GATE 2021...For 1‐mark MCQ, 1/3 mark will be deducted for a wrong answer. Likewise for, 2‐marks...

Electronics & Communication Engineering

(Volume - II)

GATE 2021

TOPIC WISE GATE SOLUTIONS1987-2020

Umesh Dhande

TM

GATE examination has been emerging as one of the most prestigious competitive exam for engineers.

Earlier it was considered to be an exam just for eligibility for pursuing PG courses, but now GATE exam

has gained a lot of attention of students as this exam open an ocean of possibilities like :

1. Admission into IISc, IITs, IIITs, NITs

A good GATE score is helpful for getting admission into IISc, IITs, IIITs, NITs and many other

renowned institutions for M.Tech./M.E./M.S. An M.Tech graduate has a number of career

opportunities in research fields and education industries. Students get 12,400 per month as

stipend during their course.

2. Selection in various Public Sector Undertakings (PSUs)

A good GATE score is helpful for getting job in government‐owned corporations termed

as Public Sector Undertakings (PSUs) in India like IOCL, BHEL, NTPC, BARC, ONGC, PGCIL, DVC,

HPCL, GAIL, SAIL & many more.

3. Direct recruitment to Group A level posts in Central government, i.e., Senior Field Officer (Tele),

Senior Research Officer (Crypto) and Senior Research Officer (S&T) in Cabinet Secretariat,

Government of India, is now being carried out on the basis of GATE score.

4. Foreign universities through GATE

GATE has crossed the boundaries to become an international level test for entry into

postgraduate engineering programmes in abroad. Some institutes in two countries Singapore

and Germany are known to accept GATE score for admission to their PG engineering

programmes.

5. National Institute of Industrial Engg. (NITIE)

NITIE offers PGDIE / PGDMM / PGDPM on the basis of GATE scores. The shortlisted

candidates are then called for group Discussion and Personal Interview rounds.

NITIE offers a Doctoral Level Fellowship Programme recognized by Ministry of HRD

(MHRD) as equivalent to PhD of any Indian University.

Regular full time candidates those who will qualify for the financial assistance will receive

25,000 during 1st and 2nd year of the Fellowship programme and 28,000 during 3rd, 4th and 5th year of the Fellowship programme as per MHRD guidelines.

6. Ph.D. in IISc/ IITs

IISc and IITs take admissions for Ph.D. on the basis of GATE score.

Earn a Ph.D. degree directly after Bachelor’s degree through integrated programme.

A fulltime residential researcher (RR) programme.

7. Fellowship Program in management (FPM)

Enrolment through GATE score card

Stipend of 22,000 – 30,000 per month + HRA

It is a fellowship program

Application form is generally available in month of sept. and oct.

Note : In near future, hopefully GATE exam will become a mandatory exit test for all engineering

students, so take this exam seriously. Best of LUCK !

IMPORTANCE of GATE

Section Question No. No. of Questions Marks Per Question Total Marks

General Aptitude

1 to 5 5 1 5

6 to 10 5 2 10

Technical

+

Engineering Mathematics

1 to 25 25 1 25

26 to 55 30 2 60

Total Duration : 3 hours Total Questions : 65 Total Marks : 100

Note : 40 to 45 marks will be allotted to Numerical Answer Type Questions

Pattern of Questions :

(i) Multiple Choice Questions (MCQ) carrying 1 or 2 marks each in all the papers and

sections. These questions are objective in nature, and each will have a choice of four

answers, out of which the candidate has to select (mark) the correct answer.

Negative Marking for Wrong Answers : For a wrong answer chosen in a MCQ, there will

be negative marking. For 1‐mark MCQ, 1/3 mark will be deducted for a wrong answer.

Likewise for, 2‐marks MCQ, 2/3 mark will be deducted for a wrong answer.

(ii) Numerical Answer Type (NAT) Questions carrying 1 or 2 marks each in all the papers

and sections. For these questions, the answer is a signed real number, which needs to

be entered by the candidate using the virtual numeric keypad on the monitor (Keyboard

of the computer will be disabled). No choices will be shown for these type of questions.

The answer can be a number such as 10 or – 10 (an integer only). The answer may be in

decimals as well, for example, 10.1 (one decimal) or 10.01 (two decimal) or –10.001

(three decimal). These questions will be mentioned with, up to which decimal places,

the candidates need to make an answer. Also, an appropriate range will be considered

while evaluating the numerical answer type questions so that the candidate is not

penalized due to the usual round‐off errors. Wherever required and possible, it is better

to give NAT answer up to a maximum of three decimal places.

Note : There is NO negative marking for a wrong answer in NAT questions.

GATE Exam Pattern

GATE ACADEMY Team took several years’ to come up with the solutions of GATE examination. It is

because we strongly believe in quality. We have significantly prepared each and every solution of the

questions appeared in GATE, and many individuals from the community have taken time out to proof

read and improve the quality of solutions, so that it becomes very lucid for the readers. Some of the

key features of this book are as under :

This book gives complete analysis of questions chapter wise as well as year wise.

Video Solution of important conceptual questions has been given in the form of QR code and

by scanning QR code one can see the video solution of the given question.

Solutions has been presented in lucid and understandable language for an average student.

In addition to the GATE syllabus, the book includes the nomenclature of chapters according to

text books for easy reference.

Last but not the least, author's 10 years experience and devotion in preparation of these

solutions.

Steps to Open Video solution through mobile.

Note : For recent updates regarding minor changes in this book, visit ‐

www.gateacademy.co.in. We are always ready to appreciate and help you.

(1) Search for QR Code scanner

in Google Play / App Store.

(2) Download & Install any

QR Code Scanner App.

(3) Scan the given QR Code

for parcular queson.

(4) Visit the link generated & you’ll

be redirect to the video soluon.

What is special about this book ?

It is our pleasure, that we insist on presenting “GATE 2021 Electronics & Communication

Engineering (Volume ‐ II)” authored for Electronics & Communication Engineering to all of the

aspirants and career seekers. The prime objective of this book is to respond to tremendous

amount of ever growing demand for error free, flawless and succinct but conceptually empowered

solutions to all the question over the period 1987 ‐ 2020.

This book serves to the best supplement the texts for Electronics & Communication

Engineering but shall be useful to a larger extent for Electrical Engineering and Instrumentation

Engineering as well. Simultaneously having its salient feature the book comprises :

Step by step solution to all questions.

Complete analysis of questions, i.e. chapter wise as well as year wise.

Detailed explanation of all the questions.

Solutions are presented in simple and easily understandable language.

Video solutions for good questions.

It covers all GATE questions from 1987 to 2019 (33 years).

The authors do not sense any deficit in believing that this title will in many aspects, be different

from the similar titles within the search of student.

We would like to express our sincere appreciation to Mrs. Sakshi Dhande Mam

(Co‐Director, GATE ACADEMY Learning Pvt. Ltd.) for her constant support and constructive

suggestions and comments in reviewing the script.

In particular, we wish to thank GATE ACADEMY expert team members for their hard work

and consistency while designing the script.

The final manuscript has been prepared with utmost care. However, going a line that, there

is always room for improvement in anything done, we would welcome and greatly appreciate the

suggestions and corrections for further improvement.

Umesh Dhande

(Director, GATE ACADEMY Learning Pvt. Ltd.)

PREFACE

S. No. Topics Page No.

1. Electronic Devices & Circuits

1. Basic Semiconductor Physics

2. p‐n Junction Diode

3. Basics of BJT

4. JFET & MOSFET

5. IC‐Technology

6. Special Purpose Diodes

2. Analog Electronics

1. Diode Circuits & Applications

2. BJT Biasing & Region of Operation

3. Low Frequency BJT Amplifier

4. JFET & MOSFET Amplifier with Biasing

5. Feedback Amplifiers

6. Operational Amplifiers

7. Frequency Response of Amplifier

8. Oscillator Circuits

9. Power Amplifiers

10. 555 Timer

3. Electromagnetic Theory

1. Basics of Electromagnetics

2. Plane Wave Propagation

3. Polarization

4. Transmission Lines

5. S‐parameters

6. Waveguides

7. Antennas

8. Optical Fiber & Radar

CONTENTS

4. Communication Systems

1. Random Variables & Random Processes

2. Amplitude Modulation

3. Angle Modulation

4. Noise in AM & FM

5. AM Transmitters & Receivers

6. Baseband Transmission

7. Bandpass Transmission

8. Noise in Digital Communication

9. Information Theory & Coding

10. Basics of TDMA, FDMA & CDMA

5. General Aptitude (GA)

1. Verbal Ability

2. Numerical Ability

1.1 Consider two energy levels : 1,E E eV

above the Fermi level and 2 ,E E eV below the Fermi level. 1P and 2P are respectively the probabilities of 1E being occupied by an electron and 2E being empty. Then

(A) 1 2P P>

(B) 1 2P P=

(C) 1 2P P<

(D) 1P and 2P depend on number of free electrons.

1.2 In an intrinsic semiconductor the free electron concentration depends on

(A) effective mass of electrons only. (B) effective mass of holes only. (C) temperature of the semiconductor. (D) width of the forbidden energy band of

the semiconductor. 1.3 According to the Einstein relation, for

any semiconductor the ratio of diffusion constant to mobility of carriers

(A) depends upon the temperature of the semiconductor.

(B) depends upon the type of the semi-conductor.

(C) varies with life time of the semi-conductor.

(D) is a universal constant.

1.4 Due to illumination by light, the electron

and hole concentration in a heavily doped n-type semiconductor, increases by nΔ and pΔ respectively. If in is the intrinsic concentration then

(A) n pΔ < Δ (B) n pΔ > Δ (C) n pΔ = Δ (D) in p nΔ × Δ = 1.5 The concentration of ionized acceptors

and donors in a semiconductor are ,A DN N respectively. If A DN N> and in

is the intrinsic concentration, the position of the fermi level with respect to the intrinsic level depends on

(A) A DN N− (B) A DN N+

(C) 2A D

i

N Nn

(D) in

1.6 Under high electric fields, in a

semiconductor with increasing electric field,

(A) the mobility of charge carriers decreases.

(B) the mobility of the carriers increases. (C) the velocity of the charge carriers

saturates. (D) the velocity of the charge carriers

increases.

1Basic

Semiconductor Physics

1987 IIT Bombay 1989 IIT Kanpur

1990 IISc Bangalore

1.2 Topic Wise GATE Solutions [EC] GATE ACADEMY ®

1.7 A silicon sample is uniformly doped with

1610 phosphorous 3atom/cm and 162 10× boron 3atom/cm . If all the dopants are fully ionized, the material is

(A) n-type with carrier concentration of 16 310 /cm .

(B) p-type with carrier concentration of 16 310 /cm .

(C) p-type with carrier concentration of 16 32 10 cm× .

(D) none of the above.

1.8 A semiconductor is irradiated with light

such that carriers are uniformly generated throughout its volume. The semiconductor is n-type with 19 310 /cmDN = . If the excess electron concentration in the steady state is 15 310 /cmnΔ = and if 10 secpτ = μ

(minority carries life time) the generation rate due to irradiation

(A) is 20 310 - pairs/cm /se h .

(B) is 24 310 - pairs/cm /se h .

(C) is 10 310 - pairs/cm /se h .

(D) cannot be determined, the given data is insufficient.

1.9 The drift velocity of electrons in silicon (A) is proportional to the electric field for

all values of electric field. (B) is independent of the electric field. (C) increases at low values of electric

field and decreases at high values of electric field exhibiting negative differential resistance.

(D) increases linearly with electric field at low values of electric field and gradually saturates at higher values of electric field.

1.10 The probability that an electron in a metal occupies the Fermi-level at any temperature 0( 0 K)>

(A) 0 (B) 1 (C) 0.5 (D) None 1.11 In a p-type silicon sample, the hole

concentration is 15 32.25 10 /cm .× If the intrinsic carrier concentration is

10 31.5 10 /cm ,× the electron concentration is

(A) Zero (B) 10 310 /cm (C) 5 310 /cm (D) 25 31.5 10 /cm× 1.12 A small concentration of minority carrier

is injected into a homogeneous semiconductor crystal at one point. An electric field of 10 V/cm is applied across the crystal and this moves the minority carriers by a distance of 1 cm in 20 sec.μ The mobility (in 2cm /V-s ) will be

(A) 1,000 (B) 2,000 (C) 5,000 (D) 500,000

1.13 The intrinsic carrier density at 0300 K is

10 31.5 10 /cm× , in silicon. For n-type silicon doped to 152.25 10× atom 3/cm , the equilibrium electron and hole densities are

(A) 15 31.5 10 /cm ,n = ×

10 31.5 10 /cmp = ×

(B) 10 31.5 10 /cm ,n = ×

15 32.25 10 /cmp = ×

(C) 15 32.25 10 /cm ,n = ×

5 31.0 10 /cmp = ×

1991 IIT Madras

1992 IIT Delhi

1995 IIT Kanpur

1997 IIT Madras

1.3GATE ACADEMY ® Electronic Devices & Circuits : Basic Semiconductor Physics

(D) 10 31.5 10 /cmn = × 10 31.5 10 /cmp = × 1.14 An n-type silicon bar 0.1 cm long and

2100 mμ in cross-sectional area has a majority carrier concentration of

20 35 10 /m× and the carrier mobility is 20.13 m /V-s at 0300 K . If the charge of

an electron is 191.6 10−× Coulomb, then the resistance of the bar is

(A) 610 Ω (B) 410 Ω

(C) 110− Ω (D) 410− Ω

1.15 The electron concentration in a sample of

uniformly doped n-type silicon at 0300 Kvaries linearly from 17 310 /cm at x = 0 to

16 36 10 /cm× at 2 mx = μ . Assume a situation that electrons are supplied to keep this concentration gradient constant with time. If electronic charge is

191.6 10−× Coulomb and the diffusion constant 235 cm /snD = , the current density in the silicon, if no electric field present is

(A) zero (B) 120 A/cm2 (C) +1120 A/cm2 (D) –1120 A/cm2

1.16 n-type silicon is obtained by doping silicon with

(A) Germanium (B) Aluminum (C) Boron (D) Phosphorus 1.17 The band gap of silicon at 0300 K is (A) 1.36 eV (B) 1.10 eV (C) 0.80 eV (D) 0.67 eV 1.18 The intrinsic carrier concentration of

silicon sample at 3000 K is 16 31.5 10 / m .× If after doping, the number of majority carriers is 20 35 10 /m ,× then minority carrier density is

(A) 11 34.50×10 /m (B) 4 33.333 10 /m×

(C) 20 35.00 10 /m× (D) 5 33.00 10 /m−×

1.19 The resistivity of a uniformly doped

n-type silicon sample is 0.5 -cmΩ . If the electron mobility (μ )n is

21250 cm /V-sec and the charge of an

electron is 191.6 10−× Coulomb, the donor impurity concentration ( )DN in the sample is

(A) 16 32 10 /cm× (B) 16 31 10 /cm×

(C) 15 32.5 10 /cm× (D) 15 35 10 /cm× 1.20 The longest wavelength that can be

absorbed by silicon, which has the band gap of 1.12 eV, is 1.1 μm . If the longest wavelength that can be absorbed by another material is 0.87 mμ , then the band gap of this material is

(A) 1.416 eV (B) 0.886 eV (C) 0.854 eV (D) 0.706 eV

1.21 The bandgap of silicon at room

temperature is (A) 1.3 eV (B) 0.7 eV (C) 1.1 eV (D) 1.4 eV 1.22 A silicon sample A is doped with 8101

atoms/ 3cm of boron. Another sample of identical dimension is doped with

18 310 atoms/cm of phosphorus. The ratio of electron to hole mobility is 3. The ratio of conductivity of the sample A to B is

(A) 3 (B) 13

(C) 23

(D) 32

2003 IIT Madras

2004 IIT Delhi

2005 IIT Bombay


1.23 The concentration of minority carriers in

an extrinsic semiconductor under equilibrium is

(A) directly proportional to the doping concentration.

(B) inversely proportional to the doping concentration.

(C) directly proportional to the intrinsic concentration.

(D) inversely proportional to the intrinsic concentration.

1.24 Under low level injection assumption, the injected minority carrier current for an extrinsic semiconductor is essentially the

(A) diffusion current. (B) drift current. (C) recombination current. (D) induced current. 1.25 The majority carriers in an n-type

semiconductor have an average drift velocity v in a direction perpendicular to a uniform magnetic field B. The electric field E induced due to Hall effect acts in the direction

(A) v B× (B) B v× (C) along v (D) opposite to v 1.26 A heavily doped n-type semiconductor

has the following data Hole-electron mobility ratio = 0.4 Doping concentration 8 34.2 10 atoms/m= ×

Intrinsic concentration 4 31.5 10 atoms/m= × The ratio of conductance of the n-type

semiconductor to that of the intrinsic semiconductor of same material and at same temperature is given by

(A) 0.00005 (B) 2,000 (C) 10, 000 (D) 20,000

1.27 The electron and hole concentration in an

intrinsic semiconductor are 3/cmin at0300 K . Now, if acceptor impurities are

introduced with a concentration 3/cmAN

(where A iN n>> ), the electron

concentration per 3cm at 0300 K will be (A) in (B) i An N+

(C) A iN n− (D) 2i

A

nN

1.28 Silicon is doped with boron to a

concentration of 17 34 10 atom/cm× . Assume the intrinsic carrier concentration

of silicon to be 10 31.5 10 /cm× and the

value of kTq

to be 25 mV at 0300 K .

Compared to undoped silicon, the Fermi level of doped silicon

(A) goes down by 0.13 eV. (B) goes up by 0.13 eV. (C) goes down by 0.427 eV. (D) goes up by 0.427 eV.

1.29 In an n-type silicon crystal at room

temperature, which of the following can have a concentration of 19 34 10 cm−× ?

(A) Silicon atoms (B) Holes (C) Dopant atoms (D) Valence electrons 1.30 The ratio of the mobility to the diffusion

coefficient in a semiconductor has the unit

(A) 1V− (B) 1cm-V−

(C) 1V-cm− (D) V-sec

2006 IIT Kharagpur 2007 IIT Kanpur

2008 IISc Bangalore

2009 IIT Roorkee


The silicon sample with unit cross-sectional area shown below is in thermal equilibrium. The following information is given : T = 0300 K , electronic charge 191.6 10−= × C, thermal voltage = 26 mV and electron mobility = 21350 cm /V-s.

1.31 The magnitude of the electric field at

0.5 mx = μ is

(A) 1 kV/cm (B) 5 kV/cm (C) 10 kV/cm (D) 26 kV/cm 1.32 The magnitude of the electron drift current

density at 0.5 mx = μ is

(A) 4 22.16 10 A/cm×

(B) 4 21.08 10 A/cm×

(C) 3 24.32 10 A/cm×

(D) 2 26.48 10 A/cm×

1.33 Drift current in semiconductors depends

upon (A) only the electric field. (B) only the carrier concentration

gradient. (C) both the electric field and the carrier

concentration. (D) both the electric field and the carrier

concentration gradient.

1.34 A silicon bar is doped with donor

impurities 15 32.25 10 atoms/cmDN = × . Given the intrinsic carrier concentration of silicon at 0300 KT = is 10 31.5 10 cmin −= × Assuming complete impurity ionization, the equilibrium electron and hole concentrations are [Set - 02]

(A) 16 30 1.5 10 cmn −= × ,

5 30 1.5 10 cmp −= ×

(B) 10 30 1.5 10 cmn −= × ,

15 30 1.5 10 cmp −= ×

(C) 15 30 2.25 10 cmn −= × ,

10 30 1.5 10 cmp −= ×

(D) 15 30 2.25 10 cmn −= × ,

5 30 1 10 cmp −= ×

1.35 Assume electronic charge 191.6 10 C,q −= × / 25 mVkT q = and electron mobility

21000cm /V-snμ = . If the concentration gradient of electrons injected into a p-type silicon sample is 21 41 10 /cm× , the magnitude of electron diffusion current density (in 2A/cm ) is _____. [Set - 02]

1.36 A thin p-type silicon sample is uniformly illuminated with light which generates excess carriers. The recombination rate is directly proportional to [Set - 03]

(A) the minority carrier mobility. (B) the minority carrier recombination

life time. (C) the majority carrier concentration. (D) the excess minority carrier concentration.

16 310 /cmDN

1 V

0x 1 mx

2010 IIT Guwahati

Statement for Linked Answer Questions 1.31 & 1.32

2011 IIT Madras

2014 IIT Kharagpur


1.37 At 0300 KT = , the hole mobility of a semiconductor 2μ 500 cm /V-sp = and

26 mVkTq

= . The hole diffusion constant

pD in 2cm /s is ________. [Set - 03]

1.38 At 0300 KT = , the band gap and the intrinsic carrier concentration of GaAs are 1.42 eV and 6 310 cm− , respectively. In order to generate electron hole pairs in GaAs, which one of the wavelength ( )Cλ

ranges of incident radiation, is most suitable? (Given that : Planck’s constant is 346.62 10−× J-s, velocity of light

103 10 cm/s× and charge of electron is 191.6 10 C−× ) [Set - 04]

(A) 0.42 m 0.87 mCμ < λ < μ

(B) 0.87 m 1.42 mCμ < λ < μ

(C) 1.42 m 1.62 mCμ < λ < μ

(D)1.62 m 6.62 mCμ < λ < μ

1.39 In the figure, ln ( )iρ is plotted as a

function of 1T

, where iρ is the intrinsic

resistivity of silicon, T is the temperature, and the plot is almost linear.

The slope of the line can be used to

estimate [Set - 04] (A) band gap energy of silicon ( ).gE

(B) sum of electron and hole mobility in silicon ( ).n pμ + μ

(C) reciprocal of the sum of electron and hole mobility in silicon 1( ) .n p

−μ + μ (D) intrinsic carrier concentration of

silicon ( ).in

1.40 The cut-off wavelength (in mμ ) of light that can be used for intrinsic excitation of a semiconductor material of bandgap

1.1eVgE = is _______. [Set - 04]

1.41 Consider a silicon sample doped with 15 31 10 /cmDN = × donor atoms. Assume

that the intrinsic carrier concentration10 31.5 10 /cmin = × . If the sample is

additionally doped with 18 31 10 /cmAN = × acceptor atoms, the approximate number of electrons 3/cm in the sample, at 0300 K,T = will be _______. [Set - 04]

1.42 An n-type semiconductor having uniform doping is biased as shown in the figure. [Set - 04]

If CE is the lowest energy level of the

conduction band, VE is the highest energy level of the valance band and FE is the Fermi level, which one of the following represents the energy band diagram for the biased n-type semiconductor?

(A)

(B)

1/T

ln ( )i

n-type semiconductor

V

FE

VE

CE

FE

VE

CE


(C)

(D)

1.43 A silicon sample is uniformly doped with

donor type impurities with a concentration of 16 310 /cm . The electron and hole mobilities in the sample are 21200cm /V-s and 2400cm /V-s respectively. Assume complete ionization of impurities. The charge of an electron is 191.6 10 C−× . The resistivity of the sample (in Ω -cm) is _______. [Set - 01]

1.44 A piece of silicon is doped uniformly with phosphorous with a doping concentration of 16 310 /cm . The expected value of mobility versus doping concentration for silicon assuming full dopant ionization is shown below. The charge of an electron is 191.6 10 C−× . The conductivity (in 1S cm− ) of the silicon at

0300 K is _______. [Set - 02]

1.45 An n-type silicon sample is uniformly illuminated with light which generates

2010 electron-hole pairs per 3cm second. The minority carrier lifetime in the sample is 1 sμ . In the steady state, the hole concentration in the sample is approximately 10x , where x is an integer. The value of x is _______. [Set - 02]

1.46 The energy band diagram and the electron density profile n(x) in a semiconductor are shown in the figures.

Assume that 15 3( ) 10 cmq xkTn x eα

− = , with 0.1 V/cmα = and x expressed in cm.

Given 2 10.026 V, 36 cm snkT Dq

−= = and

D kTq

=μ

. The electron current density (in

2A/cm ) at 0x = is [Set - 02]

(A) 24.4 10−− × (B) 22.2 10−− ×

(C) 0 (D) 22.2 10−×

1.47 A DC voltage of 10 V is applied across an n-type silicon bar having a rectangular cross-section and length of 1cm as shown in figure. The donor doping concentration DN and the mobility of electrons nμ are 16 – 310 cm and 1000

2 1 1cm V s− − , respectively. The average time (in μs ) taken by the electrons to move from one end of the bar to other end is _______. [Set - 02]

FE

VE

CE

FE

VE

CE

1. 13 1. 14 1. 15 1. 16 1. 17 1. 18 1. 19 1. 20E E E E E E E E -3Doping concentration (cm )

200

400

600

800

1000

1200

1400

Mo

bil

ity

(cm

Vs

)2

-1-1

Hole and Electron Mobility in Silicon at 300 K0

Electron

Hole

(eV)E

Slope 0.1eV/cm

CE

VE

0x 0x x x

log( ( ))n x

2015 IIT Kanpur


1.48 A small percentage of impurity is added

to an intrinsic semiconductor at 0300 K . Which one of the following statement is true for the energy band diagram shown in the following figure? [Set - 01]

(A) Intrinsic semiconductor doped with

pentavalent atoms to form n-type semiconductor.

(B) Intrinsic semiconductor doped with trivalent atoms to form n-type semiconductor.

(C) Intrinsic semiconductor doped with pentavalent atoms to form p-type semiconductor.

(D) Intrinsic semiconductor doped with trivalent atoms to form p-type semiconductor.

1.49 Consider a silicon sample at 0300 KT = , with a uniform donor density DN =

165 10× cm−3, illuminated uniformly such that the optical generation rate is = 1.5

2010× cm−3s−1 throughout the sample. The incident radiation is turned off at

0t = . Assume low-level injection to be valid and ignore surface effects. The carrier lifetimes are

0pτ = 0.1 µs and

00.5 s.nτ = μ

The hole concentration at t = 0 and the hole

concentration at 0.3 µst = , respectively, are [Set - 01]

(A) 13 3 11 31.5 10 cm and 7.47 10 cm− −× × .

(B) 13 3 11 31.5 10 cm and 8.23 10 cm− −× × .

(C) 13 3 11 37.5 10 cm and 3.73 10 cm− −× × .

(D) 13 3 11 37.5 10 cm and 4.12 10 cm− −× × . 1.50 The I-V characteristics of three types of

diodes at the room temperature, made of semiconductors X, Y and Z, are shown in the figure. Assume that the diodes are uniformly doped and identical in all respects except their materials. If

,gX gYE E and gZE are the band gaps of X,

Y and Z, respectively, then [Set - 03]

(A) gX gY gZE E E> >

(B) gX gY gZE E E= =

(C) gX gY gZE E E< <

(D) No relationship among these band gaps exists.

1.51 A bar of Gallium Arsenide (GaAs) is

doped with Silicon such that the Silicon atoms occupy Gallium and Arsenic sites in the GaAs crystal. Which one of the following statements is true? [Set - 01]

-type Siliconn1 cm

10 V

0.01 eV

New Energy Level

Conduction Band

Valence Band

EC

EV

n-type Si

I

V

X Y Z

01V 02V 03V0

2016 IISc Bangalore

2017 IIT Roorkee


(A) Silicon atoms act as p-type dopants in Arsenic sites and n-type dopants in Gallium sites.

(B) Silicon atoms act as n-type dopants in Arsenic sites and p-type dopants in Gallium sites.

(C) Silicon atoms act as p-type dopants in Arsenic as well as Gallium sites.

(D) Silicon atoms act as n-type dopants in Arsenic as well as Gallium sites.

1.52 The dependence of drift velocity of electrons on electric field in a semiconductor is shown below. The semiconductor has a uniform electron

concentration of 16 31 10 cmn −= × and

electronic charge 191.6 10 Cq −= × . If a

bias of 5 V is applied across a 1 mμ

region of this semiconductor, the resulting current density in this region, in kA/cm2, is _________. [Set - 01]

1.53 As shown, a uniformly doped Silicon (Si) bar of length L = 0.1 mμ with a donor

concentration 16 310 cmDN −= is illuminated

at x = 0 such that electron and hole pairs are generated at the rate of

0

1 , 0L LxG G x LL

= − ≤ ≤

,

where 0

17 310 /cm sLG = .

Hole life time is 410 s− electronic charge 191.6 10 C,q −= × hole diffusion coefficient

2100 cm /spD = and low level injection condition prevails. Assuming a linearly decaying steady state excess hole concentration that goes to 0 at x = L, the magnitude of the diffusion current density at x = L/2, in A/cm2, is ________. [Set - 01]

1.54 Which one of the following options

describes correctly the equilibrium band diagram at T = 300 K of a Siliconpnn p+ ++ configuration shown in the

figure?

p n n+ p++

710

55 10

Drift velocity (cm/s)

constant

linea

r

Electric field (V/cm)0

16 3Si ( 10 cm )DN

0.1 mL 0x

Light

CE

VE

FE(A)

CE

VE

FE(B)

2019 IIT Madras


1.55 A single crystal intrinsic semiconductor

is at a temperature of 300 K with effective density of states for holes twice that of electrons. The thermal voltage is 26 mV. The intrinsic Fermi level is shifted from mid-bandgap energy level by

(A) 9.01 meV (B) 13.45 meV (C) 18.02 meV (D) 26.90 meV

CE

(C) FE

VE

CE

VE

FE

(D)

2020 IIT Delhi


1.1 B 1.2 C 1.3 A 1.4 C 1.5 A

1.6 A and C 1.7 B 1.8 A 1.9 D 1.10 B

1.11 C 1.12 C 1.13 C 1.14 A 1.15 D

1.16 D 1.17 B 1.18 A 1.19 B 1.20 A

1.21 C 1.22 B 1.23 B 1.24 A 1.25 B

1.26 D 1.27 D 1.28 C 1.29 C 1.30 A

1.31 C 1.32 A 1.33 C 1.34 D 1.35 4000

1.36 D 1.37 13 1.38 A 1.39 A 1.40 1.127

1.41 225 1.42 D 1.43 0.52 1.44 1.92 1.45 14

1.46 C 1.47 100 1.48 A 1.49 A 1.50 C

1.51 A 1.52 1.6 1.53 16 1.54 B 1.55 A

Given : (i) 1E is above the fermi level, 1 eVFE E E− = (ii) 2E is below the fermi level, 2 eVFE E E− = − (iii) 1P = Probability that 1E is occupied by

electron (iv) 2P = Probability that 2E is empty . Method 1 : Fermi dirac probability distribution function is given by,

( )/1( )

1 Fn E E kTf Ee −=

+

where, E = Given energy level FE = Fermi level energy Probability that 1E is occupied by electron is given by,

11 1 ( )/1( )

1 Fn E E kTP f Ee −= =

+

1 /1

1 E kTPe

=+

… (i)

Probability that 2E is not occupied by electron is given by,

22 2 ( )/11 ( ) 1

1 Fn E E kTP f Ee −= − = −

+

/

2 / /1

1 1

E kT

E kT E kT

ePe e

−

−= =+ +

… (ii)

From equations (i) and (ii), 1 2P P=

Hence, the correct option is (B). . Method 2 : Fermi dirac probability distribution function is given by,

( )/1( )

1 Fn E E kTf Ee −=

+

kTq

= 26 mV at 0300 KT =

At 00 KT = : (i) Case 1 : When ,FE E>

1 1( ) 01nf E

e∞= = =+ ∞

Answers Basic Semiconductor Physics

Explanations Basic Semiconductor Physics

1.1 (B)


When given energy level is more than fermi energy level then probability of finding electron in given energy level at

00 K will be zero. (ii) Case 2 : When ,FE E<

1 1( ) 11 1 0nf E

e−∞= = =+ +

When energy level is less than fermi energy level then probability of finding electron in given energy level at 00 K will be 100%.

(iii) Case 3 : When ,FE E=

01 1( ) 0.5

1 2nf Ee

= = =+

Since, 1 FE E> , probability of finding electron

i.e. 1P in 1E will be close to zero. Since,

2 ,FE E< probability of finding electron i.e. 3P

in 2E will be close to one. Hence, probability of

not finding electron in 2E will be,

2 31 1 1 0P P= − ≈ − ≈

So, 1 2P P= Hence, the correct option is (B). Key Point 1. Fermi dirac distribution function describes the probability of existence of electron as a function of energy. 2. ( )nf E denotes the ratio of filled quantum state to the total quantum state at any energy level .E 3. Fermi dirac distribution function at 00 K is shown below,

This rectangular distribution implies that at 00 K, every available quantum state up to fE are filled with electron and all the

quantum state above FE are empty.

1;

( )0;

Fn

F

E Ef E

E E<

= >

4. Fermi dirac distribution function at is shown below,

(i) There is some non-zero probability

that some energy state above FE will be occupied by electron and some energy state below FE will be empty.

(ii) Fermi energy Level FE is defined as energy level at which fermi dirac distribution function ( )f E will be 50% of its maximum value.

(iii) In case of metal or conductor, ( ) 1nf E =

% ( ) 100%nf E =

5. The Fermi level represents the energy state with 50% probability of being filled if no forbidden band exists.

From Mass Action Law, 2

inp n=

where, in = Intrinsic carrier concentration

n = Free electron concentration p = Hole concentration E

( )nf E

FE

1

0

E

( )nf E

FE

1

0.5

0

1.2 (C)


0/2 3

0gE kT

in A T e−=

2 3in T∝

3/2in T∝

For intrinsic semiconductor, in p n= =

Hence, 3/2n T∝ Therefore, for intrinsic semiconductor, the free electron concentration depends on temperature of the semiconductor. Hence, the correct option is (C).

Einstein’s relation is given by,

11600

pnT

n p

DD TV= = =μ μ

… (i)

where, nD = Electron diffusion constant

pD = Hole diffusion constant

nμ = Electron mobility

pμ = Hole mobility

TV = Thermal voltage

T = Temperature in 0 K So, from equation (i), ratio of diffusion constant to mobility of carrier will depend upon temperature of semiconductor. Hence, the correct option is (A). Key Point For room temperature, 300KT = From Einstein’s relation,

300 0.0258611600

pn

n p

DD = = =μ μ

Therefore, 39n nDμ =

39p pDμ =

Given : nΔ = Excess electron concentration

due to illumination of light.

pΔ = Excess hole concentration due to illumination of light.

Due to illumination by light, electron and hole pair generation will occur. Number of generated hole and generated electrons are same because holes are absence of electrons. Therefore, n pΔ = Δ Hence, the correct option is (C). Key Point (i) Law of mass action is only valid for

thermal equilibrium. (ii) In case of excess carrier concentration,

law of mass action is not applicable.

Given : (i) The concentration of ionized acceptors

and donors in a semiconductor are ,A DN N respectively.

(ii) A DN N>

(iii) in = intrinsic concentration For intrinsic semiconductor, Number of electrons 0( )n = Number of holes 0( )p

When donor atom with concentration DN is added in an intrinsic semiconductor then semiconductor becomes n-type semiconductor. Hence, concentration of electron in this n-type semiconductor will increase by DN and concentration of hole will not change. Therefore, number of electrons 0 Dn N= +

number of holes 0p= When above obtained n-type semiconductor is doped with acceptor atom having concentration

AN then concentration of hole will increase by

AN and concentration of electron will not change.

1.3 (A)

1.4 (C)

1.5 (A)


Hence, number of holes 0 Ap N= +

number of electrons 0 Dn N= +

Case 1 : When A DN N= ,

Concentration of electron 0 Dn N= +

Concentration of hole 0 Ap N= +

Since, A DN N= and 0 0n p= Hence, Number of electrons = Number of holes Hence, semiconductor is intrinsic. Case 2 : When D AN N> ,

Concentration of hole 0( )Ap N+ <

Concentration of electron 0( )Dn N+

Since, semiconductor is of n-type ( )D AN N> Hence, effective number of electrons,

effD D AN N N= −

For a n-type semiconductor, the fermi level with respect to intrinsic level is given by,

ln eff

n i

DF F

i

NE E kT

n

= +

Case 3 : When A DN N> ,

Concentration of hole 0( )Ap N+ >

Concentration of electron 0( )Dn N+

Since, semiconductor is of p-type ( )A DN N> Hence, effective number of holes,

effA A DN N N= −

For a p-type semiconductor, the fermi level with respect to intrinsic level is given by,

ln eff

p i

AF F

i

NE E kT

n

= −

Energy band diagram of a p-type semiconductor is shown below,

Hence, the correct option is (A).

Drift velocity is given by,

dV E= μ

where, μ = Mobility

E = Applied electric field Waveform of mobility :

Waveform of drift velocity :

From above graphs : 1. Under high electric field, with

increasing electric field : (i) The mobility of charge carriers decreases as electric field increases. (ii) The drift velocity of charge carriers

saturates.

(eV)E

CE

FiE

FpE

VE

lnAeff

i

NkT

n

310 410 710(V/m)E

1/2E

1

E

0

2(m /V-sec)

310 410(V/m)E

(m/s)dV

710

Linear

Saturation

0

Sub-linear

1.6 (A) and (C)


2. For small electric field, with increasing electric field :

(i) Mobility of carrier is constant. (ii) Drift velocity increases linearly with

the applied electric field. Hence, the correct options are (A) and (C).

Given : (i) Donor concentration, DN = Phosphorus atoms 16 310 /cm= (ii) Acceptor concentration, AN = Boron atoms 16 32 10 /cm= ×

Since, A DN N>> hence it is p-type semiconductor. Therefore, majority carrier will be hole and minority carrier will be electron. Concentration of hole will be equal to the effective concentration of acceptor atom. Effective concentration of acceptor atom, 16 162 10 10

effA A DN N N= − = × −

16 310 /cmeffAN =

Therefore, concentration of hole, 16 310 /cm

effAp N= =

Hence, the correct option is (B).

Given : (i) Irradiation is a process of exposing

radiation. (ii) Electron concentration, 19 310 /cmDN = (iii) Excess electron concentration, 15 310 /cmnΔ = (iv) Minority carrier life time, 10 secpτ = μ

For n-type semiconductor generation rate is given by,

n p

n pg Δ Δ= =τ τ

For illumination of light, n pΔ = Δ

So, 15

610

10 10p

ng −

Δ= =τ ×

20 310 - pairs/cm /sg e h=


Refer Solution 1.6

Energy band diagram for metal is shown below,

At 00 KT > , conduction and valence band overlap each other in case of metal. Therefore, probability that an electron in a metal occupies Fermi level is 1. Hence, the correct option is (B).

Key Point (i) For insulator, fermi dirac probability

distribution function,

( ) 0

% ( ) 0%f E

f E=

=

(ii) For semiconductor, fermi dirac probability distribution function,

( ) 0.5

% ( ) 50%f E

f E=

=

(iii) For conductor (metal), fermi dirac probability distribution function,

( ) 1

% ( ) 100%f E

f E=

=

EC

EVCB

VB

Overlapping ofconduction bandand valance band

1.7 (B)

1.8 (A)

1.9 (D)

1.10 (B)


Given : (i) p-type silicon sample. (ii) Hole concentration, 15 32.25 10 /cmp = ×

(iii) Intrinsic concentration, 10 31.5 10 /cmin = ×

From mass action law, 2inp n=

Electron or minority carrier concentration is given by,

2 10 2 20

15 15(1.5 10 ) 2.25 102.25 10 2.25 10

innp

× ×= = =× ×

5 310 /cmn =

Hence, the correct option is (C). Key Point Mass action law is used to find the minority carrier concentration.

Given : (i) Electric field = 10 V/cm (ii) Distance = 1 cm, Time = 20 secμ

Since, velocity DistanceTime

=

So, 61 50000 cm/sec

20 10dV −= =×

Drift velocity is given by, dV E= μ

where, μ = Mobility, E = Electric field

5000010

dVE

μ = = 25000 cm /volt-sec=

Hence, the correct option is (C).

Given : (i) Intrinsic carrier density, 10 31.5 10 /cmin = ×

(ii) Concentration of doped atom 15 32.25 10 atom/cmDN = ×

Since, to create n-type semiconductor, intrinsic semiconductor is doped with donor atom. Hence, donor atom concentration, 15 32.25 10 / cmDN = ×

Concentration of majority carrier i.e. electron is, 15 32.25 10 atoms/cmDn N= = ×

From mass action law, 2

inp n=

Hole or minority carrier concentration is given by,

2 2 10 2

15(1.5 10 )2.25 10

i i

D

n npn N

×= = =×

5 310 /cmp =


Given : (i) n-type silicon bar. (ii) Majority carrier concentration, 20 35 10 /mn = ×

(iii) Length, 30.1 cm 10 ml −= =

(iv) Electron mobility, 20.13 m /V-secnμ =

(v) Cross-sectional area, 2 6 2 2100 m 100 (10 ) mA −= μ = ×

12 2100 10 mA −= ×

(vi) Electronic charge, 191.6 10 Cq −= ×

Resistance is given by, lRAρ=

where, resistivity, 1 1nnq

ρ = =σ μ

n

lRnq A

=μ

1.11 (C)

1.12 (C)

1.13 (C)

1.14 (A)


3

20 19 1210

(5 10 ) (1.6 10 ) (0.13) (100 10 )R

−

− −=× × ×

8

610 10104

R = ≈ Ω


Given : (i) Uniformly doped n-type Silicon. (ii) Diffusion constant, 235 cm /snD =

(iii) Electronic charge, 191.6 10 Cq −= × (iv) Electric field, 0 V/cmE =

Concentration gradient is given by,

2 1

2 1

n ndndx x x

−=−

16 17

20 44

6 10 10 2 10 / cm2 10 0

dndx −

× −= = − ×× −

Current density for n-type semiconductor is given by, (drift) (diffusion)n n nJ J J= +

n n ndnJ nq E qDdx

= μ +

n ndnJ qDdx

= [Given: 0E = ]

19 2035 1.6 10 ( 2) 10nJ −= × × × − ×

21120 A/cmnJ = − Hence, the correct option is (D).

n-type semiconductor is obtained by doping of pentavalent atoms. From given options, phosphorus is a pentavalent atom. Hence, the correct option is (D).

Key Point

Energy band gap in silicon is given by, 4( ) 1.21 3.60 10 eVgE T T− = − ×

At 0300 K,T = 0 4(300 K) 1.21 3.60 10 300 eVgE − = − × ×

0(300 K) 1.102 eVgE =

Hence, the correct option is (B). Key Point For Ge, energy band gap is given by, 4( ) 0.785 2.23 10 eVgE T T− = − ×

Given : (i) Intrinsic carrier concentration, 16 31.5 10 /min = × (ii) Majority carrier concentration 20 35 10 /m= × From mass action law, 2

inp n=

2Majority carrier Minority carrier concentration concentration in

× =

Minority carriers 2

Majority carriersin=

1 0x 2 2 mx

16 3

2 6 10 /cmn 17 3

1 10 /cmn

Donor orn-type impurities

Nitrogen (N )7

Phosphorus (P )15

Arsenic (As )33

Antimony (Sb )51

Bismuth (Bi )83

Acceptor orp-type impurities

Boron (B )5

Aluminium (Al )13

Gallium (Ga )31

Thallium (Tl )81

Indium (In )49

Semiconductor

Ge

Si

0.72 eV

1.1 eV

Eg(0 K)0

0.785 eV

1.21 eV

Eg(300 K)0

1.15 (D)

1.16 (D)

1.17 (B)

1.18 (A)


Minority carriers 16 2

20(1.5 10 )

5 10×=×

Minority carriers 11 34.5 10 /m= × Hence, the correct option is (A).

