FOR MARCH 2018

EXAM

PHYSICS

thStrictly Based on the Latest CBSE Syllabus Dated 4 April 2017 for Academic Year 2017-18

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07

04

l Syllabus 5 – 8

l Solved Paper, 2017 (KVS) – Agra Region 9 – 19

l Solved Paper, 2017 (KVS) – Silchar Region 20 – 31

l Solved Paper, 2016 (KVS) – Agra Region ix – xix

l Solved Paper, 2016 (KVS) – Guwahati Region xx – xxix

l Solved Paper, 2016 (KVS) – Ahmedabad Region xxx – xi

l Solved Paper, 2015 (KVS) 9 – 24

Unit–I : Physical World and Measurement

1. Physical World 1 – 9

2. Units and Measurement 10 – 30

Unit–II : Kinematics

3. Motion in a Straight Line 31 – 53

4. Motion in a Plane 54 – 75

Unit–III : Laws of Motion

5. Laws of Motion 76 – 102

Unit–IV : Work, Energy and Power

6. Work, Energy and Power 103 – 123

Unit–V : Motion of System of Particles and Rigid Body

7. System of Particles and Rotational Motion 124 – 147

Unit–VI : Gravitation

8. Gravitation 148 – 167

Unit–VII : Properties of Bulk Matter

9. Mechanical Properties of Solids 168 – 182

10. Mechanical Properties of Fluids 183 – 207

11. Thermal Properties of Matter 208 – 225

Unit–VIII : Thermodynamics

12. Thermodynamics 226 – 244

Unit–IX : Behaviour of Perfect Gases and Kinetic Theory of Gases

13. Kinetic Theory 245 – 263

Unit–X : Oscillations and Waves

14. Oscillations 264 – 282

15. Waves 283 – 303

CONTENTS

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PREFACE

CBSE always believes in Global Trends of Educational Transformation. The CBSE

curriculum gets its lead from National Curriculum Framework – 2005 and Right to Free and

Compulsory Education Act – 2009. The aim of CBSE Curriculum is not just to let learners

obtain basic knowledge but to make them life-long learners. CBSE always updates and

reviews the syllabus to make it more relevant with educational transformation and in last few

years the chapters and topics which CBSE has added are very interesting and increase

practical knowledge.

Oswaal Question Banks are designed to nurture individuality and thus enhance one's

innate potentials which helps in increasing the self-study mode for students. This book

strengthens knowledge and attitude related to subject. It is designed in such a way that students

can set their own goals and can improve their problem solving and thinking skills.

The journey of this book is never ending as this book is reviewed every year and

new questions, previous year's examination questions, new HOTS or any change in

syllabus is updated time to time. Also regular review and readers’ feedback increases the

efficiency of this book gradually.

Moreover every Question Bank strictly follows the latest syllabus and pattern, and

contains more than sufficient questions and brief description of chapters, which help students in

practicing and completing the syllabus. Questions incorporated in this Question Bank

encompass all the ‘Typologies’ mentioned by CBSE namely Remembering, Understanding,

Application, High Order Thinking Skills and Evaluation. Solutions for these have been checked

twice and efforts have been made to align them closely to the Marking Scheme. Practically, this

book provides students everything they need to learn and excel.

At last we would like to thank our authors, editors, reviewers and specially students who

regularly send us suggestions which helps in continuous improvement of this book and makes

this book stand in the category of “One of the Best”. Wish you all Happy Learning.

