Sample PDF of NEET UG, JEE Main Challenger Physics Vol 2 …

Challenger

PhysicsVol. II

NEET – UG & JEE (Main)

Printed at: Print to Print, Mumbai

© Target Publications Pvt. Ltd. No part of this book may be reproduced or transmitted in any form or by any means, C.D. ROM/Audio Video Cassettes or electronic, mechanicalincluding photocopying; recording or by any information storage and retrieval system without permission in writing from the Publisher.

P.O. No. 1591TEID: 1884

For all Medical and Engineering Entrance Examinations held across India.

Salient Features

Concise theory for every topic. Eclectic coverage of MCQs under each sub-topic. ‘1908’ MCQs including questions from previous NEET and JEE examinations. ‘78’ Numerical Value Type (NVT) questions including selected questions from

JEE (Main) 2020 Includes selective solved MCQs upto NEET-Phase I 2020, JEE (Main) 2020. Includes NEET-Phase II 2020 and JEE (Main) 2020 2nd September (Shift - II)

Question Paper and Answer Key along with Hints. Multiple Study Techniques to Enhance Understanding and Problem Solving. Inclusion of ‘Problems To Ponder’ to engage students in scientific enquiry. Hints provided wherever deemed necessary.

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The journey to create a complete book is strewn with triumphs, failures and near misses. If you think we’ve nearly missed something or want to applaud us for our triumphs, we’d love to hear from you.

Please write to us on: [email protected]

Disclaimer This reference book is based on the NEET-UG and JEE (Main) syllabus prescribed by National Testing Agency (NTA). We the publishers are making this reference book which constitutes as fair use of textual contents which are transformed by adding and elaborating, with a view to simplify the same to enable the students to understand, memorize and reproduce the same in examinations. This work is purely inspired upon the course work as prescribed by the National Council of Educational Research and Training (NCERT). Every care has been taken in the publication of this reference book by the Authors while creating the contents. The Authors and the Publishers shall not be responsible for any loss or damages caused to any person on account of errors or omissions which might have crept in or disagreement of any third party on the point of view expressed in the reference book. © reserved with the Publisher for all the contents created by our Authors. No copyright is claimed in the textual contents which are presented as part of fair dealing with a view to provide best supplementary study material for the benefit of students.

PREFACE

‘Challenger Physics Vol - II’ is a compact guidebook, extremely handy for preparation of various competitive exams like NEET, JEE (Main). This edition provides an unmatched comprehensive amalgamation of theory with MCQs. The chapters are aligned with the syllabus for NEET (UG) and JEE (MAIN) examinations and runs parallel to NCERT curriculum. The book provides the students with scientifically accurate context, several study techniques and skills required to excel in these examinations.

In this book the Theoretical Concepts are presented in the form of pointers, tables, charts and diagrams that form a vital part of preparation any competitive examination.

Multiple Choice Questions have been specially created and compiled with the following objective in mind – to help students solve complex problems which require strenuous effort and understanding of multiple-concepts. The assortment of MCQs is a beautiful blend of questions based on higher order thinking, theory, and multiple concepts.

MCQs in each chapter are segregated into following sections. • Concept Building Problems section is designed to boost prerequisite understanding of concepts.

• Practice Problems section contains questions crafted for thorough revision .

• Numerical Value Type section cater to newly added NVT questions in JEE (MAIN).

• Problems to Ponder section offers questions of diverse pattern created to instill the attitude ofconcentrating on the problems and to understand the application of various concepts in Physics.

All the features of this book pave the path of a student to excel in examination.The features are designed keeping the following elements in mind: Time management, easy memorization or revision and non-conventional yet simple methods for MCQ solving. To keep students updated, selected questions from most recent examinations of NEET 2020 and JEE (Main) 2020 are covered exclusively.

NEET-UG 2020 (Phase II) and JEE (Main) 2020 2nd SEPTEMBER (Shift - II) Question Papers and Answer Keys have been provided to offer students glimpse of the complexity of questions asked in entrance examination. The paper has been split unit-wise to let the students know which of the units were more relevant in the latest examination.

We hope the book benefits the learner as we have envisioned.

A book affects eternity; one can never tell where its influence stops. From, Publisher Edition: Fourth

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FEATURES

‘Smart tips’ comprise importanttheoretical or formula based shorttricks considering their usage insolving MCQ. This is our attempt to highlightcontent that would come handywhile solving questions.

Smart tip

If both inputs of NAND gates are shorted, then itbecomes NOT gate (similar is applicable for NOR gate).

Smart tip - 4

NOT gate

YA

‘Time Saver’ illustrates quickmethod over lengthy conventionalmethod of solving numerical.This is our attempt to benefitstudents by saving their timewherever possible.