Given : (i) Uniformly doped n-type Si sample. (ii) Resistivity, 0.5 -cmnρ = Ω (iii) Electron mobility, 2μ 1250 cm /V-secn =

(iv) Electronic charge, 191.6 10 Cq −= × Resistivity of n-type semiconductor is given by,

1 1n

n D nN qρ = =

σ μ

1D

n n

Nq

=μ ρ 19

11.6 10 1250 0.5−=

× × ×

16 310 /cmDN = Hence, the correct option is (B).

Given : (i)

11 1.1 m when 1.12 eVgEλ = μ =

(ii) 22 0.87 m when ?gEλ = μ =

. Method 1 : Band gap (in eV) and wavelength (in m)μ are related by following equation,

1.24 1.24eV eV( m) 0.87gE = =

λ μ

1.425 eV 1.416 eVgE = ≈

Hence, the correct option is (A). . Method 2 :

Since, 1( m)gE ∝

λ μ

So, 1

2

2

1

g

g

EE

λ=λ

2

1.12 0.871.1gE

=

2

1.416 eVgE =

Hence, the correct option is (A). . Method 3 :

Since, 1( m)gE ∝

λ μ

Hence, for 1 2λ > λ ,

2 1g gE E>

Only option (A) satisfies this condition. Hence, the correct option is (A).

Refer Solution 1.17

Given : (i) Sample A is doped with boron, hence,

it is p-type semiconductor. So, 18 310 atom/cmAN = (ii) Sample B is doped with phosphorus

hence, it is -typen semiconductor.

So, 18 310 atom/cmDN = (iii) Ratio of electron to hole mobility,

3n

p

μ =μ

Conductivity of sample A is given by, A p A ppq N qσ = μ = μ [ ]Ap N=

Conductivity of sample B is given by, B n D nnq N qσ = μ = μ [ ]Dn N=

Ratio of conductivity of sample A to B is given by,

18

1810 110 3

A p A pA

B D n D n

N q NN q N

μ μσ ×= = =σ μ μ ×

13

pA

B n

μσ = =σ μ


1.19 (B)

1.20 (A)

1.21 (C)

1.22 (B)


Let us consider an n-type semiconductor, where electrons are the majority carriers and holes are the minority carriers. Dn N≈ (i.e. Donor ion concentration) From mass action law, 2

inp n=

Minority carriers, 2 2i i

D

n npn N

= ≈

Minority carriers, 1D

pN

∝

Hence, the correct option is (B). Key Point Minority carriers in an extrinsic semiconductor under equilibrium is, (i) Directly proportional to the square of

intrinsic concentration. (ii) Inversely proportional to the doping

concentration.

Low level injection is a process which injects minority carrier concentration into an extrinsic semiconductor. Generally, doping is used to increase majority carrier concentration and low level injection is used to increase minority carrier concentration. Assume uniformly doped n-type semiconductor, hence Dn N=

Let minority carriers be np , then

0

2

0i

n x nD

np pN= = =

When light is illuminated at one side of semiconductor, the covalent bonds are broken. So, the electron-hole ( - )e h− pair are generated.

If Δ number of e− and holes are generated, then total number of e− becomes , where D DN N+ Δ is very large compared to Δ .

So, D DN N+ Δ ≈

So, the change in majority carrier is negligible. Compared to minority carrier Δ is very large. Hence, we cannot neglect Δ as compared to minority carrier. At 0,x =

total number of minority carrier = np + Δ .

As x increases, the minority carrier will be reduced to

0.np

To establish equilibrium condition, the holes will move from higher concentration to the lower concentration, which is called diffusion. Because of low level injection, large number of minority carrier will be created and they will move from higher concentration to lower concentration due to diffusion process and the

DN DN DN DN DN

0np

0x x

0np0np

0np0np

0x x

DN DN DN

0np 0np

0np

Light

x

0np

0np

( )np x

0x x

DN DN DN

0np 0np

0np

holes

1.23 (B)

1.24 (A)


current resulting due to diffusion process is called diffusion current. Hence, the correct option is (A).

According to Lorentz’s law, at equilibrium ( ) 0qE q v B+ × =

0E v B+ × =

E v B= − ×

E B v= ×

Hence, the correct option is (B). Key Point 1. If a specimen (metal or semiconductor)

carrying a current I is placed in a transverse magnetic field, an electric field is induced in the direction perpendicular to both I and B. This phenomenon is known as the Hall effect.

Hall voltage is given by,

HBJd BIV E d Bvd

w= × = = =

ρ ρ

Hall coefficient is given by,

1 HH

V wRBI

= =ρ

2. Hall effect is used to determine, (i) The type of semiconductor (ii) Carrier concentration of specimen (iii)Conductivity

Given :

(i) Hole-electron mobility ratio, 0.4p

n

μ=

μ

(ii) Doping concentration, 8 34.2 10 atoms/mDN = × (iii) Intrinsic concentration, 4 31.5 10 atoms/min = × For n-type semiconductor conductivity is given by, n n D nn q N qσ = μ = μ …(i) For intrinsic semiconductor conductivity is given by, ( )i i n pn qσ = μ + μ …(ii)

From equations (i) and (ii),

( )

1

n D n D

pi i n pi

n

N q Nn q

n

σ μ= =μσ μ + μ

+ μ

8

44.2 10

1.5 10 (1 0.4)n

i

σ ×=σ × +

42 10 20000n

i

σ = × =σ

Hence, the correct option is (D).

According to law of electrical neutrality, D AN p N n⊕ −+ = + For p-type semiconductor, or 0A D DN N N>> = and p n>>

Hence, Ap N≈ From mass action law, 2

inp n=

2A inN n=

Electron concentration, 2i

A

nnN

=


Scan for

Video Solution

2

Y

Z

B

I

X

d

w

1

E

1.25 (B)

1.26 (D)

1.27 (D)


Given : (i) Boron is an acceptor atom so, silicon

doped with boron will be of p-type semiconductor.

(ii) Doping concentration, 17 34 10 atom/cm

effA AN N= = ×

(iii) Intrinsic carrier concentration, 10 31.5 10 cmin −= ×

(iv) 025 mV at 300 KkTq

=

Energy band diagram of a p-type semiconductor is shown below,

Position of Fermi level in p-type semiconductor with respect to intrinsic semiconductor is given by,

ln eVeff

i p

AF F

i

NE E kT

n− =

17

310

4 1025 10 ln1.5 10i pF FE E − ×− = × ×

0.427 eVi pF FE E− =

0.427p iF FE E= − eV

Hence, from above expression we can see that compared to undoped silicon, Fermi level of doped silicon goes down by 0.427 eV. Hence, the correct option is (C). Key Point 1. If we increase the doping level, fermi

level moves away from intrinsic fermi level.

2. If we increase the temperature, fermi level moves towards the intrinsic fermi level.

3. Shift in the position of fermi level with respect to intrinsic fermi level in n-type semiconductor is given by,

ln eff

n i

DF F

i

NE E kT

n

− =

4. Probability of electron occupying donor

energy level = 111

D FE EkTe

g

− +

5. Probability of hole occupying acceptor

energy level 111

F AE EkTe

g

−

=+

where, g = Degeneracy factor

DE = Donor energy level

FE = Fermi energy level

k = Boltzmann constant

5 08.625 10 eV/ K−= ×

AE = Acceptor energy level

Given : (i) n-type Si crystal. (ii) Intrinsic carrier concentration, 10 31.5 10 /cmin = ×

For option (A), Number of atoms per 3cm 225 10= × Hence, option (A) is wrong. For option (B), Let 19 34 10 /cmAp N≈ = ×

(eV)E

CE

FiE

FpE

VE

lneffA

i

NkT

n

1.28 (C)

1.29 (C)


From mass action law, Electron or minority concentration is given by,

2 10 2

319

(1.5 10 ) 5.625 /cm4 10

i

A

nnN

×= = =×

n p<< (Not possible for a n-type material) Hence, option (B) is wrong. For option (C), Let 19 34 10 /cmDn N≈ = ×

From mass action law, Hole or minority concentration is given by,

2 10 2

319

(1.5 10 ) 5.625 /cm4 10

i

D

npN

×= = =×

So, Dn N p≈ >>

Here, condition for a n-type material is satisfied. For option (D), Number of valence electron in one Silicon atom = 4 Density of Silicon atom in Silicon crystal 22 35 10 /cm= × Density of valence electron in silicon crystal 22 3 22 34 5 10 /cm 20 10 /cm= × × = × Hence, option (D) is wrong. Hence, the correct option is (C).

The Einstein’s relation is given by,

p nT

p n

D D V= =μ μ

Unit of 11unit of (Volt)n

n TD V−μ = =


Given : (i) 0300 KT = (ii) Electronic charge, 191.6 10q −= × C

(iii) Thermal voltage, TV = 26 mV (iv) Electron mobility, nμ = 21350 cm /V-s

(v) Donor concentration, DN = 16 310 /cm

. Method 1 : The sample is in thermal equilibrium. Hence, at any distance from x = 0 (end of the bar) electric field will be constant.

66

1 10 V/m1 10

VEd −= = =

×

10 kV/cmE = Hence, the correct option is (C). . Method 2 : Conductivity in n-type semiconductor is given by, n D nN qσ = μ 16 1910 1.6 10 1350−= × × ×

2.16 mho/cmnσ =

Length, 41 m = 10 cml −= μ Area of cross-section, unitA = Resistance is given by,

41 1 10

2.16lRA A

−

= = ×σ

Current is given by,

42.16

10V AIR −

×= = [ ]1 VV =

Current density is given by,

4 24

2.16 2.16 10 A/cm10

I AJA A−

×= = = ××

From ohm’s law, J E= σ

4

42.16 10 10 V/cm2.16

JE ×= = =σ

10 kV/cmE =

16 310 /cmDN

1 V

0x 1 mx

1.30 (A)

1.31 (C)


Hence, electric field is constant throughout the bar. Hence, the correct option is (C).

Electron drift current density is given by, n nJ E= σ

where, nσ = Electron conductivity

n D nJ N q E= μ [ ]n D nN qσ = μ 16 19 310 (1.6 10 ) 1350 (10 10 )nJ −= × × × × ×

4 22.16 10 A/cmnJ = × Hence, the correct option is (A).

Drift current density is given by,

IJ EA

= σ =

Total drift current density is given by, p nJ J J= +

p nJ p qE n qE= μ + μ

( )p nJ p n qE= μ + μ

So, ( )p nIJ p n qEA

= = μ + μ

( )p nI p n qEA= μ + μ

Since, Ap N= and Dn N=

Hence, ( )A p D nI N N qEA= μ + μ

Therefore, drift current depends upon, (i) Carrier concentration (ii) Doping concentration (iii) Cross sectional area (iv) Electric field (v) Mobility Hence, the correct option is (C).

Given : (i) Donor concentration, 15 32.25 10 atoms/cmDN = ×

(ii) Intrinsic carrier concentration, 10 31.5 10 /cmin = ×

Let concentration of electrons and holes are 0p

and 0n respectively at thermal equilibrium.

Total positive charge 0 Dp N= +

Total negative charge 0n= For complete impurity ionization, 0 0 Dn p N= +

0 Dn N≈ [ ]0 0n p>>

15 30 2.25 10 atoms/cmn = ×


0 0 ip n n=

2 2 10 2

0 150

(1.5 10 )2.25 10

i i

D

n npn N

×= = =×

5 30 10 /cmp =


Key Point Concept of complete impurity ionization : (i) For n-type semiconductor, the complete

impurity ionization is a process in which all donor atoms present in donor states ( )DE donate an electron to conduction band.

Fig. Energy band diagram of complete

ionization of donor states

Scan for

Video Solution

E

CE

DE

VE

Electrons are donated toconduction band

1.32 (A)

1.33 (C)

1.34 (D)


(ii) For p-type semiconductor, the complete impurity ionization is a process in which all acceptor atoms present in acceptor states ( )AE accept an electron from valence band.

Fig. Energy band diagram of complete

ionization of acceptor states

Given :

(i) 191.6 10 Cq −= × , 25 mVkTq

=

(ii) 21000 cm /V-secnμ =

(iii) 21 41 10 /cmdndx

= ×


nT

n

D V=μ

31000 25 10n n TD V −= μ = × ×

225 cm /secnD = Electron diffusion current density is given by,

n ndnJ q Ddx

=

19 211.6 10 25 10nJ −= × × ×

3 24 10 A/cmnJ = × Hence, the magnitude of electron diffusion current density is 4000 / 2A cm .

Let 0n and 0p are concentrations of electrons and holes in equilibrium.

When light is exposed then more covalent bond will be broken. Let, after illumination of light, number of generated electron is nΔ and number of generated hole is pΔ . n pΔ = Δ Hence, electron concentration becomes, 0n n n= + Δ and hole concentration becomes, 0p p p= + Δ Recombination rate is given by,

n p

n pg Δ Δ= =τ τ

… (i)

where, nΔ = excess minority carrier concentration pΔ = excess majority carrier concentration

nτ = mean life time of excess minority electrons in p-type semiconductor.

pτ = mean life time of holes.

Hence, the recombination rate is directly proportional to the excess minority carrier concentration. Hence, the correct option is (D). Key Point Recombination rate is, (i) Directly proportional to excess minority

carrier concentration. (ii) Inversely proportional to mean life time

of excess minority carrier concentration. (iii) Directly proportional to excess majority

carrier concentration. (iv) Inversely proportional to mean life time

of excess majority carrier concentration.

Given : (i) Hole mobility, 2500 cm /V-s,pμ =

(ii) 26 mVkTq

=

E

CE

AE

VE

Holes are accepting electronsfrom valence band

1.35 4000

1.36 (D)

1.37 13



pnT

n p

DD kTVq

= = =μ μ

Hole diffusion coefficient, p T pD V= μ

500 0.026pD = × 213 cm /sec=

Hence, the hole diffusion constant pD is 213 cm /sec .

Given : (i) 0300 KT = (ii) Energy band gap, 1.42 eVgE =

(iii) Intrinsic carrier concentration, 6 310 /cmin =

(iv) Plank’s constant 346.62 10 J-s−= ×

(v) Velocity of light 83 10 m/s= ×

(vi) Electronic charge, 191.6 10 Cq −= × Energy band gap of GaAs is shown below,

Minimum energy required to break the covalent bond (or for valence electron to jump from valence band to conduction band) is given by, 1.42 eVgE =

191.42 1.6 10 JoulesgE −= × ×

192.272 10 JoulesgE −= ×

Energy of each photon in radiation is given by,

hcE h= υ =λ

To break covalent bond, gE E≥ .

Hence, minmax

ghcE E= =

λ

34 8

max 196.62 10 3 10

2.272 10g

hcE

−

−

× × ×λ = =×

7max 8.74 10 m 0.874 m−λ = × = μ

0.874 mλ < μ

OR

minmax

1.24(eV)( m)

E =λ μ

maxmin

1.24 1.24( m) 0.874 m(eV) 1.42E

λ μ = = = μ


Intrinsic concentration of a semiconductor is given by, 0

/2 30

gE kTin A T e−=

0/23/2

0gE kT

in A T e−= … (i)

Intrinsic conductivity of a semiconductor is given by, ( )i i n pq nσ = μ + μ

Intrinsic resistivity of a semiconductor is given by,

1 1

( )ii i n pq n

ρ = =σ μ + μ

… (ii)

From equation (i) and equation (ii),

0

/23/20

1( ) gi E kT

n pq A T e−ρ =μ + μ

0

3/22

0

1( )

gE kTi

n p

T eq A

−

ρ =μ + μ

0

0

1 3ln ( ) ln ln2 2( )

gi

n p

ET

kTq A

ρ = − +

μ + μ

In above expression, with variation in

temperature, variation in term 3 ln2

T is very

small. Hence, it is considered as constant.

Conduction band

Valence bandEV

EC

E (eV)

1.42 eVgE

1.38 (A)

1.39 (A)


Let, 0

1 3ln ln2( )n p

T Cq A

− =

μ + μ

Then, 0 1ln ( )2

gi

EC

k T ρ = +

Compare above equation with y mx C= + ,

ln ,iy = ρ 0 1,2

gEm x

k T= =

Slope 0

2gEdym

dx k= = =

Hence, slope can be used to estimate the band gap gE .

Hence, the correct option is (A). Key Point (i) As temperature increases, gE decreases

and electron hole pair generation increases and hence σ increases, therefore iρ decreases.

(ii) Similarly, as temperature decreases, gE

increases and electron hole pair generation decreases, hence σ decreases and therefore iρ increases.

So, iρ increase as 1T

increases.

Given : 1.1 eVgE =

Relationship between energy band gap and wavelength is given by,

1.24(eV)( m)gE =

λ μ

1.24 1.24 1.1272 m1.1gE

λ = = = μ

1.127 mλ ≈ μ Hence, the cut-off wavelength of light is 1.127 mμ .

Given : (i) Intrinsic carrier concentration, 10 31.5 10 /cmin = × (ii) Donor concentration, 15 31 10 /cmDN = × (iii) Acceptor concentration, 18 31 10 /cmAN = ×

. Method 1 : If concentration of electrons and holes are n and p respectively then according to charge electrical neutrality, D AN p N n⊕ −+ = +

Since, ,A DN N>> hence it is p-type semiconductor. So, p n>>

Hence, 18 31 10 /cmAp N≈ = ×


inp n=

2 10 2

18(1.5 10 ) 225

1 10innp

×= = =×

Hence, the approximate number of electrons3/cm in the sample, at 0300 KT = is 225.

. Method 2 : Since, A DN N> Hence, doped Si sample will be of p-type. Effective acceptor ion doping is given by,

effA A DN N N= −

18 15 1810 10 10effAN = − ≈

1/T

ln ( )i

1.40 1.127

1.41 225


Majority carrier i.e. hole concentration is given by,

18 310 cmeffAp N −= =

From mass action law,

2inp n=

Electron or minority carrier concentration,

2 10 2

18(1.5 10 )

10innp

×= ≈

3225cmn −≈

Hence, the approximate number of electrons3/cm in the sample, at 0300 KT = is 225.

Given circuit is shown below,

Let 5 VV = , then terminal A will be at less positive voltage i.e. 0 volt and terminal B will be at higher positive voltage i.e. 5 volt. Hence, it looks like, terminal A will be less reverse biased and terminal B will be more reverse biased. As reverse bias will increase, energy levels will move down. Therefore, height of energy level towards terminal A will be higher and height of energy level toward terminal B will be lower.

Energy band diagram, before application of voltage V, is shown below,

Energy band diagram after application of voltage V is shown below,


Key Point An open circuited p-n diode with its energy band diagram is shown below,

where,

p nC CE E− and p nV VE E− are height of

potential barrier.


V

A B


5 VV

A B

Less positivevoltage

Higher positivevoltage

CE

FE

VE

CE

FE

VE

Scan for

Video Solution

p n

p nC CE E

nCE

pCE

p nV VE E

nVE

pVE

pFEnFE

Width of depletionlayer i.e. W

1.42 (D)


( )p nC CE E− ∝ built-in potential i.e. biV

( )p nV VE E− ∝ built-in potential i.e. biV

When forward biased is applied to p-n diode, then height of potential barrier and width of depletion layer will decrease therefore,

pCE will

move down and nCE will move up. Hence

p nC CE E− will decrease and in same way

p nV VE E− will also decrease. Therefore, electron

from nCE will jump to

pCE and hole from pVE

will jump to nVE .

Given : (i) Donor concentration, 16 310 /cmDN = (ii) Electron mobility, 21200 cm /V-snμ = (iii) Hole mobility, 2400 cm /V-spμ =

(iv) Electronic charge, 191.6 10 Cq −= × Given sample is an n-type semiconductor due to donor concentration. Resistivity of an n-type semiconductor is given by,

1 1n

n D nN qρ = =

σ μ

16 191

10 1.6 10 1200n −ρ =× × ×

0.52 -cm= Ω

Hence, the resistivity of the sample is 0.52 -cm .Ω

Given :

(i) Donor concentration, 16 310 /cmDN =

(ii) Electronic charge, 191.6 10 Cq −= × From the given graph, at concentration

16 310 /cm (i.e. 1. 16E + ), the mobility of electron 21200 cm /V-snμ = . Hence, conductivity of n-type semiconductor is given by, n n D nn q N qσ = μ = μ

16 1910 1.6 10 1200n−σ = × × ×

11.92 Scmn−σ =

Hence, the conductivity of the silicon at 0300 Kis 11.92 Scm− .

Given : (i) n-type Si sample. (ii) Electron-hole pairs generation rate, 20 310 /cm -secg = (iii) Minority carrier life time (i.e. hole), pτ 61 s 10 sec−= μ =

(iv) Number of hole generated in 1 sec 20 310 /cm=

Electron-hole pairs generation rate is given by,

p n

p ng Δ Δ= =τ τ

Hence, number of holes left after 610 sec− (i.e. under steady state condition) is given by, 20 6 14 310 10 10 /cmpp g −Δ = × τ = × =

1410 10x = [Given : 10xpΔ = ] x = 14 Hence, the value of x is 14.

Given :

(i) 15 3( ) 10 cmq xkTn x eα

− = 1. 13 1. 14 1. 15 1. 16 1. 17 1. 18 1. 19 1. 20E E E E E E E E

-3Doping concentration (cm )

200

400

600

800

1000

1200

1400

Mo

bil

ity

(cm

Vs

)2

-1-1

Hole and Electron Mobility in Silicon at 300 K0

Electron

Hole

1.43 0.52

1.44 1.92

1.45 14

1.46 (C)


(ii) 0.1 V/cmα = , 0.026 VkTq

=

(iii) 2 136 cm snD −= , D kTq

=μ

(iv) 236 cm /V-sec/ 0.026n n

nT

D DkT q V

μ = = =

15 3( ) 10 cmq xkTn x eα

− = … (i)

15 /( ) 10 q x kTdn x q edx kT

αα= ×

15 0

0

( ) 0.1100.026x

dn x edx =

= × ×

17

15

0

( ) 100 101026 26x

dn xdx =

= × =

For non-uniform semiconductor, the electron are moving from higher concentration to lower concentration. Hence, electron diffusion current density is given by,

1 0

0

( )( )n nxx

dn xJ x qDdx=

=

=

1

1719 10(0) 1.6 10 36

26nJ −= × × ×

1

21.6 36(0) 1026nJ −×= ×

1

2 2(0) 2.21 10 A/cmnJ −= × … (ii)

In above figure, slope of energy bands indicates applied electric field. Hence,

Electric field, ( ) 0.1 V/cmE x = − Electron drift current density is given by,

2 00( ) ( ) ( )n n xx

J x n x q E x==

= μ

From equation (i), 15 3(0) 10 /cmn =

Hence, 2

15 19 36(0) 10 1.6 10 ( 0.1)0.026nJ −= × × × × −

2

2 2(0) 2.21 10 A/cmnJ −= − × … (iii)

From equation (ii) and (iii), total electron current density at 0x = is given by,

1 2(0) (0) (0)n n nJ J J= +

2 2(0) 2.21 10 ( 2.21 10 )nJ − −= × + − ×

2(0) 0 A/cmnJ = Hence, the correct option is (C). Key Point If diagram is given in this way,

In that case, vedndx

= −

1( ) venJ x = −

1

2 2( ) 2.2 10 A/cmnJ x −= − ×

Given : (i) Length = 1 cm (ii) 16 310 cmDN −=

(iii) 2 1 11000 cm V sn− −μ =

( )Slope ve

dn x

dx

x

log ( ( ))n x

(eV)E

Slope 0.1eV/cm

CE

VE

0x x

log( ( ))n x

0x x

-Sin

1 cm

10 V

1.47 100


Electric field is given by,

VEd

= 10 10 V/cm1

= =

Drift velocity is given by, 1000 10dV E= μ = × 410 cm/sec=

The average time (in sμ ) taken by the electrons to move from one end of the bar to other end is given by,

41

10d

LtV

= =

410 sec 100 sect −= = μ Hence, the average time taken by the electrons to move from one end of the bar to other end is 100 secμ .

Given :

When a pentavalent impurity is introduced in an intrinsic semiconductor to form n-type semiconductor then an energy level is created just below the conduction band, known as donor energy level. Hence, the correct option is (A). Key Point When a trivalent impurity is introduced in an intrinsic semiconductor to form p-type semiconductor then an energy level is created just above the valence band, known as acceptor energy level.

Given : (i) 0300 KT =

(ii) Donor density, DN = 5× 1610 /cm3,

(iii) Optical generation rate,

optG = 1.5 × 2010 /cm3-s,

(iv) Hole carrier life time, 0pτ = 0.1 µs

(v) Electron carrier life time, 0 0.5 snτ = μ

Consider a n-type semiconductor having ‘0np ’

hole concentration at equilibrium and illuminated by light.

0(0) ' (0)n n np p p= + … (i)

where, (0)np is total hole concentration at

0,t = after illumination.

0np is thermally generated hole concentration

before illumination. ' (0)np is excess hole concentration at 0,t = due

to illumination. ( )np t = Hole concentration at any time t, after

illumination is stopped and it is given by,

0

/( ) '(0) ptn n np t p p e− τ= + … (ii)

Scan for

Video Solution

0.01 eV

New (Donor) Energy Level

Conduction Band

Valence Band

EC

EV

ED

0.01 eV

Acceptor Energy Level

Conduction Band

Valence Band

EC

EV

EA

time( )t

0

/( ) ' (0) pt

n n np t p p e

0np' (0)np

(0)np ( )np t

0np

1.48 (A)

1.49 (A)


Since, 0

' (0)n opt pp G= × τ 20 6 3' (0) (1.5 10 0.1 10 ) /cmnp −= × × × ' (0)np 13 31.5 10 /cm= × From mass action law, 2

inp n= where,

0np p= and Dn N=

Hence, 0

2i

nD

npN

=

0

2 10 23

16(1.5 10 ) /cm

5 10i

nD

npN

×= = ×

0

4 30.45 10 /cmnp = ×

(a) From equation (i), At 0t = sec, hole concentration is given by,

0

(0) '(0)n n np p p= +

13 4 3(0) (1.5 10 0.45 10 ) /cmnp = × + ×

13 3(0) (1.5 10 ) /cmnp × (b) From equation (ii), At 0.3 sec,t = μ hole

concentration is given by,

6

0

0

0.3 10 /(0.3) '(0) pn n np p p e

−− × τ= +

4 13(0.3) [0.45 10 1.5 10np = × + ×

6

60.3 10

0.1 10 3] / cme−

−

− × × ×

11 3(0.3) 7.46 10 /cmnp ≈ ×


Given :

Contact potential is given by,

0 2ln A D

i

N NV kTn

= … (i)

0/2 3

0gE kT

in A T e−= … (ii)

where, 0gE = Energy band gap

From equation (i) and (ii),

0 /0 3

0

ln gE kTA DN NV kT eA T

=

From the above equation, when 0V increases then

0gE will increase, hence

00 gV E∝ … (iii)

From the graph, 03 02 01V V V> >

Hence, from equation (iii), gZ gY gXE E E> >

Hence, the correct option is (C). Key Point gE Vγ∝

(Ge) (Si) (GaAs)x y z

V V Vγ γ γ< <

(Ge) (Si) (GaAs)g g gE E E< <

where, gE is energy band gap, and Vγ is cut-in

voltage.

Given : GaAs is doped with Si i.e. Si atom is doped into GaAs. Case 1 : Si+Ga (G-III) : Since Ga is G-III element, So number of valence electron in Ga is 3 but it requires 4 valence electrons to be as a part of compound semi–conductor. Ga will take one electron from Si. Therefore, Si will donate one electron to GaAs crystal. Since Si is donating the electron, Hence Si is acting as n-type dopant.

Scan for

Video Solution

I

V

X Y Z

01V 02V 03V0

1.50 (C)

1.51 (A)


Case 2 : Si+As (G-V) : Since As is G-V element, So number of valence electron in As is 5 but it requires only 4 valence electron to be as a part of compound semi-conductor. So, As will loose one electron and Si accepts one electron. Due to this there is a vacancy created in crystal that is treated as hole. Since, Si is able to create a hole in crystal it means Si is acting as p–type dopant. [Characteristic of compound semiconductor is each elements present in compound semiconductor should have 4 valence electrons.] Hence, the correct option is (A).

Given : (i) Electron concentration, 16 31 10 atoms/cmn = × ,

(ii) Electronic charge, 191.6 10 Cq −= × , (iii) Applied bias voltage, 5 volts,V =

(iv) 6 410 m 10 cmd − −= =

Electric field 45 V/cm

10VEd −= =

Drift velocity is given by, dV E= μ

From the given waveform, 55 10 V/cm,E = × 710 cm/secDV =

Hence, mobility is given by,

7

25

10 20 cm /V-sec5 10

dVE

μ = = =×

Current density is given by, n D nJ N q E= μ

16 194

510 20 1.6 1010nJ −

−= × × × ×

21.6 kA/cmnJ = Hence, the resulting current density is 21.6 kA/cm

Given : (i) Electron-hole pairs generation rate,

0

1L LxG GL

= −

(ii) 0

17 310 /cm sLG =

(iii) Length of the bar, 40.1 10 cmL −= × (iv) Hole diffusion coefficient, 2100cm /secpD =

(v) Hole life time, 410 sec,p−τ =

(vi) Electronic charge, 191.6 10 Cq −= ×

Hole diffusion current density is given by,

/2

p px L

dpJ qDdx =

= − … (i)

Concentration of hole in Si bar is given by, L pp G= × τ

0

4 17 41 10 10 1 10Lx xp GL L

− − = − × = − ×

13 310 1 /cmxpL

= −

710

55 10

Drift velocity (cm/s)

constant

linea

r

Electric field (V/cm)0

16 3Si ( 10 cm )DN

0.1 mL 0x

Light

0LG

PLx

( )LG x

0

1710

1.52 1.6

1.53 16


Hence, 13

13 1 1010 0dpdx L L

− = − =

13

18 44

10 10 /cm0.1 10

dpdx −

−= = −×

… (ii)

Since, hole concentration gradient is independent of length of Si bar hence, current density will be independent of the length of the silicon bar. From equations (i) and (ii), 19 181.6 10 100 ( 10 )pJ −= − × × × −

216 A/cmpJ =

Hence, the magnitude of the diffusion current density at x = L/2 is 216 A/cm .

Fermi energy level is constant in equilibrium and FE lies just above VE in p-type

semiconductor while FE lies just below CE in n-type semiconductor. As the concentration increases, Fermi energy level penetrates in case of n-type and p-type both CE and VE respectively. Hence, the correct option is (B).

Given : At T = 300 K, the effective density of states for holes is twice that of electrons i.e.,

2V CN N= and thermal voltage is 26 mV. Location of Fermi level is given by,

ln2 2

C V CF

V

E E NKTEN

+= −

The difference of Fermi level from mid band gap energy level is

1ln 9.01 meV2 2dKTE − = =


1.54 (B)

1.55 (A)

Topics : Characteristics of Diode, Clipper & Clamper Circuits, Rectifier and Zener Diode.

1.1 The 6 V Zener diode shown below has

zero Zener resistance and a knee current of 5 mA. The minimum value of R so that the voltage across R does not fall below 6 V is

(A) 1.2 kΩ (B) 50 Ω (C) 80 Ω (D) 0 Ω

1.2 The wave shape of 0V in figure is

(A)

(B)

(C)

(D)

1.3 A Zener diode in the circuit shown in

below figure, has a knee current of 5 mA, and a maximum allowed power dissipation of 300 mW. What are the minimum and maximum load currents that can be drawn safely from the

1Diode

Circuits & Applications

50

10 V R6 V

I

6 V

10k10sin 314t0V~

4.1V 4.1V

0V

5.9

5.9

0 t

0V

4.1

4.1

0 t

0V

4.1

4.1

0 t

0V

5.9

0 t

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1993 IIT Bombay

1996 IISc Bangalore


circuit, keeping the output voltage 0V constant at 6 V?

(A) 0 mA, 180 mA (B) 5 mA, 110 mA (C) 10 mA, 55 mA (D) 60 mA, 180 mA

1.4 For full wave rectification, a four diode

bridge rectifier is claimed to have the following advantages over a two diode circuit.

1. Less expensive transformer 2. Smaller size transformer 3. Suitability for higher voltage

application of these (A) Only (1) and (2) are true (B) Only (1) and (3) are true (C) Only (2) and (3) are true (D) (1), (2), as well as (3) are true 1.5 For small signal a.c. operation, a

practical forward biased diode can be modeled as

(A) a resistance and a capacitance in series.

(B) an ideal diode and resistance in parallel.

(C) a resistance and an ideal diode in series.

(D) a resistance.

1.6 A dc power supply has a no-load voltage

of 30 V, and a full-load voltage of 25 V at a full-load current of 1 A. Its output resistance and load regulation, respectively, are

(A) 5 Ω and 20% (B) 25 Ω and 20% (C) 5 Ω and 16.7% (D) 25 Ω and 16.7%

1.7 For the circuit in the figure, the voltage

0v is

(A) 2 V (B) 1 V (C) – 1 V (D) None of these

1.8 A Zener diode regulator in the figure is

to be designed to meet the specifications

LI = 10 mA, 0V = 10 V and inV varies from 30 V to 50 V. The Zener diode has

ZV = 10 V and ZKI (knee current) = 1 mA. For satisfactory operation

(A) 1800R ≤ Ω (B) 2000 2200RΩ ≤ ≤ Ω (C) 3700 4000RΩ ≤ ≤ Ω (D) 4000R > Ω

1.9 The circuit shown in the figure is best

described as a

50

9 V

Load

0V

2 2

2 0v2 V4 V

R

ZI 10mALI

inV 0V

ZV LR

~Output

iV

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1999 IIT Bombay

2000 IIT Kharagpur

2002 IISc Bangalore

2003 IIT Madras

2.3GATE ACADEMY ® Analog Electronics : Diode Circuits & Applications

(A) bridge rectifier

(B) ring modulator

(C) frequency discriminatory

(D) voltage doubler

1.10 In the voltage regulator shown in the

figure, the load current can vary from 100 mA to 500 mA. Assuming that the Zener diode is ideal (i.e., the Zener knee current is negligibly small and Zener resistance is zero in the breakdown region), the value of R is

(A) 7 Ω (B) 70 Ω

(C) 703

Ω (D) 14 Ω

1.11 In a full-wave rectifier using two ideal diodes, dcV and mV are the dc and peak values of the voltage respectively across a resistive load. If PIV is the peak inverse voltage of the diode, then the appropriate relationships for this rectifier are

(A) , PIV 2mdc m

VV V= =π

(B) 2 , PIV 2mdc m

VV V= =π

(C) 2 , PIVmdc m

VV V= =π

(D) , PIVmdc m

VV V= =π

1.12 The Zener diode in the regulator circuit

shown in the figure has a Zener voltage of 5.8 volts and a Zener knee current of 0.5 mA. The maximum load current drawn from this circuit ensuring proper functioning over the input voltage range between 20 volts and 30 volts, is

(A) 23.7 mA (B) 14.2 mA (C) 13.7 mA (D) 24.2 mA

1.13 For the circuit shown below, assume

that the Zener diode is ideal with a breakdown voltage of 6 volts. The waveform observed across R is

(A)

(B)

(C)

5V12 V

R

Variable 100to 500 mA

1k

inV (20 V -30 V) LoadVZ = 5.8 V

RV

6 V

~12sin t R

6 V

6 V

12 V

12 V

6 V

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2005 IIT Bombay

2006 IIT Kharagpur


(D)

1.14 The correct full wave rectifier circuit is (A)

(B)

(C)

(D)

1.15 For the Zener diode shown in the figure, the Zener voltage at knee is 7 V, the knee current is negligible and the Zener dynamic resistance is 10 Ω . If the input voltage ( )iV ranges from 10 V to 16 V, the output voltage 0( )V ranges from

(A) 7.00 to 7.29 V (B) 7.14 to 7.29 V (C) 7.14 to 7.43 V (D) 7.29 to 7.43 V

1.16 In the following limiter circuit, an input

voltage 10sin100 π ViV t= is applied. Assume that the diode drop is 0.7 V when it is forward biased. The Zener breakdown voltage is 6.8 V.

The maximum and minimum values of

the output voltage respectively are (A) 6.1 V, − 0.7 V (B)0.7 V, − 7.5 V (C) 7.5 V, − 0.7 V (D)7.5 V, − 7.5 V

1.17 In the circuit below, the diode is ideal.

The voltage V is given by

6 V

Outp

utIn

put

Ou

tpu

tInp

ut

Outp

utIn

put

Ou

tpu

tInp

ut

iV

200

0V

iV

1k

0V2D

1D

Z 6.8V

iV 1AD

1 1

V

2007 IIT Kanpur

2008 IISc Bangalore

2009 IIT Roorkee


(A) min ( ,1)iV (B) max ( ,1)iV

(C) min ( ,1)iV− (D) max ( ,1)iV−

In the circuit shown below, assume that the voltage drop across a forward biased diode is 0.7 V. The thermal voltage / 25 mVTV kT q= = The small signal input voltage cos ( )i pV V t= ω

where 100 mVpV = .

1.18 The bias current DCI through the diodes

is

(A) 1 mA (B) 1.28 mA

(C) 1.5 mA (D) 2 mA

1.19 The ac output voltage acV is

(A) 0.25 cos( ) mVtω

(B) 1cos( ) mVtω

(C) 2cos( ) mVtω

(D) 22cos( ) mVtω

1.20 The diodes and capacitors in the circuit

shown are ideal. The voltage V(t) across the diode 1D is

(A) cos( ) 1tω − (B) sin( )tω (C) 1 cos( )t− ω (D) 1 sin( )t− ω 1.21 The i-v characteristics of the diode in the

circuit given below are

0 0.7 V0.7 A 0.7 V

500

vi v v

<= − ≥

The current in the circuit is (A) 10 mA (B) 9.3 mA (B) 6.67 mA (D) 6.2 mA

1.22 In the circuit shown below, the knee

current of the ideal Zener diode is 10 mA. To maintain 5 V across LR , the minimum

value of LR in Ω and the minimum power rating of the Zener diode in mW, respectively, are

(A) 125 and 125 (B) 125 and 250 (C) 250 and 125 (D) 250 and 250

12.7 V

DC acI i DC acV V

iV

9900

cos( )t ~

1D

2C

1C 2D( )V t

i

10 V

1k

v

10 V

5VZ

V L

R

100LoadI

2011 IIT Madras


2012 IIT Delhi

2013 IIT Bombay


1.23 A voltage 1000sin tω Volts is applied across YZ. Assuming ideal diodes, the voltage measured across WX in Volts, is

(A) sin tω (B) (sin sin )/2t tω + ω

(C) (sin sin )/2t tω − ω

(D) 0 for all t

1.24 The figure shows a half-wave rectifier.

The diode D is ideal. The average steady state current (in Amperes) through the diode is approximately ________. [Set - 03]

1.25 Two silicon diodes with a forward

voltage drop of 0.7 V are used in the circuit shown in the figure. The range of input voltage iV for which the output

voltage 0 iV V= , is [Set - 04]

(A) 0.3V 1.3ViV− < <

(B) 0.3V 2.0 ViV− < <

(C) 1.0V 2.0ViV− < <

(D) 1.7 V 2.7 ViV− < < 1.26 In the figure, assume that the forward

voltage drops of the PN diode 1D and

Schottky diode 2D are 0.7 V and 0.3 V, respectively. If ON denotes conducting state of the diode and OFF denotes nonconducting state of the diode, then in the circuit,

(A) Both 1D and 2D are ON.

(B) 1D is ON and 2D is OFF.

(C) Both 1D and 2D are OFF.

(D) 1D is OFF and 2D is ON. 1.27 The diode in the circuit shown has

0.7VonV = but is ideal otherwise. If

5sin( )ViV t= ω , the minimum and

maximum values of 0V (in Volts) are, respectively,

(A) –5 and 2.7 (B) 2.7 and 5 (C) –5 and 3.85 (D) 1.3 and 5

W

Y

1k

Z

X

1k

R C

D

100~

4 mF10sint

= 50 Hzf

0ViV

R

1V

1D 2D

2 V

2D1D

1k 20k

10 V

0V

2R1R

1k

1k

2 V

iV

2014 IIT Kharagpur


1.28 For the circuit with ideal diodes shown

in the figure, the shape of the output ( )outV for the given sine wave input ( )inV will be [Set - 01]

(A)

(B)

(C)

(D)

1.29 In the circuit shown below, the Zener diode is ideal and the Zener voltage is 6 V. The output voltage 0V (in Volts) is _______. [Set - 01]

1.30 If the circuit shown has to function as a

clamping circuit then which one of the following conditions should be satisfied for the sinusoidal signal of period T ?

[Set - 02]

(A) RC T<< (B) 0.35RC T= (C) RC T≈ (D) RC T>> 1.31 The diode in the circuit given below has

0.7VONV = but is ideal otherwise. The current (in mA) in the 4 kΩ resistor is _______. [Set - 02]

1.32 In the circuit shown, assume that diodes

1D and 2D are ideal. In the steady-state condition, the average voltage abV (in Volts) across the 0.5 Fμ capacitor is ______. [Set - 03]

1.33 In the circuit shown, assume that the

diodes 1D and 2D are ideal. The average value of voltage abV (in Volts), across terminals ‘a’ and ‘b’ is ______.

[Set - 03]

1.34 An AC voltage source V = 10 sin(t)

volts is applied to the following network. Assume that 1R = 3 kΩ, 2R = 6 kΩ and 3R = 9 kΩ, and that the diode is ideal. [Set - 01]

0.5 T T0 Vin Vout

++

−−

0.5 T T0

0.5 T T0

0.5 T T0

0.5 T T0

10 V 0V

1k

1k

C

V ~ R

1mA

2k 3k

6k4k

1kD

2D~50sin t

1 F

abVab

0.5 F

1D

10k

1D 2D

~ b6 sin( )t

20k10k

a

abV

2015 IIT Kanpur

2016 IISc Bangalore


RMS current rmsI (in mA) through the

diode is ________. 1.35 The figure shows a half-wave rectifier

with a 475 Fμ filter capacitor. The load draws a constant current 0 1AI = from the rectifier. The figure also shows the input voltage iV , the output voltage CV and the peak-to-peak voltage ripple u on

CV . The input voltage iV is a triangle-wave with an amplitude of 10 V and a period of 1 ms. [Set - 02]

The value of the ripple u (in volts) is

_______.

1.36 Assume that the diode in the figure has 0.7VonV = , but is otherwise ideal.

The magnitude of the current 2I is (in

mA) is equal to _______. [Set - 02] 1.37 The diodes 1D and 2D in the figure are

ideal and the capacitors are identical. The product RC is very large compared to the time period of the ac voltage. Assuming that the diodes do not breakdown in the reverse bias, the output voltage 0V (in volt) at the steady state is [Set - 03]

1.38 The output 0V of the diode circuit

shown in the figure is connected to an averaging DC voltmeter. The reading on the DC voltmeter in volts, neglecting the voltage drop across the diode, is _______. [Set - 02]

R1

R2

R1 R2

d c

ba

R2

R3

R2 R3

e h

gf

R3

R2

R1

R2

rmsI

10sin( )V t

iV

CV475 F

0 1 AI

iV

10 V

10 V

0 V t

0 V t

CV

u

6k2R

1R

2I

2 V

2k

R~ 0V

2D

C10sin t

ac voltageC

1D

1 k10sin50 Hz

t

f

0V

2017 IIT Roorkee


1.39 In the figure, 1D is a real silicon p-n junction diode with a drop of 0.7 V under forward bias condition and 2D is a Zener diode with breakdown voltage of 6.8V− . The input ( )inV t is a periodic square wave of period T , whose one period is shown in the figure.