–Publisher

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Latest Syllabus for Academic Year 2017-18

Time : 3 Hours Max. Marks : 70

No. of Periods Marks� �

Unit-I Physical World & Measurement

Chapter-1: Physical World 10

Chapter-2: Units and Measurements

Unit-II Kinematics 23 Chapter-3: Motion in a Straight Line 24

Chapter-4: Motion in a Plane

Unit-III Laws of Motion 14 Chapter-5: Laws of Motion

Unit-IV Work, Energy and Power 12

Chapter-6: Work, Energy and Power

Unit-V Motion of System of particles and Rigid Body 17 18 Chapter-7: System of particles and Rotational Motion

Unit-VI Gravitation 12

Chapter-8: Gravitation

Unit-VII Properties of Bulk Matter

Chapter-9: Mechanical Properties of Solids 24

Chapter-10: Mechanical Properties of Fluids

Chapter-11: Thermal Properties of Matter

Unit-VIII Thermodynamics 2012

Chapter-12: Thermodynamics

Unit-XI Behaviour of Perfect Gases and Kinetic

Theory of gases 08

Chapter-13: Kinetic Theory

Unit-X Oscillations and Waves

Chapter-14: Oscillations 26 10

Chapter-15: Waves

Total 160 70

Unit-I : Physical World and Measurement 10 Periods

Chapter-1: Physical World : Physics-scope and excitement; nature of physical laws; Physics, technology and society.

Chapter-2: Units and Measurements : Need for measurement: Units of measurement; systems of units; SI units, fundamental and derived units. Length, mass and time measurements; accuracy and precision of measuring instruments; errors in measurement; significant figures.

Dimensions of physical quantities, dimensional analysis and its applications.

Unit-II : Kinematics 24 Periods

Chapter-3: Motion in a Straight Line : Frame of reference, Motion in a straight line: Position-time graph, speed and velocity.

}}

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Elementary concepts of differentiation and integration for describing motion, uniform and non-uniform motion, average speed and instantaneous velocity, uniformly accelerated motion, velocity-time and position-time graphs.

Relations for uniformly accelerated motion (graphical treatment).

Chapter-4 : Motion in a Plane : Scalar and vector quantities; Position and displacement vectors, general vectors and their notations; equality of vectors, multiplication of vectors by a real number; addition and subtraction of vectors, relative velocity. Unit vector; resolution of a vector in a plane - rectangular components. Scalar and Vector product of vectors.

Motion in a plane, cases of uniform velocity and uniform acceleration-projectile motion, uniform circular motion.

Unit-III : Laws of Motion 14 Periods

Chapter-5: Laws of Motion : Intuitive concept of force. Inertia, Newton’s first law of motion; momentum and Newton’s second law of motion; impulse; Newton’s third law of motion.

Law of conservation of linear momentum and its applications.

Equilibrium of concurrent forces. Static and kinetic friction, laws of friction, rolling friction, lubrication.

Dynamics of uniform circular motion : Centripetal force, examples of circular motion (vehicle on a level circular road, vehicle on a banked road).

Unit IV : Work, Energy and Power 12 Periods

Chapter-6: Work, Energy and Power : Work done by a constant force and a variable force; kinetic energy, work-energy theorem, power.

Notion of potential energy, potential energy of a spring, conservative forces: conservation of mechanical energy (kinetic and potential energies); non-conservative forces: motion in a vertical circle; elastic and inelastic collisions in one and two dimensions.

Unit V : Motion of System of Particles and Rigid Body 18 Periods

Chapter-7: System of Particles and Rotational Motion : Centre of mass of a two-particle system, momentum conservation and centre of mass motion.

Centre of mass of a rigid body; centre of mass of a uniform rod.

Moment of a force, torque, angular momentum, laws of conservation of angular momentum and its applications.

Equilibrium of rigid bodies, rigid body rotation and equations of rotational motion, comparison of linear and rotational motions.

Moment of inertia, radius of gyration, values of moments of inertia, for simple geometrical objects (no derivation). Statement of parallel and perpendicular axes theorems and their applications.

Unit-VI : Gravitation 12 Periods

Chapter-8: Gravitation : Keplar’s laws of planetary motion, universal law of gravitation.

Acceleration due to gravity and its variation with altitude and depth.

Gravitational potential energy and gravitational potential, escape velocity, orbital velocity of a satellite, Geo-stationary satellites.