Time Saver ⏰ Time saver - 1

An unpolarised beam of light is incident on a group of ‘n’polarising sheets which are arranged in such a way that thecharacteristic direction of each polarising sheet makes anangle of with that of the preceding sheet then, fraction ofincident unpolarised light emerging out of last sheet hasintensity

0I2

( cos2 )n−1 Example: Two polaroids P1 and P2 are placed with their axisperpendicular to each other. Unpolarised light I0 is incident onP1. A third polaroid P3 is kept in between P1 and P2 such thatits axis makes an angle 45 with that of P1. The intensity oftransmitted light through P2 is [NEET (UG) 2017] (A) 0I

4 (B) 0I

8 (C) 0I

16(D) 0I

2

Conventional Method: The intensity of light

after passing through P1 is half the intensity of unpolarised light.

I1 = 0I2

According to the law of Malus’, intensity of light after passing through P2 is,

I2 = 0I2

cos2 (45) = 0I4

Similarly, the intensity oflight through P3 is,I3 = 0I

4cos2 (45) = 0I

8 Ans: (B)

Quick method: n = 3 and θ = 45° I3 = 0I

2[cos2 (45)]3−1

I3 = 0I2

212

= 0I8

Ans: (B)

Sample

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FEATURES

‘Caution’ apprises students aboutmistakes which are made whilesolving an MCQs. This is our attempt to make a studentaware of possible common mistakes.

Caution

Two terms ‘deuteron’ and ‘deuterium’ are different and not to be mixed with. Deuteron is the nucleus of deuterium atom which contains one proton and one neutron.

!

‘Formulae’ includes all of the keyformulae in the chapter. This is our attempt to offer studentstools of formulae accessible whilesolving problems and last minuterevision at a glance.

Formulae

1. Charge: q = It 2. Electrostatic force: F =

0

14

1 22

q qr

3. Forces between multiple charges:

net n1 2 n 1F F F ....... F F

Formulae

‘Smart Code’ showcases simpleand smart mnemonic created forselected concepts. This is our attempt to offer studentsa memory technique that facilitateseasy recollection.

Smart Code

EVLMHVUSE Every Version of Learning Module Has Various UseSin Education

SMART CODE - 1

Classification of radio waves in communication: With reference to the frequency range, the radio

waves are classified as under:

‘Q.R. code’ provides access to avideo in order to boost understandingof a concept or activity. This is our attempt to facilitatelearning with visual aids.

Q.R. Codes

Students can scan the adjacent QR code to get demonstration of eddy current with the aid of a linked video.

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FEATURES

‘Learning Pointers’ providecompilation of additional notes whichinvolve fusion of two concepts fromdifferent chapters or come as indirectextensions of subtopics. This is our attempt to offer gist ofknowledge required from examinationpoint of view as a snippet.

Learning Pointers 1. If a piece of metal and a piece of non-metal are dropped

from the same height near the surface of the earth, thenthe non-metallic piece will reach the ground firstbecause there will be no induced current in it.

2. Skin Effect: A direct current flows uniformlythroughout the cross-section of the conductor. Analternating current, on the other hand, flows mainlyalong the surface of the conductor.

Learning Pointers

‘Mind Over Matter’ offers insight in solving brain-racking questions. This is our attempt to build a framework for focused, guided thinking.

Mind Over Matter

The key to crack this question lies in comprehending that,on passing current through the coil, it will act as a magneticdipole. Torque acting on magnetic dipole will be counterbalanced by the moment of additional weight aboutposition O.

Mind over Matter

3. A beam of light from a source L is incident normallyon a plane mirror fixed at a certain distance x fromthe source. The beam is reflected back as a spot on ascale placed just above the source L. When themirror is rotated through a small angle , the spot ofthe light is found to move through a distance y on thescale. The angle is given by [NEET (UG) 2017]

(A) yx

(B) x2y

(C) xy

(D) y2x

‘Clock Symbol’ instructs students that given MCQ can be solved apace by applying either smart tips, learning pointers, timesaver or thinking hatke. This is our attempt to make students attentive towards their perception of approaches possible for solving an MCQ.

Clock Symbol

Miscellaneous MCQs covers conceptof different sub-topics of same chapter or from different chapters. This is our attempt to developcognitive thinking in the studentsessential to solve questions involvingfusion of multiple key concepts.

Miscellaneous Miscellaneous 1. A tube of uniform square cross sectional side l has

mercury of density in it as shown below. Whencurrent I is passed through sealed electrodes andmagnetic field B is applied across horizontal sectionperpendicular to plane of paper. Then difference inmercury level in two arms is

(A) BI

l (B) BI

gl (C) 2

BIgl

(D) BIgl

+

–

I

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FEATURES

‘Memory Map’ summarizes the keypoints in the chapter. This is our attempt to providechapter overview succinctly.

Memory Map MEMORY MAP

MAGNETIC EFFECT OF ELECTRIC CURRENT

Motion in amagnetic field

1. Force on Charge: i. Moving in a uniform

magnetic field:

F = q V B q vB sin

Biot-Savart’s law, Magnetic field due to a

current element

1. Biot –Savart’s Law: dB ∝ 2

Id sinr

l

2. Magnetic field on the axis of a circular loop:

B = 2

02 2 3/2

2 Ir4 (r x )

dB

sin

r

C I

dl

a

x P

‘Thinking Hatke’ reveals quick witted approach to crack thespecific question. This is our attempt to develop skillof lateral thinking in students.