Assuming 10 Tτ << , where τ is the

time constant of the circuit, the maximum and minimum values of the output waveform are respectively.

[Set - 02] (A) 7.5 V and – 20.5 V (B) 6.1 V and – 21.9 V (C) 7.5 V and – 21.2 V (D) 6.1 V and – 22.6 V

1.40 The circuit shown in the figure is used

to provide regulated voltage (5V) across the 1kΩ resistance. Assume that the Zener diode has a constant reverse breakdown voltage for a current range, starting from a minimum required Zener current

min2 mAZI = to its maximum

allowable current. The input voltage IV may vary by 5% from its nominal value of 6 V. The resistance of the diode in the breakdown region is negligible.

The value of R and the minimum

required power dissipation rating of the diode, respectively, are

(A) 186 and 10 mWΩ

(B) 100 and 40 mWΩ

(C) 100 and 10 mWΩ

(D) 186 and 40 mWΩ

1.41 A dc current of 26 Aμ flows through the circuit shown. The diode in the circuit is forward biased and it has an ideality factor of one. At the quiescent point, the diode has a junction capacitance of 0.5 nF. Its neutral region resistances can be neglected. Assume that the room temperature thermal equivalent voltage is 26 mV.

For 62 10 rad/secω = × , the amplitude

of the small-signal component of diode current (in Aμ , correct to one decimal place) is _________.

t (sec)

14 V

– 14 V

( )in

V t

T

10

10 F

1D

2D

( )outV t( )

inV t

R

1kIV 5V

100 5sin( ) mVt

V

2018 IIT Guwahati


1.42 In the circuit shown, sV is a square wave

of period T with maximum and minimum values of 8 V and – 10 V, respectively. Assume that the diode is ideal and 1 2 50R R= = Ω . The average

value of LV is_______ volts (rounded off to 1 decimal place).

1.43 In the circuit shown, the breakdown

voltage and maximum current of Zener diode are 20 V and 60 mA, respectively. The values of 1R and LR are 200 Ω and 1 kΩ , respectively.

What is range of iV that will maintain

the Zener diode in ‘on state’? (A) 18 V to 24 V (B) 20 V to 28 V (C) 22 V to 34 V (D) 24 V to 36 V

1.44 In the circuit shown below, all the

components are ideal and the input voltage is sinusoidal. The magnitude of the steady - state output 0V (rounded off to two decimal places) is ______V.

LR

1R

10−

8

T2T sV

+

−

LV0

sV

t

1R

+_LRiV

230 V (rms)

1 0.1 FC = μ2D

1D2 0.1 FC = μ 0V

2019 IIT Madras

2020 IIT Delhi


1.1 C 1.2 A 1.3 C 1.4 D 1.5 D

1.6 A 1.7 D 1.8 A 1.9 D 1.10 D

1.11 B 1.12 C 1.13 B 1.14 C 1.15 C

1.16 C 1.17 A 1.18 A 1.19 B 1.20 A

1.21 D 1.22 B 1.23 D 1.24 0.1 1.25 D

1.26 A 1.27 C 1.28 C 1.29 5 1.30 D

1.31 0.6 1.32 100 1.33 5 1.34 1 1.35 2.105

1.36 0.25 1.37 0 1.38 3.18 1.39 A 1.40 B

1.41 6.4 1.42 – 3 1.43 D 1.44 650.538

Concept of Clamper & Peak Detector Circuits

Multiplier Circuit by using Clamper & Peak Detector Circuits

Given : 6VZV = , (min) 5 mAknee ZI I= =

The regulator circuit is shown below,

Concept of load regulation : (i) Variable load resistor. (ii) Fixed input voltage (or fixed input

current).

For variable load R, input current I is given by, (min) (max)Z LI I I= + … (i)

(max) (min)Z LI I I= + … (ii)

Current through 50 Ω resistor is given by,

10 6 80 mA50

in zV VIR− −= = =

From equation (i), (max)80 5 LI= + (min)Given : 5 mAZI = (max) 75 mALI = For minimum value of R, current LI has to be maximum.

min(max) (max)

Z R

L L

V VRI I

= =

min 36 80

75 10R −= = Ω

×



Scan for Video

Explanation

Scan for Video

Explanation

50

10 V

ZI

LI

R6 V

I

6 V

ZV

10k10sin 314iV t0V~

1D 2D

4.1V 4.1V

Answers Diode Circuits & Applications

Explanations Diode Circuits & Applications

1.1 (C)

1.2 (A)


Source voltage is sinusoidal hence, consider two cases for positive and negative half cycles. Case 1 : During positive half cycle, Diode 1D will be in forward bias, hence it acts as a short circuit. (a) 0 4.1 ViV< < : When input voltage is less than

breakdown voltage of 2D , 2D will act as normal diode in reverse bias without breakdown and therefore, 2D can be replaced by open circuit.

Hence, 0 0 10 0 VV = × =

(b) 4.1 ViV > : When input voltage will be higher than

breakdown voltage of 2D , 2D will work in reverse bias breakdown region. Therefore, 2D can be replaced by its breakdown voltage.

Applying KVL in above figure, 04.1 0iV V− + + =

0 4.1 10sin ω 4.1iV V t= − = −

Case 2 : During negative half cycle, Diode 2D will be in forward bias, hence acts as a short circuit. (a) 4.1V 0 ViV− < < : When input voltage will be less than

breakdown voltage of 1D , 1D will act as a normal diode in reverse bias. Therefore, 1D can be replaced by open circuit.

Hence, 0 0 10 0 VV = × =

(b) 4.1 ViV < − : When input voltage will be higher than

breakdown voltage of 1D , 1D will work in reverse bias breakdown region. Therefore, 1D can be replaced by its breakdown voltage.

Applying KVL in above figure, 04.1 0iV V− − + =

0 4.1iV V= + Hence, from case (1) and case (2), expression of output voltage can be written as given below,

0

0V ; 0V 4.1V4.1 ; 4.1V

0V ; 4.1V 0V4.1 ; 4.1V

i

i i

i

i i

VV V

VV

V V

< < − >= − < < + < −

0V~ iV

O.C.

10 k

S.C. 0 A

0V~ iV 10 k

4.1V

S.C.

0V

5.9 V

t

0V~ iV

O.C.

10 k

S.C.

0 A

0V~ iV 10 k

S.C.

4.1V


Hence, the output waveform is shown below,


Given : (i) Knee current, (min) 5mAZI =

(ii) Maximum allowed power dissipation = 300 mW Regulator circuit is shown below,


current). For variable load, input current inI is given by,

(min) (max)in Z LI I I= + … (i)

(max) (min)in Z LI I I= + … (ii)

Input current through 50 Ω resistor is given by,

9 6 350 50inI −= = 60 mA=

Maximum power dissipation in Zener diode is given by, (max) (max) 300 mWZ Z ZP I V= =

3

(max)300 10

6ZI−×= 50 mA=

From equation (ii), (min)60 50 LI= + (min) 10 mALI = From equation (i), (max)60 5 LI= + (max) 55 mALI = Hence, the correct option is (C).

A four diode bridge rectifier uses a smaller size of transformer, which is less expensive transformer and these rectifiers are suitable for higher voltage applications because of low peak inverse voltage (PIV) rating requirement of each diode. Hence, the correct option is (D).

Fig. (a) Equivalent circuit representation of above practical diode is series combination of a forward resistance and a cut-in voltage, as shown below,

Fig. (b)

In small signal ac analysis, dc voltage source is replaced by short circuit. Hence, ac equivalent circuit of practical diode is shown below,

Fig. (c)

Therefore, from above circuit, it is clear that in ac analysis any practical diode is represented by a forward resistance only. Hence, the correct option is (D).

Given : (i) 30 VNo loadV = , 25V,Full loadV = (ii) 1 AFull load DCI I= =

t

iV

0V

10

5.94.1

4.1

5.9

10

/ 2

3 / 2

0V

2 1

50

50

9 V

ZI

ZI

0 6 VV

inI

Z

V

A B

V

A BRf

A BRf

S.C.

1.3 (C)

1.4 (D)

1.5 (D)

1.6 (A)


Voltage Regulation (in %) is given by,

% regulation 100%No load Full load

Full load

V VV

−= ×

% regulation 30 25 100%25−= ×

% regulation 1 100% 20%5

= × =

Output resistance, No load Full load

DC

V VR

I−

=

30 25 51

R −= = Ω



(i) Assume diode is in forward bias i.e. ON. (ii) For diode to be ON, current I should be

greater than zero.

Apply KCL at node AV ,

2 4 02 2 2A A AV V V+ −+ + =

4 3 2AV= +

2 V3AV =

From figure, 1 A2 3AVI = =

Since, I is positive, so our assumption is correct i.e. diode is ON.

From figure,

022 V3Av V I= = =


Given : (i) Load current, 10mALI =

(ii) Output voltage, 0 10VV = (iii) (min) 30 VinV = , (max) 50 VinV = ,

(iv) 10VZV = , ZKI (knee current) = 1 mA Regulator circuit is shown below,

Concept of line regulation : (i) Fixed load resistor. (ii) Variable input voltage (or variable input

current). Minimum input current is given by, min (min)Z LI I I= +

min 1 10 11 mAI = + = From figure, input current I is given by,

in ZV VIR−=

in ZV VRI−=

Maximum value of R is given by,

maxmin

in ZV VRI−=

For 50 VinV = ,

max 350 10 3636.4

11 10R −

−= = Ω×

For 30 VinV = ,

max 330 10 1818.2

11 10R −

−= = Ω×

2 2

2 0v2 V4 V

2 2

2 0v

2 V4 V

AV

I

S.C.

R

ZI 10mALI

inV

ZV LR

I

1.7 (D)

1.8 (A)


The value of R should be selected in such a way that circuit works satisfactorily for all the value of input voltage between 30 V to 50 V. If

max 3636.4R = Ω is selected then value of I falls below 11 mA which is not required since,

min 11 mA.I = Hence, for max 1818.2R = Ω , the regulator circuit works satisfactory for the whole range of input. Hence, the correct option is (A).


Output voltage 0V is given by,

2 10 C CV V V= −

Case 1 : During 1st positive half cycle, Diode 1D is in reverse bias, hence OFF.

Diode 2D is in forward bias, hence ON.

Capacitor 2C will start charging from zero volt

to maximum of input voltage i.e. mV .

In this case, charging time constant, 0 secCτ =

Therefore, 2C mV V=

Case 2 : During 1st negative half cycle, Diode 1D is in forward bias, hence ON.

Diode 2D is in reverse bias, hence OFF.

Capacitor 1C starts charging in negative

direction from zero volt to minimum of input voltage i.e. mV− . In this case, charging time constant,

0 secCτ =

Therefore, 1C mV V= −

Case 3 : During 2nd positive half cycle, The positive terminal of diode 1D is at potential

mV− . Diode 1D will be forward biased when the potential of negative terminal of 1D will be lower or more negative than mV− . Since, the potential of negative terminal of 1D can not be less than mV− , hence 1D will be always reverse biased. The negative terminal of diode 2D is at potential

mV . Diode 2D will be forward biased when the potential of positive terminal of 2D will be higher than mV . Since, the potential of positive terminal of 2D can not be higher than mV , hence 2D will be always reverse biased.

At steady state, diodes will be OFF and capacitors will not discharge. From figure,

2 10 C CV V V= −

0 ( )m mV V V= − − 0 2 mV V=

iV ~

0V2D1D

2CV

1CV1C 2C

~iV 0V

1CV

2CV

O.C.S.C.

~iV 0V

1CV

2CV

O.C.S.C.

~iV 0V

1CV2CV

O.C.O.C.

1.9 (D)


OR At steady state :

Applying KVL in loop shown in figure, 0 0m mV V V+ − + =

0 2 mV V=

Thus, the circuit behaves like a voltage doubler circuit. Hence, the correct option is (D). Key Point In the above question, polarity of 0V is not

given. So if we assume polarity of 0V as shown in figure below,

Then, 0 2 mV V= − (Voltage doubler circuit)

Given : (i) (min) 100 mALI = , (max) 500 mALI =

(ii) 12ViV = , 5 VzV = , (min) 0 AZI =

Regulator circuit is shown below,


current). For variable load, input current I is given by, (min) (max)Z LI I I= + … (i)

(max) (min)Z LI I I= + … (ii)

Note : In any voltage regulator circuit, Zener diode will always work in breakdown region to provide a constant output voltage. Current through resistor R is given by,

12 12 5 7zVIR R R− −= = = … (iii)

From equation (i), 0 500 500 mA 0.5 AI = + = = From equation (i),

70.5R

=

14R = Ω Hence, the correct option is (D).

Circuit of centre tapped full wave rectifier is shown below,

~iV 0V

O.C.O.C.

mV mV

iV~

0V

2D1D

2C

V1C

V1C 2C

5V12 V

IZI LI

R

0 ( )V tsinmV t

1D

2D

( )i

V t LR

2

t

mV

( )i

V t

0 ( )V t

mV

2 3

3

4

4

mV

0

0 t

1.10 (D)

1.11 (B)


(i) DC or average value of output voltage ( )dcV :

DC or average value of output voltage is given by,

00

1 ( ) (ω )T

dcV V t d tT

=

0

1 sin ω (ω )dc mV V t d tπ

=π

0

1 cosdc mV V t π= − ωπ

2 mdc

VV =π

(ii) Peak inverse voltage (PIV) : Peak inverse voltage (PIV) or peak

reverse voltage (PRV) is the maximum value of reverse voltage which occurs at the peak of the input cycle when the diode is in reverse-bias.

OR PIV represents maximum reverse bias

voltage across diode for which diode can operate safely without breakdown.

Case 1 : During positive half cycle,

1D will be forward biased and 2D will be reverse biased.

Applying KVL in loop shown in figure, PIV 0m mV V+ − =

PIV 2 mV=

Case 2 : During negative half cycle, Diode 2D will be forward biased and Diode 1D will be reverse biased.

Applying KVL in loop shown in figure, PIV 0m mV V+ − = PIV 2 mV= Hence, the correct option is (B). Key Point (i) Half wave rectifier :

, PIVmdc m

VV V= =π

(ii) Full wave rectifier :

2 , PIV 2mdc m

VV V= =π

(iii) Full wave bridge rectifier :

2 , PIVmdc m

VV V= =π

Given : (i) 5.8 VoltszV = (ii) Knee current, (min) 0.5mAZI =

(iii) (min) 20 VinV = , (max) 30 VinV =


Concept of line regulation : (i) Fixed load resistor. (ii) Variable input voltage (or variable input

current). For variable input, current inI is given by, (min) (min) (fixed)in Z LI I I= + … (i)

(max) (max) (fixed)in Z LI I I= + … (ii)

0VLR

mV

mV

PIV

iV

S.C.

0VLR

mV

mV

iV

PIV

S.C.

inV

inI

ZI L

I

1k

(20-30 V)L

RZ

V

inR

1.12 (C)


Current through resistor inR is given by,

(min)(min)

in Zin

in

V VI

R−

= … (iii)

From equation (i) and (iii),

(min)(min) (fixed)

in ZZ L

in

V VI I

R−

= +

(fixed)20 5.8 0.5

1 LI− = +

(fixed) (20 5.8) 0.5LI = − −

(fixed) 14.2 0.5 13.7 mALI = − = This fixed load current may be referred as maximum load current. Hence, the correct option is (C). Key Point Consider the condition that input voltage and load resistance both are variable as shown below,

In this case, the conditions for minimum and maximum Zener currents are given as (i) (min)ZI occurs when the load current is

maximum and the source voltage is minimum.

(ii) (max)ZI occurs when the load current is minimum and the source voltage is maximum.



(a) 0 6 ViV< < :

When voltage across Zener diode will be less than its break down voltage, Zener diode will act as a normal diode in reverse bias. Therefore, Zener diode can be replaced by open circuit.

From figure, 0 0VRV R= × =

(b) 6 ViV > : When voltage across Zener diode will be higher than its break down voltage, Zener diode will start to work in reverse bias break down region. Therefore, Zener diode can be replaced by its break down voltage.

Applying KVL in above figure, 6 0i RV V− + + =

6R iV V= −

Hence, from case 1(a) and case 1(b),

0V ; 0V 6V

6 ; 6Vi

Ri i

VV

V V< <

= − >

inV

inI

ZI

LI

LR

ZV

inR

0V

RV

6 V

~

ZI

12sin t R

RV

6 V

~ RiV

RV~ Ri

V

O.C.

0 A

RV

6 V

~ RiV

1.13 (B)


Case 2 : During negative half cycle, Zener diode will be forward bias and hence it will work as a normal diode. Therefore, Zener diode can be replaced by short circuit.

From figure, R iV V= Hence, from case 1 and case 2, the expression of output voltage can be summarized as given below,

; 0 V0 V ; 0 V 6 V

6 ; 6 V

i i

R i

i i

V VV V

V V

<= < < − >

The transfer characteristics is shown below,

The waveform observed across R is shown below,


For full wave bridge rectifier : Arrangement of diode in opposite arm should be in same face. From observation option (A) and option (B) are wrong.

From option (C),

Case 1 : During positive half cycle, Diode 2D and 3D will be forward biased, hence these diodes can be replaced by short circuits. Diode 1D and 4D will be reverse biased, hence these diodes can be replaced by open circuits.

Hence, 0 iV V= [ 0V = Voltage across resistor] Case 2 : During negative half cycle, Diode 2D and 3D will be reverse biased, hence these diodes can be replaced by open circuit. Diode 1D and 4D will be forward biased, hence these diodes can be replaced by short circuit.

Hence, 0 0iV V− − = 0 iV V= − Hence, from case 1 and case 2 the expression of output can be written as given below,

0

; 0V; 0V

i i

i i

V VV

V V>

= − <

~ RiV

S.C.

RV

6 V

12 V

12 V6 V

6 V

RV

iV

12 V

slop

e=

1

slop

e=

1

Z = R

B/B

D

Z = F

B/O

N

Z = RB/OFF

t

6 V

12 V

RV

iV

RV

0

12 V

/ 23 / 2

1D2D

3D4D

iV

0V

iV

0V

O.C.

O.C.

S.C.

S.C.

iV

0V

S.C. O.C.

S.C.O.C.

1.14 (C)


Above waveform is output of full wave bridge rectifier. From option (D),

Case 1 : During positive half cycle, All four diodes will be reverse biased. Hence, these diodes can be replaced by open circuit.

Hence, 0 0 VV = Case 2 : During negative half cycle, All four diodes will be forward biased. Hence, these diodes can be replaced by short circuit.

0 ViV = [i.e. not possible]

Above circuit does not exist, since it violates the KVL. Therefore, option (D) is incorrect. Hence, the correct option is (C).

Given : (i) Zener knee voltage = 7 V (ii) Zener dynamic resistance 10= Ω (iii) (min) 10 VinV = , (max) 16 VinV = Regulator circuit is shown below.

Replacing Zener diode with dynamic resistance and Zener voltage in the above circuit as shown below,

Output voltage 0V is given by,

0 Z ZV V IR= +

where, i Z

Z

V VIR R

−=+

(max)max

i Z

Z

V VI

R R−

=+

16 7 9 A200 10 210

−= =+

(min)

(min)i Z

iZ

V VI

R R−

=+

10 7 3 A200 10 210

−= =+

Output voltage 0V will be maximum when I is maximum. 0 max max( ) Z ZV V I R= +

0 max9( ) 7 10

210V = + × 7.43 V=

2

t

mV

iV

0V

mV

2 3

3

4

4t

mV

0

0

1D2D

3D4D

iV

0V

iV

0V

O.C.

O.C.

O.C.

O.C.

iV

0V

S.C. S.C.

S.C.S.C.

iV

200

0V

iV

200R

0V

10zR

7 VZV

I

1.15 (C)


Output voltage 0V will be minimum when I is minimum. 0 min min( ) Z ZV V I R= +

0 min3( ) 7 10

210V = + ×

0 min( ) 7.14 VV =

Hence, 0V lies between 7.14 V and 7.43 V.


Given : (i) Input voltage, 10sin100 ViV t= π

(ii) Diode drop = 0.7 V (iii) Zener breakdown voltage, 6.8 VZV =

Limiter circuit is shown below,

. Method 1 : Case 1 : During positive half cycle,

2D is reverse biased and Zener diode Z is also reverse biased. (a) 0 0.7 ViV< < :

Diode 1D , 2D and Zener diode Z will be OFF.

Hence, 0 iV V=

(b) 0.7 V 7.5 V (i.e. + )i ZV V Vγ< < :

Diode 2D will be in reverse bias i.e.

OFF. Diode 1D will be ON and Zener diode Z will be in reverse bias without breakdown because input voltage is less than ZV Vγ+ .

Hence, 0 iV V=

(c) 7.5 V (i.e . )i ZV V Vγ> + :

When output voltage across diode 1D

will be greater than cut-in voltage, 1D will be forward biased and therefore, it can be replaced by its cut-in voltage

0.7VVγ = .

Since, voltage across Zener diode Z is higher than breakdown voltage, hence Zener diode Z will work in breakdown region and therefore, Zener diode Z can be replaced by its breakdown voltage.

Hence, 0 7.5 VZV V Vγ= + =

Case 2 : During negative half cycle,

Diode 1D will be reverse biased and hence, it can be replaced by open circuit. Zener diode Z will be forward biased and hence, it will act as normal diode and therefore, it can be replaced by short circuit.

iV

1k

0V2D

1D

Z 6.8V

0.7

0ViV O.C.

O.C.

O.C.1k

0 A

0ViV O.C.

O.C.

S.C.1kΩ

0ViV O.C.

1kV

0.7 V

ZV 6.8V

1.16 (C)


(a) 0.7 V (i.e. )iV Vγ> − : When voltage across 2D is less than Vγ ,

2D will be reverse biased and therefore,

2D will act as open circuit.

Hence, 0 iV V= (b) 0.7 V (i.e. )iV Vγ< − : When voltage across diode 2D will be

higher than Vγ , 2D will be forward biased and therefore, 2D can be replaced by its cut-in voltage Vγ . Zener diode will be in forward bias and acts as a normal diode.

Hence, 0 0.7 VV = − Hence, from case 1 and case 2, expression of 0V can be written as given below,

0

; 0 7.5 V7.5 V ; 7.5 V

; 0.7 0 V0.7 V ; 0.7V

i i

i

i i

i

V VV

VV V

V

< < >= − < <− <

The transfer characteristics of given circuit is shown below,

Hence, the output waveform of given circuit is shown below,

Hence, the correct option is (C). . Method 2 : Case 1 : (max) 10 ViV =

Since, voltage across Zener diode Z is higher than breakdown voltage, hence Zener diode Z will work in breakdown region and therefore, Zener diode Z can be replaced by its breakdown voltage.

Hence, 0(max) 7.5 VZV V Vγ= + =

Case 2 : (min) 10 ViV = −

When voltage across diode 2D will be higher

than Vγ , 2D will be forward biased and

therefore, 2D can be replaced by its cut-in voltage Vγ . Zener diode will be in forward bias

and acts as a normal diode.

Hence, 0(min) 0.7 VV = −


0ViV O.C.

O.C.1k

S.C.

0Vi

V

O.C.1k

V 0.7 V

0V

inV

7.5V 10 V

0.7 V10 V

0.7 V

7.5V

10 V

0.7 V

7.5 V

0V

t

10 V

0

max

min

0(max)V(max) 10 Vi

V O.C.

1kV

0.7 V

ZV 6.8V

0(min)V(min) 10 V

iV

O.C.1k

V 0.7 V



Case 1 : Assume diode is ON, Hence, diode can be replaced by short circuit.

Applying KVL in the loop shown, 0 0iV V− + + =

iV V= … (i) where, ( 1) 1 1V I I= − − × = − 1I V= − From equation (i), 1 iI V= − … (ii) For diode to be ON, current I must be positive i.e. 0I > . 1 0iV− >

1ViV <

Hence, for diode to be ON, the voltage iV must be less than 1 V. If 0.5ViV = , then 0.5VV =

Option V Status (A) min( ,1)iV 0.5 V Correct

(B) max( ,1)iV 1 V Wrong

(C) min( ,1)iV− –0.5 V Wrong

(D) max( ,1)iV− 1 V Wrong

Case 2 : Assume diode is OFF, Hence, diode can be replaced by open circuit.

From figure, 1 1 1VV = × = Applying KVL in the loop shown, 0i DV V V− + + =

D iV V V= −

1D iV V= −

For diode to be OFF, DV must be positive i.e.

0VDV > .

1 0iV − >

1ViV >

Hence, for diode to be OFF, iV must be greater than 1 V and V must be equal to 1 V. Let, 1.5ViV = and 1 VV =

Option V Status (A) min( ,1)iV 1 V Correct

(B) max( ,1)iV 1.5 V Wrong

(C) min( ,1)iV− –1.5 V Wrong

(D) max( ,1)iV− 1 V Correct

Hence, from the Case 1 and Case 2 it is clear that option (A) is common for both cases. Hence, the correct option is (A).

Given : (i) Forward voltage drop of diode = 0.7 V

iV 1AD

1 1

V

iV 1AS.C.

1 1

V

( 1)I

I

iV 1A

1 1

V

1 A 1 A

DV

n

p

Scan for

Video Solution

1.17 (A)

1.18 (A)


(ii) 25 mVTkTVq

= =

(iii) Input signal cos( )i pV V t= ω where, 100 mVpV = For DC analysis, short circuit the AC voltage source.

By replacing all diodes by their cut-in voltage, we get the modified circuit as given below,

4 0.7 2.8VDCV = × =

Current, 12.7 2.8 1 mA9.9DCI −= =


For AC analysis, short circuit the DC voltage source.

Dynamic resistance or ac resistance of diode is given by,

Tac

VrI

η=

Tac

VrI

= (Assume 1η = )

25mV 251mAacr = = Ω

By replacing all diodes by their ac resistance, we get the modified circuit as given below,

4

4ac

ac iac

rV VR r

= ×+

[By VDR]

4 259900 4 25ac iV V×= ×

+ ×100

10,000 iV=

acV 1100 iV= 1 100cos

100t= × ω

cos mVacV t= ω Hence, the correct option is (B). Key Point If ideality factor, η is not given in question then we will consider 1η = . [For Ge as well as Si]

The given circuit can be redrawn as given below,

12.7 VDC

V

9900

DCI

S.C.

12.7 VDC

V

9900

DCI

S.C.

0.7 V

0.7 V

0.7 V

0.7 V

iV

acV

9900

I

S.C.

iV

acV

9.9 kR

acr

acr

acr

acr

~ 2C

1C 2D

0 ( )V t1D ( )V t

Circuit 1 Circuit 2

iVcos( )t

1.19 (B)

1.20 (A)


In above figure, circuit (1) represents ideal negative clamper circuit and circuit (2) represents ideal negative peak detector circuit. Hence, ( )V t is the output of negative clamper circuit and hence, it will be shifted form of input in downward direction. ( ) cos 1i mV t V V t= − = ω −

Hence, the correct option is (A). Output of negative peak detector circuit will be maximum negative value of ( )V t .

0 ( ) 2 VV t = −

Working : Case 1 : During 1st positive half cycle, Diode 1D will be forward biased and diode 2D will be reverse biased. Hence, charging of capacitor 1C will start and it will charge upto maximum value of input because charging time constant is zero.

Hence,

1 maxcos 1 VCV t= ω =

Case 2 : During 1st negative half cycle, Diode 1D will be reverse biased and diode 2D will be forward biased. Hence, charging of capacitor 2C will start and

capacitor 2C will charge upto maximum value

of input, which appear across capacitor 2C .

Applying KVL in the loop shown in figure,

1 20 0i C CV V V− + + + =

2 1C i CV V V= −

2

cos 1CV t= ω −

Case 3 : During 2nd positive half cycle, Positive terminal of 1D is at 1V− and positive

terminal of 2D is at 2 V− . Now, diode 1D and

2D will be forward biased when voltage at their negative terminal will be less than that of the positive terminal, which is never possible beyond first negative half cycle and first positive half cycle. Hence, in this case, 1D and 2D will be reverse biased.

From above circuit,

20 ( ) cos 1CV t V t= = ω −



. Method 1 : Applying KVL in the loop shown, 10 0iR v− + + =

0.7 , 1k500

vi R−= = Ω [Given]

( 0.7)10 1000500

v v−= × +

~ 2C

1C

0 ( )V ti

V

1CV

( )V t

O.C.

cos( )t S.C.

~ 2C

1C

0 ( )V ti

V

1CV

2CV

( )V tO.C.cos( )t

S.C.

~ 2C

1C

0 ( )V ti

V

1CV

2CV

( )V tO.C.

O.C.

cos( )t

i

10 V

1k

v

1.21 (D)


10 2 1.4v v= − + 3 11.4v = 3.8 Vv =

3.8 0.7 1000 mA500

i − = ×

6.2 mA=

Hence, the correct option is (D). . Method 2 : From given i-v characteristics,

0.7500

vi −= … (i)

i-v characteristics of forward bias practical diode is given by,

f

v Vi

Rγ−

= … (ii)

fR = forward resistance of diode Vγ = cut-in voltage (offset voltage) From equation (i) and (ii), 500 0.5 kfR = Ω = Ω , 0.7 VVγ = The given circuit is replaced by its equivalent as shown below,

Applying KVL in the above loop, 10 1 0.5 0.7 0i i− + × + × + = 1.5 9.3i× =

9.3 6.2 mA1.5

i = =


Given : (i) Knee current of Zener diode, (min)ZI = 10 mA

(ii) 5 VoltZV =


For variable load, input current inI is given by,

(max) (min)in Z LI I I= + … (i)

(min) (max)in Z LI I I= + … (ii)

(i) Calculation for minimum load resistance :

Input current inI is given by,

S Zin

V VIR−=

10 50.1inI −= 50 mA=

From equation (ii), (max) (min) 50 10 40 mAL in ZI I I= − = − =

From figure, 5V

LR L L ZV I R V= = =

(max) (min) 5 VL LI R =

(min)(max)

5 5 12540L

L

RI

= = = Ω

(ii) Calculation for minimum power rating :

Power dissipation across Zener diode is given by,

Z Z ZP V I=

V

1

fR

i

v0

10 V

1k

0.7 VV

0.5 kfR

i

Scan for

Video Solution

10VSV 5VZ

V

inI

LR

100

ZI

LI

LR

V

1.22 (B)


Minimum power rating of Zener diode represents maximum allowed power dissipation across Zener diode

(max) (max)Z Z ZP V I= … (iii)

For no load condition LR = ∞ , (min) 0 ALI =

From equation (i),

(max) 50 mAZI I= =

From equation (iii), (max) 5 50 250 mWZP = × =


Avoid This Mistake (i) The maximum power rating of the Zener

diode is (max) (max)Z Z ZP I V= × . [Correct]

(ii) The minimum power rating of the Zener diode is (min) (min)Z Z ZP I V= × . [Incorrect]

(iii) The maximum power dissipation across Zener diode is (max) (max)Z Z ZP I V= × [Correct]

(iv) The minimum power dissipation across Zener diode is (min) (min)Z Z ZP I V= × [Correct]

Given : (i) 1000sinYZV t= ω

(ii) All diodes are ideal. Given circuit is shown below,


All the four diodes are reverse biased, so all the diodes will be open circuited. Applying KVL in the loop shown, 1 0 0WXV− + × = Hence, 0 VWXV = … (i) Case 2 : During negative half cycle,

All the four diodes are forward biased, so all the diodes will be short circuited. Applying KVL in the loop shown, 1 0 0WXV− + × = 0 VWXV = … (ii) From equation (i) and (ii), 0WXV = for all t Hence, the correct option is (D).

Given : (i)

W

Y

1k

Z

X

1k

W

Y

1k

Z

X

1k

O.C.O.C.

O.C.O.C.

0 A

0 A

0 A

0 A

0 A

0 A

0 A

0 A

0 A

0 A

WXV

W

Y

1k

Z

X

1k

S.C.S.C.

S.C.S.C.

0 V

WXV

I

D

100~

4 mF10sint

= 50 Hzf( )

CV t

1.23 (D)

1.24 0.1


(ii) This circuit is used in AM receiver to detect the envelope of the amplitude modulated carrier waveform. The name of the circuit is peak detector (Envelope detector).

(iii) In envelope detector circuit, charging time should be small and discharging time should be large.

When diode is ON charging time constant is given by,

Charging time constant, C THR Cτ =

0 secCτ =

Therefore, capacitor will charge immediately. When diode is OFF, discharging time constant is given by,

D THR C RCτ = =

3100 4 10 secD−τ = × ×

400 msecDτ =

Time period of input signal is given by,

1 1 20 msec

50T

f= = =

[Given : f = 50 Hz] Since, DT << τ

[Condition for envelope detector]

At steady state, capacitor voltage varies between B and C,

2

r p pdc m

VV V −= −

Where, 2

dcr p p

IVf C− =

4

dcdc L m

II R Vf C

= −

14dc L mI R V

f C

+ =

3

101 1100

4 4 50 4 10

mdc

L

VIR

f C −

= =+ +

× × ×

0.09 0.1 AdcI = ≈

Hence, the average steady state current is 0.1 A.

Given : Forward voltage drop of both diodes = 0.7 V. Circuit is shown below,

Fig. Parallel diode clipper circuit

S.C.

R CS.C.

0THR

O.C.

R CS.C.

THR R

t

smV

t

CV

smV

1t 2t

sV

sV

exceeds VC

at point BB

0

0

A C

B

1t 2t

-r p pV

Transient Steady state

0ViV

R

1V

1D 2D

2 V

1.25 (D)


Case 1 : During positive half cycle, Diode 1D will be reverse biased and hence, 1Dwill act as open circuit. (a) 0 2.7ViV< < : Diode 2D will be

reversed biased and hence, it acts as open circuit.

Hence, 0 iV V=

(b) 2.7ViV > : Diode 2D will be forward biased and hence, it can be replaced by cut-in voltage Vγ .

Hence, 0 2.7VV = Case 2 : During negative half cycle, Diode 2D will be reverse biased and hence, 2Dwill act as open circuit. (a) 1.7V 0ViV− < < : Diode 1D will be

reverse biased and hence it acts as open circuit.

Hence, 0 iV V=

(b) 1.7 ViV < − : Diode 1D will be forward biased and hence it can be replaced by cut-in voltage Vγ .

Hence, 0 1.7VV = − Hence, from Case 1 and Case 2, expression of output voltage can be written as given below,

0

; 0V 2.7V2.7V ; 2.7V

; 1.7V 0V1.7V ; 1.7V

i i

i

i i

i

V VV

VV V

V

< < >= − < <− < −

The transfer characteristics of given circuit is shown below,

Hence, the output waveform is shown below,

Therefore when 1.7 2.7iV− < < , then both 1D and 2D are OFF and hence 0 iV V= . Hence, the correct option is (D).

0Vi

V

R

1.7 V

1(ideal)D

2.7 V

2 (ideal)D

0Vi

V

R

1.7 V

2.7 V

O.C. O.C.

0Vi

V

R

1.7 V

2.7 V

O.C. S.C.

0Vi

V

R

1.7 V

2.7 V

O.C. O.C.

0Vi

V

R

1.7 V 2.7 V

O.C.S.C.

0V

inV

2.7 V

1.7 V

1.7 V

2.7 V

1.7 V

0V

t

2.7 V

Scan for

Video Solution


Given : (i) Voltage drop of PN diode = 0.7 V. (ii) Voltage drop of Schottky diode = 0.3 V. . Method 1 : (i) Assume both diodes 1D and 2D are ON.

(ii) For both diodes to be ON, 1DI and

2DI

should be greater than zero from p to n.

Applying KVL in the loop (1), 10 1 0.7 0I− + × + = 9.3 mAI = Applying KVL in the loop (2),

20.7 20 0.3 0DI− + × + =

2

0.4 0.02 mA20DI = =

1 2

9.3 0.02 9.28 mAD DI I I= − = − =

Since, 1

0DI > and 2

0DI > therefore, this

assumption is correct and both 1D and 2D are ON. Hence, the correct option is (A). . Method 2 : (i) Assume both diodes 1D and 2D are OFF.

(ii) For both diodes to be OFF, 1DV should

be less than 0.7 V and 2DV should be

less than 0.3 V (from p to n).

From figure,

1 210 VD DV V= =

Hence, 1

0.7 VDV > , 2

0.3 VDV >

Therefore, this assumption is incorrect and both

1D and 2D are ON. Hence, the correct option is (A). . Method 3 : (i) Assume diode 1D is OFF and diode 2D

is ON. (ii) For diode 1D to be OFF,

1DV should be

less than 0.7 V from p to n. (iii) For diode 2D to be ON,

2DI should be

greater than 0 from p to n.

Applying KVL in the above loop, 10 20 0.3 0I I− + + + =

10 0.3 0.461 mA21

I −= =

2

0.461 mADI I= =

Hence, 2

0 ADI >

Therefore, this assumption is correct and 2D is ON.

120 0.3DV I= × +

1

0.461 20 0.3 9.53 VDV = × + =

Hence, 1

0.7 VDV >

Therefore, this assumption is incorrect and diode 1D is ON. Hence, the correct option is (A).

1k 20k

10 V 0.3V0.7 V

1DI

2DI

(1) (2)

I

1k 20k

10 V O.C.1D

V O.C.2D

V

0 A 0 A

0 A

1k 20k

10 V 0.3V

I

1DV

2DI0 A

Scan for

Video Solution

1.26 (A)


Key Point (i) When there are two or more diodes in

the circuit then our assumption about the diodes should be based on observation.

(ii) In all assumptions, direction of current and polarity of voltage across diode is taken from p to n.

For diode to be ON, DV and DI should

be positive.

Given : (i) 0.7VonV = , 5sin( ) ViV t= ω

(ii) (max) max5sin 5 ViV t= ω = +

(iii) (min) min5sin 5 ViV t= ω = −

Case 1 : (max) 5 ViV = + , Diode will be forward biased, hence it can be replaced by its cut-in voltage i.e. onV .

Applying KVL in above figure,

5 0.7 2 2.32 2

I − −= =

02.3 1 0.7 2 3.85 V2

V = × + + =

Case 2 : 5 ViV = − , Diode will be reverse biased, hence it can be replaced by open circuit.

From the above circuit, 0 5 ViV V= = −

From Case 1 and Case 2,

0(max)0

0(min)

3.85 V ; 5 V5 V ; 5V

i

i

V VV

V V= = +

= = − = −


Given :

Diode circuit can be simplified as shown below,

Case 1 : During positive half cycle, Both diodes 1D and 2D are ON.

DV

DI

p n

0V1k

1k

2 V

0.7 V

I

5V

0V

1k

1k

2 V

5 V

O.C.

0 A

0 A

Scan for

Video Solution

0.5 T T0

D1

Vout

D2

inVoutV

1D

2D

T0.5T0

inV out

V

S.C.

S.C.

1.27 (C)

1.28 (C)


From figure, out inV V= −

Case 2 : During negative half cycle, Both diodes 1D and 2D are OFF.

Hence, 0 VoutV =

Hence, from case 1 and case 2, the output waveform can be drawn as shown below,



For the given regulator circuit, check whether Zener diode is in breakdown or not, by finding open circuit voltage across its terminal. Let us assume Zener diode is not in reverse breakdown region, hence it can be replaced by open circuit.

0

1k 10(1 1) k

V Ω= ×+ Ω

5 V=

For breakdown of Zener diode, 0 ZV V≥ 0(max) 5 VV = Therefore, Zener diode will be in reverse bias region without breakdown. Given, 6 VZV = and terminal voltage of Zener diode is less than ZV , hence our assumption is correct i.e., Zener diode is not in breakdown region. Hence, the output voltage 0V is 5 V.

Avoid This Mistake (i) Zener voltage, 6 VZV = Input voltage, 10 ViV = i ZV V> (Zener diode will breakdown) (ii) Always check the voltage across Zener

diode for condition of breakdown by finding open circuit voltage across Zener diode.

For an ideal clamping circuit, once the capacitor is charged, it should not discharge. Hence, discharging time constant (RC) must be much larger than the time period of input signal. i.e. RC T>> Hence, the correct option is (D). Key Point (i) Negative clamper :

00.5T

outV

t

inV out

V

O.C.

O.C.

00.5T

outV

tT

00.5T

outV

tT

10 V 0V

1k

1k

10 V 0V

1k

1k

C

V ~ R

1.29 5 1.30 (D)


(ii) Positive clamper :

For proper operation of clamping circuit (Both negative and positive clamper) RC T>> . where, T = Time period of input signal


. Method 1 : From figure, 2 kABR = Ω , 6 kCDR = Ω 4 kBCR = Ω , 3 kADR = Ω Since, AB CD BC ADR R R R× = × hence, bridge is balanced and hence, no current will flow through the branch BD. Hence, branch BD will act as open circuit.

Current 1I is given by,

19 1

9 6I = ×

+ [By CDR]

19 mA

15I = 0.6 mA=

Hence, the current in the 4 kΩ resistor is 0.6 mA.

. Method 2 : In case you have no idea about balanced bridge then assume either diode is ON or diode is OFF. If we assume diode is ON then the calculations become complicated so let us assume diode is OFF. For diode to be OFF, voltage across diode D should be less than 0.7 V from p to n.

From figure, (2 4) || (3 6) 6 || 9R = + + =

6 9 3.6 k6 9

R ×= = Ω+

3 3 31 10 3.6 10 10 3.6 VV R − −= × × = × × = From figure,

4 4 3.6 2.4 V2 4 6a

VV × ×= = =+

[By VDR]

6 6 3.6 2.4 V3 6 9b

VV × ×= = =+

2.4 2.4 0 VD a bV V V= − = − =

0.7 VDV < Therefore, our assumption is correct and the diode is OFF. From figure, 4 2.4 VaV I= =

2.4 0.6 mA4

I = =

Hence, the current in the 4 kΩ resistor is 0.6 mA.

C

V ~ R

1mA

2k 3k

6k4k

1kD

A

D

C

B

1mA

2k3k

6k4k

2I

2I1I

1I

O.C.

A

C

B D

1mA

2k 3k

6k4k

DV

aV

bV

1k

0 A

1mA

V

R

I

Scan for

Video Solution

1.31 0.6


The given circuit can be redrawn as given below,

Circuit 1 Circuit 2 In above figure, circuit (1) represents ideal positive clamper circuit and circuit (2) represents ideal positive peak detector circuit. Hence, ( )V t is the output of positive clamper circuit and so, it will be shifted form of input in upward direction. ( ) 50sin 50i mV t V V t= + = ω + Output of positive peak detector circuit will be maximum positive value of ( ).V t

100 VabV =

Hence, the average voltage abV is 100 V. Working : As input to the circuit is a periodic sinusoidal signal, we can take 50sininV t= ± ω i.e. we can consider any of the positive or negative half first. Here we are taking 50sininV t= − ω and Considering the negative half cycle first.

Case 1 : During 1st negative half cycle, Diode 1D will be forward biased and diode 2D will be reverse biased.