Unit-VII : Properties of Bulk Matter 24 Periods

Chapter-9: Mechanical Properties of Solids : Elastic behaviour, Stress-strain relationship, Hooke’s law, Young’s modulus, bulk modulus, shear modulus of rigidity, Poisson’s ratio; elastic energy.

Chapter-10: Mechanical Properties of Fluids : Pressure due to a fluid column; Pascal’s law and its applications (hydraulic lift and hydraulic brakes), effect of gravity on fluid pressure.

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Viscosity, Stokes’ law, terminal velocity, streamline and turbulent flow, critical velocity. Bernoulli’s theorem and its applications.

Surface energy and surface tension, angle of contact, excess of pressure across a curved surface, application of surface tension ideas to drops, bubbles and capillary rise.

Chapter-11 : Thermal Properties of Matter : Heat, temperature, thermal expansion; thermal expansion of solids, liquids and gases, anomalous expansion of water; specific heat capacity; Cp, Cv - calorimetry; change of state - latent heat capacity.

Heat transfer-conduction, convection and radiation, thermal conductivity, Qualitative ideas of Blackbody radiation, Wein’s displacement Law, Stefan’s law, Green house effect.

Unit-VIII : Thermodynamics 12 Periods

Chapter-12: Thermodynamics : Thermal equilibrium and definition of temperature (zeroth law of thermodynamics), heat, work and internal energy. First law of thermodynamics. Isothermal and adiabatic processes.

Second law of thermodynamics: reversible and irreversible processes, heat engine and refrigerator.

Unit-IX : Behaviour of Perfect Gases and Kinetic Theory of Gases 08 Periods

Chapter-13: Kinetic Theory : Equation of state of a perfect gas, work done in compressing a gas.

Kinetic theory of gases - assumptions, concept of pressure. Kinetic interpretation of temperature; rms speed of gas molecules; degrees of freedom, law of equi-partition of energy (statement only) and application to specific heat capacities of gases; concept of mean free path, Avogadro’s number.

Unit-X : Oscillations and Waves 26 Periods

Chapter-14: Oscillations : Periodic motion - time period, frequency, displacement as a function of time, periodic functions. Simple harmonic motion (S.H.M) and its equation; phase; oscillations of a loaded spring-restoring force and force constant; energy in S.H.M. Kinetic and potential energies; simple pendulum derivation of expression for its time period.

Free, forced and damped oscillations (qualitative ideas only), resonance.

Chapter-15: Waves : Wave motion. Transverse and longitudinal waves, speed of wave motion, displacement relation for a progressive wave, principle of superposition of waves, reflection of waves, standing waves in strings and organ pipes, fundamental mode and harmonics, Beats, Doppler effect.

PRACTICALS Total Periods : 60

The record, to be submitted by the students, at the time of their annual examination, has to include :l Record of at least 15 Experiments [with a minimum of 6 from each section] to be performed by

the students.l Record of at least 5 Activities [with a minimum of 2 each from section A and section B], to be

demonstrated by the teachers.l Report of the project to be carried out by the students.

EVALUATION SCHEME

Time Allowed : Three hours Max. Marks : 30

Two experiments one from each section 8+8 Marks

Practical record (experiment and activities) 6 Marks

Investigatory Project 3 Marks

Viva on experiments, activities and project 5 Marks

Total 30 Marks

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Time 3 Hours Max. Marks: 70

S. Typology of Questions Very Short Short Value Long Total % No. Short Answer-I Answer -II based Anser Marks Weigh- Answer (SA-I) (SA-II) question (L.A.) tage (VSA) (2 marks) (3 marks) (4 marks) (5 marks) (1 mark) 1 Remembering- (Knowledge based Simple recall questions, to know specific facts, terms, 2 1 1 - - 7 10% concepts, principles, or theories, Identify, define, or recite, information) 2 Understanding- (Comprehension -to be familiar with meaning and to understand - 2 4 - 1 21 30% conceptually, interpret, compare, contrast, explain, paraphrase information)