Thinking Hatke 1. An infinite number of capacitors, having capacitances

1 F, 2 F, 4 F, 8 F, ….. are connected in series.The equivalent capacitance of the system is

(A) infinite (B) 4.5 F (C) 0.5 F (D) 2 F 1. (C) 1. When number of capacitors having capacitance

1 F, 2 F, 4 F, 8 F are connected in series, theneffective capacitance is given by

1C

= 11

+ 12

+ 14

+ 18

+ ……

= 1

112

=

12 1

2 = 2

C = 12

= 0.5 F.

Thinking Hatke - Q. 1 In series combination, resultant capacitance is

lesser than every individual capacitance. Amongst givenoptions, only option (C) satisfies this condition andhence must be correct option.

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Why Challenger Series? Gradually, every year the nature of competitive entrance exams is inching towards conceptual

understanding of topics. Moreover, it is time to bid adieu to the stereotypical approach of solving a problem using a single conventional method. To be able to successfully crack the NEET examination, it is imperative to develop skills such as data interpretation, appropriate time management, knowing various methods to solve a problem, etc. With Challenger Series, we are sure, you’d develop all the aforementioned skills and take a more holistic approach towards problem solving. The way you’d tackle advanced level MCQs with the help of hints, Smart tips, Smart codes and Thinking Hatke section would give you the necessary practice that would be a game changer in your preparation for the competitive entrance examinations.

What is the intention behind the launch of Challenger Series? The sole objective behind the introduction of Challenger Series is to severely test the student’s

preparedness to take competitive entrance examinations. With an eclectic range of critical and advanced level MCQs, we intend to test a student’s MCQ solving skills within a stipulated time period.

What do I gain out of Challenger Series? After using Challenger Series, students would be able to: a. assimilate the given data and apply relevant concepts with utmost ease. b. tackle MCQs of different pattern such as match the columns, diagram based questions,

multiple concepts and assertion-reason efficiently. c. garner the much needed confidence to appear for competitive exams. d. easy and time saving methods to tackle tricky questions will help ensure that time consuming

questions do not occupy more time than you can allot per question. Can the Questions presented in Problems to Ponder section be a part of the NEET

Examination? No, the questions would not appear as it is in the NEET Examination. However, there are fair chances that these questions could be covered in parts or with a novel question construction.

Best of luck to all the aspirants!

Frequently Asked Questions

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No. Topic Name Page No. 1 Electrostatics 1

2 Capacitors 65

3 Current Electricity 105

4 Magnetic Effect of Electric Current 160

5 Magnetism 215

6 Electromagnetic Induction and Alternating Current 253

7 Electromagnetic Waves 307

8 Ray Optics 327

9 Wave Optics and Interference of light 391

10 Diffraction and Polarisation of light 434

11 Dual Nature of matter and radiation 465

12 Atoms and Nuclei 496

13 Electronic Devices 538

14 Communication system 588

15 NEET (UG) Phase-II 2020 Question Paper & Answer Key 610

16 JEE (Main) - 2020 : Question Paper & Answer Key 2nd SEPTEMBER (Shift - II) 613

CONTENTS

Note: ** marked subtopic in a chapter is unlisted in the JEE syllabus. However, since questions on the same have appeared in the recent examinations, subtopic is covered judiciously in the book.

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2.1 Dielectrics, Electric polarisation and Electric susceptibility

Dielectrics

Dielectrics Dielectrics are (non-conducting) material that transmit electric effect without conducting. Types of dielectrics Polar dielectrics Non polar dielectrics

i. It has permanent electric dipole momenteven if the electric field is absent.

ii. Net dipole moment of polar dielectric iszero because polar molecules are randomlyoriented in absence of electric field.

iii. In presence of electric field, polarmolecules align themselves in thedirection of the field.Examples: Alcohol, Water, NH3, HCl.

i. Every non-polar molecule has zerodipole moment in its normal state.

ii. When electric field is applied,molecule acts like electric dipole.Examples: N2, O2, Methane, Benzene.

Electric polarisation:i. In presence of an external electric field, a dielectric develops a net dipole moment. The dipole moment per

unit volume is called polarization.It is denoted by P

and expressed in Cm2.

ii. Relation between polarisation vector and surface density of induced charges:When a dielectric slab is placed in an electric field, the charges are induced on the surfaces of the dielectric.The negative induced surface chaitairge is equal to the positive induced surface charge. So, dielectric as awhole remains electrically neutral. Thus, polarisation of a dielectric involves only a relative displacement ofcharges in it.