Hence, charging of capacitor 1C will start and it will charge upto maximum value of input.

Hence,

1 max50sin 50VCV t= ω =

and 2

0 VCV =

Case 2 : During 1st positive half cycle, Diode 1D will be reverse biased and diode 2D will be forward biased. Hence, charging of capacitor 2C will start and capacitor 2C will charge upto maximum value of input, which appear across capacitor 2C .

Hence,

2 1 max50sinC CV V t= + ω

2

50 50 100 VCV = + =

Case 3 : During 2nd positive half cycle, Negative terminal of 1D is at 50 V and negative terminal of 2D is at 100 V. Now, diode 1D and

2D will be forward biased when voltage at their positive terminal will be higher than that of the negative terminal, which is never possible beyond first negative half cycle and first positive half cycle. Hence, in this case, 1D and 2D will be reverse biased.

~ 1D2C

1C 2D

abV( )V t

iV50sin t

a

b

inV

50

50

t

~ 2C

1C

abV

iV

1CV

( )V t

O.C.

S.C.50sin t 2CV

a

b

~ 2C

1C

iV

1CV

2CV

( )V tO.C.

S.C.

50sin tab

V

a

b

~ 2C

1C

iV

1CV

2CV

( )V tO.C.

O.C.

50sin tab

V

a

b

1.32 100


Hence, from above circuit

1( ) 50sin 50 50sinCV t V t t= + ω = + ω

and 2max 100 Vab CV V= =

Hence, the average voltage across 0.5 Fμ capacitor is 100 V. Key Point (1)

Circuit 1 Circuit 2 Circuit 1 : It is an ideal negative clamper circuit. Output of this circuit will be shifted form of input in downward direction. Hence, output

( )V t of this circuit is given by, ( ) sini m m mV t V V V t V= − = ω − Circuit 2 : It is an ideal negative peak detector circuit. Output of this circuit will be maximum negative value of ( )V t . Hence, output 0 ( )V t of this circuit is given by,

0 min( ) sin 2m m m m mV t V t V V V V= ω − = − − = −

(2)

Circuit 1 Circuit 2 Circuit 1 : It is an ideal positive clamper circuit. Output of this circuit will be shifted form of input in upward direction. Hence, output ( )V t of this circuit is given by,

( ) sini m m mV t V V V t V= + = ω + Circuit 2 : It is an ideal negative peak detector circuit. Output of this circuit will be maximum negative value of ( )V t . Hence, output 0 ( )V t of this circuit is given by,

0 min( ) sin 0Vm m m mV t V t V V V= ω + = − + =

(3)

Circuit 1 Circuit 2 Circuit 1 : It is an ideal negative clamper circuit. Output of this circuit will be shifted form of input in downward direction. Hence, output

( )V t of this circuit is given by,

( ) sini m m mV t V V V t V= − = ω −

Circuit 2 : It is an ideal positive peak detector circuit. Output of this circuit will be maximum positive value of ( )V t . Hence, output 0 ( )V t of this circuit is given by,

0 max( ) sin 0Vm m m mV t V t V V V= ω − = − =

(4)

Circuit 1 Circuit 2

~ 1D2C

1C 2D

0ViV sinm

V t LRLR( )V t

~ 1D2C

1C 2D

0Vsini mV V t ( )V t

~ 1D2C

1C 2D

0Vsini mV V t ( )V t

~ 1D2C

1C 2D

0V( )V tsini mV V t


Circuit 1 : It is an ideal positive clamper circuit. Output of this circuit will be shifted form of input in upward direction. Hence, output ( )V t of this circuit is given by, ( ) sini m m mV t V V V t V= + = ω +

Circuit 2 : It is an ideal positive peak detector circuit. Output of this circuit will be maximum positive value of ( )V t . Hence, output 0 ( )V t of this circuit is given by,

0 max( ) sin 2m m m m mV t V t V V V V= ω + = + =

Given : Input voltage = 6 sin tπ ω



Diode 1D is ON and 2D is OFF.

10 6 sin10 20abV t= × π ω

+ [By VDR]

2 sin VabV t= π ω

Case 2 : During negative half cycle, Diode 1D is OFF and 2D is ON.

10 6 sin 3 sin10 10 2

iab

V tV t− × − π ω= = = − π ω+

[By VDR] Hence, from case 1 and case 2 expression of

abV is given by,

2 sin ; 03 sin ; 2ab

t tV

t tπ ω < ω < π

= − π ω π < ω < π

Average value of abV is given by,

0

1( ) ( )T

ab avg abV V d tT

= ω

2

0

1( ) ( )2ab avg abV V d t

π

= ωπ

0

iV

t 2

6

10k

1D 2D

~ b6 sin( )t

20k10k

a

abV

10k~ b6 sin( )

iV t

20k10k

a

O.C.S.C.

iV

20k

10k

iV

abV

10k~ bi

V

20ki

V10k

a

O.C. S.C.

10k

10k

iV

abV

a b

1.33 5


0

1( ) 2 sin ( )2ab avgV t d t

π= π ω ω π

2

( 3 sin ) ( )t d tπ

π

+ − π ω ω

[ ]0

1( ) 2 cos2ab avgV t π= π − ω

π

[ ]23 cos2

t π

π

π+ ωπ

0 3( ) cos cos02ab avgV π = − π + + π

[ ]cos 2 cosπ − π

3( ) (1 1) (1 1)2ab avgV = + + +

( ) 5 Voltab avgV =

Hence, the average value of voltage abV is 5 V.

Given : 1 2 33 k , 6 k , 9 kR R R= Ω = Ω = Ω

Case 1 : During positive half cycle, Diode will be ON. Applying KVL in the above loop,

1 2 33 6 3I I IV R R R= × + × + ×

3 6 9 53 6 3I I IV I= × + × + × =

10sin 2sin mA5 5V tI t= = =

Case 2 : During negative half cycle, Diode will be OFF. Hence, 0 AI =

RMS value of current I is given by,

2

0

1 ( )T

rmsI I t dtT

= 2

2

0

1 ( )2

I t dtπ

=π

2

2 2

0

1 (2sin ) 02rmsI t dt dt

π π

π

= + π

0

1 1 cos 242 2rms

tI dtπ − = π

[ ]0

1 sin 2 1 02rms

tI tπ

= − = π − π π

1 mArmsI =

Hence, the RMS current rmsI through the diode is 1 mA. Key Point (i) For half wave rectifier,

RMS current, 2m

rmsII =

(ii) For full wave rectifier,

RMS current,

2m

rmsII =

R1

R2

R1 R2

R2

R3

R2 R3

R3

R2

R1

R2

rmsI

10sin( )V t

/ 3I

/ 3I

/ 6I

/ 3I

I

/ 3I

/ 3I

/ 3I

/ 6I

/ 6I

/ 6I/ 6I

/ 6I

0

I

t 2 3 4

2

0

I

t 2 3 4

mI

0

I

t 2 3 4

mI

1.34 1


Given : 0 1 A, 475 F, 1 msecDCI I C T= = = μ =

From the above waveform of the capacitor

voltage, it is clear that capacitor charging time

is very less as compared to total time period i.e.

total time period is almost equal to discharging

time. Because of this, we can neglect charging

time of the capacitor.

During discharging time, diode will be OFF and

circuit is shown below,

0CdVI C

dt= − [Ohm’s law]

0C

IdV dtC

−=

0C

I TV uC

Δ = =

where, (max) (min)C C CV u V VΔ = = −

3

61 1 10 2.105 Volts475 10CV u

−

−

× ×Δ = = =×

Hence, the value of the ripple u is 2.105 V.


Assume diode D is ON, hence it can be replaced by cut-in voltage i.e. onV .

From the above figure, 2V, 0.7 1.3 VB A BV V V= = − = Applying KCL at node A,

2 6

A B AD

V V V I− + =

1.3 2 1.32 6 DI− + =

0.13 mADI = − For diode to be ON, DI must be positive from p to n. From the above analysis, DI is negative. So, diode will be OFF. Hence, our assumption is wrong. Since, diode is OFF, hence it can be replaced by open circuit. Hence, the modified figure is shown below,

0 V t

CV

CV u

Dischargingof capacitor

Chargingof capacitor

iV

CV475 F

0 1 AI

O.C.

0I

6k2R

1R

2I

2 V

2k

6k2R

DI

2I

2 V

2k

0.7

AB

6k2R

1R2I

2 V

2k

1.35 2.105 1.36 0.25


Applying KVL in above figure,

22 1 0.25 mA8 4

I = = =

Hence, the magnitude of the current 2I is equal to is 0.25 mA.


The circuit can be redrawn as shown below,

Case 1 : 02

t π< ω <

Since, both the diodes will be forward biased, hence both diodes can be replaced by short circuits.

From figure, 0 0 VV =

At 2

t πω = , 10 ViV =

Charging time constant for 1C , 0 secCτ =

Charging time constant for 2C , 0 secCτ = From figure,

1 210 VC C iV V V= = =

Case 2 : 2

tπ < ω < π

10 ViV < , 1 2

andi C i CV V V V< < Since, both diode will be reverse biased, hence both diode can be replaced by open circuit. 10 V, 10 Vi CV V< =

Discharging time constant for 1C , Dτ = ∞ Discharging time constant for 2C , Dτ = ∞ Hence, both capacitor voltage fixed at 10 V. Applying KVL in loop shown in figure, 010 10 0V− + + = 0 0V =

Hence, the output voltage 0V at the steady state is 0 V.

R~ 0V

2D

C10sin t

ac voltageC

1D

R~

0V

10sin t

1C 2C

1D 2D

R~

0V

10sin t

1C 2C1C

V

ON ON

2CV

R~

0V

10sin t

10 V 10 V

OFF OFF

inV

t

2

3

2

2

1CV

t

10 V

2CV

t

10 V

1 20 C CV V V

t0

0

0

1.37 0


Given circuit is shown in below figure,

Case 1 : During positive half cycle, Diode is ON, hence it can be replaced by short circuit.

Hence, 0 iV V= Case 2 : During negative half cycle, Diode is reverse biased, hence it is OFF and therefore it can be replaced by open circuit.

Hence, 0 0 VV = Hence, from case 1 and case 2, 0V can be expressed as, given below.

0

; 00 ; 0

i i

i

V VV

V>

= <

Average value of 0V is given by,

2

0 (avg) 00

1 ( )2

V V t d tπ

= ωπ

0 (avg)0

1 10sin2

V t d tπ

= ω ωπ

[ ]0 (avg) 0

10 cos2

V t π= − ωπ

[ ]0 (avg)

10 1 1 102

V+

= =π π

0 (avg) 3.18 voltV = Hence, the reading on the DC voltmeter is 3.18 V. Key Point Average DC voltmeter reading i.e. average output voltage of half wave rectifier is given by,

0 (avg)mVV =π

Given : (i) 0.7 V, 6.8 V,ZV Vγ = = − (ii) 10 Tτ << where τ is time constant. (iii)

Case 1 : When 0 ,2Tt< <

( ) 14 VinV t = Diode 1D is in forward bias and hence it can be replaced by 0.7 V. Zener diode 2D is in reverse bias and because voltage across 2D is more than breakdown voltage, hence it is operating in breakdown region. Therefore, it can be replaced by its breakdown voltage.

10 sint

f = 50 Hz1 k V0

+

–

10 sint

f = 50 HzV0

+

–

iV

S.C.

R

10 sint

f = 50 HzV0

+

–

iV

O.C.

R

0 A

t

iV

10

10

t

0V

10

2

2

0

0

t (sec)

14 V

– 14 V

( )in

V t

T/2T

1.38 3.18

1.39 (A)


Equivalent circuit is shown in below figure,

Hence, (max) 0.7 6.8 7.5VoutV = + =

Apply KVL in the loop shown in figure, 14 7.5 0cV− + + =

6.5VcV =

At 2Tt = , capacitor will charge upto 6.5 V due

to small time constant (10 )Tτ << .

Case 2 : When 2T t T< < ,

( ) 14 VinV t = −

Diode 1D is in reverse bias and hence it can be replaced by open circuit. Zener diode 2D is forward biased and therefore, it will act as normal diode. Hence it can be replaced by short circuit. Equivalent circuit is shown in below figure.

Apply KVL in the loop shown in figure, 0in C outV V V− + + =

14 6.5 0outV+ + =

(min) 20.5VoutV = −


Avoid This Mistake Zener break down voltage is given either

6.8 VZV = or 6.8 VZV = − it is represented by, 1. Correct form :

2. Incorrect form :

Given : 0 5V, 2 mA,

KZ ZV V I= = =

6 6 V of 5%iV = ±

6 0.3 ViV = ±

Concept : Line Regulation (i) Input voltage = Variable (ii) Load = Fixed Voltage regulated output = 5 volts Load current is given by,

5 5 mA1L

L

VIR

= = =

and input voltage 5.7, 6.3 ViV =

in Zin

V VIR−=

(min)(min)

min

in Zin

V VI

R−

= and (max)(max)

max

in Zin

V VI

R−

=

10

10 F0.7 V

6.8V

( )outV t

+

–

( ) 14Vin

V t

cV

+ –

+–

+–

10

10 F

( )outV t( ) 14V

inV t

6.5Vc

V + –

S.C.

6.8 V

n

p

6.8 V

n

p

R

0 5VZV V= =iV

inI

ZI

1kΩ

LI

1.40 (B)


(min) (min) ( ixed)in Z L fI I I= +

(min) 2 mA 5 mA 7 mAinI = + =

min5.7 5 1007 mA

R −= = Ω

Minimum value of R is 100 Ω . (max) (max) (fixed)in Z LI I I= + … (i)

(min) (min) (fixed)in Z LI I I= + … (ii)

(max)(max)

6.3 5 13 mA100

in Zin

V VI

R− −= = =

From equation (i), (max)13 5ZI= +

(max) 8 mAZI =

Therefore, minimum power rating of diode i.e. maximum power dissipation is (max) 5 8 40 mWZ ZV I× = × =

Hence, the correct option is (B). Note : Power rating or minimum power rating of zener diode represent maximum power dissipation. Maximum power dissipation (max)Z ZV I= × 5 8mA= × = 40mW

Given : 26 AdcI = μ , Ideality factor 1η = ,

0.5nF,dC = 62 10 rad/secω = × , 26 mVTV = .

To find small signal component of diode current deactivating dc source and replacing diode by its equivalent small signal model :

Here, 3

61 26 10

26 10T

fVrI

−

−

η × ×= =×

310= Ω

5sin ( ) mVeq

tiZω=

where, 1|| 100eq fd

Z rsC

= +

/100

(1 ) /f d

eqf d d

r sCZ

sr C sC

= + +

1001

feq

f d

rZ

j r C= +

+ ω

6 3 91000 100

1 2 10 10 0.5 10eqZj −

= + + × × × × ×

1000 1100 1001001 1eq

jZj j

+= + =+ +

2 21 1

2 2

(1100) (100) 100 1tan tan1100 1(1) (1)

eqZ − −+ = ∠ − +

0781.02 39.805eqZ = ∠ −

05sin ( ) mV

781.02 39.805ti ω=

∠ − Ω

06.401sin ( 39.805 ) Ai t= ω + μ

Hence amplitude of small signal current is 6.401 μA .

Given circuit and input waveform are shown below,

Fig. (a)

Where, 1 50LR R= = Ω

For 0 / 2t T< < , 8VsV =

100 5sin( ) mVt

fr

dC

i

LR

1R

10−

8

T2T sV

+

−

LV0

sV

t

1.41 6.4

1.42 – 3


So, diode will be off and can be replaced with open circuit

Fig. (b)

∴ 50 8 4 V50 50LV = × = +

For / 2T t T< < , 10VsV = − . So, diode will be

ON and hence it will be short circuited and 1R can be neglected as it comes across short circuit.

Fig. (c)

10 VLV = −

Therefore, equivalent output waveform will be

Fig. (d)

Average value of ( )LV t is given as

0

1 ( )T

av LV v t stT

=

/2

0 /2

1 4 ( 10)av

T T

LT

V dt dtT

= ⋅ + −

1 4 102 2avLT TV

T = × − ×

1 ( 6) 3 V2avLTV

T= × − × = −

Hence, the average value of LV is – 3 V

Given : 1 200 0.2 kΩR = Ω =

1 kΩLR =

(max) 60 mAzI =

20Vzv =

?iV = (For zener diode in ON state)

(Fixed)in z LI I I= +

( LI is fixed since LR is fixed)

(max) (max)in z LI I I= +

(min) (min)in z LI I I= +

20VzV =

0 20 20 mA1kL

L

VIR

= = =

(max)(max)

1

(60 20) min zin

V VI

R−

= = +

(max) 2080 m

0.2 kinV −

=

(max) 20 16inV − =

(max) 36 VinV =

So, for (min)inV

(min) 0zI =

(min) 0 20 minI = +

(min)

1

20min zV VR

−=

50

50 Ω

8 V

+

−

LV+

−

LR10 V

LV

10 V

4 V

( )v t

t/2T

T

3 /2T

2T

zI LI

inI1R

+_LRiV (Fixed)

1.43 (D)


(min) 2020m

0.2kinV −

=

(min) (4 20) VinV = +

(min) 24 VinV =

So, (min) (max)24 V, 36 Vin inV V= =


..Method 1.. Given circuit is shown below,

2m

rmsVV =

2 2 230Vm rmsV V= = ×

230 2 VmV =

Also given input voltage is sinusoidal. sinin mV V t= ω

230 2 sininV t= ω

Assume that negative half cycle occurs first,

Case 1 : During 1st negative half cycle, Diode 1D will be forward biased and diode 2Dwill be reverse biased. Hence, charging of capacitor 1C will start and it will charge upto maximum value of input.

Hence, 230 2V=

1230 2 sin 230 2 VCV t= ω = and

20VCV =

Case 2 : During 1st positive half cycle, Diode 1D will be reverse biased and diode 2D will be forward biased. Hence, charging of capacitor 2C will start and

capacitor 2C will charge upto maximum value

of input, which appear across capacitor 2C .

2 1 max230 2 sinC CV V t= + ω

2

230 2 230 2CV = +

2

460 2 VCV =

Case 3 : During 2nd positive half cycle, negative terminal of diode 1D is at 230 2 V and negative terminal

of 2D is at 460 2 V . Now, diode 1D and 2D will be forward bias when voltage at their positive terminal will be higher than that of the negative terminal, which is never possible beyond first negative half cycle and first positive half cycle. Hence, in this case, 1 2andD D will be reverse biased.

230 V (rms)

1 0.1 FC = μ2D

1D2 0.1 FC = μ 0V

inV

0

230 2

t

230 2

1C

( )V t 2C 0V1CV

230 2 sin tω 2CV

O.C.

1C

O.C. 2C 0V1CV

230 2 sin tω 2CV

S.C.

( )V t

1.44 650.538


Hence, from above circuit

20 460 2 VCV V= =

0 650.538VV =

Hence, the magnitude of the steady-state output

0V is 460 2 V .

..Method 2.. Given circuit is shown below,

The clamper will do the shift of input voltage to 2 .mV

Given rms value of sinusoidal input

2302mV = 2 230mV = ×

Peak detector will detect the maximum value 2 2 230( 2)mV= = × 650.53=

..Method 3.. Define the circuit as a voltage doubler. 0 2 2 230 2 650.4 VmV V= = × =

1C

( )V t 2C 0V230 2 sin tω 2CVO.C.

O.C.

1CV

230 V (rms)

1 0.1 FC = μ2D

1D 2 0.1 FC = μ 0V

Positive clamper Positive peak detector

mV2 mV

0mV−

ClamperPositive

1.1 An electrostatic field is said to be

conservative when (A) the divergence of the field is equal to

zero. (B) the curl of the field is equal to zero.

(C) the curl of the field is equal to Bt

∂−∂

(D) the Laplacian of the field is equal to 2

2E

t∂με∂

.

1.2 Two long parallel wires in free space are separated by a distance R and carry currents of equal magnitude by opposite in direction. At any general point, the z-component of

(A) The magnetic vector potential is

20 2

21

ln4

I dd

μ π

.

(B) The magnetic induction is 0 2

1

.2

I dd

μ π

(C) The magnetic induction is zero. (D) The magnetic vector potential is

20 1

224

I dd

μ π

.

1.3 On either side of a charge-free interface

between two media (A) the normal components of the electric

field are equal. (B) the tangential components of the

electric field are equal. (C) the normal components of the electric

flux density are equal. (D) the tangential components of the

electric flux density are equal. 1.4 Vector potential is a vector (A) whose curl is equal to the magnetic

flux density. (B) whose curl is equal to the electric

field intensity. (C) whose divergence is equal to the

electric potential. (D) which is equal to the vector product

E H×

.

1Basics of

Electromagnetics

z

y

x

d2d1

P

II

R

1987 IIT Bombay

1988 IIT Kharagpur


1.5 The electric field strength at a far-off

point P due to a point charge, q+ located at the origin, O is 100 millivolts/metre. The point charge is now enclosed by a perfectly conducting hollow metal sphere with its centre at the origin O. The electric field strength at the point, P

(A) remains unchanged in its magnitude and direction.

(B) remains unchanged in its magnitude but reverse direction.

(C) would be that due to a dipole formed by the charge q+ at O and q− induced.

(D) would be zero.

1.6 Which of the following field equations

indicates that the free magnetic charges do not exist?

(A) 1H A= ∇×μ

(B) 34Idl RH

R×=

π

(C) 0H∇⋅ =

(D) H J∇× =

1.7 Given 2 2 2cos sinzV x y i x e j z y k= + +

and S the surface of unit cube with one corner at the origin and edges parallel to the coordinate axis, the value of the

integral C

V n dS⋅ is ___________.

1.8 For a uniformly charged sphere of radius R and charge density ρ , the ratio of magnitude of electric fields at distances

/2R and 2R from the center i.e., ( /2)( 2 )

E r RE r R

==

is ___________.

1.9 A long solenoid of radius R, and having N turns per unit length carries a time dependent current 0( ) cos ( )I t I t= ω . The magnitude of induced electric field at a distance R/2 radially from the axis of the solenoid is

(A) 0 0 sin ( )2R NI tμ ω ω

(B) 0 0 cos ( )4R NI tμ ω ω

(C) 0 0 sin ( )4R NI tμ ω ω

(D) 0 0 sin ( )R NI tμ ω ω

1.10 Match List-I with List-II and select the

correct answer using the code given below the lists :

List-I A. H J∇× =

B. C s

BE dl dSt

∂⋅ = − ⋅∂

C. Jt

∂ρ∇ ⋅ = −∂

List-II 1. Continuity equation 2. Faraday’s law 3. Ampere’s law 4. Gauss’s law 5. Biot-Savart law Codes : A B C (A) 3 2 1 (B) 2 1 3 (C) 2 3 1 (D) 1 2 3

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3.3GATE ACADEMY ® Electromagnetic Theory : Basics of Electromagnetics

1.11 In the infinite plane, y = 6 m, there exists

a uniform surface charge density of 2(1/6000 ) C/mπ μ . The associate electric

field strength is (A) ˆ30 V/mi (B) ˆ3 V/mj

(C) ˆ30 V/mk (D) ˆ60 V/mj 1.12 The electric field strength at distant point

P due to a point charge, q+ located at the origin is 100 V/mμ . The point charge is now enclosed by a perfectly conducting metal sphere whose centre is at the origin. Then the electric field strength at the point P, outside the sphere becomes

(A) Zero (B) 100 V/mμ (C) 100 V/m− μ (D) 50 V/mμ

1.13 A metal sphere with 1 m radius and a

surface charge density of 10 Coulombs/ 2m is enclosed in a cube of 10 m side. The total outward electric displacement normal to the surface of the cube is

(A) 40π Coulombs (B) 10π Coulombs (C) 5π Coulombs (D) None of the above

1.14 A loop is rotating about the y-axis in a

magnetic field 0 cos( ) TxB B t a= ω + φ .

The voltage in the loop is (A) zero. (B) due to rotation only. (C) due to transformer action only. (D) due to both rotation and transformer

action.

1.15 The electric field on the surface of a

perfect conductor is 2 V/m. The conductor is immersed in water with

080ε = ε . The surface charge density on the conductor is

(A) 20 C/m

(B) 22 C/m

(C) 11 21.8 10 C/m−×

(D) 9 21.41 10 C/m−×

1.16 The unit of H∇×

is

(A) Ampere (B) Ampere/meter (C) Ampere/meter2 (D) Ampere-meter 1.17 Medium 1 has the electrical permittivity

1 01.5ε = ε F/m and occupies the region to the left of 0x = plane. Medium 2 has the electrical permittivity 2 02.5ε = ε F/m and occupies the region to the right of x = 0 plane. If 1E in medium 1 is

1 (2.0 3.0 1.0 )x y zE a a a= − + V/m, then

2E

in medium 2 is

(A) (2.0 7.5 2.5 ) V/mx y za a a− +

(B) (2.0 2.0 0.6 ) V/mx y za a a− +

(C) (1.2 3.0 1.0 ) V/mx y za a a− +

(D) (1.2 2.0 0.6 ) V/mx y za a a− +

1.18 If the electric field intensity is given by, ( )x y zE xa ya za= + +

volt/m,

The potential difference between x(2, 0, 0) and y (1, 2, 3) is

(A) + 1 volt (B) – 1 volt (C) + 5 volt (D) + 6 volt

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1.19 A parallel plate air-filled capacitor has

plate area of 4 210 m− and plate separation

of 310 m− . It is connected to a 0.5 V , 3.6 GHz source. The magnitude of the

displacement current is 9

010 F/m36

− ε = π

(A) 10 mA (B) 100 mA (C) 10 A (D) 1.59 mA

1.20 A medium is divided into regions I and II

about x = 0 plane, as shown in the figure below. An electromagnetic wave with electric field 1 4 3 5x y zE a a a= + +

is

incident normally on the interface from region I. The electric field 2E

in region

II at the interface is

(A) 2 1E E=

(B) 4 0.75 1.25x y za a a+ −

(C) 3 3 5x y za a a+ +

(D) 3 3 5x y za a a− + −

1.21 If C is a closed curve enclosing a surface

S, then the magnetic field intensity H

, the current density J

and the electric flux

density D

are related by

(A) S C

DH dS J dlt

∂⋅ = + ⋅ ∂

(B) C S

DH dl J dSt

∂⋅ = + ⋅ ∂

(C) S C

DH dS J dlt

∂⋅ = + ⋅ ∂

(D) C S

DH dl J dSt

∂⋅ = + ⋅ ∂

1.22 If a vector field V

is related to another

vector field A

through V A= ∇×

. Which of the following is true? Note : C and CS refer to any closed contour and any surface whose boundary is C.

(A) CC S

V dl A dS⋅ = ⋅

(B) CC S

A dl V dS⋅ = ⋅

(C) CC S

V dl A dS∇× ⋅ = ∇× ⋅

(D)

CC S

A dl V dS∇× ⋅ = ⋅

1.23 Two infinitely long wires carrying current are as shown in the figure below. One wire is in the y-z plane and parallel to the y-axis. The other wire is in the x-y plane and parallel to the x-axis. Which component of the resulting magnetic field are non-zero at the origin?

0x 0x 0x

1 0

1

1

3

0

r

Region I

2 0

2

2

4

0

r

Region II

1E

2E

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(A) x, y, z components (B) x, y components (C) y, z components (D) x, z components 1.24 A magnetic field in air is measured to be

0 2 2 2 2y xx yB B a a

x y x y

= − + +

.

What current distribution leads to this field?

[Hint : The algebra is trivial in cylindrical co-ordinate.]

(A) 02 2

0

1 , 0zB aJ rx y

= − ≠ μ +

(B) 0

2 20

2 , 0zB aJ rx y

= − ≠ μ +

(C) 0, 0J r= ≠

(D) 02 2

0

1 , 0zB aJ rx y

= ≠ μ +

1.25 A current sheet 10 A/myJ a=

lies on the

dielectric interface 0x = between two dielectric media with 1 5,rε = 1 1rμ = in

region-1 ( 0)x < and 2 2rε = , 2 2rμ = in region-2 ( 0)x > . If the magnetic field in

region-1 ( 0 )x −= is 1 3 30x yH a a= +

A/m, the magnetic field in region-2 at 0x += is

(A) 2 1.5 30 10 A/mx y zH a a a= + −

(B) 2 3 30 10 A/mx y zH a a a= + −

(C) 2 1.5 40 A/mx yH a a= +

(D) 2 3 30 10 A/mx y zH a a a= + +

An infinitely long uniform solid wire of radius a carries a uniform dc current of density J

.

1.26 The magnetic field at a distance r from the center of the wire is proportional to

(A) r for r < a and l/ 2r for r > a (B) 0 for r < a and 1/r for r > a (C) r for r < a and 1/r for r > a

(D) 0 for r < a and l/ 2r for r > a 1.27 A hole of radius b (b < a) is now drilled

along the length of the wire at a distance d from the center of the wire as shown below.

The magnetic field inside the hole is (A) uniform and depends only on d. (B) uniform and depends only on b. (C) uniform and depends on both b and d. (D) non uniform.

1.28 The force on a point charge +q kept at a

distance d from the surface of an infinite grounded metal plate in a medium of permittivity ε is [Set - 01]

x

y

2 20 Region-2 : 2, 2 r rx

1 10 Region-1 : 5, 1 r rx

x = 0J

a

d

b

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Common Data for Questions 1.26 & 1.27

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(A) 0

(B) 2

216q

dπε away from the plate

(C) 2

216q

dπε towards the plate

(D) 2

24q

dπε towards the plate

1.29 Given the vector

(cos )(sin ) (sin )(cos )x yA x y a x y a= +

where xa , ya denote unit vectors along

,x y directions, respectively.

The magnitude of curl of A

is _________. [Set - 03]

1.30 A region shown below contains a perfect conducting half-space and air. The surface current sK

on the surface of the

perfect conductor is 2s xK a= amperes

per meter. The tangential H

field in the air just above the perfect conductor is [Set - 03]

(A) ( )x za a+ amperes per meter.

(B) 2 xa amperes per meter.

(C) 2 za− amperes per meter.

(D) 2 za amperes per meter.

1.31 The electric field (assumed to be one-dimensional) between two points A and B is shown. Let Aψ and Bψ be the

electrostatic potentials at A and B, respectively. The value of B Aψ −ψ in volts is________. [Set - 04]

1.32 If 3 2 2 2(2 3 ) (6 3 )x yE y yz a xy xz a= − − − −

(6 ) zxyz a+ is the electric field in a source free region, a valid expression for the electrostatic potential is [Set - 04]

(A) 3 2xy yz− (B) 3 22xy xyz−

(C) 3 2y xyz+ (D) 3 22 3xy xyz−

1.33 Consider a straight, infinitely long,

current carrying conductor lying on the z-axis. Which one of the following plots (in linear scale) qualitatively represents the dependence of Hφ on r, where Hφ is the

magnitude of the azimuthal component of magnetic field outside the conductor and r is the radial distance from the conductor? [Set - 01]

(A)

(B)

y

xsK

Air

Perfect conductorPerfect conductor

20 kV/cm

0 kV/cm

5μm

40 kV/cm

A B

H

r

H

r

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(C)

(D)

1.34 A vector P

is given by 3 2 2 2

x y zP x ya x y a x yza= − − . Which of

the following statements is TRUE? [Set - 01]

(A) P

is solenoidal, but not irrotational. (B) P

is irrotational, but not solenoidal.

(C) P

is neither solenoidal nor irrotational. (D) P

is both solenoidal and irrotational.

1.35 In a source free region in vaccum, if the electrostatic potential 2 2 22x y czφ = + + , the value of constant c must be ______. [Set - 02]

1.36 A vector field 22 zD a zaρ= ρ + exists

inside a cylindrical region enclosed by the surfaces 1,ρ = 0z = and 5.z = Let S be the surface bounding of this cylindrical region. The surface integral of this field on ,

SS D dS⋅

is _______.

[Set - 03]

1.37 Concentric spherical shells of radii 2 m,

4m and8m carry uniform surface charge densities of 20 nC/m2, − 4 nC/m2 and sρ

respectively. The value of sρ (nC/m2 ) required to ensure that the electric flux density 0D =

at radius 10 m is_______.

[Set - 01]

1.38 The current density in a medium is given by

22

400sin Am2 ( 4) rJ a

r−θ=

π +

The total current and the average current density flowing through the portion of a

spherical surface r = 0.8 m, ,12 4π π≤ θ ≤

0 2≤ φ ≤ π are given, respectively, by [Set - 01]

(A) 15.09 A, 12.86 Am 2− (B) 18.73 A, 13.65 Am 2− (C) 12.86 A, 9.23 Am 2− (D) 10.28 A, 7.56 Am 2− 1.39 A uniform and constant magnetic field

ˆB z=

B exists in the z direction in vaccum. A particle of mass m with a small charge q is introduced into this region with an initial velocity ˆv x= xv +

z zv . Given that , , , xB m q v and zv are all non-zero, which one of the following describes the eventual trajectory of the particle? [Set - 02]

(A) Helical motion in the z -direction. (B) Circular motion in the xy -plane. (C) Linear motion in the z -direction. (D) Linear motion in the x -direction. 1.40 The parallel-plate capacitor shown in the

figure has movable plates. The capacitor is charged so that the energy stored in it is E when the plate separation is d. The capacitor is then isolated electrically and the plates are moved such that the plate separation becomes 2d. [Set - 02]

H

r

H

r

d

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At this new plate separation, what is the energy stored in the capacitor, neglecting fringing effects?

(A) 2E (B) 2E (C) E (D) /2E 1.41 A positive charge q is placed at x = 0

between two infinite metal plates placed at x = −d and at x = +d respectively. The metal plates lie in the yz plane. [Set - 02]

The charge is at rest at t = 0, when a

voltage +V is applied to the plate at – d and voltage –V is applied to the plate at x = +d. Assume that the quantity of the charge q is small enough that it does not perturb the field set up by the metal plates. The time that the charge q takes to reach the right plate is proportional to

(A) /d V (B) /d V

(C) /d V (D) /d V 1.42 Faraday’s law of electromagnetic

induction is mathematically described by which one of the following equations? [Set - 03]

(A) 0B∇⋅ =

(B) vD∇ ⋅ = ρ

(C) BEt

∂∇× = −∂

(D) DH Et

∂∇× = σ +∂

1.43 Two conducting spheres 1S and 2S of

radii a and b (b a> ) respectively, are

placed far apart and connected by a long, thin conducting wire, as shown in the figure. [Set - 02]

For some charge placed on this structure,

the potential and surface electric field on 1S are aV and aE , and that on 2S are bV

and bE , respectively. Then, which of the following is CORRECT?

(A) anda b a bV V E E= <

(B) anda b a bV V E E> >

(C) anda b a bV V E E= >

(D) anda b a bV V E E> = 1.44 If the vector function

1 2 3(3 ) ( 2 ) ( )x y zF y k z a k x z a k y z a= − + − − +

is irrotational, then the values of the constants 1 2,k k and 3k respectively, are

[Set - 02] (A) 0.3, 2.5, 0.5− (B) 0.0, 3.0, 2.0 (C) 0.3, 0.33, 0.5 (D) 4.0, 3.0, 2.0 1.45 An electron 1( )q is moving in free space

with velocity 510 m/s towards a stationary electron 2( )q far away. The closest distance that this moving electron gets to the stationary electron before the repulsive force diverts its path is _______ 810 m−× . Given :

Mass of electron 319.11 10 kgm −= × .

Charge of electrons 191.6 10 Ce −= − ×

and permittivity 90

1 10 F/m36

−ε = ×π

[Set - 02]

at=

–x

d

at=

+x

d

+ q

x = 0

Wire

S2

S1

Radius rb

Radius ra

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1.46 What is the electric flux ( )ˆE da⋅

through a quarter-cylinder of height H (as shown in the figure) due to an infinitely long line charge along the axis of the cylinder with a charge density of Q ?

(A) 04

HQε

(B) 0

4H

Qε

(C) 0

4HQε

(D) 0

HQε

1.47 In the table shown. List I and List II, respectively, contain terms appearing on the left-hand side and the right-hand side of Maxwell's equations (in their standard form). Match the left-hand side with the corresponding right-hand side.

List-I List-IĪ 1. D∇⋅

P. 0

2. E∇×

Q. ρ

3. B∇ ⋅

R. Bt

−∂∂

4. H∇×

S. DJt

∂+∂

(A) 1 - P, 2 - R, 3 - Q, 4 - S (B) 1 - R, 2 - Q, 3 - S, 4 - P (C) 1 - Q, 2 - R, 3 - P, 4 - S (D) 1 - Q, 2 - S, 3 - P, 4 - R 1.48 Two identical copper wires 1W and 2W ,

placed in parallel as shown in the figure, carry currents I and 2I. respectively, in

opposite directions. If the two wires are separated by a distance of 4r, then the magnitude of the magnetic field B

between the wires at a distance r from 1W is

(A) 065

Ir

μπ

(B) 056

Ir

μπ

(C) 0

6Ir

μπ

(D) 2 20

22Ir

μπ

1.49 For a vector field A

, which one of the

following is FALSE ?

(A) A

is solenoid if 0A∇ ⋅ =

.

(B) 2( ) ( )A A A∇× ∇× = ∇ ∇ ⋅ − ∇

(C) A∇ ×

is another vector field.

(D) A

is irrotational if 2 0.A∇ =

0ε

QH

1W

2W

r

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1.1 B 1.2 A 1.3 B, C 1.4 A 1.5 D

1.6 C 1.7 1 1.8 2 1.9 C 1.10 A

1.11 B 1.12 A 1.13 A 1.14 D 1.15 D

1.16 C 1.17 C 1.18 C 1.19 A 1.20 C

1.21 D 1.22 B 1.23 D 1.24 C 1.25 A

1.26 C 1.27 A 1.28 C 1.29 0 1.30 D

1.31 – 15 1.32 D 1.33 C 1.34 A 1.35 – 3

1.36 78.54 1.37 – 0.25 1.38 * 1.39 A 1.40 A

1.41 C 1.42 C 1.43 C 1.44 B 1.45 5.063

1.46 A 1.47 C 1.48 B 1.49 D

An electrostatic field with electric field, E

is said

to be conservative, if the closed line integral of the field is zero i.e.,

0E dl⋅ =

…(i)

Applying Stokes theorem,

( )s

E dl E dS⋅ = ∇× ⋅

Equation (i) becomes 0E∇× =

, i.e., the curl of the field, E

is equal to zero.


Key Point

(i) BEt

∂∇× = −∂

for time varying fields.

(ii) Laplacian of the field E

is given by 2

22EEt

∂∇ = με∂

, which is the wave

equation of an electromagnetic wave. (iii) If electrostatic field is divergenceless

then divergence of field is zero i.e. 0E∇⋅ =

.

Magnetic vector potential A, at point P in x-y plane is shown below,

Let us take current I of equal magnitude but opposite in direction through the wires AB and CD. Magnetic vector potential 1A due to wire AB is given by,

01 4

L

L

I dzAr−

μ=π

01 2 2

1

14

L

L

IA dzz d−

μ=π +

[From figure, 2 21r z d= + ]

A

y

x

d1

P

I

I

R

d2

00

z r

z

C

B D

Answers Basics of Electromagnetics

Explanations Basics of Electromagnetics

1.1 (B) 1.2 (A)


01 2 2

0 1

12

LIA dzz d

μ=π +

2 201 1

0ln( )

2

LIA z z dμ = + + π

2 2

2 2

1 lndx x x ax a

= + ± ±

2 2 201 1 1ln ( ) ln (0 0 )

2IA L L d dμ = + + − + + π

2 201 1 1ln ( ) ln

2IA L L d dμ = + + − π

Since 1,L d>>

[ ]01 1ln( ) ln

2IA L L dμ= + −

π

[ ]01 1ln (2 ) ln

2IA L dμ= −

π … (i)

Similarly, magnetic vector potential 2A due to wire CD is given by,

[ ]02 2ln (2 ) ln

2IA L dμ= − −

π … (ii)

[Negative sign is due to direction of current through the wire CD] Hence, total magnetic vector potential A due to both wires will be, 1 2A A A= +


[ ]01ln (2 ) ln

2IA L dμ = − π

[ ]02ln (2 ) ln

2I L dμ + − − π

[ ]02 1ln ln

2IA d dμ= −

π0 2

1

ln2

I dd

μ=π

2

0 22

1

ln4

I dAd

μ=π

log log me em a a =


Key Point (i) Vector magnetic potential, A exists when

J is present. (ii) It has applications to obtain radiation

characteristics of antenna, apertures and also to obtain radiation leakage from transmission lines, waveguide and microwave ovens.

This question is regarding the boundary conditions at the interface between two media (1) and (2). For a charge free interface between two media with dielectric constants 1ε and 2ε :

(i) The tangential components of the electric field are equal. i.e.,

1 2t tE E= or tE is

continuous at the boundary. (ii) The normal components of electric flux

density D are equal, i.e., 1 2n nD D= and

nD is continuous at the boundary.

Hence, the correct options are (B) and (C). Key Point For interface carrying charge, 0sρ ≠ . In this case

1 2n n sD D− = ρ .

Vector magnetic potential, A

is a vector whose

curl is equal to the magnetic flux density, B

A B H∇× = = μ

Hence, the correct option is (A). Key Point Some points about vector potential : (i) The unit of A

is Wb/meter.

(ii) AE Vt

∂= −∇ −∂

1.3 (B) and (C)

1.4 (A)


(iii) Magnetic vector potential is used to find near and far fields in antenna.

(iv) VAt

∂∇⋅ = −με∂

It is called the Lorentz condition for potential.

(v) For magnetostatic field, 0A∇⋅ =

(vi) 2

22

vVVt

ρ∂∇ − με = −∂ ε

2

22AA J

t∂∇ − με = −μ∂

The above two equation are called wave equation.

Given : Electric field due to point charge q is 100 mV/m .

According to Faraday’s law, the point charge +Q will induce a Q− charge on hollow conducting sphere.

As point is very far, conducting sphere will act as a point charge of Q− for point P. Electric field intensity due to point charge Q− will be 100 mV/m− . Therefore, the net electric field strength at point P will be (100 100) mV/m 0netE = − = Hence, the correct option is (D).

Gauss’s law for magnetic fields states that the magnetic flux flowing through the closed surface is equal to zero, as free magnetic charges do not exist and magnetic flux lines are closed.

i.e. 0B∇⋅ =

From divergence theorem, we have

( )s vB dS B dv⋅ = ∇ ⋅

Since, 0s

B dS⋅ =

or 0B∇ ⋅ =

As B H= μ

, 0H∇ ⋅μ =

In homogeneous medium μ is independent of position.

Since, 0H∇⋅ =


Given : 2 2 2cos sinzV x yi x e j z yk= + +

From divergence theorem,

S v

V n dS V dv⋅ = ∇ ⋅

Divergence of any vector is given by,

yx zVV VVx y z

∂∂ ∂∇ ⋅ = + +∂ ∂ ∂

2 2 2( cos ) ( ) ( sin )zV x y x e z yx y z

∂ ∂ ∂∇ ⋅ = + +∂ ∂ ∂

2 2cos 0 sin 1V y y∇ ⋅ = + + =

Hence, the value of the integral is 1.

For a uniformly charged sphere of radius ‘R’, electric field intensity inside the sphere is given by,

30

14

QrER

=πε

For ,2Rr = point P will be inside the sphere.