3 Application (Use abstract information in concrete situation, to apply knowledge to new situations, Use given - 2 4 - 1 21 30% content to interpret a situation, provide an example, or solve a problem)

4 High Order Thinking Skills ( Analysis & Synthesis- Classify, compare, contrast, or differentiate between 2 - 1 - 1 10 14% different pieces of information, Organize and/or integrate unique pieces of information from a variety of sources)

5 Evaluation - (Appraise, judge, and /or justify the value or worth of a decision 1 - 2 1 - 11 16% or outcome, or to predict outcomes based on values)

TOTAL 5×1=5 5×2=10 12×3=36 1×4=4 3×5=15 70(26) 100%

QUESTION WISE BREAK UP Type of Question Mark per Total No. of Total Question Questions Marks

VSA 1 5 05

SA-I 2 5 10

SA-II 3 12 36

VBQ 4 1 04

LA 5 3 15

Total 26 70

1. Internal Choice: There is no overall choice in the paper. However, there is an internal choice in one question of 2 marks weightage, one question of 3 marks weightage and all the three questions of 5 marks weightage.

2. The above template is only a sample. Suitable internal variations may be made for generating similar templates keeping the overall weightage to different form of questions and typology of questions same.

Question Paper Design (2017-18)Physics (Code No. 042)

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KENDRIYA VIDYALAYA SANGATHAN[AGRA REGION]

SESSION ENDING EXAMINATION 2016-17SUBJECT : PHYSICS [042]

CLASS–XI(SOLVED PAPER)

Time : 3 Hrs. M.M. : 70

General Instructions : (i) All the questions are compulsory.

(ii) Q.No. 1 to 5 are very short answer type questions, carrying one mark each.

(iii) Q. No. numbers 6 to 10 are short answer type questions, carrying two marks each.

(iv) Q.No. 11 to 22 are also short answer type questions, carrying three marks each.

(v) Q.No. 23 is a value based questions, carrying four marks.

(vi) Q.No. 24 to 26 are long answer type questions, carrying five marks each.

(vii) There is no overall choice. However, an internal choice has been provided in one question of 2 marks, one question of 3 marks and all their questions of 5 marks each. You have to attempt only one of the given choices in such questions.

(viii) Use of calculators is not permitted.

(ix) You may use the following values of physical constants wherever necessary.

Earth's radius Re = 6.38 × 106 m, c=3 × 108 m/s, h=6.6 × 10–34 Js, e=1.6 × 10–19 C, NA = 6.023 × 1023 /mole mn = 1.67 × 10–27 kg, me = 9 × 10–31 kg, k(Boltzmann constant) = 1.38 × 1023 JK–1 and g = 9.8 m/s2

1. Name the physical Quantity associated with the dimension M1 L2 T–2.

2. Why does a gun recoils when bullet is fired from it.

3. Define efficiency of a heat engine.

4. A particle is moving with a uniform velocity. What is the net force on it.

5. The slope of an adiabatic process is greater than an isothermal process. Give reason.

6. Why a cricket player lower his hands while catching a cricket ball, Explain.

7. State and prove work energy theorem.

8. A person goes to a height equals to radius of earth above earth's surface. What would be the change in the weight of the person.

9. On what principle working of hydraulic brakes is based. State that principle.

OR

What do you mean by terminal velocity. Derive its expression.

10. A ball mass 5 kg moving with a velocity of 15 m/s under goes a head-on collision with another ball of unknown mass at rest after collision it rebounds with a velocity of 5 m see. Find the mass of the other ball.

11. Obtain an expression for maximum speed with which a vehicle can negotiate a curve on a banked circular road without skidding. Draw suitable diagram showing various forces acting on the body negotiating the curve.

12. A bullet of mass .012 kg and horizontal speed 70 m/s strikes a block of wood of mass 0.4 kg using and instantly comes to rest with respect to the block is suspended from the celling by means of thin wires. Calculate the height to which the block rises. Also estimate the amount of heat produced in the block.

13. Sate :

(i) Theorem of parallel axis.