Let, ± qi = induced surface charges, ± i = surface density of induced charges,

x = thickness of slab and A = area of the slab.Dipole moment of slab = qi x,Volume of slab = Ax

Polarisation, P = i ii

q x qAx A

So, when P

is normal to the surface of the dielectric, the polarisation is numerically equal to i.

i = ˆP n

2 Capacitors

–++

– + –

–+– + +

–++

– ––

+– +–+–+– +–


2.2 Capacitance and Capacitors 2.3 Combination of capacitors

2.4 Capacitance of parallel plate capacitor with and without dielectric medium between the plates

2.5 Energy stored in a capacitor

**2.6 Van de Graaff ganerator

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Challenger Physics Vol - II (Med. and Engg.)

66

Here, n̂ is a unit normal vector at the surface. iii. Electric field gets modified in presence of dielectric, E = E Ei Where E is induced field, E is main field, Ei is field due to dielectrics. Dielectric constant: i. ii. Dielectric constant,

K =

electric field betweenthe plates with airE

electric field betweenEthe plates with medium

iii. K is also called as relative permittivity of the material or SIC (specific inductive capacitance). iv. For vacuum, K = 1 and though for air its slightly greater than 1, for practical purposes, it’s taken as 1. Electric susceptibility of dielectric: i. It has been observed that the polarisation P is directly proportional to the resultant electric field E in the

dielectric.

P

E

or P

= 0 E

, where is known as the electric susceptibility of the dielectric. ii. It describes the behaviour of a dielectric. Larger the value of , greater will be the polarisation of the

dielectric in an electric field. In the case of vacuum, the value of is zero because there are no molecules in vacuum.

2.2 Capacitance And Capacitors Capacitance:

Formula: C = qV

Units: farad in SI system; statfarad in CGS system

Conversion factors: 1farad = 9 1011 statfarad Dimensions: [M1 L2 T4 A2] 1 F = 106 F Quantity: scalar

1 nF = 10–9 F 1 pF = 1012 F Capacitance of conductor: i. Capacitance of a conductor is small and limited. ii. The capacity of conductor depends on its shape and size and nature of surrounding medium. iii. Also, if a conductor is placed near a charged conductor, the potential of charged conductor decreases and

hence it can store more charge i.e., nearness of a conductor increases the capacitance of a charged conductor. Redistribution of charges and concept of common potential: Consider two insulated conductors P and Q of capacitances C1 and C2 carrying charges q1 and q2

respectively. Let V1 and V2 be their respective potentials. When the two conductors are joined by a thin metallic wire, positive charge begins to flow from a conductor of higher potential to a conductor of lower potential till their potentials are equalised. Thus, the charges are redistributed. But the total quantity of charge remains (q1 + q2).

If we neglect the capacitance of the connecting wire and assume that the conductors are so much spaced apart that they do not electrically affect each other, then the combined capacitance of the two conductors is C1 + C2.

E

Ei

++++++

–––––

– +– + – +– + – +

– +

– + – +– +– +

Dielectric constant of dielectric medium is the ratio of the strength of the applied electric field to the strength of the reduced value of the electric field on placing dielectric between the plates of the capacitor.

The ability of a conductor to store charge is called as capacitance of conductor.

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Chapter 2: Capacitors

Common potential, V = TotalchargeTotalcapacitance

or V = 1 2

1 2

q qC C

or V = 1 1 2 2

1 2

C V C VC C

Let 1q and 2q be the charges on P and Q respectively after the redistribution of charges has taken place. Then 1q = C1V and 2q = C2V

1 1

2 2

q C Vq C V

or 1 1

2 2

q Cq C

Capacitance of an isolated spherical conductor: Consider an isolated spherical conductor of radius r metre placed in a medium of relative

permittivity r. Suppose Q is the charge on the surface of the sphere.

Potential at the surface of the sphere is given by: V = 0 r

1 Q4 r

Let C be the capacitance, then

C = QV

=

0 r

Q1 Q

4 r

= 40rr

For vacuum, r = 1, C = 40r

Capacitors

Capacitor Capacitor is a device which increases charge storing capacity of a conductor at a relatively low potential. A capacitor acts as a small reservoir of energy.

Principle of a capacitor

The capacity of a charged conductor increases if another conductor connected to earth is kept near it.

Types of capacitor Parallel plate capacitor Spherical capacitor Cylindrical capacitor Diagram

Formula of Capacitance

C = 0K Ad C = 4K0 1 2

2 1

r rr r

C = 0

2

1

2 Kr2.303logr

l

Applications of capacitors

i. For tuning in radio circuit. ii. For smoothing rectified current in power supplies. iii. For eliminating sparkling of points, when they are open or close in an ignition

system of automobile engine. iv. For storage of large amount of charge in nuclear reactors.

r ++

++++++

++++ + + + + +

+ + + + + + +

+

r2r1

l

+ + + +

+ + + + +

++

++

r2 r1

+

++

+ +

In a capacitor, it is possible to deposit still larger charges on first plate in presence of induced negative charge on the second plate.

Students can scan the adjacent QR code to understand basics of capacitor and its uses with the aid of a linked video.

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68

2.3 Combination of capacitors

Capacitors in series Capacitors in parallel Diagram

Explanation i. Potential across capacitors is V1, V2 and V3. ii. Total potential across the capacitors is V, V = V1 + V2 + V3

iii. V1 = 1

qC

, V2 = 2

qC

, V3 = 3

qC

,

S

qV=C

where, CS is equivalent capacitance in series.