RP

Q 100 mV/mE

Origin

+Q

–Q––––––

––––––– – – –

––– R

––

P

+ ++

+ +++

+++

++

++ ++

R r

P

1.5 (D)

1.6 (C)

1.7 1

1.8 2


Since, 3 30 0

1 24 8

RQQRE

R R

= =

πε πε

3

0643

Q RER

=ε π

Volume charge density, ρ is defined as charge per unit volume.

34

3

Q

Rρ =

π

So, 06

RE ρ=ε

... (i)

For a uniformly charged sphere of radius ‘R’, electric field intensity outside the sphere is given by,

20

14

QEr

=πε

For 2 ,r R= point P will be outside the sphere.

2 20 0

14 (2 ) 4 (4 )

Q QER R

= =πε πε

3 0 04 3 (4) 12

3

Q R RER

ρ= =ε ε π

... (ii)


( )

0

0

at122 2

at 2 6

RE rR

E r R R

= ε ρ = × = = ε ρ

Hence, the ratio of magnitude of electric fields at distances /2R and 2R from the center is 2. Key Point Electric field inside sphere is given by,

30

14

QrER

=πε

…. (i)

where, r = distance from center to the point R = radius of sphere Volume charge density, ρ is given by,

34

3

Q

Rρ =

π …. (ii)


03rE ρ=

ε

Given : A long solenoid of radius R, and having N turns per unit length carries a time dependent current 0( ) cos ( ).I t I t= ω

A solenoid consists of a long conducting wire made up of many loops packed closely together. For coil that are closely together, the magnetic field is approximately uniform towards the center. If the current flows through the conducting wire then it generates magnetic field and it is uniform.

+ ++

+ ++

+

+++

++

++ ++

RP

2R++

+

II

B

Magneticfield is zero

c d

ab

Amperian path

l

current out of the page

current inside the page

1.9 (C)


From Ampere’s law, encH dl I⋅ =

0 encB dl I⋅ = μ

… (i)

00 90

b c d

a b c

B

B dl B dl B dl B dl

= θ=

⋅ = ⋅ + ⋅ + ⋅

090

a

d

B dl

θ=

+ ⋅

0cos 0d

c

B dl B dl B l⋅ = =

… (ii)

If I is the current in the wire of the solenoidal then the enclosed I is given by, nI where n is number of loops. From equation (i) and (ii), 0Bl nI= μ

00

nIB NIlμ= = μ

where, nl

= number of loops per unit length

0B NI= μ [ ]( ) andH NI t B H= = μ

0 0( ) ( )B H NI t= μ = μ

0 0 cos( )B NI t= μ ω According to Faraday’s law,

BE dl dSt

∂⋅ = − ⋅∂

2

0 02 sin ( )2 4R RE NI t π⋅ π = μ ω ω

0 0 sin ( )4RE NI t= μ ω ω


H J∇× =

: Ampere’s Law

C s

BE dl dSt

∂⋅ = − ⋅∂

: Faraday’s Law

Jt

∂ρ∇ ⋅ = −∂

: Continuity equation

Hence, the correct option is (A). Key Point Gauss’s law : vD∇ ⋅ = ρ

Biot-Savart’s law : 24RIdl aH

R×=

π

Given : 21/6000 C/msρ = π μ

The electric field strength is given by,

2

snE aρ=

ε

where, na is the unit vector normal to the plane 6 my = .

Case 1 : To determine the electric field strength to the left of the plane, n ya a= −

Electric field is,

2

syE aρ= −

ε

6 910 36 106000 2 yE a

− × π×= −π×

ˆ3 V/m 3 V/myE a j= − = −

y

x

z

y = 6

ya

ya

1.10 (A)

1.11 (B)


Case 2 : To determine electric field strength to the right of the plane. n ya a=

Electric field is,

2

syE aρ=

ε

ˆ3 V/m 3 V/myE a j= =

Hence from case 1 and case 2, electric field strength will be either ˆ3 V/mj or ˆ3 V/mj− . Hence, the correct option is (B).

Refer Solution 1.5

Given : A metal sphere with radius = 1 m and surface charge density 210 C/msρ = The total charge on the sphere is given by, 2 24 10 4 (1)sQ r= ρ π = × π× (40 )CQ = π This sphere is enclosed in a cube of side = 10 m. According to Gauss’s law, the net electric flux flowing out through a closed surface (normal to the surface) is equal to the charge enclosed. enc

s

D dS Q⋅ =

Total outward electric flux (displacement) (40 )= π C normal to the surface of the cube.


Given : 0 cos( ) TxB B t a= ω + φ

The flux is defined by, m B dSψ = ⋅

Thus voltage, md dBV dSdt dtψ= = ⋅

Thus, if magnetic field is a function of time, then it will be transformed into voltage. Hence, time varying magnetic field generates voltage. Also the loop is rotating with a certain velocity thus it will also induce some voltage in the loop. According to equation,

( )V v B dl= × ⋅

Thus, voltage in the loop will be due to both rotation and transformer action. Hence, the correct option is (D).

Given : 02V/m, 80E = ε = ε

This is a special case of conductor-dielectric boundary condition. Under static condition, following conclusion can be made about a perfect conductor. 1. No electric field may exist within a

conductor. 0, 0s Eρ = =

2. Since 0E V= −∇ =

, i.e. there can not be potential difference between any two points in the conductor. So conductor is an equipotential body.

3. Electric field can be external to the conductor and normal to the surface.

0 0t r tD E= ε ε =

0n r n sD E= ε ε = ρ

Hence total electric field of a perfect conductor is given by only normal component.

0

sn

r

E ρ=ε ε

12280 8.85 10

s−

ρ=× ×

9 21.41 10 C/ms−ρ = ×


1.12 (A)

1.13 (A)

1.14 (D)

1.15 (D)


. Method 1 : From Maxwell’s equation, C DH J J∇× = +

CJ

is conduction current density and DJ

is displacement current density. Both have unit

2A/m . So, Unit of H∇×

is 2A/m . Hence, the correct option is (C). . Method 2 :

(del)∇ operator has unit per meter and H (magnetic field intensity) has unit A/m . Hence, the correct option is (C).

Given : 1 2 3x y zE a a a= − +

From the above figure, n xa a=

1 1( )n n nE E a a= ⋅

1

(2 3 ) 2n x y z x x xE a a a a a a = − + ⋅ =

1 11t nE E E= −

1

(2 3 ) 2 3t x y z x y zE a a a a a a= − + − = − +

Since it is not given that plane is having charge or not, assuming plane is charge free i.e. 0sρ =

From the boundary conditions : 1.

1 2n nD D=

1 21 2n nE Eε = ε

2 1

1

2n nE Eε=

ε

2

0

0

1.5 2 1.22.5n x xE a aε= × =

ε

2. 1 2t tE E=

2

3t y zE a a= − +

2 22 t nE E E= +

2 1.2 3x y zE a a a= − +


Given : x y zE xa ya za= + +

x y zdl dxa dya dza= + +

Potential difference is given by, V E dl= − ⋅

( ) ( )x y z x y zV xa ya za dxa dya dza= − + + ⋅ + +

2 0 0

1 2 3V x dx y dy z dz = − + +

2 0 02 2 2

1 2 32 2 2x y zV

= − − −

1 [3 4 9] 5 V2

V = − − − =


Given : Area of air filled parallel plate capacitor, 4 210 mA −=

Separation between plates, 310 md −= Source voltage, 0.5 VsV = at 3.6 GHzf =

1E

2E

Medium 1

1 01.5 Medium 2

2 02.5

0x 0x 0x

Scan for

Video Solution

Scan for

Video Solution

1.16 (C)

1.17 (C)

1.18 (C)

1.19 (A)


Displacement current density is given by, dJ j E= ωε

Magnitude of displacement current density is dJ E= ωε

where, 2 fω = π , 0 rε = ε ε , 1rε = (for air) Electric field is given by,

sVEd

=

02 sd

VJ fd

= π ε

9

93

10 0.52 3.6 1036 10dJ

−

−= π× × × ×π

2100 A/mdJ =

Since, displacement current, d dI J A= ×

4 2100 10 10 AdI − −= × =

10 mAdI = Hence, the correct option is (A).

Given : 1 4 3 5x y zE a a a= + +

From the above figure, n xa a=

1 1( )n n nE E a a= ⋅

1

(4 3 5 ) 4n x y zz x x xE a a a a a a = + + ⋅ =

1 11t nE E E= −

1

(4 3 5 ) 4 3 5t x y zz x y zE a a a a a a= + + − = +

Since it is not given that plane is having charge or not, assuming plane is charge free i.e. 0sρ =

From the boundary conditions :

1. 1 2n nD D=

1 21 2n nE Eε = ε

….(i)

2 1

1

2n nE Eε=

ε

2

0

0

3 4 34n x xE a aε= × =

ε

2. 1 2t tE E=

2

3 5t y zE a a= +

2 22 t nE E E= +

2 3 3 5x y zE a a a= + +


According to Maxwell’s 4th equation,

DH dl J dS dSdt∂⋅ = ⋅ + ⋅

S

DH dl J dSdt

∂⋅ = + ⋅


Given : V A= ∇×

Taking surface integral in both the sides,

( )S SV dS A dS⋅ = ∇× ⋅

… (i)

From Stokes theorem,

( )S C

A dS A dl∇× ⋅ = ⋅

… (ii)


S CV dS A dl⋅ = ⋅


0x 0x 0x

1 0

1

1

3

0

r

Region I

2 0

2

2

4

0

r

Region II

1E

2E

Scan for

Video Solution

1.20 (C) 1.21 (D)

1.22 (B)


. Method 1 :

By applying right thumb rule it is observed that field due to wire in y-z plane is in –x-direction and field due to wire in x-y plane is in –z-direction. Therefore, x and z components of the resulting magnetic field are non-zero at the origin.


. Method 2 :

Direction of magnetic field can be found as,

H la a aρ= ×

where, la = unit vector in the direction of current

aρ = unit vector along perpendicular

distance from conductor to point of interest

Due to wire placed in y-z plane,

( )H y z xa a a a= × − = −

Due to wire placed in x-y plane,

( )H x y za a a a= × − = −


Given : 0 2 2 2 2y xx yB B a a

x y x y

= − + +

Convert to cylindrical coordinates and put cosx r= φ

ˆ cos sinx ra a aφ= φ − φ

siny r= φ

ˆ sin cosy ra a aφ= φ + φ

Magnetic field in cylindrical coordinate is given below,

2

0

2

cos (sin cos )

sin (cos sin )

r

r

r a arB Br a a

r

φ

φ

φ φ + φ =

φ − φ − φ

0 [ (cos sin sin cos )rBB ar

= φ φ − φ φ

2 2(cos sin )]aφ+ φ + φ

0BB ar φ=

0BBH ar φ= =

μ μ

But H J∇× =

.… (i)

where, 0r zH H= = and 0BHrφ =

μ

Hence, from equation (i),

0

1

0 / 0

r za ra a

Jr r z

rB r

φ

∂ ∂ ∂ =∂ ∂φ ∂

μ

0 0

0 0

1r z

B BJ r a r ar z r r r ∂ ∂= − × + × ∂ μ ∂ μ

0J =

(for 0r ≠ ) Hence, the correct option is (C).

y

x

xHa

z

Scan for

Video Solution

1.23 (D) 1.24 (C)


Given : 10 A/myJ a=

The dielectric interface, 0x = or -y z plane with a current sheet is shown in figure.

n xa a=

1 3 30x yH a a= +

1 1( )n n nH H a a= ⋅

1

(3 30 ) 3n x y x x xH a a a a a = + ⋅ =

1 11t nH H H= −

1

(3 30 ) 3 30t x y x yH a a a a= + − =

From the boundary conditions : 1.

1 2n nB B=

1 21 2n nH Hμ = μ

….(i)

2 1

1

2n nH Hμ=

μ

2

0

0

3 1.52n x xH a aμ= × =μ

2. 1 2

( )t t n sH H a J− = ×

2

(30 ) 10 10y t x y za H a a a− = × =

2

30 10t y zH a a= −

2 22 t nH H H= +

2 1.5 30 10x y zH a a a= + −


According to Ampere’s law,

H dl⋅

encJ dS I= ⋅ =

Case I : r a< 2(2 )H r J rπ = π

2JH r=

H r∝ Case II : r a> 22H r J a⋅ π = × π

2

2JH ar

=

1Hr

∝

Hence, the correct option is (C). Key Point For a long solid cylindrical conductor carrying uniform current, magnitude of magnetic field intensities at different points are given by,

22

2

2

Ir r aa

IH r ar

I r aa

< π= > π = π

x

y

2 20 Region-2 : 2, 2 r rx

1 10 Region-1 : 5, 1 r rx

x = 0J

Scan for

Video Solution

rr

y

z

x

a

1.25 (A) 1.26 (C)


Given figure is shown below,

The magnetic field at a point within the hole is the sum of the fields due to current distributions. The first is that of the solid cylinder obtained by filling the hole and has a current density that is same as that in the original cylinder (with the hole). The second is the solid cylinder that fills the hole and it has the current density with same magnitude as that of the original cylinder but is in the opposite direction. If these two situations are superimposed the total current in the region of the hole is zero. Now, a solid cylinder carrying current density ‘J’ which is uniformly distributed over a cross-section, produces a magnetic field with magnitude.

22IrBR

μ=π

At a distance ‘r’ from its axis inside the cylinder. where, R = the radius of the cylinder. For the cylinder of this problem, current density ‘J’ is

2 2( )I IJA a b

= =π −

where 2 2( )A a b= π − is the cross sectional area of the given cylinder (with hole).

Hole current in the cylinder without the hole is

2

21 2 2 2 2( ) ( )

I IaI JA aa b a b

= = × π =π − −

and the magnetic field it produces inside, at a distance 1' 'r from the axis is

2

1 1 11 2 2 2 22 2 ( )

I r Ir aBa a a b

μ μ= =π π −

11 2 22 ( )

IrBa bμ=

π −

The current in the cylinder that fills the hole is

2

22 2 2( )

IbI Jba b

= π =−

and the field it produces at a point inside the hole, at a distance 2' 'r from its axis has magnitude,

2

2 2 22 2 2 2 22 2 ( )

I r Ir bBb b a b

μ μ= =π π −

22 2 22 ( )

IrBa bμ=

π −

Net magnetic field, 1 2B B B= −

1 22 2 ( )2 ( )

IB r ra bμ= −

π −

Now consider the point at the axis of the hole i.e. 2 0r = and 1r d=

2 22 ( )IB d

a bμ= ×

π −

2JB dμ =

B d∝ Hence, for a given current density, B is independent of ‘b’ and it just depends on ‘d’. Hence, the correct option is (A).

Given : The point charge is kept at a distance ‘d’ from the surface of an infinite grounded metal plate. This is shown in figure below.

(Straight line)H r

1(Hyperbolic)H

r

H

2

I

a

0 ar

a

db

1.27 (A)

1.28 (C)


The image theory states that a given charge configuration above an infinite grounded perfect conducting plane may be replaced by a charge configuration itself, its image and equipotential surface in place of conducting plane.

From Coulomb’s law,

2 2

2 20 04 (2 ) 16q qF

d d− −= =

πε πε

Here negative sign indicates that force is attractive i.e. towards the plate. Hence, the correct option is (C).

Given : (cos )(sin ) (sin )(cos )x yA x y a x y a= +

From above expression, cos sin , sin cosx yA x y A x y= =

Curl of A

is given by,

x y z

x y z

a a a

Ax y z

A A A

∂ ∂ ∂∇× =∂ ∂ ∂

cos sin sin cos 0

x y za a a

Ax y z

x y x y

∂ ∂ ∂∇× =∂ ∂ ∂

A∇×

0 (sin cos )xa x yz

∂ = − ∂

(sin cos ) 0ya y xz

∂ + − ∂

0A∇× =

Hence, the magnitude of curl of A

is 0.

Key Point 0A∇× =

A

is irrotational vector.

Given : 2s xK a=

Magnetic field inside the perfect conductor is zero. From magnetic boundary conditions,

1 2( )t t n sH H a K− = ×

[0 ( )] 2x x y y z z y xH a H a H a a a− + + = ×

( ) 2x x y y z z zH a H a H a a− + + = −

On comparing both sides, 2zH =

The tangential H

field in air just above the conductor is 2 A/mzH a=

. Hence, the correct option is (D).

Given figure is,

20 kV/cm 2 V/ mAE = = μ

40 kV/cm 4 V/ mBE = = μ

Assume that 0x = at A and 5 mx = μ at B. Equation of line between A and B is given by,

2 11 1

2 1

( )y yy y x xx x

−− = −−

4 22 ( 0)5 0

E x−− = −−

+q

dGroundedmetal plate

+q

d

–q

d

V = 0

40 kV/cm

BA

20 kV/cm

5 m

0 kV/cm

(5 m, 4 V/ m) y

(0, 2 V/ m)

x

1.29 0

1.30 (D)

1.31 –15


0.4 2E x= +

(0.4 2) xE x a= + and xdl dx a=

Potential difference is given by,

B

AB AE dlψ = − ⋅

(0.4 2)B

AB x xAx a dx aψ = − + ⋅

(0.4 2)B

B A Ax dxψ − ψ = − +

52

0

0.4 22B Ax x

ψ − ψ = − +

250.4 2 5 15V2B A

ψ − ψ = − × + × = −

Hence, the value of B Aψ −ψ is – 15 V.

Given :

3 2 2 2(2 3 ) (6 3 )x yE y yz a xy xz a= − − − − (6 ) zxyz a+

. Method 1 : Let electrostatic potential at any point ( , , )P x y z is ( , , )V x y z i.e. scalar function.

E V= −∇

x y zV V VE a a ax y z

∂ ∂ ∂= − + + ∂ ∂ ∂

Compare above equation with the given expression of electric field,

3 22 3V y yzx

∂ = −∂

….(i)

2 26 3V xy xzy

∂ = −∂

…(ii)

6V xyzz

∂ = −∂

…(iii)

Using equation (i),

3 21(2 3 ) ( , )V y yz dx f y z= − +

3 212 3 ( , )V xy xyz f y z= − + …(iv)

Similarly from (ii) and (iii),

3 222 3 ( , )V xy xyz f z x= − + …(v)

233 ( , )V xyz f x y= − + …(vi)

where, 1( , )f y z is the function of y and z only.

2 ( , )f z x is the function of z and x only. 3( , )f x y is the function of x and y only. Using equation (iv), (v) and (vi),

3 22 3V xy xyz C= − +

where, C is constant. Hence, the correct option is (D).

. Method 2 : Checking from option : In option (A),

3 2V xy yz= −

3 2 3 2

3 2

( ) ( )

( )

x y

z

xy yz a xy yz ax yE

xy yz az

∂ ∂ − + − ∂ ∂ = −∂ + − ∂

3 2 2[ (3 ) 2 ]x y zE a y xy z a yz a= − + − −

(Not matched) In option (B),

3 22V xy xyz= −

3 2 3 2

3 2

(2 ) (2 )

(2 )

x y

z

xy xyz a xy xyz ax yE

xy xyz az

∂ ∂ − + − ∂ ∂ = −∂ + − ∂

( 2 ) ]zxyz a+ − (Not matched)

Scan for

Video Solution

3 2 2 2[(2 ) (6 )x yE y yz a xy xz a= − − + −

1.32 (D)


In option (C), 3 2V y xyz= +

3 2 3 2

3 2

( ) ( )

( )

x y

z

y xyz a y xyz ax yE

y xyz az

∂ ∂ + + + ∂ ∂ = −∂ + + ∂

2 2 2[ (3 ) (2 ) ]x y zE yz a y xz a xyz a= − + + +

(Not matched) In option (D),

3 22 3V xy xyz= −

3 2 3 2

3 2

(2 3 ) (2 3 )

(2 3 )

x y

z

xy xyz a xy xyz ax yE

xy xyz az

∂ ∂ − + − ∂ ∂ = −∂ + − ∂

3 2 2 2[(2 3 ) (6 3 )x yE y yz a xy xz a= − − + −

(6 ) ]zxyz a− 3 2 2 2(2 3 ) (6 3 )x yE y yz a xy xz a= − − − −

(6 ) zxyz a+ (Matched) Hence, the correct option is (D).

Magnetic field at any distance r from the conductor can be calculated using Ampere’s law. encH dl I⋅ =

0cos0H dl Iφ =

2H r Iφ π =

2

IHrφ =

π

1Hrφ ∝


Given : 3 2 2 2

x y zP x ya x y a x yza= − −

For a vector to be solenoidal,

0P∇⋅ =

yx zPP PPx y z

∂∂ ∂∇ ⋅ = + +∂ ∂ ∂

2 2 23 2P x y x y x y∇ ⋅ = − −

0=

So, P

is solenoidal. For a vector to be irrotational,

0P∇× =

x y z

x y z

a a a

Px y z

A A A

∂ ∂ ∂∇× =∂ ∂ ∂

3 2 2 2

x y za a a

Px y z

x y x y x yz

∂ ∂ ∂∇× =∂ ∂ ∂

− −

2( ) (2 )x yP a x z a xyz∇ × = − +

2 3( 2 ) 0za xy x+ − − ≠

So, P

is not irrotational. Hence, the correct option is (A).

Given : 2 2 22x y czφ = + +

Electric field is given by,

E V= −∇ = −∇φ

Conductor

y

x

p

I

z

rdl

H

Amperianpath

H

r

1.33 (C)

1.34 (A)

1.35 –3


Now x y zE a a ax y z

∂φ ∂φ ∂φ= − + + ∂ ∂ ∂

2 2 2

2 2 2

2 2 2

(2 )

(2 )

(2 )

x

y

z

x y cz ax

x y czE ay

x y cz az

∂ + + ∂

∂ + + = − + ∂ ∂ + ++ ∂

(4 2 2 )x y zE xa ya cza= − + +

In source-free region, 0vρ =

0D∇⋅ =

(If 0vρ ≠ then vD∇ ⋅ = ρ

) ( ) 0Eε ∇ ⋅ =

0E∇⋅ =

0yx zEE Ex y z

∂∂ ∂+ + =∂ ∂ ∂

( 4 ) ( 2 ) ( 2 ) 0x y czx y z

∂ ∂ ∂− + − + − =∂ ∂ ∂

4 2 2 0c+ + = 3c = − Hence, the value of constant c must be –3.

. Method 1 : Given : Vector field 22 zD a zaρ= ρ +

in the region enclosed by the surfaces 1, 0zρ = = and 5z = S be the surface bounding this cylindrical region. So surface integral on S is,

s

I D dS= ⋅

S is the closed surface of cylindrical. Using Gauss divergence theorem, ( )

vs

I D dS D dv= ⋅ = ∇ ⋅

…. (i)

Divergence of D

is given by,

1 1 zD D Ddiv D Dz

ρ φ∂ ∂ ∂= ∇⋅ = + + ρ ∂ρ ρ ∂φ ∂

22 , 0D Dρ φ= ρ = and zD z=

So, 21 ( 2 ) 0 ( )D zz

∂ ∂∇ ⋅ = ρ⋅ ρ + +ρ ∂ρ ∂

22 3 1 6 1D × ρ∇ ⋅ = + = ρ +

ρ

V is the volume bounded by closed surface ‘S’. ( )dv d d dz d d dz= ρ ρ φ = ρ ρ φ …. (ii) From equation (i) and (ii),

5 2 1

0 0 0

(6 1)z

z

I d d dz= π

= φ= ρ=

= ρ + ⋅ρ ρ φ

5 2 1

2

0 0 0

(6 )z

z

I d d dz= π

= φ= ρ=

= ρ + ρ ρ φ

15 2 2

3

0 0 0

22

z

z

I d dz= π

= φ=

ρ= ρ + φ

5 2

0 0

12 12

z

z

I d dz= π

= φ=

= × + φ

5 2

0 0

52

z

z

I d dz= π

= φ=

= φ

5 2 5 25 78.542

I = × π× = π =

Hence, the surface integral of this field on S is 78.54. . Method 2 : Closed surface of cylindrical region,

sS D dS= ⋅

cylindrical top bottom

S D dS D dS D dS= ⋅ + ⋅ + ⋅

y

x

z

z = 5

= 1

S

1.36 78.54


2 5 2

0 0, 1(2 )zz

S a za d dz aπ

ρ ρφ= = ρ== ρ + ⋅ρ φ

2 5 2

0 0, 5(2 )z zz

a za d d aπ

ρφ= ρ= =+ ρ + ⋅ρ ρ φ

2 1 2

0 0, 0(2 ) ( )z zz

a za d d aπ

ρφ= ρ= =+ ρ + ⋅ −ρ ρ φ

2 5 3

0 0, 12

zS d dz

π

φ= = ρ== ρ φ +

2 1 2 1

0 0, 5 0 0, 0z zz d d z d d

π π

φ= ρ= = φ= ρ= =ρ ρ φ − ρ ρ φ

1(2 2 5) 2 52

S = × π× + × π×

25 78.54S = π = Hence, the surface integral of this field on S is 78.54.

Given : 1 2 32 m, 4 m, 8 mr r r= = =

2 21 220 nC/m , 4 nC/ms sρ = ρ = −

From Gauss law,

encD dS Q⋅ =

If 0D =

, then 0encQ =

1 2 3 0enc enc encQ Q Q+ + =

Since, enc sQ ds= ρ

Hence, 2 2 21 1 2 2 3 3(4 ) (4 ) (4 ) 0s s sr r rρ π + ρ π + ρ π =

2 2 220(4 2 ) 4(4 4 ) (4 8 ) 0sπ× − π× + ρ π× =

320 256 256 0sπ − π + ρ π =

2256 320 0.25 nC/m256s−ρ = = −

Hence, the value of sρ is 20.25 nC/m− .

Given : 22

400sin Am2 ( 4) rJ a

r−θ=

π +

, 0.8mr =

Total current is given by, I J dS= ⋅

/4 2

22

012

400sin sin2 ( 4)

I r d dr

π π

π φ=θ=

θ= ⋅ θ θ φπ +

[ ]/4

22 22 0

/12

400 sin2 ( 4)

I r dr

ππ

π

= × × φ × θ θπ +

/42

2/12

400 2 1 cos 22 ( 4) 2

rI dr

π

π

× π − θ = θ π +

/42

2/12

400 sin 24 12( 4) 2 4

rIr

π

π

π π − θ = − +

2

2

11400 2( 4) 12 4

rIr

− π= − +

2400 (0.8) 0.136 7.56 A

4 64I ×= × =

⋅

Total area, S 2 sins

dS r d d= = θ θ φ

/4

/42 2/12

12

2 sin 2 cosS r d rπ

π

ππθ=

= π θ θ = π − θ

S

S 2

0.82 0.259

rr

== π ×

S 2 2(0.8) 2 0.259 1.041 m= × π× = Average current is given by,

2 27.56 7.26 A/m 7.56 A/m1.041

IS

= = ≈

No option is correct as per key provided by IISc Bangalore.

Scan for

Video Solution

/42

12

2 sinr dπ

πθ=

= π θ⋅ θ

1.37 – 0.25

1.38 *


Given : Magnetic field ˆB zB=

and

Initial velocity, ˆ ˆx zv x v z v= +

By Lorentz’s force equation,

( )F Q v B= × ˆ ˆ ˆ( )x zQ v x v z Bz= + ×

ˆ ˆ( )x xF Q v B y Qv By= − = −

Since, we have force in y-direction, hence we will have velocity in y-direction as well, with changing magnitude. In the xy plane, it is exactly what causes a circular motion. The z-component of velocity is constant. Hence the combined effect will result in helical motion in z-direction. Hence, the correct option is (A). Note : When a motion has two components, one of them constant and other changing with changing direction, then the trajectory so formed will be helical. ˆxF Qv By= −

F ma=

ˆxQv BFa ym m

−= =

… (i)

Also da vdt

=

Let ˆ ˆ ˆx y zv v x v y v z= + +

ˆ ˆ ˆyx zdvdv dva x y z

dt dt dt= + + … (ii)


y xdv Qv Bdt m

−=

xy

Qv Bv t Km

−= + … (iii)

At 0t = , initial velocity vector is given as 0 ˆ ˆx zv v x v z= +

Hence, 0

0y tv

==

From equation (iii),

0 0xQv B Km

−= × +

0K =

Hence, xy

Qv Bv tm

−=

We can see that velocity will be changing in y-direction with respect to time but velocity in z-direction will be constant as there is no z-component of acceleration. Hence, the eventual trajectory will be helical motion in z-direction.

Energy stored in the capacitor 212

CV=

QVC

=

2

2QEC

=

Let when 1d d= then 1E E=

i.e. 2

11

12QEC

= … (i)

For electrically isolated capacitor charge stored is same. 2 1Q Q=

When 2 2d d= then 12 2

CC =

Hence, 2 2 22 1 1

212 1

22 22

2

Q Q QE CC C

= = =

Using equation (i) 2 12 2E E E= =


Scan for

Video Solution

1.39 (A)

1.40 (A)


In equilibrium, 212

mv qV=

Velocity is given by,

2d qVvt m

= =

dtV

∝


Faraday’s law of electromagnetic induction, rate of change of magnetic field results in induced voltage,

BEt

∂∇× = −∂


Given :

Potential on sphere 1S is given by,

4

aa

a

qVr

=πε

Potential on sphere 2S is given by,

4

bb

b

qVr

=πε

When two sphere are connected through conducting wire, then charge flows from higher potential to lower potential sphere until both spheres attains same potential.

So, a ba b

a b

kq kqV Vr r

= =

a b

a b

q qr r

= …. (i)

Electric field outside sphere is given by,

2kqEr

=

For sphere A, 2a

aa

kqEr

=

For sphere B, 2b

bb

kqEr

=

1 1b a

b a

r rr r

> < …. (ii)


2 2a b

a b

kq kqr r

>

a bE E>


Given :

1 2(3 ) ( 2 )x yF y k z a k x z a= − + −

3( ) zk y z a− +

and F

is irrotational i.e. 0F∇× =

1 2 3(3 ) ( 2 ) ( 1)

x y za a a

Fx y z

y k z k x z k y

∂ ∂ ∂∇× =∂ ∂ ∂− − − +

3 2( 1) ( 2 )xF a k y k x zy z

∂ ∂∇× = − + − − ∂ ∂

3 1( 1) (3 )ya k y y k zx z

∂ ∂ − − + − − ∂ ∂

2 1( 2 ) (3 )za k x z y k zx y

∂ ∂+ − − − ∂ ∂

3 1 2( 2) ( ) ( 3)x y zF k a k a k a∇× = − + − + −

Wire

S2

S1

Radius rb

Radius ra

Scan for

Video Solution

1.41 (C)

1.42 (C)

1.43 (C)

1.44 (B)


For F∇×

to be zero, 1 20, 3k k= = and 3 2k =


Given : Mass of electron 319.11 10 kgm −= ×

Charge of electrons 191.6 10 Ce −= − ×

Velocity of electron 51 ( ) 10 m/secq v =

Initial kinetic energy and potential energy is,

2initial

1(KE)2

mv=

1 2initial

0

1(PE) 04 r

q qr

→∞

= =πε

Final kinetic energy and potential energy is,

2final

0

1(KE) 02 v

mv=

= =

1 2final

0

1(PE)4

q qr

=πε

From energy conservation theorem,

initial initial final final(KE) (PE) (KE) (PE)+ = +

2 1 2

0

1 12 4

q qmvr

=πε

1 22

0

24

q qrmv

=πε

191 2 1.6 10q q e −= = = − ×

9 19 2

31 5 22 9 10 ( 1.6 10 )

9.1 10 (10 )r

−

−

× × × − ×=× ×

2 29

2118 1.6 10

9.1 10r

−

−

× ×=×

28 818 1.6 10 5.063 10 m

9.1r − −×= × = ×

Hence, the closest distance is 85.063 10 m−× .

Given : Infinite line charge density, A/mL Qρ =

Flux is given by, E daψ = ⋅ …(i)

From diagram, ˆda d dz aρ= ρ φ

(Taking the differential surface of ρ direction as on the quarter cylindrical surface, only ρ is constant while φ and z are varying).

02π< φ < and 0 z H< <

Due to infinite line charge electric field intensity is given by,

0 0

ˆ ˆ2 2

L QE a aρ ρρ= =

πε ρ πε ρ

From equation (i),

0

ˆ ˆ( )2

QE da a d dz aρ ρ

ψ = ⋅ = ⋅ ρ φ πε ρ

/2

00 0

( )2

H

z

Q d dzπ

φ= =

ψ = ⋅ ρ φπε ρ

ˆ ˆ 1a aρ ρ ⋅ =

[ ] [ ]/2

/2

0 00 00 0 2 2

HH

z

Q d dz Q zπ

π

φ= =

φψ = = φπε πε

04

QHψ =ε


Correct Maxwell’s equation in point form for time varying fields are given as vD∇ ⋅ = ρ

Scan for

Video Solution

1q 510 m/sv 2q

rMoving Stationary

1.45 5.063

1.46 (A)

1.47 (C)


0B∇ ⋅ =

BEt

∂∇× = −∂

DH Jt

∂∇× = +∂


Given copper wires 1W and 2W are carrying

currents I and 2I in opposite directions and having a total separation of 4r between them.

Magnetic field intensity due to infinitely long current carrying conductor is given as

ˆ2

IH aφ= ⋅πρ

00 2

IB H μ= μ =πρ

Due to current carrying conductor 1W ,

01 2

IBr

μ=π

[ ]1 rρ =

Due to current carrying conductor 2W ,

02

22 3

IBr

μ=π×

[ ]2 3rρ =

Using right hand rule, intensities or densities due to both current carrying conductors will be in same direction at the point between two conductors, hence the total magnetic flux density is given as

1 2B B B= +

0 0 0 02 3 22 6 6

I I I IBr r r

μ μ μ + μ= + =π π π

∴ 056

IBr

μ=π


Key point : 1. 0A∇⋅ =

Then it is solenoidal or divergenceless.

2. 2( )A A A∇ ×∇ × = ∇ ∇ ⋅ − ∇

[identity] referred to as vector triple product.

3. A∇×

is an another vector because when we have curl of vector we will get new vector.

4. For irrotational vector 0A∇× =

. Hence, the correct option is (D).

4r2 3r

1 r r

I

2I2W

1W

1.48 (B)

1.49 (D)

1.1 The variance of a random variable X is

2Xσ . Then the variance of KX− (where

K is an positive constant) is (A) 2

Xσ (B) 2XK− σ

(C) 2XKσ (D) 2 2

XK σ

1.2 Zero mean Gaussian noise of variance N

is applied to a half wave rectifier. The mean squared value of the rectifier output will be

(A) Zero (B) N/2 (C) / 2N (D) N

1.3

For a random variable ‘X’ following the

probability density function (PDF) as shown in the above figure, the values of the mean and the variance are, respectively

(A) 1 2and2 3

(B) 41 and3

(C) 21 and

3 (D) 42 and

3

1.4 A probability density function is given

by 2 /2( ) ,xP x Ke x−= −∞ < < ∞ . The

value of K should be

(A) 12π

(B) 2π

(C) π

2 (D) 1

π 2

1.5 The spectral density of a real valued

random process has (A) An even symmetry (B) An odd symmetry (C) A conjugate symmetry (D) No symmetry

1.6 The PSD and the power of a signal g(t)

are respectively, ( )gS ω and gP . The PSD and the power of the signal ag(t) are, respectively

(A) ( )2 2andg ga S a Pω

(B) ( )2 andg ga S aPω

(C) ( ) 2andg gaS a Pω

(D) ( ) andg gaS aPω

1Random Variables

& Random Processes

( )XP x

1/4

1 1 2 3x

0

1987 IIT Bombay

1989 IIT Kharagpur

1992 IIT Delhi

1997 IIT Madras

1998 IIT Delhi

2001 IIT Kanpur


1.7 The PDF of a Gaussian random variable

X is given by ( )( )24

1813 2

x

XP x e− −

=π

.

The probability of the event X = 4 is

(A) 12

(B) 13 2π

(C) 0 (D) 1

4

1.8 If the variance 2

dσ of ( ) ( ) ( 1)d n x n x n= − − is one-tenth the

variance 2Xσ of stationary zero-mean

discrete-time signal ( )x n , then the normalized autocorrelation function

2( )XX

X

R kσ

at 1k = is

(A) 0.95 (B) 0.90 (C) 0.10 (D) 0.05

1.9 Let X and Y be two statistically

independent random variable uniformly distributed in the ranges (– 1, 1) and (–2, 1) respectively. Let Z = X + Y, then the probability ( 1)Z ≤ − is

(A) Zero (B) 16

(C) 13

(D) 112

( )X t is a random process with a constant mean

value of 0 and the autocorrelation function 0.2( ) 4( 1)XXR e− ττ = + .

1.10 Let X be the Gaussian random variable obtained by sampling the process at

it t= and let

( )2

212

y

Q e dyα

−∞α =

π

The probability that [ ]1X ≤ is

(A) ( )1 0.5Q− (B) ( )0.5Q

(C) 12 2

Q

(D) 112 2

Q −

1.11 Let Y and Z be the random variable obtained by sampling X(t) at t = 2 and t = 4 respectively. Let W = Y – Z. The variance of W is

(A) 13.36 (B) 9.36 (C) 2.64 (D) 8.00

1.12 The distribution function ( )XF x of a

random variable X is shown in the figure. The probability that X = 1 is

(A) Zero (B) 0.25 (C) 0.55 (D) 0.30 1.13 A random variable X with uniform

density in the interval 0 to 1 is quantized as follows :

If 0 0.3,X≤ ≤ 0qx =

If 0.3 1,X≤ ≤ 0.7qx =

where qx is the quantized value of X.

The root-mean square value of the quantization noise is

(A) 0.573 (B) 0.198 (C) 2.205 (D) 0.266

( )XF x

1.0

0.55

0.25

2 0 1 3x

2002 IISc Bangalore

2003 IIT Madras

Common Data for Questions 1.10 & 1.11

2004 IIT Delhi

4.3GATE ACADEMY ® Communication Systems : Random Variables & Random Processes

1.14 An output of a communication channel

is a random variable V with the probability density function as shown in the figure. The mean square value of V is

(A) 4 (B) 6 (C) 8 (D) 9

1.15 A probability density function is of the

form | |( ) ,xP x Ke−α= ( , )x −∞ ∞

The value of K is (A) 0.5 (B) 1 (C) 0.5 α (D) α 1.16 A uniformly distributed random variable

X with probability function

1( ) [ ( 5) ( 5)]10Xf x u x u x= + − −

where ( )u ⋅ is the unit step function is passed through a transformation given in the figure below. The probability density function of the transformed random variable Y would be

(A) ( ) ( )1( ) [ 2.5 2.5 ]5Yf y u y u y= + − −

(B) ( ) ( )( ) 0.5 0.5 1Yf y y y= δ + δ −

(C) ( )( ) 0.25 2.5Yf y y= δ +

( ) ( )0.25 2.5 5y y+ δ − + δ

(D) ( ) ( )( ) 0.25 2.5 0.25 2.5Yf y y y= δ + + δ −

( ) ( )1 2.5 2.510

u y u y+ + − −

1.17 A zero mean white Gaussian noise is passed through an ideal low pass filter of bandwidth 10 kHz. The output is then uniformly sampled with sampling period

0.03st = msec. The samples so obtained would be

(A) Correlated (B) Statistically independent (C) Uncorrelated (D) Orthogonal

The following two questions refer to wide-sense stationary stochastic processes. 1.18 It is desired to generate a stochastic

process (as voltage process) with power spectral density

216( )

16S ω =

+ ω

By driving a linear time invariant system by zero mean white noise (as voltage process) with power spectral density being constant equal to 1. The system which can perform the desired task could be

(A) First order lowpass R-L filter (B) First order highpass R-C filter (C) Tuned L-C filter (D) Series R-L-C filter

( )P v

0 4

K

v

2.5 2.5

1

y

x

2005 IIT Bombay

2006 IIT Kharagpur



1.19 The parameters of the system obtained in above question would be

(A) first order R-L lowpass filter would have 4R = Ω , L = 1 H.

(B) first order R-C highpass filter would have 4R = Ω , 0.25 FC = .

(C) tuned L-C filter would have L = 4 H, C = 4 F.

(D) series R-L-C lowpass filter would have 1R = Ω , L = 4 H, C = 4 F.

1.20 If ( )R τ is the autocorrelation function of

a real, wide-sense stationary random process, then which of the following is not true?

(A) ( ) ( )R Rτ = − τ

(B) ( ) ( )0R Rτ ≤

(C) ( ) ( )R Rτ = − − τ

(D) Mean square value of the process is R(0).

1.21 If ( )S f is the power spectral density of a real, wide-sense stationary random process, then which of the following is always true?

(A) ( ) ( )0S S f≥

(B) ( ) 0S f ≥

(C) ( ) ( )S f S f− = −

(D) ( ) 0S f df∞

−∞

=

1.22 If E denotes expectation, the variance of a random variable X is given by

(A) 2 2[ ] [ ]E X E X−

(B) 2 2[ ] [ ]E X E X+

(C) 2[ ]E X (D) 2[ ]E X

1.23 The probability density function (PDF)

of a random variable X is as shown below.

The corresponding cumulative distribution

function (CDF) has the form (A)

(B)

(C)

(D)

1.24 ( ) exp( 2 | |) exp( 3 | |)XP x M x N x= − + −

is the probability density function for the real random variable X, over the entire x-axis. M and N are both positive real numbers. The equation relating M and N is

(A) 2 13

M N+ = (B) 12 13

M N+ =

(C) 1M N+ = (D) 3M N+ =

1 0 1x

1PDF

CDF

1 0 1x

1

CDF

1 0 1x

1

CDF

1 0 1x

1

CDF

1

1 0

1

1 x

2007 IIT Kanpur

2008 IISc Bangalore


1.25 If the power spectral density of

stationary random process is a sinc-squared function of frequency, the shape of its autocorrelations is

(A) (B)

(C) (D)

1.26 A white noise process X(t) with two-sided power spectral density 101 10−×W/Hz is input to a filter whose magnitude squared response is shown below.

The power of the output process Y(t) is

given by (A) 75 10 W−× (B) 61 10 W−× (C) 62 10 W−× (D) 51 10 W−× 1.27 Consider two independent random

variables X and Y with identical distributions. The variables X and Y take values 0, 1 and 2 with probabilities 1 1 1, and2 4 4

respectively. What is the

conditional probability ( )2 | 0P X Y X Y+ = − = ?

(A) 0 (B) 116

(C) 16

(D) 1

1.28 A discrete random variable X takes value from 1 to 5 with probabilities as shown in the table. A student calculates the mean X as 3.5 and her teacher calculates the variance of X as 1.5. Which of the following statements is true?

K 1 2 3 4 5 ( )P X K= 0.1 0.2 0.4 0.2 0.1

(A) Both the student and the teacher are right.

(B) Both the student and the teacher are wrong.

(C) The student is wrong but the teacher is right.

(D) The student is right but the teacher is wrong.

1.29 ( )X t is a stationary process with the

power spectral density ( ) 0XS f > for all f. The process is passed through a system shown below.

Let ( )YS f be the power spectral density

of Y(t). Which one of the following statements is correct?