(ii) Theorem of perpendicular axis.

14. Three bodies a ring, a solid cylinder and a solid sphere roll down the same inclined plane without slipping. The start from rest, the radii of the bodies are identical. Which of the body reaches the ground with maximum velocity. Explain.

10 ] OSWAAL CBSE Question Bank Chapterwise Solutions, Physics-XI

15. Prove that a gun will shoot three times as high when its angle of elevation is 60 degree as when it is shoot at 30 degree, but cover the same horizontal range.

16. Draw strain stress curve for an elastic metal wire. Explain (i) Proportional limit (ii) Permanent set.

17. Discuss elastic collision in one dimension. Obtain expression for velocities of the two bodies after such a collision.

18. Prove that the average kinetic energy of a molecule of an ideal gas is directly proportional to the absolute temperature of the gas.

OR

Write the expression for pressure exerted by a gas containing n particles each of mass m in a container of volume V. Using this relation show that root mean square velocity of molecules of a gas is directly proportional to the square root of absolute temperature of gas.

19. A 400 kg. satellite is in a circular orbit of radius 2 RE about the earth. How much energy is required to transfer it to a circular orbit of radius 4Re ? What are the changes in the kinetic and potential energy.

20. What is heat engine ? Explain the principle and working of a heat engine. Define its efficiency.

21. Show that the ratio of the frequency of first three harmonics in open pipe is 1 : 2 : 3. A pipe 30 cm, long, is open at both ends. Which harmonic mode of the pipe resonates 1.1 KHz sources.

22. Two parallel rail tracks run north south. Train A moves north with a speed of 54 km/h, and train B moves south with a speed of 90 km/h. What is the

(a) velocity of B with respect to A

(b) velocity of ground with respect to B

(c) velocity of a monkey running on the roof of the train ‘A’ against its motion with a velocity of 18 km/h with respect to train A as observed by a man standing on the ground.

23. Once Arun was going to his house. He was listening music on mobile with earphone while crossing the railway line he was engaged in the music that he did not heard the sound of approaching train though the train was blowing horn. A person nearby ran towards him and post away just as the train reached there. Arun realized his mistake and thanked the person.

(a) Write the values possessed by the person.

(b) Name the phenomenon of change in frequency of sound when there is relative motion between the observer and source of sound.

(c) Apparent frequency is more or less than the original frequency when source comes closer to the observer.

24. What is projectile ? Obtain an expression for :

1. Maximum height

2. Time of flight

3. Horizontal range when a projectile is fired at an angle θ with the horizontal.

OR

State Parallelogram law of vector addition. Find analytically the magnitude and direction of resultant vector when.

1. Two vectors are parallel to each other.

2. Two vectors are perpendicular to each other.

25. State and prove Bernoulli’s theorem. Give two applications.

OR

What is the surface energy ? Find the relation between surface tension and surface energy.

Explain why

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

(b) Detergents should have small angle of contact.

26. Derive an expression for the kinetic and potential energies of a harmonic oscillator.

Hence, show that total energy is conserved in SHM.

OR

Obtain Discuss Newton's formula for velocity of sound in air medium and apply Laplace's correction.

Solved Paper - 2017 [ 11

SOLUTIONS 1. Work / Energy / Moment of force.

2. When a gun is fired, it exerts a forward force on the bullet. The bullet exerts an equal and opposite reaction force on the gun. This results in the recoil of the gun.

3. Efficiency of a heat engine is defined as the ratio of net work done per cycle by the engine to the total amount of heat absorbed per cycle by the working substance from the source. It is denoted by h.

h = 1 –

QQ

2

1

4. As body is moving with uniform velocity, its, acceleration is zero. Force acting on the body F = ma, where m is mass and a is the acceleration. Since, the acceleration is zero, the net force acting on the body is zero.