S

qC

= 1

qC

+ 2

qC

+ 3

qC

i. Potential across each capacitor is V. ii. Charges on each plates of capacitors are q1,

q2, q3. iii. Total charge q = q1 + q2 + q3

q1 = C1V, q2 = C2V, q3 = C3V and q = CpV where, CP is equivalent capacitor in

parallel. p 1 2 3C V C V C V C V

Formula S

1C

= 1

1C

+ 2

1C

+ 3

1C

CP = C1 + C2 + C3

+ V K

C1

C2

C3

+q1

+q2

+q3

q1

q2

q3q q

+q q +q q +q q

V1 V2 V3

E F C D

()K +

V

A B

⏰ Time saver - 1

When the difference in the effective capacity of two identical capacitors when joined in series and then

in parallel is given as C, then capacity of each capacitor say C is given by, 2CC3

Example: The difference in the effective capacity of two similar capacitors when joined in series and then in parallel is6 μF. The capacity of each capacitor is (A) 2 μF (B) 4 μF (C) 8 μF (D) 16 μF

Conventional Method: CP = C1 + C2

S

1C

= 1

1C

+ 2

1C

= 1 2

1 2

C CC C

CS = 1 2

1 2

C CC C

According to given condition CP – CS = 6 F C1 + C2 – 1 2

1 2

C CC C

= 6

As C1 = C2 = C 2C –

2C2 C

= 6

4C2 – C2 = 12C 3C2 = 12C C = 4 F Ans: (B)

Quick method: Capacity of each capacitor,

2C 2 6C3 3

= 4 μF

Ans: (B)

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Wheatstone’s bridge of capacitors: i. When potential of B is equal to potential of D, there is no flow of

charge in the diagonal arm BD. So, C5 becomes ineffective. In this situation, C1 and C2 are in series.

ii. Let q1 be the charge on each capacitor C1 and C2. Similarly, C3 and C4 are in series.

iii. Let q2 be the charge on each capacitor C3 and C4.

VA VB = 1

1

qC

VA VD = 2

3

qC

= VA VB ….( VB

= VD)

Also, VB VC = 1

2

qC

VD VC = 2

4

qC

= VB VC

Clearly, 1 2 1 2

1 3 2 4

q q q q;C C C C

1 1 2

2 3 4

q C Cq C C

1 3

2 4

C CC C

2.4 Capacitance of parallel plate capacitor with and without

dielectric medium between the plates

Parallel Plate Capacitor Conditions Diagram Formula

Parallel plate capacitor without dielectric medium.

C = 0Ad E

A

d

+

+

+

+

+

+

+

+–

+ V

BC1

D

q1

q2

C2 +–

C5 C A

C4 + –

q2

q1

C3 +

–

1

4 2

3

6

5 7

i. If n capacitors each of same capacitance C, when are connected in a. series then equivalent capacitance of combination is CS = (C/n) b. parallel then equivalent capacitance is CP = nC. ii. If n identical plates are arranged such that even numbered of plates are connected together and odd

numbered plates are connected together, then (n 1) capacitors will be formed in parallel grouping. For such a circuit, equivalent capacitance C = (n 1)C

where C = capacitance of a capacitor = 0Ad

Smart tip - 1

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70

Parallel plate capacitor completely filled with dielectric medium (K) i.e., dielectric slab

C = 0KAd

Parallel plate capacitor partially filled with dielectric medium (K) i.e., dielectric slab.

C = 0Atd tK

Parallel plate capacitor partially filled with a metallic dielectric medium i.e., metallic slab

C =

0A

d t

Here, C = Capacity of capacitor, 0 = permittivity of free space, A = Area of each plate, d = distance between two plates, t = thickness of dielectric slab.

2.5 Energy stored in a capacitor i. When battery is connected across the two plates of the capacitor, work is done by the battery to charge the

capacitor. ii. Charging of capacitor increases the potential across plates of the capacitor. iii. The work done by the battery in charging a capacitor is stored in the capacitor in the form of electric

potential energy. iv. Electric potential energy is given by,

U = 12

2qC

Where, q is charge on plates of the capacitor C is capacitance U is electrical potential energy. v. Energy stored can also be given as,

U = 12

qV = 12

CV2

vi. Total energy stored per unit volume of the capacitor, u = U 1volume 2

0E2

t

E

A

d

+

+

+

+

+

+

+

K

E

A

d

+

+

+

+

+

+

+

K

If more than one dielectric slabs are inserted between the plates of a capacitor, value of capacitance will getdecided on the basis of arrangement of slabs (series/parallel/mixed), thickness and area occupied by the slabs.