(A) ( )YS f > 0 for all f

(B) ( ) 0YS f = for | | 1kHzf >

(C) ( ) 0YS f = for 0 0, 2 kHz,f nf f= = n any integer

(D) ( ) 0YS f = for 0(2 1)f n f= + ,

0 1kHzf = , n any integer

( )R

( )R

( )R

( )R

( )X t

10 kHz 10kHz

( )Y t

2( )H f

f

d

dt( )Y t( )X t

Delay = 0.5 ms

+

2009 IIT Roorkee

2010 IIT Guwahati


1.30 X(t) is a stationary random process with

autocorrelation function ( ) expXR τ =2( )− πτ . This process is passed through

the system shown below. The power spectral density of the output process Y(t) is

(A) 2 2 2(4 1) exp( )f fπ + − π (B) 2 2 2(4 1) exp( )f fπ − − π (C) 2 2(4 1) exp( )f fπ + − π (D) 2 2(4 1) exp( )f fπ − − π

1.31 The power spectral density of a real

process ( )X t for positive frequencies is shown below. The values of E[ 2X (t)] and [ ]( )E X t , respectively, are

(A) 6000 / , 0π (B) 6400 / , 0π

(C) 6400 / , 20 / ( 2)π π

(D) 6000 / , 20 / ( 2)π π

1.32 Let U and V be two independent zero

mean Gaussian random variables of

variances 14

and 19

respectively. The

probability (3 2 )P V U≥ is (A) 4/9 (B) 1/2 (C) 2/3 (D) 5/9

1.33 Consider two identically distributed zero-mean random variables U and V. Let the cumulative distribution functions of U and 2V be F(x) and G(x) respectively. Then, for all values of x

(A) ( ) ( ) 0F x G x− ≤

(B) ( ) ( ) 0F x G x− ≥ (C) ( ( ) ( )) 0F x G x x− ⋅ ≤ (D) ( ( ) ( )) 0F x G x x− ⋅ ≥

1.34 Let 1,X 2X and 3X be independent and

identically distributed random variables with the uniform distribution on [ ]0, 1 .

The probability P 1X is the largest is _________. [Set - 01]

1.35 Let X be a real-valued random variable with [ ]E X and 2E X denoting the

mean values of X and 2X , respectively. The relation which always holds [Set - 01]

(A) 2 2( [ ]) [ ]E X E X>

(B) 2 2[ ] ( [ ])E X E X≥

(C) 2 2[ ] ( [ ])E X E X=

(D) 2 2[ ] ( [ ])E X E X>

1.36 Consider a random process ( )X t =

2 sin(2 ),tπ + φ where the random phase φ is uniformly distributed in the interval [0, 2 ]π . The auto-correlation

1 2[ ( ) ( )]E X t X t is [Set - 01]

(A) 1 2cos (2 ( ))t tπ +

(B) 1 2sin (2 ( ))t tπ −

(C) 1 2sin (2 ( ))t tπ +

(D) 1 2cos (2 ( ))t tπ −

( )X t ( )Y t

( ) 2H f j f

0 9 10 11

6

( )XS 4400 ( 10 )

3(10 rad/s)

2011 IIT Madras

2012 IIT Delhi

2013 IIT Bombay

2014 IIT Kharagpur


1.37 Let X be a random variable which is uniformly chosen from the set of positive odd numbers less than 100. The expectation [ ]E X is ______. [Set - 02]

1.38 The power spectral density of a real stationary random process ( )X t is given by

1

( ) '0,

X

f WS f W

f W

≤= >

The value of the expectation 1( )

4E X t X t

W π −

is ________.

[Set - 02] 1.39 Let 1 2,X X and 3X be independent and

identically distributed random variables with the uniform distribution on [0, 1]. The probability 1 2 3 P X X X+ ≤ is _________. [Set - 03]

1.40 Let ( )X t be a wide-sense stationary (WSS) random process with power spectral density ( )XS f . If ( )Y t is the process defined as ( ) (2 1),Y t X t= − the power spectral density ( )YS f is

[Set - 03]

(A) 1( )2 2

j fY X

fS f S e− π =

(B) /21( )2 2

j fY X

fS f S e− π =

(C) 1( )

2 2Y XfS f S =

(D) 21( )2 2

j fY X

fS f S e− π =

1.41 A real band-limited random process( )X t has two-sided power spectral

density 610 (3000 ) W/Hz for 3kHz( )

0 otherwiseXf f

S f− − ≤

=

where f is the frequency expressed in Hz. The signal ( )X t modulates a carrier cos16000 tπ and the resultant signal is passed through an ideal band-pass filter of unity gain with centre frequency of 8 kHz and bandwidth 2 kHz. The output power (in Watts) is _________.

[Set - 03]

1.42 nn n

X = ∞

= −∞ is an independent and

identically distributed (i.i.d.) random process nX equally likely to be 1+ or

–1. nn n

Y = ∞

= −∞ is another random process

obtained as 10.5n n nY X X −= + . The

autocorrelation function of nn n

Y = ∞

= −∞

denoted by [ ]YR k , is [Set - 02]

(A)

(B)

(C)

(D)

3 2 1 0 1 2 3k

1

...

[ ]YR k

3 2 1 0 1 2 3k

0.5

...

[ ]YR k

1.25

0.5

3 2 1 0 1 2 3k

0.5

...

[ ]YR k

1.25

0.50.25 0.25

3 2 1 0 1 2 3k

...

[ ]YR k 1.25

0.25 0.25

2015 IIT Kanpur


1.43 Let the random variable X represent the number of times a fair coin needs to be tossed till two consecutive heads appear for the first time. The expectation of X is ________. [Set - 02]

1.44 Let 0, 1X ∈ and 0, 1Y ∈ be two independent binary random variables. If

( 0)P X p= = and ( 0)P Y q= = , then ( 1)P X Y+ ≥ is equal to [Set - 02]

(A) (1 )(1 )pq p q+ − − (B) pq (C) (1 )p q− (D) 1 pq− 1.45 A zero mean white Gaussian noise

having power spectral density 0

2N is

passed through an LTI filter whose impulse response ( )h t is shown in the figure. The variance of the filtered noise at 4t = is [Set - 02]

(A) 2

032

A N (B) 20

34

A N

(C) 20A N (D) 2

012

A N

1.46 The variance of the random variable X with probability density function

1( )2

xf x x e −= is ______. [Set - 03]

1.47 A random binary wave ( )y t is given by

( ) ( )nn

y t X p t nT∞

=−∞

= − − φ ,

where ( ) ( ) ( ), ( )p t u t u t T u t= − − is the unit step function and φ is an independent random variable with uniform distribution in [ ]0, T . The sequence consists of independent and identically distributed binary valued random variables with

[ ] [ ]1 1 0.5n nP X P X= + = = − = for each n.

The value of the autocorrelation 3 3( )4 4YYT TR E y t y t Δ −

equals

______. [Set - 03] 1.48 A fair dice with faces [1, 2, 3, 4, 5, 6] is

thrown repeatedly till ‘3’ is observed for the first time. Let X denote the number of times the dice is thrown. The expected value of X is ____. [Set - 03]

1.49 Consider a random process

( ) 3 ( ) 8X t V t= − , where ( )V t is a zero mean stationary random process with autocorrelation 5( ) 4vR e− ττ = . The power in X(t) is __________. [Set - 02]

1.50 Two random variables X and Y are distributed according to

,

( ), 0 1, 0 1( , )

0, otherwise.X Y

x y x yf x y

+ ≤ ≤ ≤ ≤=

The probability ( 1)P X Y+ ≤ is ______. [Set - 02]

1.51 A wide-sense stationary random process ( )X t passes through the LTI system

shown in the figure. If the autocorrelation function of ( )X t is ( )XR τ , then the autocorrelation function ( )YR τ of the output ( )Y t is equal to [Set - 03]

(A) 0 02 ( ) ( ) ( )X X XR R T R Tτ + τ − + τ +

(B) 0 02 ( ) ( ) ( )X X XR R T R Tτ − τ − − τ +

(C) 02 ( ) 2 ( 2 )X XR R Tτ + τ −

(D) 02 ( ) 2 ( 2 )X XR R Tτ − τ −

A

A

( )h t

1 2 3t

0

( )Y t

Delay = T0

( )X t

2016 IISc Bangalore


1.52 Let X(t) be a wide-sense stationary

random process with the power spectral

density ( )XS f as shown in figure (a),

where f is in Hertz (Hz). The random

process X(t) is input to an ideal lowpass

filter with the frequency response

11, Hz2( )10, Hz2

fH f

f

≤= >

as shown in figure (b). The output of the

lowpass filter is Y(t),

Fig. (a)

Fig. (b)

Let E be the expectation operator and

consider the following statements :

I. [ ( )] [ ( )]E X t E Y t=

II. 2 2[ ( )] [ ( )]E X t E Y t=

III. 2[ ( )] 2E Y t =

Select the correct option : [Set - 01] (A) only I is true

(B) only II and III are true

(C) only I and II are true

(D) only I and III are true

1.53 Consider the random process ( )X t U Vt= + , where U is a zero mean

Gaussian random variable and V is a random variable uniformly distributed between 0 and 2. Assume that U and V are statistically independent. The mean value of the random process at 2t = is ______. [Set - 02]

1.54 Passengers try repeatedly to get a seat reservation in any train running between two stations until they are successful. If there is 40% chance of getting reservation in any attempt by a passenger, then the average number of attempts that passengers need to make to get a seat reserved is _________.

[Set - 02]

1.55 Consider a white Gaussian noise process

( )N t with two-sided power spectral

density ( ) 0.5NS f = W/Hz as input to a

filter with impulse response 2 /20.5 te−

(where t is in seconds) resulting in output ( ).Y t The power in ( )Y t in watts

is

(A) 0.11 (B) 0.22

(C) 0.33 (D) 0.44

1.56 If X and Y are random variable such that

[ ]2 0E X Y+ = and [ ]2 33E X Y+ = ,

then [ ] [ ]E X E Y+ = _________.

( )XS f

exp( )f

0f

Ideal lowpassfilter ( )h t

cutoff = 1/2 Hz

X t( ) Y t( )

2017 IIT Roorkee

2018 IIT Guwahati

2019 IIT Madras


1.57 Let a random process Y(t) be described

as ( ) ( ) ( ) ( )*Y t h t X t Z t= + , where X(t)

is a white noise process with power

spectral density ( ) 5 W/HzXS f = . The

filter h(t) has a magnitude response

given by ( ) 0.5H f = for 5 5f− ≤ ≤ ,

and zero elsewhere. Z(t) is a stationary

random process, uncorrelated with X(t),

with power spectral density as shown in

the figure. The power in Y(t), in watts, is

equal to ______W (rounded off to two

decimal places).

1.58 The random variable

( ) ( ) ,Y W t t dt∞

−∞

= φ

where 1, 5 7

( )0, otherwise

tt

≤ ≤φ =

and ( )W t is a real white Gaussian noise process with two-sided power spectral density ( ) 3 W/HzWS f = , for all .f The variance of Y is ________.

1.59 X is a random variable with uniform probability density function in the interval [ ]2, 10 .− For 2 6,Y X= − the

conditional probability ( 7 5)P Y X≤ ≥

(rounded off to three decimal places) is _______.

5− 5

( ) (W/ Hz)zS f1

(Hz)f

2020 IIT Delhi


1.1 D 1.2 B 1.3 B 1.4 A 1.5 A

1.6 A 1.7 C 1.8 A 1.9 C 1.10 D

1.11 C 1.12 D 1.13 B 1.14 C 1.15 C

1.16 B 1.17 A 1.18 A 1.19 A 1.20 C

1.21 B 1.22 A 1.23 A 1.24 A 1.25 B

1.26 B 1.27 C 1.28 B 1.29 D 1.30 A

1.31 B 1.32 B 1.33 D 1.34 0.33 1.35 B

1.36 D 1.37 50 1.38 4 1.39 0.1667 1.40 C

1.41 2.5 1.42 B 1.43 1.5 1.44 D 1.45 A

1.46 6 1.47 0.25 1.48 6 1.49 100 1.50 0.33

1.51 B 1.52 A 1.53 2 1.54 2.5 1.55 B

1.56 11 1.57 17.5 1.58 6 1.59 0.3

Given : Variance of a random variable X is 2 .Xσ Variance of random variable is given by, 2 2 2[ ] ( )X E X Xσ = − …(i)

Mean square value is given by,

2 2[ ] ( )XE X x f x dx∞

−∞

=

2 2[( ) ] ( ) ( )XE KX Kx f x dx∞

−∞

− = −

2 2 2[( ) ] ( )XE KX K x f x dx∞

−∞

− =

2 2 2[( ) ] [ ]E KX K E X− = …(ii) Mean value is given by,

[ ] ( )XX E X x f x dx∞

−∞

= =

[ ] ( )XE KX Kx f x dx∞

−∞

− = −

[ ]E KX K X− = − …(iii) Variance is given by, 2 2MSV (Mean)σ = − From equation (ii) and (iii), 2 2 2 2[ ] ( )K E X KXσ = − −

2 2 2 2 2[ ] ( )K E X K Xσ = − 2 2 2 2 [ ] ( ) K E X Xσ = − Using equation (i), 2 2 2

XKσ = σ Hence, the correct option is (D).

Half wave rectification is represented as, for 0Y X x= ≥

0 for 0x= <

The PDF of zero mean Gaussian random variable is given by,

2

22

2

1( )2

X

x

X

f x e dx−

σ=πσ

…(i)

Answers Random Variables & Random Process

Explanations Random Variables & Random Process

1.1 (D)

1.2 (B)


where, 2Xσ = variance of Gaussian random

variable X. The PDF of random variable Y with zero mean and variance N is given by,

2 21( )

2y Nf y e

N−=

π …(ii)


2 2

0

( ) ( )E Y y f y dy∞

=

2

22 2

0

( )2

y NyE Y e dyN

∞−=

π

Let, 2

2y tN

=

2 2y dy N dt=

2

N dt Ndtdyy Nt

= =

1/2

2Ndy t dt−=

2 1/2

0

2( )2 2

tNt NE y e t dtN

∞− −=

π

2 1/2

0

( ) tNE y t e dt∞

−=π …(iii)

Definition of gamma function :

1

0

t ne t dt n∞

− − = …(iv)

On comparing equation (iii) and (iv),

112

n − =

32

n =

1/2

0

3 1 1 112 2 2 2 2

tt e dt∞

− π= = + = = …(v)

11 and2

n n n

+ = = π

From equation (iii) and (v),

2( )2 2

N NE Y π= × =π


. Method 1 : The given PDF is shown in below figure,

Mean value is given by,

[ ] ( )XX E X x P x dx∞

−∞

= =

3

2 31

1

1 1[ ]4 8

X x dx x −−

= × =

1X = …(i) Mean square value is given by,

2 2[ ] ( )XE X x P x dx∞

−∞

=

3

2 2 3 31

1

1 1[ ] [ ]4 12

E X x dx x −−

= × =

2 1 28[ ] [27 1]12 12

E X = + =

2 7[ ]3

E X =

Variance is given by, 2 2 2[ ] ( )X E X Xσ = −

2 7 413 3Xσ = − = …(ii)


( )XP x

1/4

1 1 2 3x

0

1.3 (B)


1X = and 2 43Xσ =

Hence, the correct option is (B). . Method 2 : Mean value is given by,

1 3 12

X − += =

Variance is given by,

[ ]22 3 ( 1) 4

12 3X

− −σ = =


Key Point PDF of uniformly distributed function is shown below,

Mean value for UDF is given by,

2

a bX +=

Variance for UDF is given by,

2

2 ( )12X

b a−σ =

Given : Probability density function,

2

2( )x

P x Ke−

= , x−∞ < < ∞

. Method 1 : From property of PDF, area under the PDF is always equal to one.

( ) 1P x dx∞

−∞

=

2

2 1x

Ke dx∞ −

−∞

=

2

2

0

2 1x

K e dx∞ −

=

Let, 2

2x t=

22

x dx dt⋅ =

x dx dt=

1/2

2 2dt dt t dtdxx t

−

= = =

1/2

0

2 12

t tK e dt∞ −

− =

1/2

0

2 1tK e t dt∞

− − = …(i)

Definition of gamma function :

1

0

t ne t dt n∞

− − = …(ii)

On comparing equation (i) and (ii),

112

n − = −

12

n =

1/2

0

12

te t dt∞

− − = = π

From equation (i), 2 1K π =

12

K =π

Hence, the correct option is (A). . Method 2 : For Gaussian random variable, probability density function is given by,

2

2( )

2

2

1( )2

X

x X

X

P x e−−σ=

πσ …(i)

2

2( )x

P x Ke−

= …(ii) On comparing equation (i) and (ii),

0 a

( )f x

xb

1

b a

1.4 (A)


2 1Xσ = , 0X = and 12

K =π


The power spectral density of a real-valued random process is an even function of frequency, i.e. ( ) ( )X XS f S f− =


Given : PSD and power of a signal ( )g t are respectively ( )gS ω and gP .

Power of a signal ( )g t is given by,

2

2

2

1lim ( )

T

g TT

P g t dtT→∞

−

=

For signal ( )ag t , power is given by,

2

2

2

1' lim ( )

T

g TT

P a g t dtT→∞

−

=

2

22

2

1' lim ( )

T

g TT

P a g t dtT→∞

−

=

2

22

2

1' lim ( )

T

g TT

P a g t dtT→∞

−

=

2'g gP a P= …(i)

Power spectral density is defined as power per unit bandwidth. If power spectral density of ( )g t is ( )gS ω then

( ) gg

PS

BWω = …(ii)

Power spectral density of ( )a g t is

2

' ( ) gg

a PS

BWω =

From equation (ii), 2' ( ) ( )g gS a Sω = ω The PSD and the power of the signal ( )a g t are

2 ( )ga S ω and 2ga P respectively.


Given : PDF of a Gaussian random variable X is given by,

2( 4)

181( )3 2

x

XP x e− −

=π

. Method 1 : We denote the random variable of the probability density function with a subscript, so we may write ( ) ( )X

A

P X A P x dx∈ =

The definition of this probability using an integral gives one important consequence for continuous random variables. If the set A contains just a single element, we can immediately see that probability that X is equal to that one value is exactly zero, as the integral over a single point is zero. So,

( 4) 0P X = = . Hence, the correct option (C). . Method 2 :

Using the property of PDF,

2

1

1 2( ) ( )x

Xx

P x X x f x dx< < =

For Gaussian PDF,

1( ) ( )2

P X m P X m< = ≥ = ;

x4 44

( )Xf x

1

3 2

1.5 (A)

1.6 (A)

1.7 (C)


where, m = mean value

1( 4) ( 4)2

P X P X< = ≥ =

Here, m = 4

4

4

( 4) (4 4 ) ( )XP X P X f x dx+

−

− += = < < =

4 4

44

( 4) ( ) ( )X XP X f x dx f x dx+

−

= = +

Since, ( ) ( )X Xf x F x=

4 4

4 4( 4) ( ) ( )X XP X F x F x

+

−= = +

( 4) (4) (4 ) (4 ) (4)X X X XP X F F F F− += = − + −

1 1( 4) (4) (4)2 2X XP X F F= = − + −

( 4) 0P X = = Hence, the correct option (C).

Given : 2 2110d Xσ = σ , [ ( )] 0E x n =

( ) ( ) ( 1)d n x n x n= − − Variance of ( )d n is given by,

2 2 2[ ( )] [ ( )]d E d n E d nσ = − …(i)

2 2[ ( )] [ ( ) 2 ( ) ( 1)E d n E x n x n x n= − −

2 ( 1)]x n+ −

2 2[ ( )] [ ( )] 2 [ ( ) ( 1)]E d n E x n E x n x n= − −

2[ ( 1)]E x n+ −

2 2[ ( )] [ ( )] 2 (1)XXE d n E x n R= −

2[ ( 1)]E x n+ − Autocorrelation function is defined as, ( ) [ ( ) ( )]XXR k E x n x n k= −

Since this process is stationary, the mean is independent of time, so [ ( )] [ ( 1)]E x n E x n= −

2 2[ ( )] 2 [ ( )] 2 (1)XXE d n E x n R= − …(ii)

Mean of ( )d n can be written as,

[ ( )] [ ( ) ( 1)]E d n E x n x n= − − [ ( )] [ ( )] [ ( 1)] 0E d n E x n E x n= − − =

…(iii) The variance of ( )x n can be written as,

2 2 2[ ( )] [ ( )]X E x n E x nσ = − …(iv)

On substituting the value of equation (ii) and (iii) in equation (i), 2 22 [ ( )] 2 (1)d XXE x n Rσ = − …(v)

Given, Mean of ( )x n is zero.

[ ( )] 0E x n = From equation (iv), 2 2[ ( )]X E x nσ =

From equation (v), 2 22 2 (1)d X XXRσ = σ −

2

22 2 (1)10

XX XXRσ = σ −

2

22 (1) 210

XXX XR σ= σ −

21

( ) 19 0.9520

XX

X k

R k

=

= =σ


X is uniformly distributed random variable in the range ( 1,1)− then density function is given by,

1 1 1

( ) 20 otherwise

Xx

f x − < <=

1/2

1 1x

( )Xf x

1.8 (A)

1.9 (C)


Y is also uniformly distributed random variable in the range ( 2,1)− then density function is given by,

1 2 1

( ) 30 otherwise

Yx

f y − < <=

According to Central limit theorem,

( ) ( ) ( )Z X Yf z f x f y= ⊗

Convolution of two unsymmetrical rectangular signal is trapezoidal signal.

Equation of line AB :

1 03( ) 0 ( 3)1 3Zf z z−

− = +− +

1( ) ( 3)6Zf z z= + ; 3 1z− < < −

The probability ( 1)Z ≤ − is

1

( 1) ( )ZP Z f z dz−

−∞

≤ − =

1

3

1( 1) ( 3)6

P Z z dz−

−

≤ − = +

2 1

3[( 3) ]1( 1)6 2

zP Z−−+≤ − =

1 1( 1) (4 0)12 3

P Z ≤ − = − =


Given : [ ( )] 0E X t X= =

0.2( ) 4( 1)XXR e− ττ = +

. Method 1 : From property of autocorrelation, 2(0) [ ( )]XXR E X t=

2 0[ ( )] 4( 1) 4(1 1) 8E X t e= + = + = Mean square value of ( )X t is 2[ ( )] 8E X t = For Gaussian random variable, CDF is given by,

( ) 1XX

x XF x Q −= − σ

Let X be the Gaussian random variable obtained by sampling the process with mean 0 and variance 2 2 2[ ( )] [ ( )]X E X t E X tσ = −

2 8 0 8Xσ = − =

2 2Xσ = The cumulative distribution function is defined as, ( ) [ ]XF x P X x= ≤ [ 1] (1)XP X F≤ = For Gaussian random variable,

1 0[ 1] (1) 12 2XP X F Q − ≤ = = −

So, 1[ 1] 12 2

P X Q ≤ = −


1/3

2 1y

( )Yf y

X

Y

Z X Y Summer

11,

3

( )Zf z

1/3

3 1 2z

B

A

Scan for

Video Solution

1.10 (D)


. Method 2 : [ ( )] 0E X t X= =

0.2( ) 4( 1)XXR e− ττ = +

Total power = Mean square value 2 (0)XXE X R = =

Therefore, 2 0.2 04 1 8E X e− × = + =

Variance, [ ] 22 2 8X E X E X σ = − =

PDF for Gaussian random variable is given as

2

22

2

1( )2

X

x

X

X

f x e−

σ=πσ

2

2 81( )2 8

x

Xf x e−

×=π×

[ ] [ ]1 1 1P X P X≤ = − >

[ ]1

1 1 ( )XP X f x dx∞

≤ = −

[ ]2

2 8

1

11 12 8

x

P X e dx∞

−×≤ = −

π×

[ ]2( /2 2 )

2

1

1 11 18 2

x

P X e dx∞

−≤ = −

π

…(i)

Let 2 22 2

x dx d= λ = λ

When 112 2

x → λ →

x → ∞ λ → ∞ Hence from equation (i),

[ ]2

2

1/2 2

1 11 1 2 28 2

P X e d∞ λ−

≤ = − λπ

[ ]2

2

1/2 2

11 12

P X e d∞ λ−

≤ = − λπ …(ii)

Given 2

21( )2

y

Q e dy∞

−

α

α =π …(iii)

Comparing equation (ii) and (iii),

[ ] 11 12 2

P X Q ≤ = −


[ ( 2)] [ (2)]Y E X t E X= = =

[ ( 4)] [ (4)]Z E X t E X= = = W Y Z= − Variance of W is given by, 2 2 2[ ] [ ]W E W E Wσ = − …(i)

[ ] [ ] [ ] [ ]E W E Y Z E Y E Z= − = −

[ ] [ (2)] [ (4)]E W E X E X= −

( )X t is a wide-sense stationary random process with constant mean of 0. So, [ ] 0E W = …(ii)

2 2 2 2[ ] [( ) ] [ 2 ]E W E Y Z E Y Z YZ= − = + −

2 2 2[ ] [ ] [ ] 2 [ ]E W E Y E Z E YZ= + −

2 2 2[ ] [ (2)] [ (4)]E W E X E X= + 2 [ (2) (4)]E X X− …(iii)

2 0.2 0[ (2)] (0) 4( 1) 8XXE X R e− ×= = + =

2[ (4)] (0) 8XXE X R= =

[ ( ) ( )] ( )XXE X t X t R+ τ = τ

[ (2) (4)] (2)XXE X X R=

0.2 24( 1) 6.68e− ×= + =

From equation (iii), 2[ ] 8 8 2 6.68 2.64E W = + − × =

From equation (i), 2 2.64 0 2.64Wσ = − =


Scan for

Video Solution

1.11 (C)


From the property of CDF if CDF has discontinuity then, ( ) ( ) ( )X XP X x F x F x−= = −

( 1) (1) (1 )X XP X F F −= = − ( 1) 0.55 0.25 0.30P X = = − = Hence, the correct option is (D).

Given : A random variable X with uniform density in the interval 0 to 1 has probability density function ( )f x . Hence ( )f x can be given as

( ) 1, 0 1f x x= ≤ ≤ Quantization error = Actual value – Quantized value e qq x x= −

Mean square value of quantization error is given by,

2 2[ ] ( )e eE q q f x dx∞

−∞=

2

2[ ] ( ) ( )e qE q x x f x dx∞

−∞= −

212

0[ ] ( ) ( )e qE q x x f x dx= −

0.3 12 2 2

0 0.3[ ] ( 0.7)eE q x dx x dx= + −

0.33 12 2

0.30

[ ] ( 0.49 1.4 )3exE q x x dx

= + + −

0.3 13 3 22

0 0.3

[ ] 0.49 1.43 3 2ex x xE q x

= + + −

2[ ] 0.0393333eE q =

The root mean square value 2[ ]eE q=

RMS = 0.198 Hence, the correct option is (B).

The probability density function of random variable V is shown in figure below.

Equation of line :

0( ) 0 ( 0)4 0KP v v−− = −

−

( ) , 0 44

KvP v v= < <

From property of PDF, area under PDF is unity.

( ) 1P v dv∞

−∞

=

Area of triangle = 1

1 4 12

K× × =

12

K =


2 2[ ] ( )E V v P v dv∞

−∞

=

4 4 3

2 2

0 0

[ ]4 8

Kv vE V v dv dv= =

2 4 40

1 256[ ] [ ] 832 32

E V v= =


Given : A probability density function is

,( )0 , otherwise

xKe xP x−α −∞ < < ∞=

0 1

1

( )f x

x

( )P v

(0,0) 4

(4, )KK

v

1.12 (D)

1.13 (B)

1.14 (C)

1.15 (C)


. Method 1 : From property of PDF, area under PDF is always equal to unity.

( ) 1P x dx∞

−∞

=

1xKe dx∞

−α

−∞

=

0

0

1x xKe dx Ke dx∞

α −α

−∞

+ =

00[ ] [ ] 1x xK Ke eα −α ∞

−∞ + =α −α

[1 ( 1)] 1K − − =α

0.52

K α= = α

The value of K is 0.5α . Hence, the correct option is (C). . Method 2 :

| |

0

2 1x xKe dx Ke dx∞ ∞

−α −α

−∞

= =

Key Point As

x

is an even function so the property of

even function is as follows.

0

( ) 2 ( )a a

a

f x dx f x dx−

=

02 1K e e−∞− − = α

2 1K =α

0.52

K α= = α


Given : A uniformly distributed random variable X with probability function,

1( ) [ ( 5) ( 5)]10Xf x u x u x= + − −

The probability density function of the transformed random variable Y is given as,

Random variable Y has just two levels either 0 or 1, it means that it is a discrete random variable and its PDF is given in the form,

2

1( ) ( ) ( )Y i i

if y P Y y y y

=

= = δ −

For 1i = , 1 0y = ( 0) ( 2.5) (2.5 )P Y P X P X= = < − + ≤ < ∞

2.5

2.5

( 0) ( ) ( )X XP Y f x dx f x dx− ∞

−∞

= = +

2.5 5

5 2.5

1 1( 0)10 10

P Y dx dx−

−

= = +

2.5( 0) 2 0.510

P Y = = × =

For 2i = , 2 1y = ( 1) ( 2.5 2.5)P Y P X= = − ≤ <

2.5 2.5

2.5 2.5

1( 1) ( )10XP Y f x dx dx

− −

= = =

1( 1) 5 0.510

P Y = = × =

For discrete random variable, probability density function is given by,

1

( ) ( ) ( )n

Y i ii

f y P Y y y y=

= = δ −

5 0 5x

1/10

( )Xf x

y

1

2.5 0 2.5x

1.16 (B)


2

1( ) ( ) ( )Y i i

if y P Y y y y

=

= = δ −

( ) ( 0) ( 0)Yf y P Y y= = δ − ( 1) ( 1)P Y y+ = δ −

( ) 0.5 ( ) 0.5 ( 1)Yf y y y= δ + δ −


Given : Sampling period, 0.03 secst = μ

If 1( )X t and 2( )X t are two samples obtained from a random process X(t) at 1t and 2t instances then,

1. If 1 2 1 2( ) ( ) ( ) ( )k j k jE X t X t E X t E X t =

then 1 2( ), ( )X t X t are statistically independent, for all values of and where and are positive integers.

2. If [ ]1 2 1 2( , ) ( , ) 0XXR t t E X t t= =

then 1 2( ), ( )X t X t are orthogonal.

3. 1 2 1 2( , ) ( , )XXCov t t R t t= [ ] [ ]1 2( ) ( ) 0E X t E X t− =

i.e. [ ]1 2( ) ( )E X t X t = [ ] [ ]1 2( ) ( )E X t E X t then

1( )X t and 2( )X t are uncorrelated.

Assume that the random process at the input of the LPF is ( )X t which is given zero mean white Gaussian noise and output of the LPF is

( )Y t . The spectral densities relation as shown in figure below.

According to Wiener-Khinchin theorem, . .( ) ( )F TR S fτ ←⎯⎯→

0( ) rect2 2Y

N fS fB

=

Since, . .rect sinc( )F TA AT fTTτ ←⎯⎯→

From duality property,

. .sinc ( ) rectF T fAT T AT

τ ←⎯⎯→

On comparing ( )YS f with above equation,

02 ,2

NT B A= =

0( ) sinc (2 )YR N B Bτ = τ

The mean value of random process ( )Y t is given by, [ ] [ ](0)E Y H E X=

[ ] 0 1 0E Y = × =

( )Y t is also zero mean Gaussian random process. Let ( ), (2 )s sY t Y t are two consecutive samples obtained from ( )Y t at st and 2 st instances (Since, uniform sampling) [ ] [ ]( ) (2 ) 0s sE Y t E Y t= =

[ ]( )Since, ( ) 0E Y t =

( )Yf y

0y

0.5 ( )y

1

0.5 ( 1)y

Scan for

Video Solution

k j k j

( )XS f

0 / 2N

f

1

B B

( )H f

f

( )YS f

0 /2N

fB B

1.17 (A)


[ ]( )Since, ( ) ( ) ( )YYR E Y t Y tτ = + τ

0 sinc (2 )sN B Bt=

Given : 10 kHz, 0.03 mssB t= =

30( ) 10 10 sinc(2 10YY sR t N= × × ×

3 310 0.03 10 )−× × ×

( )YY sR t 4010 sinc(0.6)N=

sinsinc( ) xxxπ = π

( ) 0YY sR t ≠ → Not orthogonal

[ ]cov ( , 2 ) ( ) ( ) [ (2 )]s s YY s s st t R t E Y t E Y t= −

cov ( , 2 ) ( ) 0.(0)s s YY st t R t= − Correlated


Given : 0 216( ) 1, ( )

16iS Sω = ω =+ ω

Relation between input and output power spectral densities is given by,

20 ( ) ( ) ( )iS H Sω = ω ω

So, 22

16 ( ) 116

H= ω ×+ ω

2 2

4 1( )16

14

H ω = =+ ω ω +

…(i)

Transfer function of first order low pass R-L can be obtained as,

0 ( )( )( )i

V j RHV j R j L

ωω = =ω + ω

1( )1

H j LR

ω = ω+

2

1( )

1/

H

R L

ω =ω +

…(ii)

( )H ω represents a transfer function of a first order low pass filter. So the desired operation can be performed with a first order low pass R-L filter.

Transfer function of given system,

2

1( )

14

H ω =ω +

Comparing equation (i) and (ii),

4RL

=

4R L= If 1 HL = then 4R = Ω . Hence, the correct option is (A).

Properties of autocorrelation : (a) ( ) (0)XX XXR Rτ ≤ (b) ( ) ( )XX XXR R−τ = τ (c) 2(0) ( )XXR E X t = = Power

(d) If [ ( )] 0E X t X= ≠ and ( )X t is ergodic with no periodic components then

2lim ( )XXR Xτ →∞

τ =

(e) If ( )X t is periodic component, then ( )XXR τ will have a periodic component

with the same period.

[ ]( ) ( ,2 ) ( ) (2 )YY s YY s s sR t R t t E Y t Y t= =

[ ]( ) (2 ) ( )s s YY sE Y t Y t R t=

0≠ →

( )iS 0 ( )S ( )H

( )iV j 0 ( )V jR

j L

1.18 (A)

1.19 (A)

1.20 (C)


(f) If ( )X t is ergodic, zero-mean and has no periodic component, then

lim ( ) 0XXRτ →∞

τ =

(g) ( )XXR τ cannot have an arbitrary shape.


Properties of power spectral density : 1. The power spectral density ( )XS f is

always real and non-negative, i.e. ( ) 0XS f ≥

2. The power spectral density of a real-valued random process is an even function of frequency, i.e.

( ) ( )X XS f S f− =

3. The total normalized power of a random process is defined in terms of power spectral density as,

( )total XP S f df∞

−∞=

2[ ( )] ( )XE X t S f df∞

−∞=

(0) ( )X XR S f df∞

−∞=

4. The zero-frequency value of the power spectral density of a stationary process equals the total area under the graph of the autocorrelation function, i.e.

(0) ( )X XS R d∞

−∞= τ τ


The variance of a random variable X is given by,

2 2 2[ ] [ ]X E X E Xσ = −


Cumulative distribution function (CDF) of a random variable in terms of its probability density function (PDF) is given by,

( )x

XCDF f d−∞

= ε ε

where, ( )Xf x is PDF of variable.

From given PDF curve,

1, 1 0

( )1, 0 1X

x xf x

x x+ − < <

= − + < <

0

1 0

CDF ( 1) ( 1)x

d d−

= ε + ε + −ε + ε

02 2

1 0

CDF2 2

x

−

ε − ε= + ε + + ε

21CDF

2 2x x= − +

At 0x = , 1CDF2

=

At 1x = , CDF 1= The corresponding CDF satisfying the values is as shown below.


1 0 1x

1

( )Xf x

1 1x

CDF

1

0.5

Scan for

Video Solution

1.21 (B)

1.22 (A)

1.23 (A)


Note : 1. CDF is always non-decreasing function.

Therefore, option (D) is eliminated. 2. CDF is integration of PDF. So,

integration of a straight line is a parabolic function. Only option (A) gives a parabolic variation of CDF.

3. Option (B) is not correct because it gives a linear variation of CDF.

4. From property of CDF, ( ) 1F ∞ = but in option (C) it is going towards infinite.

Given : Probability density function is

2 3( ) x xXP x Me Ne− −= +

From property of PDF, the area under the PDF is always equal to unity.

( ) 1XP x dx∞

−∞

=

2 3( ) 1x xMe Ne dx∞

− −

−∞

+ =

02 3 2 3

0

( ) ( ) 1x x x xMe Ne dx Me Ne dx∞

− −

−∞

+ + + =

2 0 3 0 20[ ] [ ] [ ]

2 3 2x x xM N Me e e− ∞

−∞ −∞+ +−

30[ ] 1

3xN e− ∞+ =

−

0 0 0( ) ( ) ( )2 3 2M N Me e e e e e−∞ −∞ −∞− + − + −

− 0( ) 1

3N e e−∞+ − =−

12 3 2 3M N M N+ + + =

2 13NM + =


Key Point ( )xP x is even function as x is even function.

So, by the property of even function,

2 3

0

2 ( ) 1x xMe Ne dx∞

− −+ =

2 12 3M N × + =

2 13

M N+ =

Autocorrelation function and power spectral density form Fourier transform pair. . .( ) ( )F TR S fτ ←⎯⎯→ The above relationship is known as Wiener Khinchin relationship. The Fourier transform of gate pulse is sinc function.

[ ]. .rect sincF TA AT fTTτ ←⎯⎯→

Fourier transform of triangular pulse is sinc squared function,

[ ]. . 21 sincF TA AT fTT

τ − ←⎯⎯→


Given : Two-sided power spectral density,

101 102

−η = × W/Hz

. .F T

( )R

/2T /2T

( )S f

2

T

1

T

1

T

2

T

f

A

AT

A

( )R

T T

( )S f

f

. .F T

1.24 (A)

1.25 (B)

1.26 (B)


For a LTI system, the relationship between output PSD and input PSD is given by,

20 ( ) ( ) ( )nS f H f S f=

Two-sided power spectral density of white noise process ( )X t is

The total area under power spectral density gives power of the signal. Area under 0 ( )S f can be calculated as,

0 0 ( )P S f df∞

−∞

= 3 101 20 10 102

−= × × ×

60 1 10 WP −= ×

The power of the output process ( )Y t is 61 10 W−× .


Given : X and Y are identical and independent i.e. [ ] [ ] [ ]E XY E X E Y= and [ ] [ ]P X P Y= .

Conditional probability is given by,

( )( )

A P A BPB P B

∩ =

( 2 0)P X Y X Y+ = − =

( 2 0)

( 0)P X Y X Y

P X Y+ = ∩ − ==

− = …(i)

Where, ( 2) ( 1, 1) ( 0, 2)P X Y P X Y P X Y+ = = = = + = =

( 2, 0)P X Y+ = = …(ii) ( 0) ( 0, 0)P X Y P X Y− = = = =

( 1, 1) ( 2, 2)P X Y P X Y+ = = + = = ( 0) ( 0) ( 0)P X Y P X P Y− = = = =

( 1) ( 1)P X P Y+ = = ( 2) ( 2)P X P Y+ = =

1 1 1 1 1 1( 0)2 2 4 4 4 4

P X Y− = = × + × + ×

3( 0)8

P X Y− = = …(iii)

Intersection of ( 2)P X Y+ = and ( 0)P X Y− = means the common term in both the equation,

( 2 0) ( 1, 1)P X Y X Y P X Y+ = ∩ − = = = = ( 2 0) ( 1) ( 1)P X Y X Y P X P Y+ = ∩ − = = = =

1 1 1( 2 0)4 4 16

P X Y X Y+ = ∩ − = = × =

From equation (i), 1/16 1( 2 0)3/8 6

P X Y X Y+ = − = = =


Given : 3.5X = , 2 1.5Xσ =

For a discrete random variable, mean value is given by,

1

( )m

i ii

X P x x=

=

5

1( )i i

iX P x x

=

=

(0.1 1) (0.2 2) (0.4 3)X = × + × + × (0.2 4) (0.1 5)+ × + ×

3X =

( )nS f

η / 2

f

1

10 10

2( )H f

0 ( )S f

η / 2

10 kHz 10kHzf

1

4

1

8

1

8

1

8

1

8

1

16

1

16

1

16

1

16

1( 0)

2P Y

1( 1)

4P Y

1( 2)

4P Y

1( 0)

2P X

1( 1)

4P X

1( 2)

4P X

1.27 (C) 1.28 (B)


Student calculated mean as 3.5 but by calculation it is 3. So, student is wrong. For a discrete random variable, mean square value is given by,

2 2

1[ ] ( )

m

i ii

E X x P x=

=

5

2 2

1[ ] ( )i i

iE X x P x

=

=

2 2 2 2[ ] (1) 0.1 (2) 0.2 (3) 0.4E X = × + × + × 2 2(4) 0.2 (5) 0.1+ × + ×

2[ ] 10.2E X = Variance is given by, 2 2 2[ ] ( )X E X Xσ = −

2 210.2 (3) 10.2 9Xσ = − = −

2 1.2Xσ =

Teacher calculated variance as 1.5 but by calculation variance is 1.2. So, teacher is also wrong. Hence, the correct option is (B).

If Fourier transform of ( )x t is ( )X f i.e.

. .( ) ( )F Tx t X f←⎯⎯→ Using time-shifting property, 2. .( ) ( )dj ftF T

dx t t e X f− π− ←⎯⎯→ …(i)

Delay element will have transfer function

3 32 2 0.5 10 101( ) dj ft j f j fH f e e e

− −− π − π × × − π= = =

For an LTI system,

The output and input are related as, ( ) ( ) ( )y t x t h t= ⊗ ( ) ( ) ( )Y f H f X f= ⋅ …(ii)

On comparing equation (i) and (ii), 2( ) dj ftH f e− π= Now, we have differentiator in the system as shown in figure. . .( ) ( )F Tx t X f←⎯⎯→

. .( ) 2 ( )F Td x t j f X fdt

←⎯⎯→ π …(iii)

( ) ( ) ( )y t x t h t= ⊗ ( ) ( ) ( )Y f X f H f= …(iv) On comparing equation (iii) and (iv), ( ) 2H f j f= π The system with impulse response is shown in figure below.

Solving parallel block,

Solving cascade block,

The transfer function is given by,

310( )( ) 2 (1 )( )

j fY fH f j f eX f

−− π= = π +

3( ) 2 [1 cos( 10 )H f j f f −= π + π 3sin( 10 )]j f −− π 3( ) 2 [1 cos( 10 )]H f j f f −= π + π 32 sin( 10 )f f −+ π π

2 3 2

2 2 3

(2 ) [1 cos( 10 )]( )

(2 ) sin ( 10 )f f

H ff f

−

−

π + π=

+ π π

3 2 3

2 3

1 2cos( 10 ) cos ( 10 )( ) (2 )

sin ( 10 )f f

H f ff

− −

−

+ π + π= π

+ π

3( ) (2 ) 2 2cos( 10 )H f f f −= π + π 2 2 2 3( ) 8 [1 cos( 10 )]H f f f −= π + π

( )x t ( ) ( )dy t x t t Delay, td

( )X f ( )Y fLTI system

( )H f

( )x t ( ) ( )d

y t x tdt

d

dt

( )h t

310j fe

( )X f 2j f ( )Y f

( )X f 3101 j fe ( )Y f2j f

( )X f310(1 ) 2j fe j f

( )Y f

1.29 (D)


For an LTI system, the relationship between output power spectral density and input power spectral density is given by,

2( ) ( ) ( )Y XS f H f S f=

2 2 3( ) 8 [1 cos 10 ] ( )Y XS f f f S f−= π + π

If 0(2 1)f n f= + , 0 1kHzf = , n any integer

then 3cos 10 1f−π× = − and ( ) 0YS f = .