5. Slope given by – dP/dV

For isothermal process – PV = K

Differentiating both sides, we get

P. dV + V. dP = 0

V. dP = –P. dV

dd

PV

= –P/V

...(i),

For adiabatic change

PVg = K

Differentiating both sides

gPVg–1 dV + Vg dP = 0

Vg dp = –gP/V ...(ii)

From eq. (i) & (ii)

We find that slope (dP/dV) of an adiabatic curve is g time the slope of (dP/dV) of an isothermal curve. As g is greater than 1 therefore slope of adiabatic process is greater than an isothermal process.

6. He allows a longer time for his hands to stop the ball. As impulse = Force × Time = change in linear momentum. Player has to apply smaller, force against ball to stop it. The ball in turn exerts a smaller force on his hands and his hands are not injured. Therefore, force not only depends on change in momentum but also on how fast the change is brought about. The same change in momentum brought about in shorter time needs greater force and vice-versa.

7. Work-Energy Theorem :

Statement : Work done by net force in displacing a body is equal to change in kinetic energy of the body.

Proof : Consider

m = mass of body

u = initial velocity of body

F = force applied on the body along its direction of motion.

a = acceleration produced in the body

v = final velocity produced in the body.

dW = F (ds) small amount of work done by applied force on the body.

Where ds = small distance moved by the body in the direction of the force applied

F = ma = m

dvdt

= m

dsdt

dv = mvdv dsdt

v=

Total work done by applied force on the body in increasing its velocity from u to v is

W =

u

v

∫ mv dv = m u

v

∫ vdv =

m

v2

u

v

W =

12

m (v2 – u2) =

12

2mv −

12

2mu

12

mv2 = Kf = final K.E. of body

12

mu2 = Ki = initial K.E. of body

i.e. work done on body = increase in K.E. of body.

8. At surface of earth, weight, W = mg = GMm /R2 (h = R) at height h,

W' = mg' = GMmR( )+ h 2

(R = radius of earth)

= GMmR R( )+ 2

WW

'

=

RR

2

22( )

WW

'

=

14

W' = W4

9. Pascal's Law :

Statement : If states that if gravity effect is neglected, the pressure at every point of liquid in equilibrium of rest is same.

12 ] OSWAAL CBSE Question Bank Chapterwise Solutions, Physics-XI

OR

Terminal velocity : It is the maximum constant velocity acquired by the drop while falling in a viscous medium. Terminal velocity of a drop of radius, density r moving through viscous medium of viscosity η & density r6 is given by–

V =

29

×

r2

η (r – rs) g

Thus, the value of V depends upon r, r, r6 and h. 10. m1 = 5 kg, V2 = 0, u1 = 15 m/s, V1 = 5 m/s, m2 = ?

V1 =

( )m m um m

1 2 1

1 2

−+

+

2 2 2

1 2

m um m+

5 =( )5 15

52

2

−+mm

+ 0

5 (5 + m2) = 15 (5 – m2)

25 + 5m2 = 75 – 15m2

(5 + 15) m2 = 75 – 25

20m2 = 50

m2 =

5020

= 2.5 kg.

m2 = 2.5 kg.

11. Weight (mg) of the vehicle going over the banked road acts vertically downwards. Normal reaction R of the banked road acts upwards in a direction perpendicular to OA. Therefore, R cos θ, acting vertically upwards and R sin θ acting along the horizontal towards centre of the circular track. If v is velocity of vehicle over the banked circular road of radius r, then centripetal force required = mv2/r.

R cos θ balances the weight in equilibrium –

R cos θ = mg ....(i)

Necessary centripetal force provides by R sin θ

R sin

θ =

mvr

2

...(ii)

Dividing (i) by (ii), we get

RR

sincos

θθ

=

mvrmg

2

tan θ = V2

rg ...(iii)

The speed limit at which the curve can be negoti-ated safely is :

V = rg tan θ

If µ is coefficient of friction between the tyres and the road, safe value of speed limit–

V =

rg(tan )tanθ µ

µ θ+

−1

q

q

R sin q

R cos qR

mg

A

XB

Outeredge

raisedO

12. Given that

Initial speed of bullet u = 70 m/s

Suppose the final speed of block and bullet which move together = v m/s

According to law of conservation of momentum,

m u = (m+M) v

Þ v =

mum( )M +

M = 0.4 kg, m = 0.012 kg

\ v =

0 012 700 412

..