!

t

A

d

K= ∞E

If space between the plates iscompletely filled with a conductor then,K = and hence C =

Smart tip - 2

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2.6 Van de Graaff generator i. Van de Graaff generator is an electrostatic generator which can generate very high potential of the order of

5 106 V. It is used to accelerate charged particles to carry out nuclear reactions. ii. Principle: It is based on two electrostatic phenomena: a. A sharp pointed conductor has large charge

density. Hence the surrounding air becomes conducting and produces discharge called Corona discharge.

b. When a charged conductor is brought in contact with hollow conductor, it transfers the charge to hollow conductor. The transferred charge resides on the outer surface of the hollow conductor.

Applications of Van de Graff generator: i. It is used to produce high potential of the order

of few millions of volt. ii. It is used to accelerate charged particles such as

proton, deuteron. iii. It is used in nuclear physics for studies. iv. In medicine, such beams are used to treat cancer. 1. Capacitance: C = q

V 2. Parallel plate capacitor: i. Without dielectric

C = 0Ad

ii. With dielectric C = 0A

1d t 1K

3. Spherical capacitor: C = 4K0 1 2

2 1

r rr r

4. Cylindrical capacitors: C = 0

2

1

2 Kr2.303logr

l

5. Capacitors in series: S

1C

= 1

1C

+ 2

1C

+ 3

1C

6. Capacitors in parallel: CP = C1 + C2 + C3 7. Wheatstone’s bridge of capacitors:

1

2

CC

= 3

4

CC

8. Energy stored in capacitor:

U = 12

CV2 = 12

qV = 12

2qC

9. Total energy stored per unit volume of the

capacitor:

u = U 1volume 2

0E2

Target

M1 M2 C1

C2

P1

P2

H T

I D

S Metallic comb

Metallic sphere

Insulating belt

Motor driven pulley

Steel chamber

Formulae

When gap between two plates of parallel plate capacitor is completely filled with dielectric slab of dielectric constant K and

i. a battery is connected to it then, ii. no battery is connected or a battery is disconnected from it then,

a. charge becomes K times. a. charge remains same. b. capacity becomes K times. b. capacity C becomes K times. c. becomes K times. c. remains constant. d. V remains same. d. V becomes 1/K times. e. E (electric field) remains same. e. E (electric field) becomes 1/K times. f. U (energy stored) becomes K times. f. U (energy stored) becomes 1/K times.

Smart tip - 3

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72

2.1 Dielectrics, Electric

polarisation and Electric susceptibility

1. Dielectric constant for gold is (A) zero (B) infinite (C) 1 (D) greater than 1 2. Increasing order of dielectric constant for air,

rubber and copper is (A) dair < drubber < dcopper (B) dair > drubber > dcopper (C) drubber < dair < dcopper

(D) dcopper < dair < drubber

Concept Building Problems

Learning Pointers

1. The maximum electric field that a dielectricmedium can withstand, without breakingdown of its insulating property is called itsdielectric strength.

2. Two large conducting plates X and Y keptclose to each other. The plate X is given acharge Q1 while plate Y is given a chargeQ2 (Q1 > Q2), the distribution of charge onthe four faces a, b, c, d will be as shown inthe following figure.

3. When dielectric is partially filled between the

plates of a parallel plate capacitor then itscapacitance increases but potential differencedecreases. To maintain the capacitance andpotential difference of capacitor as before,separation between the plates has to beincreased say by d. In such case

K = tt d

4. If plates are moving away with constant

velocity and have separation at any instantx then x is the function of time.

The capacitance of parallel plate condenserwhen plate separation is x is,

0AC

x

and time rate of change of capacitance isgiven as,

0

0 2

dC d 1 A dxAdt dt x x dt

02

A vx

….(As dx/dt = v)

2dC x

dt

5. For infinite network, if the effective

capacitance between two points is Ce, then even if we add/remove one pair of capacitors from the chain the remaining network would still have infinite pairs of capacitors.

X

1 2Q Q

2

YQ2

c

d

1 2Q Q2

1 2Q Q2

Y

1 2Q Q2

X Q1

a

b

Q

C1 C1 C1

C2 C2 C2

P

then, effective capacity of network is equalto effective capacity of the unit circuit ofthe ladder as shown below.

6. Capacitor actually stores electric energy inthe form of electric field between the plates.The net charge on the capacitor is zero.

7. Even if the current through capacitor iszero, a finite amount of energy can bestored in a capacitor.

8. When two charged capacitors are connectedtogether, there is a chance of loss of energy,which can be given by equation,

U = U U =

21 2 1 2

1 2

C C V V2 C C

U is positive so there is always loss ofenergy. This loss is usually in the form ofheat.

9. When n identical liquids drops, each of radiusr coalesce to form a big drop of radius R, then

R = n1/3 r. Also, if each small drop has a capacitance C

a potential V, and a charge q then: Total charge, Qbig = nqsmall

Electric field intensity, Ebig = n1/3 Esmall Electric potential, Vbig = n2/3 Vsmall Surface density, big = n1/3

small Electric potential energy, Ubig = n5/3 Usmall Electric capacitance, Cbig = n1/3 Csmall

O

C1 x

C2 Ce

P

y

Sample

Con

tent

73


3. With the rise in temperature, the dielectricconstant K of a liquid(A) increases.(B) decreases.(C) remains unchanged.(D) changes erratically.