Given : 2

( )XR e−πττ =

From Wiener-Khinchin relationship, auto-correlation and power spectral density are Fourier transform pair.

. .( ) ( )F TR S fτ ←⎯⎯→

Fourier transform of Gaussian signal is also Gaussian.

2 2. .F T fe e−πτ −π←⎯⎯→

Solving parallel path,

The transfer function is given by,

( )( ) 2 1( )

Y fH f j fX f

= = π −

2 2( ) 1 4H f f= + π

For an LTI system, the relationship between input and output PSD is given by,

2( ) ( ) ( )Y XS f H f S f=

22 2( ) (1 4 ) fYS f f e−π= + π


The power spectral density of a real process

( )X t for positive frequencies is shown below.

From property of PSD, 2[ ( )]E X t = Mean square value = Power

The total area under PSD gives power of the signal.

2 1[ ( )] ( )2 XE X t S d

∞

−∞

= ω ωπ

2

0

2[ ( )] ( )2 XE X t S d

∞

= ω ωπ

3

3

11 102

9 10

1[ ( )] ( )XE X t S d×

×

= ω ωπ

2 1[ ( )] Area under PSD ( )XE X t S= × ωπ

2 31 1[ ( )] 400 2 6 1 102

E X t = + × × × × π

[ ]2 1 6400[ ( )] 400 6000E X t = + =π π

From property of autocorrelation,

[ ]22lim ( ) (Mean) ( )XR E Xτ→∞

τ = =

Applying final value theorem,

0lim ( ) lim ( )X Xf

R s S fτ→∞ →

τ =

From figure, at 0, ( ) 0Xf S f= =

[ ]2

0( ) lim ( ) 0 0Xf

E X s S f s→

= = × =

( ) 0E X =


Scan for

Video Solution

( )X f ( 2 1)j f ( )Y f

0 9 10 11

6

( )XS 4400 ( 10 )

3(10 rad/s)

1.30 (A)

1.31 (B)


Given : 2 14Uσ = , 2 1

9Vσ = , [ ] [ ] 0E U E V= =

(3 2 ) (3 2 0)P V U P V U≥ = − ≥ Let 3 2Z V U= − The sum of two Gaussian random variable is also Gaussian. So, Z will also be a Gaussian random variable. Now, we have to find ( 0)P Z ≥ Since Z is a Gaussian random variable, its probability density function is given by,

2

2( )2

2

1( )2

Z

z Z

Z

Z

f z e− −

σ=πσ

where, Z = Mean of Gaussian random variable Z

2Zσ = Variance of Gaussian random

variable Z. [ ] [3 2 ]Z E Z E V U= = − Since U and V are independent random variables, we can write 3 [ ] 2 [ ]Z E V E U= −

3 0 2 0Z = × − × 0= Variance is given by, 2 2 2[ ] ( )Z E Z Zσ = −

2 2[ ]Z E Zσ =

Mean square value is given by, 2 2[ ] [(3 2 ) ]E Z E V U= −

2 2 2[ ] [9 4 12 ]E Z E V U UV= + −

2 2 2[ ] 9 [ ] 4 [ ] 12 [ ]E Z E V E U E UV= + −

2 1 1[ ] 9 4 12 [ ] [ ]9 4

E Z E U E V= × + × −

2[ ] 1 1 12 0E Z = + − × 2=

Hence, 2 2 2z E Z σ = =

The Gaussian random variable Z has mean 0 and variance 2. The PDF of Z is shown in figure below.

Probability for 1( 0)2

P Z ≥ = .

(Since area under the PDF is unity). Since Gaussian is symmetrical about mean, the probability will be distributed 50% about mean. Hence, the correct option is (B).

Given : Two identically distributed zero-mean random variable U and V . Let distribution function of U and V be ( )UF u and ( )VF v respectively.

( ) ( ) ( )U VF u F v F x= = CDF of U and 2V are ( )F x and ( )G x . CDF is defined as, ( ) ( ) ( )UF u P U u F x= ≤ = ( ) ( )VF v P V v= ≤

2 ( ) (2 ) ( )2VvF v P V v P V G x = ≤ = ≤ =

Since U and V are identical, u and v are also same i.e. u v x= = From property of CDF, CDF is non-decreasing function. Case I : For positive value of random variable U and V : If 1 2x x> then 1 2( ) ( )X XF x F x>

So, ( ) ( )F x G x> Since,2xx >

( ) ( ) 0F x G x− > …(i)

z

( )Zf z

1

2

0 2–2

1.32 (B)

1.33 (D)


Case II : For negative value of random variable U and V :

2v u>

Since,

2xx <

For example, 1u v x= = = −

1 1

2− > −

Hence, ( ) ( ) 0G x F x− > …(ii) Case III : For zero value of random variable U and V i.e. 0x = . ( ) ( )F x G x= So, ( ) ( ) 0F x G x− = …(iii) Combining equation (i), (ii) and (iii), [ ( ) ( )] 0F x G x x− ≥ Hence, the correct option is (D).

Given : The three random variables 1 2,X X and

3X are uniformly distributed on [0, 1]. So, we have following possible values 1 2 3X X X> >

1 3 2X X X> >

2 1 3X X X> >

2 3 1X X X> >

3 1 2X X X> >

3 2 1X X X> >

Since, all the three variables are identical, the probabilities for all the above inequalities are same. Hence, the probability that 1X is the largest,

P ( 1X is the largest) 2 1 0.336 3

= = =

Hence, the probability P 1X is the largest is 0.33.

Variance is given by,

Variance = mean square value – 2(mean) 2 2 2[ ] ( )X E X Xσ = − Variance can not be negative. So, 2 0Xσ ≥

2 2[ ] ( )E X X≥ 2 2[ ] ( [ ])E X E X≥ This relation always holds true. Hence, the correct option is (B).

Given : ( ) 2 sin(2 )X t t= π + φ where, φ is random variable uniformly distributed in the interval (0, 2 )π . The density function for UDF can be written as,

1 , 0 2

( ) 20, otherwise

fφ

< φ < πφ = π

Autocorrelation function is defined as,

1 2 1 2[ ( ) ( )] ( ) ( ) ( )E X t X t X t X t f d∞

φ−∞

= φ φ

2

1 2 10

[ ( ) ( )] [ 2 sin(2 )]E X t X t tπ

= π + φ

21[ 2 sin(2 )]

2t dπ + φ φ

π

1 2[ ( ) ( )]E X t X t = 2

1 20

2 sin(2 )sin(2 )2

t t dπ

π + φ π + φ φπ

Using the trigonometric relation,

1sin sin [cos( ) cos( )]2

A B A B A B= − − +

1

2

( )f

0 2

1.34 0.33

1.35 (B)

1.36 (D)


2

1 2 1 20

1[ ( ) ( )] cos(2 2 ) cos2

E X t X t t tπ

= π − π −π

1 2[2 ( ) 2 ]t t dπ + + φ φ

2

1 2 1 2 0

1[ ( ) ( )] cos 2 ( )2

E X t X t t t dπ

= π − φπ

2

1 20

1 cos[2 ( ) 2 ]2

t t dπ

− π + + φ φπ

1 2 1 2

1[ ( ) ( )] cos[2 ( )] 22

E X t X t t t= π − × ππ

21 2

0

sin 2 ( ) 212 2

t tπ

π + + φ − π

1 2 1 2[ ( ) ( )] cos[2 ( )]E X t X t t t= π −

Hence, the correct option is (D). Key Point Autocorrelation function for random process

cos( )A tω + φ or sin( )A tω + φ is given by,

2

( ) cos ( )2XXAR τ = ωτ

Given : The random variable X is uniformly chosen from the set of positive odd number less than 100. So, we have 1,3,5,7,.....,99X = So, the total number of values in set of X is 50n = Since, X is uniformly chosen from the set of numbers. So, the probability density function for each value of X is

1 1( )50Xf x

n= =

The expected value of X is given by,

1

[ ] ( )m

Xi

E X x f x=

=

1 3 5 99[ ] ..........50 50 50 50

E X = + + + +

1[ ] [1 3 5 ....... 99]50

E X = + + + +

1 50[ ] 2 1 (50 1)250 2

E X = × + −

[ ]Sum of AP is given by, 2 ( 1)2nnS a n d = + −

1[ ] (2 49 2) 502

E X = + × =

Hence, the expectation [ ]E X is 50.

Given : The power spectral density

1

( )0

X

f WS f W

f W

≤= >

…(i)

From Wiener-Khinchin relationship, auto-correlation function and power spectral density are Fourier transform pair. . .( ) ( )F T

X XR S fτ ←⎯⎯→ By taking inverse Fourier transform of equation (i),

2( ) ( ) j fX XR S f e df

∞π τ

−∞

τ =

21( )W

j fX

W

R e dfW

π τ

−

τ =

2[ ]1( )2

j WW

XeR

W j

πτ−τ =

πτ

2 21 [ ]( )

2

j W j W

Xe eR

W j

πτ − π τ−τ =πτ

sin(2 )( ) 2(2 )X

WRW

πττ =πτ

( ) 2sin (2 )XR c Wτ = τ

sin( )sinc( ) xxxπ = π

…(ii)

Autocorrelation is given by, ( ) [ ( ) ( )]XR E X t X tτ = − τ

1.37 50

1.38 4


The value of expectation 1( )4

E X t X tW

π −

can be written as,

1 1( )4 4XXE X t X t RW W

π − = π …(iii)

Using equation (ii),

1sin 21 42

14 24

XX

WWR

W WW

π ⋅ = π ⋅

sin

1 4224

2

XXRW

π = = π π

From equation (iii),

1 4( ) 44

E X t X tW

π − = π× = π

Hence, the value of the expectation is 4. . Short Method :

If . .rect sinc( )F TtA A f ←⎯⎯→ τ τ τ

Where, A = height of pulse τ = width of pulse From duality property,

. .sinc( ) rect rectF T f fA t A A− τ τ ←⎯⎯→ = τ τ

. . 12sinc(2 ) rect2

F T fWW W

τ ←⎯⎯→

1 1( )4 4

E X t X t RW W

π − = π

1 2sinc (2 )

4XXR WW

= τ

1sin 21 42

14 24

XX

WWR

W WW

π ⋅ = π ⋅

sin

1 4224

2

XXRW

π = = π π

1 4( ) 44

E X t X tW

π − = π× = π

Hence, the value of the expectation is 4.

1 2,X X and 3X be independent and identically

distributed random variables with the uniform distribution on [0,1] . We have to find 1 2 3( )P X X X+ ≤ or 1 2 3( 0)P X X X+ − ≤ Let 1 2 3X X X Z+ − = Z is a random variable which is sum of three independent random variables. According to central limit theorem,

1 2 3( ) ( ) ( ) ( )Z X X Xf z f x f x f x= ⊗ ⊗ −

All the random variables are identical so the density function is given by,

1 2( ) ( ) ( )X X Xf x f x f x= =

3

1, 0 1( )

0, otherwiseX

xf x

< <= =

( ) ( ) ( 1)Xf x u x u x= − −

Scan for

Video Solution

2X

3X

1X 1 2 3Z X X X

0 1

1

( )Xf x

x

1.39 0.1667


First calculate 1 2'X X X= + , Using central limit theorem,

1 2' ( ) ( ) ( )X X Xf x f x f x= ⊗

[ ] [ ]' ( ) ( ) ( 1) ( ) ( 1)Xf x u x u x u x u x= − − ⊗ − −

' ( ) ( ) ( 1) ( 1) ( 2)Xf x r x r x r x r x= − − − − + −

' ( ) ( ) 2 ( 1) ( 2)Xf x r x r x r x= − − + −

Convolution of two symmetric rectangular wave is a triangular signal. The triangular signal is shown below.

For random variable 3X− , probability density function can be drawn as,

From above figure, PDF can be expressed as mathematically

3( ) ( 1) ( )Xf x u x u x= + − …(i)

The triangular PDF ' ( )Xf x can be expressed as, ' ( ) ( ) 2 ( 1) ( 2)Xf x r x r x r x= − − + − …(ii) Now, calculate 3'Z X X= − , using central limit theorem. Property of convolution,

1 2 1 2[ ( ) ( )] ( ) ( )d dx t x t x t x tdt dt

⊗ = ⊗

3 '( ) ( ) ( )Z X X

d df x f x f xdx dx

= ⊗

From differentiating equation (i),

3( ) ( 1) ( )X

d f x x xdx

= δ + − δ …(iii)

Density function of random variable Z can be written as,

( ) [ ( 1) ( )]Zd f x x xdx

= δ + − δ

[ ( ) 2 ( 1) ( 2)]r x r x r x⊗ − − + −

( ) ( 1) 2 ( ) ( 1)Zd f x r x r x r xdx

= + − + −

( ) 2 ( 1) ( 2)r x r x r x− + − − −

( ) ( 1) 3 ( ) 3 ( 1)Zd f x r x r x r xdx

= + − + −

( 2)r x− − The derivative of ( )Zf x can be drawn for 0x ≤ as shown below.

Equation of line :

1 0( ) 0 ( 1)0 ( 1)Z

d f x xdx

−− = +− −

On integrating w.r.t. 1x + ,

2( 1)( )

2Zxf x +=

Key Point Integrating ramp signal we get parabola. But here ramp signal is starting from –1 so on integrating the equation of ramp we will get parabola that should have center at –1. Equation of parabola in vertex form is given by, 2( ) ( )y k a x h− = − where, h and k are coordinates of center of the parabola

21( ) 0 [ ( 1)]2Zf x x− = − −

0 1

1

1( )Xf x

x0 1

1

2( )Xf x

x

' ( )Xf x

0 1 2x

0–1

1

3( )Xf x

x

( )Z

df x

dx

0–1x

1


We need to find ( 0)P Z ≤

0

( 0) ( )ZP Z f x dx−∞

≤ =

0 2

1

( 1)( 0)2

xP Z dx−

+≤ =

3 0

1[( 1) ] 1( 0)6 6

xP Z −+≤ = =

( 0) 0.1667P Z ≤ = Hence, the probability is 0.1667.

Given : ( )X t is a wide-sense stationary process. So, its autocorrelation function is given by, ( ) [ ( ) ( )]XXR E X t X tτ = − τ From Wiener-Khinchin relationship, autocorrelation function and power spectral density are Fourier transform pair. . .( ) ( )F T

XX XXR S fτ ←⎯⎯→ Autocorrelation function of random process

( )Y t is given by, ( ) [ ( ) ( )]YYR E Y t Y tτ = − τ ( ) [ (2 1) (2 1 2 )]YYR E X t X tτ = − − − τ ( ) (2 )YY XXR Rτ = τ From Wiener-Khinchin relationship, ( ) [ (2 )]YY XXS f FT R= τ (From scaling property of Fourier transform)

1( )2 2YY XX

fS f S =


Given : Power spectral density of random process ( )X t is

610 (3000 ) 3kHz( )0 otherwiseX

f fS f

− − ≤=

The power spectrum can be obtained as,

Now, ( )X t modulates a carrier cos(16000 )tπ . So, the spectrum becomes (carrier frequency

8kHzCf = ).

Again this spectrum is passed through an ideal band pass filters of unity gain with centre frequency of 8 kHz and bandwidth of 2 kHz. Spectrum of bandpass filter is

So, we get output spectrum as

The total power is equal to area under power spectral density. So, the output power is given by,

3 3 3 31 10 10 10 2 102

P − − = × × + × ×

1 22

P = + 2.5 W=

Hence, the output power is 2.5 W.

Scan for

Video Solution

( )XS f

63000 10

(kHz)f3 3

(kHz)f811 5 5 8 11

' ( )XS f

63000 10

4

–9 –8 –7 7 8 9

( )H f

(kHz)f

1

–9 –8 –7 7 8 9

63000 10

4

62000 10

4

(kHz)f

" ( )XS f

Scan for

Video Solution

1.40 (C)

1.41 2.5


Given : 1 1x = + and 2 1x = −

1 21 1( 1) , ( 1)2 2

P x P x= + = = − =

Mean value of nX is given by,

[ ] ( )n i ii

E X x P x=

1 1[ ] 1 ( 1) 02 2nE X = × + − × =

The autocorrelation function is defined as,

[ ]( )Y n n kR k E Y Y +=

Now, 10.5n n nY X X −= +

So,

[ 1( ) ( 0.5 )(Y n n n kR k E X X X− += + ]10.5 )n kX + −+

( ) [(Y n n kR k E X X += 10.5 n n kX X + −+

10.5 n n kX X− ++ 1 10.25 ]n n kX X− + −+

( ) [ ]Y n n kR k E X X += 10.5 [ ]n n kE X X + −+

10.5 [ ]n n kE X X− ++ 1 10.25 [ ]n n kE X X− + −+

( ) ( ) 0.5 ( 1)Y X XR k R k R k= + −

0.5 ( 1) 0.25 ( )X XR k R k+ + +

( ) 1.25 ( ) 0.5 ( 1)Y X XR k R k R k= + −

0.5 ( 1)XR k+ + …(i)

( ) [ ]X n n kR k E X X +=

For 0k = , we obtain

2 2(0) [ ] ( )X n i ii

R E X x P x= =

2 21 1(0) 1 ( 1) 12 2XR = × + − × =

For 0k ≠ , we obtain ( ) [ ]X n n kR k E X X += [ ] [ ]n n kE X E X +=

(since, independent) ( )XR k = 0

Hence from equation (i), (0) 1.25 (0) 0.5 ( 1)Y X XR R R= + −

0.5 (1)XR+ = 1.25

(1) 1.25 (1) 0.5 (0)Y X XR R R= +

0.5 (2)XR+ = 0.5

( 1) 1.25 ( 1) 0.5 ( 2)Y X XR R R− = − + −

0.5 (0)XR+ = 0.5

( )YR k for k other than 0, 1 and –1 = 0. Thus,

the autocorrelation function ( )YR k is plotted as shown below.


Random variable X = Number of times a fair coin needs to be tossed till two consecutive heads appears for the first time.

3 2 1 0 1 2 3k

1

...

( )XR k

3 2 1 0 1 2 3k

1.25

...

1.25 ( )XR k

3 2 1 0 1 2 3k

0.5...

0.5 ( 1)XR k

3 2 1 0 1 2 3k

0.5...

0.5 ( 1)XR k

3 2 1 0 1 2 3k

0.5...

( ) 1.25 ( ) 0.5 ( 1) 0.5 ( 1)Y X X XR k R k R k R k

0.5

1.25

3 2 1 0 1 2 3k

0.5

1.25

0.5

Scan for

Video Solution

1.42 (B)

1.43 1.5


The probability of getting head in each toss

12

H =

The probability of getting tail in each toss

12

T =

The probabilities,

21 1 1( 2)

2 2 2P X HH = = = × =

31 1 1 1( 3)

2 2 2 2P X THH = = = × × =

41( 4)

2P X TTHH = = =

The expectation of X is given by, [ ] ( )i i

iE X x P x=

2 3 41 1 1[ ] (2) 3 4 .........

2 2 2E X = + +

Consider 2 3 41 1 1(2) 3 4 .........

2 2 2S = + +

…(i)

3 4 51 1 1(2) 3 4 .........2 2 2 2S = + +

…(ii)

Subtracting equation (ii) from equation (i) 2 3 41 1 1(2) [3 2] [4 3].........

2 2 2 2SS − = + − + −

2 3 4 51 1 1 1(2) .........2 2 2 2 2S = + + +

3 4 51 1 1 1 .........2 2 2 2 2S − = + +

3

21

1 1212 2 212

S − = =

−

1 12 2 4S = + 11

2= +

32

S = 1.5=

So, 3[ ] 1.52

E X = =

Hence, the expectation of X is 1.5.

Given : ( 0)P X p= = So, ( 1) 1P X p= = − ( 0)P Y q= = and ( 1) 1P Y q= = − Let, Z X Y= +

X Y Z 0 0 0 0 1 1 1 0 1 1 1 2

So, ( 1) 0 and 1P Z P X Y≥ = = = 1 and 0

1 and 1P X Y

P X Y+ = =

+ = =

( 1) 1 0 and 0P Z P X Y≥ = − = = 1 [ 0] [ 0]P X P Y= − = = Since, X and Y are independent ( 1) 1P Z pq≥ = − Hence, the correct option is (D).

. Method 1 : Given : The input noise has zero mean i.e. [ ( )] 0E W t =

Let ( )N t be the noise at the output of filter. Variance of ( )N t is given by,

2 2Var[ ( )] [ ( )] [ ( )]N t E N t E N t= − Output noise mean is also zero because

[ ( )] [ ( )] ( )E N t E W t h t dt∞

−∞

=

LTI( )h t

N t( )W t( )

0( )2

W

NS f

1.44 (D)

1.45 (A)


( )W t is white noise, so

2 2Var[ ( )] [ ( )] 0 [ ( )]N t E N t E N t= − =

Var[ ( )]N t (0)NR=

Since, ( ) ( ) ( ) ( )N WR h h Rτ = τ ⊗ −τ ⊗ τ

and 0( ) ( )2W

NR tτ = ⋅δ

0( ) [ ( ) ( )]2N

NR h hτ = τ ⊗ −τ ⋅

0( ) ( ) ( )2N

NR h k h k dk∞

−∞

τ = ⋅ ⋅ τ +

20(0) ( )2N

NR h k dk∞

−∞

=

1 2 32 2 200 1 2

(0) ( ) ( ) ( )2N

NR A dk A dk A dk = + − +

20(0) (3 )2N

NR A= 20

32

A N=

Hence, the correct option is (A). . Method 2 : Variance of output noise is

(0) ( )N NR S f df∞

−∞=

2

(0) ( ) ( )N WR S f H f df∞

−∞=

2

0(0) ( )2N

NR H f df∞

−∞=

From Parseval’s theorem,

2 2

( ) ( )h k dk H f df∞ ∞

−∞ −∞=

2

0(0) ( )2N

NR h k dk∞

−∞=

1 2 32 2 200 1 2

(0) ( ) ( ) ( )2N

NR A dk A dk A dk = + − +

2 200

3(0) (3 )2 2N

NR A A N= =

Given : Probability density function,

1( )2

xf x x e−=

Variance is given by, 2 2 2[ ] [ ]X E X E Xσ = − …(i)

Mean value is given by,

1[ ] ( )2

xE X x f x dx x x e dx∞ ∞

−

−∞ −∞

= =

Since, it is an odd function, [ ] 0E X = Mean square value is given by,

2 2[ ] ( )E X x f x dx∞

−∞

=

2 2 1[ ]2

xE X x x e dx∞

−

−∞

=

2 3

0

[ ] xE X x e dx∞

−=

2 3 22[ ] 3

1 ( 1)

x xe eE X x x− −

= − − −

3 40

6 6 6( 1) ( 1)

x xe ex∞− −

+ − = − −

From equation (i), 2 26 (0) 6Xσ = − =

Hence, the variance of the random variable X is 6.

Given : ( ) ( )nn

y t X p t nT∞

=−∞

= − − φ

where ( ) ( ) ( ), ( )p t u t u t T u t= − − is the unit step function and φ is an independent random variable with uniform distribution in [ ]0, T .

[ ] [ ]1 1 0.5n nP X P X= + = = − = Scan for

Video Solution

1.46 6

1.47 0.25


For a binary symmetric pulse,

2( ) 1YYR AT

τ τ = −

where, A = pulse amplitude Here, 1A = ±

33 414YY

TTR

T

= −

1 0.254

= =

Hence, the value of the autocorrelation is 0.25.

Probability of getting 3 on dice 16

=

Probability of not getting 3 on dice 1 516 6

= − =

Now, the random variable X represents the number of throws required for getting 3.

So, 11, ( 1)6

X P x= = =

5 12, ( 2)6 6

X P x= = = ×

5 5 13, ( 3)6 6 6

X P x= = = × ×

5 5 5 14, ( 4)6 6 6 6

X P x= = = × × ×

For discrete random variable, the mean value is given by,

1

[ ] ( )m

i ii

E X x P x=

=

21 5 1 5 1(1) (2) (3)6 6 6 6 6

X = + +

35 1(4) ..........6 6

+ +

16

X =

2 35 5 51 (2) (3) (4) ......6 6 6

+ + + +

Let 2 35 5 51 (2) (3) (4)

6 6 6S = + + +

…(i)

2 3 45 5 5 5 52 3 4

6 6 6 6 6S = + + +

…(ii) Subtracting equation (ii) from (i),

5 51 (2 1)6 6

S S − = + −

2 35 5(3 2) (4 3) .......6 6

+ − + − +

2 35 5 51 .......

6 6 6 6S = + + + +

1 6 3656 16

S S= = =−

The expected value of X is

36[ ] 66

X E X= = =

Hence, the expected value of X is 6.

Given : ( ) 3 ( ) 8, [ ( )] 0X t V t E V t= − = and

5( ) 4vR e− ττ =

Mean square value of ( )V t is given by,

2[ ( )] (0) 4VE V t R= =

Power of ( )X t = mean square value of ( )X t

2 2[ ( )] [(3 ( ) 8) ]P E X t E V t= = −

2[9 ( ) 64 48 ( )]P E V t V t= + −

29 [ ( )] [64] 48 [ ( )]P E V t E E V t= + −

9 4 64 48 0 36 64P = × + − × = + 100 WP =

Hence, the power in X(t) is 100.

1.48 6

1.49 100


Given :

,

( ), 0 1, 0 1( , )

0, otherwise.X Y

x y x yf x y

+ ≤ ≤ ≤ ≤=

Probability ( 1)P X Y+ ≤ is given by,

(1 )1

0 0

( 1) ( , )x

xyx y

P X Y f x y dx dy−

= =

+ ≤ =

(1 )1

0 0

( 1) ( )x

x y

P X Y x y dx dy−

= =

+ ≤ = +

11 2

0 0

( 1)2

x

x

yP X Y xy−

=

+ ≤ = +

1 2

0

(1 )( 1) (1 )2x

xP X Y x x dx=

−+ ≤ = − +

1 22

0

1 2( 1)2x

x xP X Y x x dx=

− ++ ≤ = − +

11 2 3

0 0

1( 1)2 2 2 6x

x x xP X Y dx=

+ ≤ = − = −

1 1 1( 1) 0.332 6 3

P X Y+ ≤ = − = =

Hence, the correct answer is 0.33.

Given : A wide sense stationary random process ( )X t passes through the LTI system as shown in the figure.

From the given block diagram, 0( ) ( ) ( )Y t X t X t T= − −

Autocorrelation function of output ( )Y t is given by,

[ ]( ) ( ) ( )YR E Y t Y tτ = − τ

0( ) ( ) ( )YR E X t X t Tτ = − −

0( ) ( )X t X t T− τ − − τ −

[( ) ( ) ( ) ( )YR E X t X t X tτ = − τ − 0( )X t T− τ −

0( ) ( )X t T X t− − − τ ]0 0( ) ( )X t T X t T+ − − τ −

[ ] [ ]0( ) ( ) ( ) ( ) ( )YR E X t X t E X t X t Tτ = − τ − − τ −

[ ]0( ) ( )E X t T X t− − − τ

[ ]0 0( ) ( )E X t T X t T+ − − τ −

0 0( ) ( ) ( ) ( )Y X X XR R R T R Tτ = τ − τ + − τ − ( )XR+ τ

0 0( ) 2 ( ) ( ) ( )Y X X XR R R T R Tτ = τ − τ + − τ −


Given :

1 11,

( ) 2 20, Otherwise

fH f

− < <=

(i) The mean value of output random process is given by,

[ ( )] (0) [ ( )]E Y t H E X t= [ ( )] 1 [ ( )]E Y t E X t= ×

y

y = – x1

x1dx

1

( )Y t

Delay = T0

( )X t

LTISystem

X t( ) Y t( )

( )H f

0f

1

2

1

2

1

Stopband

Stopband

Passband

1.50 0.33

1.51 (B)

1.52 (A)


[From figure, (0) 1H = ]

[ ( )] [ ( )]E Y t E X t=

Statement 1 is correct. (ii) The relationship between output and

input PSD is given by,

2( ) ( ) ( )YY XXS f H f S f=

( ) ( )YY XXS f S f= for 1 12 2

f− < <

Power of output random process is equal to area under power spectral density.

0 ( )YYP S f df∞

−∞=

1/2

0 1/2

fP e df−

−=

1/21/2 0

0 0

2[ ]21

ff eP e df

−−= =

−

1/20 2(1 ) 0.786 WP e−= − =

2 ( ) 0.786 WE Y t =

Statement 3 is incorrect. (iii) Power of input random process can be

calculated as,

( )i XXP S f df∞

−∞=

0

2f fiP e df e df

∞ ∞− −

−∞= =

02[ ] 2 W1

f

ieP

− ∞

= =−

2 ( ) 2 WE X t =

2 2[ ( )] [ ( )]E X t E Y t≠

Statement 2 is also incorrect. Hence, the correct option is (A).

Given :

( )X t U Vt= + , [ ] 0E U = , [0,2]V UDF=

1 , 0 2

( ) 20, Otherwise

Vv

f v < <=

Mean value of V is given by,

0 2[ ] 12 2

a bE V + += = =

[ ( )] [ ] [ ]E X t E U E Vt= + At 2 sect = , [ ( 2)] [ ] [2 ]E X t E U E V= = + [ ( 2)] [ ] 2 [ ]E X t E U E V= = + [ ( 2)] 0 2 1 2E X t = = + × = Hence, the mean value of the random process at

2t = is 2.

Let ' 'X be the number of attempts by the passenger. Probability of getting reservation, p = 0.4 Probability of not getting reservation, q = 1 – 0.4 = 0.6 In 1st time probability of getting reservation, p = 0.4 In 2nd time probability of getting reservation, qp = 0.6 × 0.4 In 3rd time probability of getting reservation, 2 20.6 0.4q p = × and so on. X 1 2 3 4 5 P 0.4 0.6 0.4× 20.6 0.4× 30.6 0.4× 40.6 0.4×

Scan for

Video Solution

1

2

0 2v

( )Vf v

1.53 2

1.54 2.5


Average number of attempts = Mean value of the attempts [ ] ( )i i

iE X x P x=

[ ]E X = 20.4 2 (0.6) 0.4 3 (0.6) 0.4+ × × + × × 3 44 (0.6) 0.4 5 (0.6) 0.4 .......+ × × + × × +

[ ]E X = 20.4(1 2 0.6 3 0.6+ × + × 34 0.6 ........)+ × +

[ ] 210.4

(1 0.6)E X =

−1

21

1(1 )

n

nn

∞−

=

α = − α

[ ]E X =1 2.5

0.4=

Hence, the average number of attempts is 2.5.

Given : Two sided power spectral density, ( ) 0.5NS f = Watt/Hz

Response of the filter, 2 /2( ) 0.5 th t e−=

Relationship between output and input power spectral density is given by,

2( ) ( ) ( )Y NS f H f S f= …(i)

Fourier transform of Gaussian function is given by,

2 2 2F.T. /at f ae e

a− −ππ←⎯⎯→

Therefore, Fourier transform of ( )h t is given by,

2 2 2F.T./2 20.5 0.5 2t fe e− − π←⎯⎯→ π

From equation (i), output power spectral density is given by,

2 2 22( ) 0.5 2 0.5f

YS f e− π= π

2 2 2 24 4( ) 0.125(2 )

4f f

YS f e e− π − ππ= π =

Area under power spectral density gives power of the process. Power of ( )Y t is given by,

2 24( )

4f

YP S f df e df∞ ∞ − π

−∞ −∞

π= = …(ii)

Area of 2fe−π is always equal to 1

i.e. 2

1fe df∞ −π

−∞= .

Note : ( )x t dt A∞

−∞=

( ) Ax kt dtk

∞

−∞=

2

1fe df∞ −π

−∞=

2(2 ) 1

2fe df

∞ −π π

−∞=

π

From equation (ii), 2 24 1 0.2215

4 4 82fe df

∞ − π

−∞

π π π= × = =π


Given : [ ]2 0E X Y+ =

[ ] [ ]2 0E X E Y+ = …(i)

Also, [ ]2 33E X Y+ =

[ ] [ ]2 33E X E Y+ = …(ii) By linear solution of equation (i) and equation (ii) we get,

∴ [ ] 11E X = −

f

0.5

( )NS f

Filter( )N t ( )Y t2 /2( ) 0.5 th t e

4 2 0E X E Y

2 33E X E Y

3E X 33

1.55 (B)

1.56 11


Substituting value of [ ]E X in equation (i),

[ ] [ ]2 22E Y E X= − =

∴ [ ] [ ] 11 22 11E X E Y+ = − + =

Hence, the value of [ ] [ ]E X E Y+ = 11.

Given : Random process ( )Y t as

( ) ( )* ( ) ( )Y t h t X t Z t= +

Let, ( )* ( ) '( )h t X t X t=

( ) '( ) ( )Y t X t Z t= +

Given ( )X t and ( )Z t are uncorrelated.

We know that, total area under PSD gives the

power of the process.

Autocorrelation of ( )Y t is given as

( ) [ ( ) ( )]yR E Y t Y tτ = ⋅ + τ

( ) [ '( ) ( ) '( ) ( )]yR E X t Z t X t Z tτ = + + τ + + τ

( ) [ '( ) '( )] [ '( ) ( )]YR E X t X t E X t Z tτ = + τ + + τ[ ( ) '( )] [ ( ) ( )]E Z t X t E Z t Z t+ + τ + + τ

' ' '( ) ( ) ( ) ( ) ( )Y X X Z ZX ZR R R R Rτ = τ + τ + τ + τ ...(i)

Where, ' ( )X ZR τ and ' ( )ZXR τ are correlation of

'( )X t and ( )Z t .

From Wiener-Khinchin relationship,

autocorrelation function and PSD are Fourier

transform pair.

Therefore, taking Fourier transform pair of

equation (i),

' ' '( ) ( ) ( ) ( ) ( )Y X X Z ZX ZS f S f S f S f S f= + + +

…(ii)

Given ( )X t and ( )Z t are uncorrelated and have

constant means,

' '( ) ( ) ' ( )X Z ZXS S X Zω = ω = δ ω …(iii)

Where, 'X an Z are means of 'X and Z

Given ( )X t is white noise with PSD

( ) 5W / HzXS f =

Impulse response ( )h t whose system response

is

0.5 ; 5 5

( )0 ; otherwise

fH f

− ≤ ≤=

PSD of ( )Z t is

Now PSD of ' ( )XS f can be drawn as

2' ( ) | ( ) | ( )X XS f H f S f=

Now taking Inverse Fourier transform of

' ( )XS f

' ( ) 12.5sin (10 )XR cτ = τ

From property of ACF,

2'lim ( ) ( ')XR X

τ→∞τ =

'lim ( ) lim[12.5sinc(10 )]XRτ→∞ τ→∞

τ = τ

2( ') 0X = limsin ( ) 0cτ→∞

τ =Θ

' 0X =

5 5

( ) (W/ Hz)zS f1

(Hz)f

5

( )xS f( )X t

f0

0 55

( )H f

f

'( )X t

0 55

' ( )XS f

f

1.250.5

1.57 17.5


Putting 'X in equation (iii),

' '( ) ( ) 0X Z ZXS f S f= =

Now equation (ii) becomes

'( ) ( ) ( )Y X ZS f S f S f= +

Power spectral density of is given as

' ( ) ( )X ZS f S f+ as shown in figure,

Power in ( )Y t is given as area under the power spectral density function. Area under the above figure

1(10 1.25) 1 10 17.52YP = × + × × =

Hence, the power in ( )y t is 17.5 W.

Given : ( ) ( )Y W t t dt∞

−∞

= φ

where 1, 5 7

( )0, otherwise

tt

≤ ≤φ =

( ) 3 W/HzWS f =

As ( )tφ exists only for 5 7t≤ ≤ , so the product of ( )W t and ( )tφ will also exist for this duration only.

∴ 7

5

( )Y W t dt=

Where, ( )W t = white Gaussian Noise with PSD

3 W/Hz2η = .

So Y represents a white Gaussian noise truncated in interval 5 to 7. As mean of WGN is zero, so the power will be equal to the variance of Y. Variance of WGN truncated in interval 1T to

1T T+ ,

Variance 1

1

( )T T

T

W t dt+

2

Tη= ×

Variance 5 2

5

( )W t dt+

= Variance [ ] 3 2 6Y = × =

Hence, the variance of Y is 6.

Given : 2 6Y X= − X is uniform probability density function in the interval [ ]2,10− . The probability function of X is shown below,

62

YX +=

minX for Y less than 7 is 5 because X has to be greater or equal to 5.

maxmax

6 7 6 6.52 2

YX + += = =

( 7 | 5)P Y X≤ ≥ =

0 55−

' ( )XS f

f

1.25 +

5− 0 5

1

f

( )zS f

5− 0 5

2.25

f

' ( ) ( )X ZS f S f+

1.25=

5 7

1

t

( )tφ

( )Xf x

2− 10x

1 /12

1.58 6

1.59 0.3


( 7 5) ( 6.5 5)( 5) ( 5)

P Y X P X XP X P X≤ ∩ ≥ ≤ ∩ ≥=

≥ ≥

… (i) (5 6.5)( 7 | 5)

( 5)P XP Y X

P X≤ ≤≤ ≥ =

≥

From property of PDF,

[ ]2

1

1 2 ( )x

Xx

P x X x f x dx≤ ≤ =

[ ]6.5

5

1 1.55 6.512 12

P X dx≤ ≤ = =

5

( 5) ( )XP X f x dx∞

≥ =

10

5

1 5( 5)12 12

P X dx≥ = =

From equation (i), 1.5 /12( 7 | 5) 0.35 /12

P Y X≤ ≥ = =

Hence, the conditional probability is 0.3.

( )Xf x

2− 5 6.5 10x

1/ 12

. One Mark .

1.1 Which of the following options is the closest in meaning to the word below?

Circuitous (A) Cyclic (B) Indirect (C) Confusing (D) Crooked 1.2 The question below consists of a pair of

related words followed by four pairs of words. Select the pair that best expresses the relation in the original pair.

Unemployed : Worker (A) Fallow : Land (B) Unaware : Sleeper (C) Wit : Jester (D) Renovated : House 1.3 Choose the most appropriate word from

the options given below to complete the following sentence :

If we manage to ________ our natural resources, we would leave a better planet for our children.

(A) uphold (B) restrain (C) cherish (D) conserve 1.4 Choose the most appropriate word from


His rather casual remarks on politics _______ his lack of seriousness about the subject.

(A) masked (B) belied (C) betrayed (D) suppressed . Two Marks .

1.5 Modern warfare has changed from large scale clashes of armies to suppression of civilian populations. Chemical agents that do their work silently appear to be suited to such warfare; and regretfully, there exist people in military establishments who think that chemical agents are useful tools for their cause.

Which of the following statements best sums up the meaning of the above passage?

(A) Modern warfare has resulted in civil strife.

(B) Chemical agents are useful in modern warfare.

(C) Use of chemical agents in warfare would be undesirable.

(D) People in military establishments like to use chemical agents in war.

. One Mark .

1.6 Choose the word from the options given below that is most nearly opposite in meaning to the given word:

Frequency (A) Periodicity (B) Rarity (C) Gradualness (D) Persistency

Verbal Ability12010 IIT Guwahati

2011 IIT Madras


1.7 Choose the most appropriate word from the options given below to complete the following sentence :

Under ethical guidelines recently adopted by the Indian Medical Association, human genes are to be manipulated only to correct diseases for which ________ treatments are unsatisfactory.

(A) similar (B) most (C) uncommon (D) available 1.8 The question below consists of a pair of

related words followed by four pairs of words. Select the pair that best expresses the relation in the original pair :

Gladiator : Arena (A) Dancer : Stage (B) Commuter : Train (C) Teacher : Classroom (D) Lawyer : Courtroom 1.9 Choose the most appropriate word from


It was her view that the country’s problems had been _______ by foreign technocrats, so that to invite them to come back would be counter-productive.

(A) identified (B) ascertained (C) exacerbated (D) analysed . Two Marks .

1.10 The horse has played a little known but very important role in the field of medicine. Horses were injected with toxins of diseases until their blood built up immunities. Then a serum was made from their blood. Serums to fight with diphtheria and tetanus were developed this way.

It can be inferred from the passage, that horses were

(A) Given immunity to diseases. (B) Generally quite immune to diseases. (C) Given medicines to fight toxins. (D) Given diphtheria and tetanus serums.

. One Mark .

1.11 Choose the most appropriate alternative from the options given below to complete the following sentence :

If the tired soldier wanted to lie down, he _____ the mattress out on the balcony.

(A) should take (B) shall take (C) should have taken (D) will have taken 1.12 Choose the most appropriate word from


Given the seriousness of the situation that he had to face, his _____was impressive.

(A) beggary (B) nomenclature (C) jealousy (D) nonchalance 1.13 Which one of the following options is

the closest in meaning to the word given below?

Latitude (A) Eligibility (B) Freedom (C) Coercion (D) Meticulousness 1.14 One of the parts (A, B, C, D) in the

sentence given below contains an ERROR. Which one of the following is INCORRECT?

2012 IIT Delhi

5.3GATE ACADEMY ® General Aptitude : Verbal Ability

I requested that he should be given the driving test today instead of tomorrow.

(A) requested that (B) should be given (C) the driving test (D) instead of tomorrow . Two Marks .

1.15 One of the legacies of the Roman legions was discipline. In the legions, military law prevailed and discipline was brutal. Discipline on the battlefield kept units obedient, intact and fighting, even when the odds and conditions were against them.

Which one of the following statements best sums up the meaning of the above passage?

(A) Thorough regimentation was the main reason for the efficiency of the Roman legions even in adverse circumstances.

(B) The legions were treated inhumanly as if the men were animals.

(C) Discipline was the armies inheritance from their seniors.

(D) The harsh discipline to which the legions were subjected to led to the odds and conditions being against them.

. One Mark .

1.16 Choose the grammatically CORRECT sentence :

(A) Two and two add four . (B) Two and two become four. (C) Two and two are four. (D) Two and two make four.

1.17 Statement : You can always give me a ring whenever you need.

Which one of the following is the best inference from the above statement?

(A) Because I have a nice caller tune. (B) Because I have a better telephone

facility. (C) Because a friend in need is a friend

indeed. (D) Because you need not pay towards

the telephone bills when you give me a ring.

1.18 Complete the sentence : Dare ________ mistakes (A) commit (B) to commit (C) committed (D) committing 1.19 They were requested not to quarrel

with others. Which one of the following options is

the closest in meaning to the word quarrel?

(A) Make out (B) Call out (C) Dig out (D) Fall out

. Two Marks .

1.20 Statement : There were different streams of freedom movements in colonial India carried out by the moderates, liberals, radicals, socialists, and so on.

Which one of the following is the best inference from the above statement?

(A) The emergence of nationalism in colonial India led to our Independence.

(B) Nationalism in India emerged in the context of colonialism.

(C) Nationalism in India is homogeneous. (D) Nationalism in India is heterogeneous.

2013 IIT Bombay


. One Mark : Set - 01 .

1.21 Choose the most appropriate phrase from the options given below to complete the following sentence.

The aircraft ______ take off as soon as its flight plan was filed.