×

= 2.04 m/s

There will be loss in kinetic energy of block and bullet, which is equal to the gain in potential energy after collision.

12

(m+M)v2 = (m+M) gh (block rises to height h)

h = v2/2g =

( . ).

2 042 9 8

2

×

= 0.212 m

Now, loss of K.E. will be equal to heat produced in the block.

Thus, heat produced = K.E. of bullet – K.E. of block and bullet

=

12

12

2mu − (m+M)v2

=

12

0 012 7012

0 412 2 042 2× × − × ×. ( ) . ( . )

= 28.543 Joule.

13. Theorem of Parallel Axes : According to this theorem, moment of inertia of a rigid body about any axis, AB is equal to moment of inertia of the body about another axis KL passing through centre of mass C of the body in a direction parallel to AB, plus the product of total mass M of the body and square of the perpendicular distance between the two parallel axes, i.e.

IAB = TKL + Mh2

Solved Paper - 2017 [ 13

A K

B L

Oh mi

Cri

Theorem of Perpendicular Axes : According to this theorem, the M.I. of plane lamina about any axis OZ perpendicular to the plane of the lamina is equal to sum of moments of inertia of lamina about any two mutually, perpendicular axes OX and OY in the plane of lamina, meeting at a point where the given axis OZ passes through the lamina, i.e.

Iz = Ix + Iy

Z

X

Y

Iy

Ix

Iz

O

14. As. we know, energy of a rolling body is conserved. So,

Potential Energy lost by the body = kinetic energy gained

As body starts from rest, the K.E. gained will be equal to final K.E. of the body

K.E. of rolling body, K =

12

mv2

1

2

2+

KR

...(i)

Here, V = final velocity of body K = radius of gyration Now, with lost potential energy,

mgh =

12

122

2mv +

KR

Þ V2 =

2

12

2

ghKR

+

For a ring, K2 = R2

\ Vring =

21 1gh+

= gh

For solid cylinder, k2 =

R2

2

\ Vcylinder =

2

112

43

gh gh

+=

For solid sphere, K2 =

25

2R

\ V sphere =

2

125

107

gh gh

+=

Thus, the sphere has the greatest velocity.

15. When θ = 60°, the max height

H1 =

ug

2 2 602

sin °

=

ug

2 23 22

( / )

=

38

2ug

Horizontal range, R1 =

ug

2 2 60sin × °

=

ug

2

sin 120° =

ug

2 3 2/

=

32

2ug

When θ = 30°, max height,

H2 =

ug

2 2 302

sin °

=

ug

2 21 22

( / )

=

ug

2

8

Horizontal range, R2 =

ug

2

sin 2 × 30°

=

ug

2

sin 60° =

u

g

2 3 2× ( )/

=

32

2ug

∴ H1 = 3H2 and R1 = R2. 16. (i) Up to point P, Hooke's law is obeyed because

the graph is a straight line from O to P. As strain is proportional to stress upto P, hence OP is called proportional limit.

Str

ess

Stress-strain curvefor metallic wire C

B

AE

P

O StrainO'

(ii) strain varies directly proportional to load up to point P, strain increases by greater amount as compared to first case i.e. from O to P, for a given increase in load. Beyond the elastic limit E, the curve does not retrace backwards as the wire is unloaded but returns along dotted line AO'. Point O' corresponds to strain at zero load, which shows that there is a permanent strain in the wire. Hence OO' is the permanent set.