4. Which of the following statements are true?(i) Polarisation of the dielectric results in net

increase in the charge of the dielectric.(ii) Polarisation of the dielectric does not

result in any net increase in the charge ofthe dielectric.

(iii) Polarisation results into relativedisplacement of charges.

(iv) Polarisation does not result into relativedisplacement of charges.

(A) Both (i) and (iv) (B) Both (ii) and (iv)(C) Both (ii) and (iii) (D) Both (i) and (iii)

5. Assertion: The value of electric susceptibilityof the dielectric in vacuum is zero.Reason: There are no molecules in vacuum.(A) Assertion is True, Reason is True; Reason is

a correct explanation for Assertion.(B) Assertion is True, Reason is True; Reason is

not a correct explanation for Assertion.(C) Assertion is True, Reason is False.(D) Assertion is False, Reason is False.

6. The electric field between two charged parallelplates reduced to one fourth when polythenesheet is inserted and reduced to one ninth whenmica sheet is inserted. The ratio of dielectricconstants of mica to polythene is(A) 4 : 9 (B) 1 : 1(C) 9 : 4 (D) 3 : 4

7. If two charged parallel plates having electricfield of 8.2 105 V/m are filled completely withPVC, the original electric field will be reducedby (Given: KPVC = 6)(A) 1.45 105 V/m (B) 5.9 105 V/m(C) 6.83 105 V/m (D) 6.28 105 V/m

8. The capacitance of a parallel plate capacitor withair as medium is 6 F. With the introduction of adielectric medium, the capacitance becomes30 F. The permittivity of the medium is: (ε0 = 8.85 1012 C2 N1 m2)

[NEET (UG) P-I 2020] (A) 1.77 1012 C2 N1 m2

(B) 0.44 1010 C2 N1 m2

(C) 5.00 C2 N1 m2

(D) 0.44 1013 C2 N1 m2

9. If electric susceptibility of dielectric is 4 and theresultant electric field in the dielectric is2.6 V/m, then what will be the polarization ofdielectric slab?(A) 83 pC/m2 (B) 92 pC/m2

(C) 76 pC/m2 (D) 62 pC/m2

2.2 Capacitance And Capacitors

1. The outer sphere of a spherical air capacitor isearthed. To increase its capacitance(A) vacuum is created between two spheres.(B) dielectric material is filled between the

two spheres.(C) the space between two spheres is increased.(D) the earthing of the outer sphere is removed.

2. The radius of earth is 6400 km. Its capacitancewill be(A) zero (B) 7.1 104 F(C) 6.4 104 F (D) 6.4 106 F

3. Two insulated conductors are shown in thefigure below, if capacitance of first conductorand second conductor is 3.2 F and 4.2 Frespectively, then find charge on secondconductor if charge on first conductor is 6 C.

(A) 3.2 C (B) 7.9 C (C) 3 C (D) 8.5 C

4. The capacitance of a spherical condenser is2 F. If the spacing between the two spheres is2 mm, then the radius of the outer sphere is(A) 30 cm (B) 6 m(C) 5 cm (D) 3 m

5. The capacity of an isolated sphere of radius9 cm is C. When it is connected to an earthedconcentric thin hollow sphere of radius R, thecapacity becomes 15 C. Then the value of R is(A) 9 cm (B) 9.6 cm(C) 90 cm (D) 100 cm

2.3 Combination of capacitors

1. Two capacitors of capacity C1 and C2 areconnected in series and potential difference V isapplied across it. Then the potential differenceacross C1 will be

(A) 2

1

CVC

(B) 1 2

1

C CVC

(C) 2

1 2

CVC C

(D) 1

1 2

CVC C

A2 = x cm2

B

A1 = 1.6x cm2 A

The key to crack this question lies incomprehending that, when two conductorsare joined by a thin metallic wire, positivecharge begins to flow from a conductor ofhigher potential to a conductor of lowerpotential till their potentials are equalised.

Mind over Matter

Sample

Con

tent

Page no. 74 to 85 are purposely left blank.

To see complete chapter buy Target Notes or Target E‐Notes

86


86

2.1 : 1. (B) 2. (A) 3. (B) 4. (C) 5. (A) 6. (C) 7. (C) 8. (B) 9. (B)

2.2 : 1. (B) 2. (B) 3. (B) 4. (B) 5. (B)2.3 : 1. (C) 2. (B) 3. (B) 4. (D) 5. (C) 6. (B) 7. (B) 8. (B) 9. (A) 10. (C)

11. (D) 12. (C) 13. (B) 14. (C) 15. (B) 16. (A) 17. (B) 18. (A) 19. (D) 20. (D)21. (B) 22. (B) 23. (B) 24. (C) 25. (B) 26. (B) 27. (D) 28. (B)

