of laminar flow in the tube is correct].

Answer:

The volumetric flow rate of glycerine flow would be given by

V=Amount of glycerine flow per second Density of glycerine
V=Mρ
V=4×1031.3×103
V=3.08×106ms1

The viscosity of glycerine is η=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 η would flow through it with a volumetric rate of

V=πPr48ηl
P=8Vηlπr4
P=8×3.08×106×0.83π×(102)4
P=9.75×102 Pa

Reynolds number is given by

R=4ρVπdη
R=2ρVπrη
R=2×1.3×103×3.08×106π×0.01×0.83R0.3

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

Q 9.14 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 s1 and 63 ms1 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 m3

Answer:

The speed of air above and below the wings are given to be v1 = 70 m s1 and v2 = 63 m s1 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

P1+ρgh+12ρv12=P2+ρgh+12ρv22P2P1=12ρ(v12v22)ΔP=12×1.3×(702632)ΔP=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=ΔP× Area of wing F=605.15×2.5F=1512.875 N

Q 9.15 Figures (a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect? Why?

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.

Q 9.16 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 min1 , what is the speed of ejection of the liquid through the holes?

Answer:

Cross-sectional area of cylindrical tube is a1=8.0 cm2
The total area of the 40 fine holes is a2

a2=40π(d24)a2=40×π×(103)24a2=3.143×105 m2

Speed of liquid inside the tube is v1=1.5 m min1
Let the speed of ejection of fluid through the holes be v2
Using the continuity equation

a1v1=a2v2v2=a1v1a2v2=8×104×1.560×3.143×105v2=0.636 m s1

Q 9.17 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 × 102 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?

Answer:

Total weight supported by the film W=1.5×102N
Since a soap film has two surfaces, the total length of the liquid film is 60 cm .
Surface Tension is T

T=WlT=1.5×10260×102T=2.5×102Nm1

Q 9.18 Figure (a) shows a thin liquid film supporting a small weight = 4.5 × 102 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.

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 4.5×102 N .

Q 9.19 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 × 101 N m1 . The atmospheric pressure is 1.01 × 105 Pa. Also give the excess pressure inside the drop.

Answer:

Surface Tension of Mercury is T=4.65×101 N m1
The radius of the drop of Mercury is r=3.00 mm
Excess pressure inside the Mercury drop is given by

ΔP=2TrΔP=2×4.65×1013×103ΔP=310 Pa

Atmospheric Pressure is P0=1.01×105 Pa
Total Pressure inside the Mercury drop is given by

PT=ΔP+P0PT=310+1.01×105PT=1.0131×105

Q 9.20 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 (20C) is 2.50×102 N m1 ? 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 ).

Answer:

Excess Pressure inside a bubble is given by

ΔP=4Tr (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

ΔP=4×2.5×1025×103ΔP=20Pa

Atmospheric Pressure is Pa=1.01×105 Pa
The density of soap solution is ρs=1.2×103 kg m3
The pressure at a depth of 40 cm (h) in the soap solution is

Ps=Pa+ρsghPs=1.01×105+1.2×103×9.8×40×102Ps=1.01×105+4704Ps=1.057×105 Pa

Total Pressure inside an air bubble at that depth

Pt=Ps+2TrPt=1.057×105+2×2.5×1025×103Pt=1.057×105+10Pt=1.0571×105 Pa

Class 11 Physics NCERT Chapter 9: Higher Order Thinking Skills (HOTS) Questions

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 attended by the particle?


Answer:

From the figure,

mg=B+kvt2mgB=kvt2vt2=mgBkvt=mgBk

Q2: For the arrangement shown in the figure, what is the density of oil?

Answer :

ρ0+ρwgl=ρ0+ρoil (l+d)gρoil =ρωll+d=1000×135(135+12.3)ρoil =916 kg/m3

Q:3 In the given figure, the velocity V3 will be

Answer:

From continuity equation

A1V1=A2V2+A3V3V3=A1V1A2V2A3V3=(0.2×4)(0.2×2)0.4V3=0.80.40.4=0.40.4V3=1 m/s

Q4: A tank filled with water (density ρw = 1000 kg/m3) and oil (density ρo = 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 :

Height of water which exerts the same pressure on the interface as whatever oil exerts:

hρwg=4ρoilg

or

h=49001000=3.6 m

Effective height of water over the hole:

H=1+3.6=4.6m

ve=2gh=2104.6=246=96ve=9.5 m/s

Q5: The velocity of liquid coming out of a small hole of a vessel containing two different liquids of densities 2ρ and ρ as shown in the figure is-

Answer : With the help of Bernoulli's equation,we get-
P0+ρg(2h)+(2ρ)gh=12(2ρ)v2+P0P0+4ρgh=ρv2+P04ρgh=ρv2v=2gh

NCERT Solutions for Class 11 Physics Chapter 9: Topics

Topics of class 11 chapter 9 physics are listed below. These topics are very important for your class 11 exam as well as for competitive exams.

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

Mechanical Properties of Fluids Class 11 NCERT Solutions: Important Formulas

Important formulae are listed below. These formulas from Class 11 Physics Chapter 9 are crucial for board exams as well as competitive exams like JEE and NEET.

Pressure and Pascal's Law

  • Pressure, P=FA
  • Pascal's Law: Pressure applied to an enclosed fluid is transmitted equally in all directions.

Variation of Pressure with Depth

  • P=P0+ρgh (where P0 : atmospheric pressure, ρ : fluid density, g : gravity, h : depth)

Bernoulli's Theorem

  • P+12ρv2+ρgh= constant
  • Speed of Efflux (Torricelli's Law) v=2gh

Viscosity

  • Stoke's Law: F=6πηrv
    (where η : viscosity, r : radius, v : velocity)

  • Terminal Velocity vt=2r2(ρσ)g9η
    ( ρ : density of sphere, σ : density of fluid)

Surface Tension and Capillarity

  • Surface tension, T=Fl
  • Capillary rise, h=2Tcosθrρg
  • Excess pressure in drop, ΔP=2Tr
  • Excess pressure in bubble, ΔP=4Tr

Approach to Solve the NCERT Class 11 Physics Chapter 9 Questions

To solve questions from Mechanical Properties of Fluids, first note what is given, like pressure, height, density, or flow speed. Then choose the correct formula, such as Bernoulli’s equation, Pascal’s law, or the formula for viscosity. Use diagrams(FBD) if needed. Keep units consistent, apply the formulas step by step, and solving becomes much easier.

What Extra Should Students Study Beyond NCERT for JEE/NEET?

To score well in JEE/NEET, students must go beyond the NCERT in some parts of the Mechanical Properties of Fluids. Below is a comparison showing what NCERT covers and what needs extra focus.

NCERT Solutions for Class 11 Physics Chapter-Wise

NCERT Solutions for Class 11 Subject-wise

Also Check NCERT Books and NCERT Syllabus here

Subject wise NCERT Exemplar solutions

Frequently Asked Questions (FAQs)

1. What is the difference between solids, liquids, and gases in terms of fluid mechanics?

Solids have a fixed shape and volume, while fluids (liquids and gases) can flow and take the shape of their container. Gases, unlike liquids, can expand to fill any available space.

2. What is Bernoulli’s principle and its applications?

Bernoulli’s principle states that an increase in fluid speed leads to a decrease in pressure or potential energy. It is used in aerodynamics, carburettors, and medical applications like blood flow analysis.

3. What is Pascal’s Law and how is it applied?

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.

4. Why does surface tension occur in liquids?

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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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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