17. Elastic collision in one dimension : When two bodies moving initially along the same straight line, striking against each other without loss of kinetic energy and continuing to move along the same straight line after collision

14 ] OSWAAL CBSE Question Bank Chapterwise Solutions, Physics-XI

A

m1

u1

Before collision

B

m2u2

A

m1

During

collision

B

m2

A

m1

v1

After collision

B

m2v2

Relative velocity of approach before collision when u1 > u2,

= u1 – u2

when v2 > v1, Relative velocity of separation after collision

= v2 – v1

Linear momentum of two balls before collision = m1 u1 + m2 u2

Linear momentum after collision = m1 v1 + m2 v2

As linear momentum is conserved m1v1 + m2v2 = m1u1 + m2u2 ...(1) m2(v2 – u2) = m1 (u1 – v1) ...(2) Total kinetic energy of two balls before collision

=

12

m1u12 +

12

m2u22 ...(3)

Total K.E. after collision

=

12

m1v12 +

12

m2v22 ...(4)

As K.E. is also conserved,

\

12

m1v12 + 1

2 m2v2

2 =

12

m1u12 + 1

2 m2u2

2

m2(v22 – u2

2) = m1 (u12 – v1

2) ...(5)

Dividing (5) by (2), we get

m v um v u

2 22

22

2 2 2

( )( )

−−

=

m u vm u v

1 12

12

1 1 1

( )( )

−−

( )( )( )

v u v uv u

2 2 2 2

2 2

+ −−

=

( )( )( )

u v u vu v

1 1 1 1

1 1

+ −−

v2 + u2 = u1 + v1

v v u u2 1 1 2− = − ...(6)

Hence, in 1-Dimensional elastic collision relative velocity of separation is equal to the relative velocity of approach before collision

v vu u

2 1

1 1

−−

= e = 1

Calculation of velocities :

From equation – (6)

v2 = u1 – u2 + v1 ...(7)

Putting this value in equation (1).

m1v1 + m2 (u1 – u2 + v1) = m1u1 + m1u2

m1v1 + m2u1 – m2u2 + m2v1 = m1u1 + m2u2

v1 (m1 + m2) = (m1 – m2) u1 + 2 m2 u2

Velocity of A

vm m um m

m um m1

1 2 1

1 2

2 2

1 2

2=

−+

++

( ) ...(8)

Put this value of v1 from (8) in (7)

v2 = u1 – u2 +

( )m m um m

m um m

1 2 1

1 2

2 2

1 2

2−+

++

= u1 1

211 2

1 22

2

1 2+

−+

+

+−

m mm m

um

m m

= u1 m m m m

m mu

m m mm m

1 2 1 2

1 22

2 1 2

1 2

2+ + −+

+

− −+

Velocity of B,

vm u

m mm m um m2

1 1

1 2

2 1 2

1 2

2=

++

−+

( )

...(9)

18. Proof : Average K.E. of a molecule of an ideal gas is directly proportional to the absolute temperature of the gas.

Consider one gram mole of an ideal gas occupying a volume V at temperature T. Let m be the mass of each molecule of the gas, then

M = m × NA

Here, NA = Avogadro's Number If C is the r.m.s. velocity of the gas molecules, then

pressure P exerted by ideal gas is

P CMV

C

PV MC

= =

=

13

13

13

2 2

2

ρ

From perfect gas equation PV = RT

13

MC2 = RT

MC2 = 3RT Dividing both sides by 2 we get

12

MC2 =

32

RT

Average translation K.E. of one mole of gas

12

MC2 =

32

RT.

12

mNAC2 =

32

RT. (M = n NA)

12

m C2 =

32

RNA

T =

32

KB T—(1)

KR

NBA

=

Where KB is Boltzmann constant. \ Average K.E. of translation per molecule of a

gas =

32

KB T.

As, KB = Boltzmann's constant

So, Average K.E. a TOR

Same as above, As m= mass = constant, KB = constant

C T Here C = r. m. s. of molecules & T is absolute

temperature.

Oswaal CBSE Chapterwise/TopicwiseQuestion Bank For Class 11 Physics

(Mar. 2018 Exam)

Publisher : Oswaal Books ISBN : 9789386339676 Author : Panel Of Experts

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