2.4: 1. (A) 2. (A) 3. (D) 4. (C) 5. (C) 6. (A) 7. (B) 8. (A) 9. (B) 10. (D)11. (C) 12. (C) 13. (B) 14. (C) 15. (D)

2.5: 1. (A) 2. (D) 3. (B) 4. (C) 5. (D) 6. (B) 7. (C) 8. (D) 9. (C) 10. (A)11. (B) 12. (C) 13. (A)

2.6: 1. (B) 2. (B) 3. (D) 4. (A)Misc.: 1. (B) 2. (C) 3. (A) 4. (C) 5. (C) 6. (D) 7. (A) 8. (B) 9. (C) 10. (A)

11. (B) 12. (C) 13. (B) 14. (D) 15. (C) 16. (C) 17. (D) 18. (D) 19. (B) 20. (B)

2.1 : 1. (D) 2. (D) 3. (D) 4. (A) 5. (B)

2.2 : 1. (C) 2. (A) 3. (A) 4. (D) 5. (A)

2.3 : 1. (C) 2. (B) 3. (C) 4. (D) 5. (B) 6. (A) 7. (B) 8. (D) 9. (D) 10. (D)11. (A)

2.4: 1. (B) 2. (B) 3. (B) 4. (C)

2.5: 1. (A) 2. (B) 3. (D) 4. (B) 5. (B) 6. (A) 7. (A)

2.6: 1. (A) 2. (B) 3. (B)

Misc.: 1. (C) 2. (D) 3. (B) 4. (B) 5. (B)


1. Dielectric constant 0

εK =ε

Permittivity of metals (gold) is assumed to be infinite.

3. As temperature increases, owing to increasedkinetic energy, the capability of the dipole toremain united decreases, and therefore,dielectric constant decreases.

6. K = without dielectric

with dielectric

EE

K polythene = EE/4

K mica = EE / 9

= 9

K mica : K polythene = 9: 4


Thinking Hatke - Q.6 As due to mica sheet electric field

reduces more than with polythene sheet,dielectric constant of mica will be greater thanpolythene. Hence, the ratio of dielectricconstants of mica to polythene will be greaterthan 1. Amongst given options, only option(C) satisfies this condition and hence must becorrect option.

Answers to MCQs

Practice Problems


Hints to MCQs

Sample

Con

tent

87


7. K = without dielectric

with dielectric

EE

6 = 5

with dielectric

8.2 10 V / mE

Ewith dielectric = 1.37 105 V/m Ereduced = (8.2 1.37) 105 = 6.83 105 V/m 8. Using Smart tip – 3 (ii - b),

K = medium

air

C 30C 6

= 5

but, K = 0

Permittivity of medium, ε = Kε0 = 5 8.85 1012

= 0.44 1010 C2 N1 m2

9. P = 0E P = 4 8.85 10–12 2.6 P = 9.2 10–11 C/m2 = 92 10–12 C/m2 = 92 pC/m2 2.2 Capacitance And Capacitors 1. In spherical capacitor, 1 2

02 1

r rC 4 Kr r

C K 2. C = Q

V

but V = 0

1 Q4 R

for a spherical body.

C = 40R C = 4 8.85 1012 6400 103 C = 7.1 104 F 3. Let q1 and q2 be the charge on conductors A and

B respectively; C1 and C2 be their respective capacitance, then

1 1

2 2

q Cq C

6

62

6 3.2 10q 4.2 10

q2 = 6

6

6 4.2 103.2 10

q2 = 7.9 C 4. Given: (r2 – r1) = 2 10–3 m ….(i)

and 61 20

2 1

r rC 4 2 10 Fr r

6 1 29 3

1 r r2 109 10 2 10

r1 r2 = 36 ….(ii)

From equations (i) and (ii), 2

2

36 2rr 1000

1000 22r – 2r2 – 36000 = 0

Solving the quadratic equation,

2

22 ( 2) 4(1000)( 36000)

r2 1000

6

22 144 10r

2000

6 m.

5. Capacity of isolated sphere, C = q

V

C = 40r1 ….(i) Capacity of earthed concentric hollow sphere is;

CH = 401 2

2 1

r rr r

15C = 401

1

r R(R r )

….(ii)

Dividing equation (ii) by equation (i),

15CC

= 1

1 1

r R 1R r r

15 = 9R 1(R 9) 9

15R 135 = R 14R = 135 R = 9.6 cm 2.3 Combination of capacitors 1. Charge flowing 1 2

1 2

C C VC C

.

So potential difference across C1

2

1 2

C VC C

2. For series combination, V1 =

1

QC

and V2 = 2

QC

2 1

1 2

V CV C

= 4 : 1 3. In series combination of capacitors, voltage

distributes on them, in the reverse ratio of their

capacitance i.e., A

B

V 8V 4

….(i)

Also VA + VB = 12 ….(ii) On solving (i) and (ii) We get, VA = 8 V, VB = 4 V

r1 r2

O 9 cm

R

r1 = 9 cm r2 = R

Sample

Con

tent

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To see complete chapter buy Target Notes or Target E‐Notes

104

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Con

tent

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