(A) is allowed to (B) will be allowed to (C) was allowed to (D) has been allowed to 1.22 Read the statements : All women are entrepreneurs. Some women are doctors. Which of the following conclusions can

be logically inferred from the above statements?

(A) All women are doctors. (B) All doctors are entrepreneurs. (C) All entrepreneurs are women. (D) Some entrepreneurs are doctors. 1.23 Choose the most appropriate word from

the options given below to complete the following sentence.

Many ancient cultures attributed disease to supernatural causes. However, modern science has largely helped ________ such notions.

(A) impel (B) dispel (C) propel (D) repel . One Mark : Set - 02 .

1.24 Choose the most appropriate word from the options given below to complete the following sentence.

Communication and interpersonal skills are _______ important in their own ways.

(A) each (B) both (C) all (D) either

1.25 Which of the options given below best completes the following sentence?

She will feel much better if she ______.

(A) will get some rest

(B) gets some rest

(C) will be getting some rest

(D) is getting some rest

1.26 Choose the most appropriate pair of words from the options given below to complete the following sentence.

She could not _____ the thought of ______ the election to her bitter rival.

(A) bear, loosing (B) bare, loosing

(C) bear, losing (D) bare, losing


1.27 “India is a country of rich heritage and cultural diversity.”

Which one of the following facts best supports the claim made in the above sentence?

(A) India is a union of 28 states and 7 union territories.

(B) India has a population of over 1.1 billion.

(C) India is home to 22 official languages and thousands of dialects.

(D) The Indian cricket team draws players from over ten states.

1.28 ‘Advice’ is ________.

(A) a verb.

(B) a noun.

(C) an adjective.

(D) both a verb and a noun.

2014 IIT Kharagpur


. Two Marks : Set - 03 .

1.29 The multi-level hierarchical pie chart shows the population of animals in a reserve forest. The correct conclusions from this information are :

(i) Butterflies are birds. (ii) There are more tigers in this forest

than red ants. (iii)All reptiles in this forest are either

snakes or crocodiles. (iv) Elephants are the largest mammals

in this forest. (A) (i) and (ii) only. (B) (i),(ii), (iii) and (iv). (C) (i),(iii) and (iv) only. (D) (i), (ii) and (iii) only.


1.30 Which of the following option is the closest in meaning to the word underlined in the sentence below?

In a democracy, everybody has the freedom to disagree with the government.

(A) dissent (B) descent (C) decent (D) decadent

1.31 After the discussion, Tom said to me, “Please revert!”. He expects me to_______.

(A) retract (B) get back to him (C) move in reverse (D) retreat 1.32 While receiving the award, the scientist

said, “I feel vindicated”. Which of the following is closest in meaning to the word “Vindicated”?

(A) Punished (B) Substantiated (C) Appreciated (D) Chastened . Two Marks : Set - 04 . 1.33 If “KCLFTSB” stands for “best of luck”

and “SHSWDG” stands for “good wishes”, which of the following indicates “ace the exam”?

(A) MCHTX (B) MXHTC (C) XMHCT (D) XMHTC

. One Mark : Set - 01 . 1.34 Choose the appropriate word/phrase, out

of the four options given below, to complete the following sentence :

Frogs ______. (A) croak (B) roar (C) hiss (D) patter 1.35 Choose the word most similar meaning

to the given word : Educe (A) Exert (B) Educate (C) Extract (D) Extend 1.36 Choose the most appropriate word from


The principal presented the chief guest with a ____, as token of appreciation.

(A) momento (B) memento (C) momentum (D) moment

Tiger

Mammal

Insect

BirdMoth

Hawk

Butterfly

Dro

ngo

Bulbul

Crocodile

Snake

Leopard

Elephant

Beetle

Red-ant

Honey

-bee

Reptile

2015 IIT Kanpur



1.37 The following question presents a sentence, part of which is underlined. Beneath the sentence you find four ways of phrasing the underlined part. Following the requirements of the standard written English, select the answer that produces the most effective sentence.

Tuberculosis, together with its effects, ranks one of the leading causes of death in India.

(A) ranks as one of the leading causes of death.

(B) rank as one of the leading causes of death.

(C) has the rank of one of the leading causes of death.

(D) are one of the leading causes of death.

1.38 Humpty Dumpty sits on a wall every day while having lunch. The wall sometimes breaks. A person sitting on the wall falls if the wall breaks.

Which one of the statements below is logically valid and can be inferred from the above sentences?

(A) Humpty Dumpty always falls while having lunch.

(B) Humpty Dumpty does not fall sometimes while having lunch.

(C) Humpty Dumpty never falls during dinner.

(D) When Humpty Dumpty does not sit on the wall, the wall does not break.

1.39 Read the following paragraph and choose the correct statement.

Climate change has reduced human security and threatened human well-being. An ignored reality of human progress is that human security largely depends upon environmental security. But on the contrary, human progress seems contradictory to environmental security. To keep up both at the required level is a challenge to be addressed by one and all. One of the ways to curb the climate change may be suitable scientific innovations, while the other may be the Gandhian perspective on small scale progress with focus on sustainability.

(A) Human progress and security are positively associated with environmental security.

(B) Human progress is contradictory to environmental security.

(C) Human security is contradictory to environmental security.

(D) Human progress depends upon environmental security.


1.40 Choose the word most similar in meaning to the given word :

Awkward (A) Inept (B) Graceful (C) Suitable (D) Dreadful 1.41 Choose the appropriate word/phase, out

of the four options given below, to complete the following sentence :

Dhoni, as well as the other team members of the Indian team ______ present on the occasion.

(A) were (B) was (C) has (D) have


1.42 What is the adverb for the given word below?

Misogynous (A) Misogynousness (B) Misogynity (C) Misogynously (D) Misogynous


1.43 Given below are two statements followed by two conclusions. Assuming these statements to be true, decide which one logically follows :

Statements : I. All film star are playback singers. II. All film directors are film stars. Conclusions : I. All film directors are playback

singers. II. Some film stars are film directors. (A) Only conclusion I follows. (B) Only conclusion I nor II follows. (C) Neither conclusion I nor II follows.

(D) Both conclusions I and II follow. 1.44 In the following sentence certain parts

are underlined and marked P, Q and R. One of the parts may contain certain error or may not be acceptable in standard written communication. Select the part containing an ERROR. Choose D as your answer if there is NO ERROR.

The student corrected all the errorsP

that

the instructor markedQ

on the

answer bookR

.

(A) P (B) Q (C) R (D) NO ERROR

1.45 Lamenting the gradual sidelining of the arts in school curricula, a group of prominent artists wrote to the Chief Minister last year, asking him to allocate more funds to support arts education in schools. However, no such increase has been announced in this year’s Budget. The artists expressed their deep anguish at their request not being approved, but many of them remain optimistic about funding in the future.

Which of the statement(s) below is/are logically valid and can be inferred from the above statements?

(i) The artists expected funding for the arts to increase this year.

(ii) The Chief Minister was receptive to the idea of increasing funding for the arts.

(iii)The Chief Minister is a prominent artists.

(iv) Schools are giving less importance to arts education nowadays.

(A) (iii) and (iv) (B) (i) and (iv) (C) (i), (ii) and (iv) (D) (i) and (iii) . One Mark : Set - 03 .

1.46 Choose the correct verb to fill in the blank below :

Let us ______. (A) introvert (B) alternate (C) atheist (D) altruist 1.47 Choose the most appropriate word from


If the athlete had wanted to come first in the race, he ______ several hours every day.

(A) should practice (B) should have practiced (C) practiced (D) should be practicing


1.48 Find the missing sequence in the letter series below :

A, CD, GHI, ?, UVWXY (A) LMN (B) MNO (C) MNOP (D) NOPQ 1.49 Choose the most suitable one word

substitute for the following expression : Connotation of a road or way (A) Pertinacious (B) Viaticum (C) Clandestine (D) Ravenous


1.50 In the following question, the first and the last sentence of the passage are in order and numbered 1 and 6. The rest of the passage is split into 4 parts and numbered as 2, 3, 4 and 5. These 4 parts are not arranged in proper order. Read the sentences and arrange them in a logical sequence to make a passage and choose the correct sequence from the given options.

1. On Diwali, the family rises early in the morning.

2. The whole family, including the young and the old enjoy doing this.

3. Children let off fireworks later in the night with their friends.

4. At sunset, the lamps are lit and the family performs various rituals.

5. Father, mother and children visit relatives and exchange gifts and sweets.

6. Houses looks so pretty with lighted lamps all around.

(A) 2, 5, 3, 4 (B) 5, 2, 4, 3 (C) 3, 5, 4, 2 (D) 4, 5, 2, 3

1.51 Ram and Shyam shared a secret and promised to each other that it would remain between them. Ram express himself in one of the following ways as given in the choices below. Identify the correct way as per standard English.

(A) It would remain between you and me.

(B) It would remain between I and you. (C) It would remain between you and I. (D) It would remain with me.


1.52 Which of the following is CORRECT with respect to grammar and usage?

Mount Everest is ________. (A) the highest peak in the world. (B) highest peak in the world. (C) one of highest peak in the world. (D) one of the highest peak in the world. 1.53 The policeman asked the victim of a

theft, “What did you _______?” (A) loose (B) lose (C) loss (D) louse 1.54 Despite the new medicine’s ________

in treating diabetes, it is not _______ widely.

(A) effectiveness –– prescribed (B) availability –– used (C) prescription –– available (D) acceptance –– proscribed . Two Marks : Set - 01 .

1.55 In a world filled with uncertainty, he was glad to have many good friends. He has always assisted them in times of need and was confident that they would reciprocate. However, the events of the last week proved him wrong.

2016 IISc Bangalore


Which of the following inference(s) is/are logically valid and can be inferred from the above passage?

(i) His friends were always asking him to help them.

(ii) He felt that when in need of help, his friends would let him down.

(iii) He was sure that his friends would help him when in need.

(iv) His friends did not help him last week.

(A) (i) and (ii) (B) (iii) and (iv) (C) (iii) only (D) (iv) only 1.56 Leela is older than her cousin

Pavithra. Pavithra's brother Shiva is older than Leela. When Pavithra and Shiva are visiting Leela, all three like to play chess. Pavithra wins more often than Leela does.

Which one of the following statements must be TRUE based on the above?

(A) When Shiva plays chess with Leela and Pavithra, he often loses.

(B) Leela is the oldest of the three. (C) Shiva is a better chess player than

Pavithra. (D) Pavithra is the youngest of the three.


1.57 Based on the given statements, select the appropriate option with respect to grammar and usage.

Statements : (i) The height of Mr. X is 6 feet. (ii) The height of Mr. Y is 5 feet. (A) Mr. X is longer than Mr. Y. (B) Mr. X is more elongated than Mr.Y. (C) Mr. X is taller than Mr. Y. (D) Mr. X is lengthier than Mr. Y.

1.58 The students _______ the teacher on teachers’ day for twenty years of dedicated teaching.

(A) facilitated (B) felicitated (C) fantasized (D) facillitated 1.59 After India’s cricket world cup victory

in 1985, Shrotria who was playing both tennis and cricket till then, decided to concentrate only on cricket. And the rest is history.

What does the underlined phrase mean in this context?

(A) history will rest in peace (B) rest is recorded in history books (C) rest is well known (D) rest is archaic . Two Marks : Set - 02 .

1.60 Social science disciplines were in existence in an amorphous form until the colonial period when they were institutionalized. In varying degrees, they were intended to further the colonial interest. In the time of globalization and the economic rise of postcolonial countries like India, conventional ways of knowledge production have become obsolete.

Which of the following can be logically inferred from the above statements?

(i) Social science disciplines have becomes obsolete.

(ii) Social science disciplines had a pre-colonial origin.

(iii) Social science disciplines always promote colonialism.

(iv) Social science must maintain disciplinary boundaries.

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



1.61 An apple costs Rs. 10. An onion costs Rs. 8.

Select the most suitable sentence with respect to grammar and usage.

(A) The price of an apple is greater than an onion.

(B) The price of an apple is more than onion.

(C) The price of an apple is greater than that of an onion.

(D) Apples are more costlier than onions.

1.62 The Buddha said, “Holding on to anger is like grasping a hot coal with the intent of throwing it at someone else; you are the one who gets burnt.”

Select the word below which is closest in meaning to the word underlined above.

(A) burning (B) igniting (C) clutching (D) flinging 1.63 M has a son Q and a daughter R. He has

no other children. E is the mother of P and daughter-in-law of M. How is P related to M?

(A) P is the son-in-law of M. (B) P is the grandchild of M. (C) P is the daughter-in law of M. (D) P is the grandfather of M. . Two Marks : Set - 03 .

1.64 The overwhelming number of people infected with rabies in India has been flagged by the World Health Organization as a source of concern. It is estimated that inoculating 70% of pets and stray dogs against rabies can lead to a significant reduction in the number of people infected with rabies.

Which of the following can be logically inferred from the above sentences?

(A) The number of people in India infected with rabies is high.

(B) The number of people in other parts of the world who are infected with rabies is low.

(C) Rabies can be eradicated in India by vaccinating 70% of stray dogs.

(D) Stray dogs are the main source of rabies worldwide.


1.65 Some tables are shelves. Some shelves are chairs. All chairs are benches. Which of the following conclusions can be deduced from the preceding sentences?

(i) At least one bench is a table. (ii) At least one shelf is a bench. (iii)At least one chair is a table. (iv) All benches are chairs. (A) Only (i) (B) Only (ii) (C) Only (ii) and (iii) (D) Only (iv) 1.66 I ___________ made arrangements

had I __________ informed earlier. (A) could have, been (B) would have, being (C) had, have (D) had been, been 1.67 She has a sharp tongue and it can

occasionally turn (A) hurtful (B) left (C) methodical (D) vital

2017 IIT Roorkee



1.68 “If you are looking for a history of India, or for an account of the rise and fall of the British Raj, or for the reason of the cleaving of the subcontinent into two mutually antagonistic parts and the effects this mutilation will have in the respective sections, and ultimately on Asia, you will not find it in these pages, for though I have spent a lifetime in the country. I lived too near the seat of events, and was too intimately associated with the actors, to get the perspective needed for the impartial recording of these matters”.

Here, the word ‘antagonistic’ is closest in meaning to

(A) impartial (B) argumentative (C) separated (D) hostile 1.69 S, T, U, V, W, X, Y and Z are seated

around a circular table. T’s neighbors are Y and V. Z is seated third to the left of T and second to the right of S. U’s neighbors are S and Y ; and T and W are not seated opposite each other. Who is third to the left of V ?

(A) X (B) W (C) U (D) T . One Mark : Set - 02 .

1.70 The ninth and the tenth of this month are Monday and Tuesday _______.

(A) figuratively (B) retrospectively (C) respectively (D) rightfully 1.71 It is ____ to read this year’s textbook

_____ the last year’s. (A) easier, than (B) most easy, than (C) easier, from (D) easiest, from


1.72 Each of P, Q, R, S, W, X, Y and Z has been married at most once. X and Y married and have two children P and Q, Z is grandfather of the daughter ‘S’ of P. Further Z and W married and are parents of R. Which one of the following must necessarily be FALSE?

(A) X is the mother-in-law to R (B) P and R are not married to each

other (C) P is son of X and Y (D) Q cannot be married to R 1.73 “If you are looking for a history of

India, or for an account of the rise and fall of the British Raj, or for the reason of the cleaving of the subcontinent into two mutually antagonistic parts and the effects this mutilation will have in the respective sections, and ultimately on Asia, you will not find it in these pages for though I have spent a lifetime in the country, I lived too near the seat of events, and was too intimately associated with the actors, to get the perspective needed for the impartial recording of these matters.”

Which of the following statement best reflects the author’s opinion?

(A) An intimate association does not allow for the necessary perspective.

(B) Matters are recorded with an impartial perspective.

(C) An intimate association offers an impartial perspective.

(D) Actors are typically associated with the impartial recording of matters.


. One Mark .

1.74 “Even though there is a vast scope for its __________, tourism has remained a/an _________ area.”

The words that best fill the blanks in the above sentence are

(A) Improvement, neglected (B) Rejection, approved (C) Fame, glum (D) Interest, disinterested 1.75 “By giving him the last __________ of

the cake, you will ensure lasting ________ in our house today.”

The words that best fill the blanks in the above sentence are

(A) peas, piece (B) piece, peace (C) peace , piece (D) peace, peas . Two Marks .

1.76 A coastal region with unparalleled beauty is home to many species of animals. It is dotted with coral reefs and unspoilt white sandy beaches. It has remained inaccessible to tourists due to poor connectivity and lack of accommodation. A company has spotted the opportunity and is planning to develop a luxury resort with helicopter service to the nearest major city airport. Environmentalists are upset that this would lead to the region becoming crowded and polluted like any other major beach resorts.

Which one of the following statements can be logically inferred from the information given in the above paragraph?

(A) The culture and tradition of the local people will be influenced by the tourists.

(B) The region will become crowded and polluted due to tourism.

(C) The coral reefs are on the decline and could soon vanish.

(D) Helicopter connectivity would lead to an increase in tourists coming to the region.

1.77 The Cricket Board has long recognized John's potential as a leader of the team. However, his on-field temper has always been a matter of concern for them since his junior days. While this aggression has filled stadia with die-hard fans, it has taken a toll on his own batting. Until recently, it appeared that he found it difficult to convert his aggression into big scores. Over the past three seasons though, that picture of John has been replaced by a cerebral, calculative and successful batsman-captain. After many years, it appears that the team has finally found a complete captain.

Which of the following statements can be logically inferred from the above paragraph?

(i) Even as a junior cricketer. John was considered a good captain.

(ii) Finding a complete captain is a challenge.

(iii)Fans and the Cricket Board have differing views on what they want in a captain

(iv) Over the past three seasons John has accumulated big scores.

(A) (i) (ii) and (iii) only (B) (iii) and (iv) only (C) (ii) and (iv) only (D) (i) (ii), (iii) and (iv)

2018 IIT Guwahati


. One Mark .

1.78 Five different books (P, Q, R, S, T) are to be arranged on a shelf. The books R and S are to be arranged first and second respectively from right side of the shelf. The number of different orders in which P, Q and T may be arranged is ________.

(A) 6 (B) 2 (C) 12 (D) 120 1.79 The boat arrived _______ dawn. (A) in (B) under (C) on (D) at 1.80 When he did not come home, she ___

him lying dead on the roadside somewhere.

(A) notice (B) looked (C) concluded (D) pictured 1.81 The strategies that the company

_______ to sell its products ________ house-to-house marketing.

(A) use, includes (B) uses, including (C) uses, include (D) used, includes . Two Marks .

1.82 “Indian history was written by British historians – extremely well documented and researched, but not always impartial. History had to serve its purpose : Everything was made subservient to the glory of the Union Jack. Latter-day Indian scholars presented a contrary picture.”

From the text above, we can infer that : Indian history written by British

Historians _______.

(A) was not well documented and researched and was sometimes biased

(B) was well documented and not researched but was always biased

(C) was not well documented and researched and was always biased

(D) was well documented and researched but was sometimes biased

1.83 Four people are standing in a line facing you. They are Rahul, Mathew, Seema and Lohit. One is engineer, one is a doctor, one a teacher and another a dancer. You are told that :

1. Mathew is not standing next to Seema.

2. There are two people standing between Lohit and the engineer.

3. Rahul is not a doctor. 4. The teacher and the dancer are

standing next to each other. 5. Seema is turning to her right to

speak to the doctor standing next to her.

Who among them is an Engineer ? (A) Lohit (B) Seema (C) Rahul (D) Mathew 1.84 Five people P, Q, R, S and T work in a

bank. P and Q don’t like each other but have to share an office till T gets a promotion and moves to the big office next to the garden. R, who is currently sharing an office with T wants to move to the adjacent office with S, the handsome new intern.

Given the floor plan, what is the current location of Q, R and T ? [O = office, WR = Wash room]

2019 IIT Madras


(A)

(B)

(C)

(D)

. One Mark .

1.85 He was not only accused of theft _____________ of conspiracy.

(A) but also (B) but even (C) rather (D) rather than

1.86 The Canadian constitution requires that equal importance be given to English and French. Last year, Air Canada lost a lawsuit and had to pay a six-figure fine to a French speaking couple after they filed complaints about formal in-flight announcements in English lasting 15 seconds, as opposed to informal 5 second messages in French.

The French - speaking couple were upset at _____.

(A) the English announcements being longer than French ones.

(B) equal importance being given to English and French.

(C) the English announcements being clearer than the French ones.

(D) the in-flight announcements being made in English.

1.87 Select the word that fits the analogy : Explicit : Implicit : : Express : _____. (A) Repress (B) Suppress (C) Compress (D) Impress 1.88 The untimely loss of life is a cause of

serious global concern as thousands of people get killed ____ accidents every year while many other die ____ diseases like cardio vascular diseases, cancer, etc.

(A) in, of (B) during, from (C) from, of (D) from, from

. Two Marks .

1.89 The global financial crisis in 2008 is considered to be the most serious world-wide financial crisis, which started with the sub-prime lending crisis in USA in 2007. The sub-prime lending crisis led to the banking crisis in 2008 with the

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collapse of Lehman Brothers in 2008. The sub-prime lending refers to the provision of loans to those borrowers who may have difficulties in repaying loans, and it arises because of excess liquidity following the East Asian crisis. Which one of the following sequences shows the correct precedence as per the given passage?

(A) East Asian crisis → subprime lending crisis → banking crisis → global financial crisis.

(B) Banking crisis → subprime lending crisis → global financial crisis → East Asian crisis.

(C) Global financial crisis → East Asian crisis → banking crisis → subprime lending crisis.

(D) Subprime lending crisis → global financial crisis → banking crisis → East Asian crisis.


1.1 B 1.2 A 1.3 D 1.4 C 1.5 D

1.6 B 1.7 D 1.8 A 1.9 C 1.10 B

1.11 C 1.12 D 1.13 B 1.14 B 1.15 A

1.16 D 1.17 C 1.18 B 1.19 D 1.20 D

1.21 C 1.22 D 1.23 B 1.24 B 1.25 B

1.26 C 1.27 C 1.28 D 1.29 D 1.30 A

1.31 B 1.32 B 1.33 B 1.34 A 1.35 C

1.36 B 1.37 A 1.38 B 1.39 B 1.40 A

1.41 B 1.42 C 1.43 D 1.44 B 1.45 C

1.46 B 1.47 B 1.48 C 1.49 B 1.50 B

1.51 A 1.52 A 1.53 B 1.54 A 1.55 B

1.56 D 1.57 C 1.58 B 1.59 C 1.60 A

1.61 C 1.62 C 1.63 B 1.64 A 1.65 B

1.66 A 1.67 A 1.68 C 1.69 A 1.70 C

1.71 A 1.72 B 1.73 C 1.74 A 1.75 B

1.76 B 1.77 C 1.78 A 1.79 D 1.80 D

1.81 C 1.82 D 1.83 D 1.84 A 1.85 A

1.86. A 1.87 A 1.88 A 1.89 A

Circuitous : Deviating from a straight course; indirect. Cyclic : Recurring in cycle. Indirect : Not leading by straight line. Confusing : Lacking clarity. Crooked : For shapes (irregular in shape). E.g. a crooked path. Hence, the correct option is (B).

A worker who is unemployed is unproductive just as a land which is fallow is unproductive. Hence, the correct option is (A).

Uphold : Cause to remain (Not appropriate). Restrain : Keep under control (not appropriate). Cherish : Be fond of (not related). Conserve : Keep in safety and protect from harm, decay, loss or destruction (most appropriate). Hence, the correct option is (D).

Masked : Hide under a false appearance (opposite). Belied : Be in contradiction with (not appropriate). Betrayed : Reveal unintentionally (most appropriate).

Answers Verbal Ability

Explanations Verbal Ability

1.1 (B)

1.2 (A)

1.3 (D)

1.4 (C)


Suppressed : To put down by force or authority (irrelevant). Hence, the correct option is (C).

(A) Modern warfare has resulted in civil

strife : There is no direct consequence of warfare given, so it is not appropriate.

(B) Chemical agents are useful in modern warfare : Passage does not say whether chemical agents are useful or not, so it is not appropriate.

(C) Use of chemical agents in warfare would be undesirable : Given that people in military think these are useful, undesirable is wrong.

(D) People in military establishment like to use chemical agents in war : Correct choice as last statement tells that military people think that chemical agents are useful tools for their cause (work silently in warfare).


Frequency : Means occurring at frequent intervals. So opposite of it is ‘rarity’ which means infrequency of occurrence. Periodicity : The equality or state of being periodic; recurrence at regular intervals. Rarity : The quality or state of being rare; infrequency of occurrence. Gradualness : The quality of change in ascending or descending in a moderate rate. Persistency : Persistent determination; the quality of being determined to do or achieve something. Hence, the correct option is (B).

Similar : Showing resemblance in qualities, characteristics, or appearance; alike but not identical.

Most : Superlative of many, much. (i) Greatest in number. (ii) Greatest in amount, extent, or degree. Uncommon : (i) Not common; rare. (ii) Wonderful; remarkable Available : Present and ready for use; at hand; accessible. (most appropriate) Hence, the correct option is (D).

Gladiator performs his action in arena, similarly dancer’s performance required stage. Gladiator : A person, usually a professional combatant, a captive, or a slave, trained to entertain the public by engaging in mortal combat with another person or a wild animal in the ancient Roman arena. Arena : An enclosed area for the presentation of sports events and spectacles. Commuter : One that travels regularly from one place to another, as from suburb to city and back. Lawyer : One whose profession is to give legal advice and assistance to clients and represent them in court or in other legal matters. Courtroom : A room in which the proceedings of a court are held. Hence, the correct option is (A).

Identified : (i) To establish the identity of. (ii) To ascertain the origin, nature, of

definitive characteristics of. Ascertained : To discover with certainty, as through examination or experimentation. Exacerbated : To increase the severity, violence, or bitterness of ; aggravate.

1.5 (D)

1.6 (B)

1.7 (D)

1.8 (A)

1.9 (C)


Analyzed : To examine in detail in order to discover meaning, essential features, etc. Hence, the correct option is (C).

The author in the paragraph wants to show the role of horses in the field of medicines for providing the immunity to various diseases. Hence, the correct option is (B).

There are a number of different ways to use 'should have been' or 'should have done' (i.e. should have + past participle). For example : You may also hear 'should have been/done' used when you expect a confirmation that something had a particular outcome in the past. This usage would be similar to 'must have been/done' Hence, the correct option is (C).

Considering the first half of the statement we can depict that “the situation was serious / intense” but the second half suggests that “his efforts / reactions were commendable” , thus it will be a “positive or a neutral word ” but definetly not a “negative word.” Beggary : (i) Extreme poverty, penury. (ii) The state of condition of being a beggar. Nomenclature : The terminology used in particular science, art, activity, etc. Jealousy : The state or quality of being jealous. Nonchalance : The trait of remaining calm and seeming not to care, a causal lack of concern. Hence, the correct option is (D).

Latitude : Freedom from normal restraints, limitations or regulations.

Eligibility : The quality or state of being eligible. Freedom : The condition of being free of restraints. Coercion : Refer to force. Meticulousness : Extremely careful and precise. Hence, the correct option is (B).

The error is ‘he should be given’. The correct way of writing the sentence is ‘he should give’. Hence, the correct option is (B).

Here, there are three key words “legacy”, “legions” and “Discipline”. Even though the paragraph starts with “legacy” but the basic idea revolves around “Strict Discipline”, the author has mentioned that the main reason because of which the legion is “obedient, intact and fighting” is “Discipline”. In option (A), we can see that “Strict Discipline” has been featured as “Regimentation.” Hence, the correct option is (A). Key Point Legion = noun 1. A division of the Roman army, usually

comprising 3000 to 6000 soldiers. 2. A military or semi-military unit. Regimentation = noun 1. The act of regimenting or the state of

being regimented. 2. The strict discipline and enforced

uniformity characteristic of military groups.

“Two and two make four”. The word “make” means “amount to”. Two and two amount to (or make) four.

1.10 (B)

1.11 (C)

1.12 (D)

1.13 (B)

1.14 (B)

1.15 (A)

1.16 (D)


“Two and two are four” suggests that they are separate numbers. Grammatically both options (C) and (D) are used in different sense. Hence, the general usage is two and two make four. Hence, the correct option is (D).

“You can always give me a ring whenever you need.” This statement, means that whenever there is any problem or need, feel free to call me. It simply means “a friend is need is a friend indeed.” Hence, the correct option is (C).

In this sentence, preposition is used to make a relation between “Dare” and “Mistakes”. So, correct sentence is “Dare to commit mistakes”. Hence, the correct option is (C).

The closest option is fall out. Make out : To cause to exist or happen; bring about; to understand. Call out : To cause to assemble ; summon. Dig out : To remove ; unearth. Fall out : To quarrel. Hence, the correct option is (D).

In the above statement, the movement in colonial India is carried out by different persons of different philosophies. For example, some are moderates, some are liberals, etc. Therefore, different philosophies shows heterogeneous character. Hence, Nationalism in India is heterogeneous. Hence, the correct option is (D).

The sentence is in past tense. So, the correct option is “was allowed to”. Hence, the correct option is (C).

Some of the entrepreneurs are doctors, as among entrepreneur some of them are women and some of women entrepreneur are doctors. Hence, the correct option is (D).

Impel : To force somebody. Dispel : To disappear something especially an idea or notion. Propel : To push. Repel : To force back. According to the sentence, ancient cultures pointed out that disease is caused due to supernatural causes. But modern science has helped disappears such notions. Hence, the correct option is (B).

‘Each’ and ‘either’ given as answer choices cannot go because the personal pronoun used in the sentence is ‘their’. Two words have been used in the sentence i.e. communication and interpersonal skills. So, ‘both’ is correct when we are referring to it. Hence, the correct option is (B).

If two events in the future is being talked about then the sentence starting with ‘if ’ takes simple present tense and the other sentence takes simple future tense. Hence, the correct option is (B).

Bear : To carry something. Bare : Without anything. Loosing : To loosen something. Losing : To stop to retain something. Hence, the correct option is (C).

1.17 (C)

1.18 (B)

1.19 (D)

1.20 (D)

1.21 (C)

1.22 (D)

1.23 (B)

1.24 (B)

1.25 (B)

1.26 (C)


Diversity is shown in terms of different languages. Hence, the correct option is (C).

Advice is a noun. It is a suggestion or recommendation. Advice is a verb also. It means to give advice and therefore refers to the act of giving advice. Hence, the correct option is (D).

Following statements can be derived from the given diagram : (i) Butterflies are shown in birds category. (ii) The area corresponding to the tiger in

pie chart is larger than that of red ants. (iii) There are only two types of reptiles i.e.

snakes and crocodiles. (iv) It is not correct as the size of animals is

not specified in the pie chart but the numbers are specified.

Elephants are in largest number in mammals. Hence, the correct option is (D).

Dissent : To fail to agree, to differ in opinion, to cause discomfort or distress. Descent : To move downward. Decent : Satisfactory. Decadent : To decline. Hence, the correct option is (A).

‘Revert’ means to come or go back (as to a former condition, period or subject). Hence, the correct option is (B).

Vindicated : To set free, to free from allegation, confirm, to provide justification. Substantiated : means to show to be true.

Chastened : To bring down. Hence, the correct option is (B).

BEST OF LUCK B _ ST _ F L_CK → KCLFTSB GOOD WISHES G _ _D W_SH_S → SHSWDG From above two we can conclude that vowels are missing and then reversed. Hence, ACE THE EXAM _C_ TH_ _X_M And reverse is MXHTC. Hence, the correct option is (B).

The sound made by Frogs is croak. Hence, the correct option is (A).

Educe : Take out or extract. Exert : Apply a force, influence or quality. Educate : Give intellectual or moral instruction to. Extract : Remove with care or effort. Extend : Make larger or longer. Hence, the correct option is (C).

Memento : A souvenir or gift. Hence, the correct option is (B).

Tuberculosis is a singular number and the work ‘ranks’ should be followed by ‘as’. Hence, the correct option is (A).

The key word to solve this question is ‘sometimes’. Hence, the correct option is (B).

1.27 (C)

1.28 (D)

1.29 (D)

1.30 (A)

1.31 (B)

1.32 (B)

1.33 (B)

1.34 (A)

1.35 (C)

1.36 (B)

1.37 (A)

1.38 (B)


Refer the second sentence “An ignored reality of human progress is that human security largely depends upon environmental security”. From above sentence, we can conclude that the correct option is (B).

Awkward : Clumsy, unskilled, inept. Graceful : A way of moving that is smooth and attractive and that is not stiff or awkward. Suitable : Proper appropriate. Dreadful : Very bad or unpleasant. Hence, the correct option is (A).

The sentence should be in simple past tense and the verb should agree with the first subject ‘Dhoni’. Hence, the correct option is (B).

Words in options (A) and (B) are nouns and (D) is an adjective. Hence, the correct option is (C).

The possible Venn diagram drawn from the given statements is :

From this diagram, conclusions I and II logically follows. Hence, the correct option is (D).

In part ‘Q’ it should be ‘the instructor had marked’ because the first action of the past should be in past perfect tense. Hence, the correct option is (B).

A close reading of the passage would show that only statements (i), (ii) and (iv) are logically valid. Hence, the correct option is (C).

Words in option (A), (C) and (D) are nouns. “Let us” is followed by a verb. Hence, the correct option is (B).

“If the athlete had” should be followed by “should have”. Hence, the correct option is (B).

In the alphabetical sequence, the missing word is as follows

Therefore, the missing word in the sequence is ‘MNOP’. Hence, the correct option is (C).

Pertinacious : Determined. Viaticum : Provisions for a journey. Clandestine : Secret. Ravenous : Very hungry Hence, the correct option is (B).

FDFD

FS

P Play-back-singer

Film director

Film star

A CD GHI

EF JKL

UVWXY

B QRST

(+1) (+2) (+3) (+4)

MNOP

1.39 (B)

1.40 (A)

1.41 (B)

1.42 (C)

1.43 (D)

1.44 (B)

1.45 (C)

1.46 (B)

1.47 (B)

1.48 (C)

1.49 (B)


Given : As per the question, the first and the last sentence of the passage are in proper order and number 1 and 6. The above question is based on time sequence. Thus, the sentence should be organized in a chronological order clockwise. Hence, the proper order of above paragraph is,

Sentence Order On Diwali, the family rises early in the morning.

1

Father, mother and children visit relatives and exchange gifts and sweets.

5

The whole family, including the young and the old enjoy doing this.

2

At sunset, the lamps are lit and the family performs various rituals.

4

Children let off fireworks later in the night with their friends.

3

Houses looks so pretty with lighted lamps all around.

6

This arrangement makes a logical sequence (5, 2, 4, 3) in clockwise order. Hence, the correct option is (B).

‘Between’ is followed by pronoun in object form. Hence, the correct option is (A).

Before superlative article ‘the’ has to be used. ‘one of ’ the expression should take plural noun and so option (C) and (D) can’t be the answer. Hence, the correct option is (A).

‘lose’ is a verb. Hence, the correct option is (B).

‘effectiveness’ is noun and ‘prescribed’ is verb. These words are apt and befitting with the word ‘medicine.’ Hence, the correct option is (A).

The words “was confident that they would reciprocate” and “last week proved him wrong” lead to statements (iii) and (iv) as logically valid inferences. Hence, the correct option is (B).

From the given data, following relations are possible. L > P (Leela is older than Pavithra) S > L (Shiva is older than Leela) So, Pavithra is youngest. Hence, the correct option is (D).

In degrees of comparison Mr. X is taller than Mr. Y. Positive degree – tall Comparative degree – taller Superlative degree – tallest Hence, the correct option is (C).

Felicitated : To compliment upon a happy event; congratulate or Honour. Facilitated : Make easy or easier. Fantasized : Imagine something desirable. Hence, the correct option is (B).

1.50 (B)

1.51 (A)

1.52 (A)

1.53 (B)

1.54 (A)

1.55 (B)

1.56 (D)

1.57 (C)

1.58 (B)


The underlined phrase mean that rest is well known. As history is permanent documentation and well known.


“Until the colonial period” means pre-colonial origin. Other options can’t be inferred.


Based on the given sentences option (C) is the correct sentence which is in the comparative degree. Options (A) and (B) convey the wrong comparison and (D) has double comparative and so they are wrong.


The underlined word ‘grasping’ means clutching or holding something tightly.


Q and R are the son and daughter of M, E is the mother of P and daughter-in-law of M means Q and E are married couples in the family

Therefore, P is the grandchild of M. Hence, the correct option is (B).

Only option ‘A’ can be logically inferred from the information “1st line of paragraph”.


. Method 1 :

Only conclusion (ii) follows. Hence, the correct option is (B). . Method 2 : Some tables are shelves, T ∈ S …(i) Some shelves are chairs, S ∈ C …(ii) All chairs are benches, C = B …(iii) Using equation (ii) and (iii), S ∈ B Therefore, at least one shelf is a bench. Hence, the correct option is (B).

Could have, been I could have made arrangements had I been informed earlier. Use of conditional sentence based on past participle form. Hence, the correct option is (A).

Hurtful means causing pain or suffering or something that is damaging or harmful. The expression “sharp tongue” defines a bitter or critical or rude manner of speaking. Hence, the correct option is (A).

T S C

BT S

C

B

T S

B

C

1.59 (C)

1.60 (A)

1.61 (C)

1.62 (C)

1.63 (B)

1.64 (A)

1.65 (B)

1.66 (A)

1.67 (A)


Antagonistic : An opponent or enemy. Impartial : Not favouring one more than another. Argumentative : Tending to argue. Separated : Divide into parts. Hostile : Showing dislike or opposition. Hence, the correct option is (C).

Following circular seating arrangement can be drawn

Only one such arrangement can be drawn. The person on third to the left of V is X. Hence, the correct option is (A).

Figuratively : It is an adjective, means “of the nature of or involving a figure of speech”. Retrospectively : With contemplation of past situations, events. Respectively : In precisely the order given. Rightfully : Having a valid or just claim as to some property or position. Hence, the correct option is (C).

Easier, than It is easier to read this year’s textbook than the last year’s. Hence, the correct option is (A).

(i) X is mother-in-law to R. Therefore

option (A) is true. (ii) P and R can marry. Therefore option (B)

is false. (iii) Q and R cannot marry. Therefore option

(C) is true. (iv) P is son of X and Y. Therefore option

(D) is true. Hence, the correct option is (B).

It is clear from the last line : “I lived too near the seat of events, and was too intimately associated with the actors, to get the perspective needed for the impartial recording of these matters.” Hence, the correct option is (C).

(A) Improvement : Thing that represents an

advancement Neglected : Not pay proper attention (B) Rejection : The act of refusing to accept Approved : To confirm or sanction

formally (C) Fame : The state of being known or

recognized by many people because of achievements, skills, etc.

Glum : Sad Glum is a word for being depressed,

bummed out or down in the dumps.

X

T

W U

S

V Y

Z

Z

W

R P Q

X Y

S

Marry

Son or daughter

=

=

1.68 (C)

1.69 (A)

1.70 (C)

1.71 (A)

1.72 (B)

1.73 (C)

1.74 (A)


(D) Interested : Wanting to give your attention to something and discover more about it.

Disinterested : Having no personal involvement and therefore free to act fairly.

From the above options we see that options (B), (C) and (D) are not appropriate to the sentence, only option (A) is logical to the question. Hence, the correct option is (A).

Peas : A spherical green seed which is eaten as a vegetable not appropriate Piece : A limited portion or quantity of something most appropriate Peace : Freedom from disturbance most suitable Hence, the correct option is (B).

(A) Culture and tradition of local people will

be influenced by the tourists : Not exactly meant by passage.

(B) Region will become crowded and polluted by tourism : This concluded most closely to the environment safety.

(C) The coral reefs are one decline and could soon vanish : Unless we change the way we live that affect the biodiversity of nature so it is not exact conclusion of the passage.

(D) Helicopter connectivity would lead to increase in tourists coming to region : Inappropriate

Hence the correct option is (B).

(i) Even as a junior cricketer, John was

considered a good captain : It means that there is no provision to make senior cricketer as captain, so it is not appropriate.

(ii) Finding a good captain is challenging : Given that to select a good captain is difficult which concluded it is relevant to the statement.

(iii) Fans and the cricket board have different views of captain : It means different people have its own concern to think so it is also inappropriate.

(iv) Over the past three seasons, John is scoring big scores : It concluded the performance of captain, so it is relevant to the passage.


. Method 1 : Given R and S are to be fixed at first and second position from the right side respectively _ _ _ S R Now, rest of the 3 books P, Q and T can be arranged as

6 ways

P Q T S RP T Q S RQ P T S RQ T P S RT P Q S RT Q P S R

Hence, total number of ways, books can be arranged is 6. Hence, the correct option is (A). . Method 2 : As positions of R and S are fixed at first and second from right, so P, Q and T can be arranged in three possible places.

1.75 (B)

1.76 (B)

1.77 (C)

1.78 (A)


Hence, total possible arrangements 3

3 3! 6P= = = Hence, the correct option is (A).

As the sentence is intended to specify particular time instant of the arrival of boat, hence ‘at’ is the most appropriate option. Also, we use ‘at’ to specify places like ‘at the market’. Hence, the correct option is (D).

The given sentence is pointing to some king of imagination, hence the most appropriate word is ‘pictured’. Hence, the correct option is (D).

‘Company’ is a singular noun, hence it will take singular verb ‘uses’ while for “strategies”, ‘include’ is correct. Hence, the correct option is (C).

It is given at the starting of the paragraph that the Indian history written by British historians was extremely well documented and researched and later it is intended to say that it was sometimes biased. The sentence ‘but not always impartial’ implies that it was sometimes biased. Hence, the correct option is (D).

Lohit

Doctor

Seema

Teacher

Rahul

Dancer

Mathew

Engineer OR

Dancer

Teacher

According to the given statements, the positions with profession of different peoples are shown above.

Only Mathew can be the engineer and Lohit can be the doctor, Seema and Rahul are teacher and dancer but we cannot surely tell about them. Hence, the correct option is (D).

From the statements given in question we can observe that (i) P and Q don’t like each other but are

sharing the same office. (ii) R is sharing an office with T (iii) S is in an adjacent office to R and T Now from the observed data we can deduce that (a) Option (D) has all the people in separate

office, hence it is rejected. (b) Option (B) has P and Q sharing an

office but R and T are not sharing an office, hence rejected.

(c) Option (C) has P and Q sharing an office but also has R and S sharing an office which is wrong according to given data.

(d) Option (A) satisfies all the given conditions, so it is correct.


The French speaking couple were upset as they were not able to get the announcement properly in such a short span of five seconds and they filed complaint as the announcement in English lasted for 10 seconds. Hence the correct option is (A).

Given words in the first relation are antonyms, the correct antonym of express is repress, which means to control an emotion or to try to prevent it from being shown.

1.79 (D)

1.80 (D)

1.81 (C)

1.82 (D)

1.83 (D)

1.84 (A)

1.85 (A)

1.86 (A)

1.87 (A)


“Suppress” is synonym of repress but it means to stop something by using force, so it is not the correct antonym for express. Hence, the correct option is (A).

According to the given paragraph, because of East Asian crisis, sub-prime lending crisis occurred due to access liquidity and hence banking crisis taken place resulting in the global financial crisis. Hence, the correct option is (A).

1.88 (A)

1.89 (A)

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