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Initial Words

Analysis on IIT-JEE... Physics

Speciality of IIT-JEE paperThe speciality of IIT-JEE Physics paper isthat the questions are always original andunique in their context, solving them oftenrequires a combination of concepts fromdifferent chapters. The knowledge of basicconcepts of all chapters is must for gettinggood score in JEE examination. Studentsmust practice apply method i.e. application ofbasic concepts, rather than recall method i.e.try to by heart or remember the solutions ofstandard model questions.

Change of pattern

One more important aspect of JEE is changeof pattern of exam. Change of pattern ofexam is very important to select the goodtalent. In IIT-JEE 2007, there was a change inexam pattern. Assertion and Reasoning basedquestions were introduced in place of subjecttype questions (fill in the box type). Therewas no change of pattern in 2008 exam butthere were major changes in IIT-JEE 2009paper. For e.g. Subjective type questionswere introduced again. There were 8subjective questions in Paper-II. After longtime "more than one choice correct" type ofquestions with negative marking wereintroduced. This will eliminate students whodo guess work in the paper. Assertion and Reason type of questionswere removed from both the papers. "Integertype questions" were introduced and therewere no options to integer type questions.Answers had value from 0 to 9. Studentsmust be ready to face any change in pattern.

Difficulty level

The difficulty of question paper varies fromyear to year. IIT-JEE 2009 Physics paper wasrelatively easy when compared to other

subjects. In Physics Paper-I around 75% ofthe questions were easy to solve and rest 25%questions were difficult. Paper-II required alot of time and patience to solve questions.Over all the papers were 'moderately'difficult.

Marking Scheme

Marking scheme is also changing from timeto time. The integer type questions wereintroduced for the first time in 2009. Therewere no options to integer type questions.The student must read instructions beforeanswering the paper.

Electricity and Magnetism topics togetherconstitute almost one third of total marks andstudents must concentrate on these two topicsand also these topics are relatively easy whencompared to Mechanics.

The next important topics are Mechanicsand General Physics. Within Mechanics, thefavourite topics seem to be Rigid bodyrotation and Motion on a Plane. It isimportant for students to realize thatMechanics though important, takes relativelymore time for preparation. They shouldrestrict their preparation of Mechanics andconcentrate on next important topics"Modern Physics" and "Thermal Physics".These topics can be covered thoroughly in arelatively less time.

'Ray Optics' and 'Physical Optics'together contributes approximately 10 to15%, with in Optics, it has been observed thatproblems on Young's Double Slit Experiment(Physical Optics) are more frequently asked.Students should concentrate on different

aspects of Young's Double Slit experiment.The last but not the least, Mechanical

waves and Sound has accounted forapproximately 10% of total marks; studentsshould concentrate some time on wavemotion.

Dr. Ch. RamakrishnaCAMBRIDGE Certified Trainer,Dr.RK's Academy, Hyderabad.

Written by

Analysis of Previous Eight Years(2000-2007) Physics Papers

TopicApproximate percentage of marks

Mechanics 25Electromagnetism 27Heat andThermodynamics 13Modern Physics 12Optics 10Waves 8Properties of Matter 5

Difficulty Level Analysis for IIT-JEE 2009Physics Paper - I

Paper - II:

Chapter wise Analysis of 2009 PaperUnit wise Distribution for Physics Paper-I

Physics Paper - II

1

2

3

0%

5%

10%

15%

20%

25%

30%

35%

40%

45%

1 2 3

Series1

0%

10%

20%

30%

40%

50%

60%

1 2 3

Optics 9%

Mechanics 24%

Modern Physics 24%

Thermodynamics 5%

Electricity & Magnetism 38%

Optics 10%

Property of matters 10%

Modern Physics 6%

Heat & Thermodynamics

13% Electricity & Magnetism 25%

Mechanics 36%

Unit wise Analysis (% of marks)

Sl.No. Units Marks Percentage1. Mechanics & Properties of Matter 73 45.6252. Electricity & Magnetism 37 23.1253. Optics 15 9.3754. Modern Physics 27 16.8755. Others 8 5

Chapter wise Analysis (No.of Questions) of IIT-JEE 2009 Physics Paper

Sl. Chapter Paper-I Paper - II Total maximum

3 4 8 3 4 8 % of marksmarks marks marks marks marks marks marks

1. General Physics & Kinematics - 1 - - 1 - 8 5

2. Laws of Motion - - - 1 - - 3 1.8753. Work, Power and

Energy - - - - 1 - 4 2.54. Centre of Mass,

Linear Momentum,Collision 2 - - - 1 - 10 6.25

5. Rotation 1 - - - 1 - 7 4.3756. Gravitation, SHM 1 - - 2 - - 9 5.6257. Properties of Matter - - - - 2 - 8 58. Waves - - - - 1 - 4 2.59. Heat &

Thermodynamics - 1 - - 2 - 12 7.510. Optics 1 - - - - 1 11 6.87511. Current Electricity - 1 - - - - 4 2.512. Electrostatics 2 - - - 2 - 14 8.7513. Magnetics - - - - 1 - 4 2.514. Electromagnetic

Induction 1 - - - 1 - 7 4.37515. Modern Physics - 6 - 1 - - 27 16.87516. Mixed type

(Multiple topics) - 1 2 - - 1 28 17.5

IIT-JEE 2006, 2007, 2008 Chapter wise Analysis

(No. of questions) Physics

Chapter 2008 '07 '061. General Physics 1 3 22. Kinematics - - -3. Laws of Motion 1 2 34. Work, Power and

Energy 4 1 -5. Centre of mass - 3 -6. Rotation 4 4 37. Gravitation 2 - 18. SHM 3 1 19. Properties of Matter 5 1 310. Waves 3 3 411. Heat and

Thermodynamics 3 5 312. Optics 4 6 413. Current electricity 2 2 114. Electrostatics 6 4 115. Magnetics 2 3 216. Electromagnetic

induction - 1 717. Modern Physics 6 5 5

easy medium difficult

easy medium difficult


3

Biot - Savart law

Torque on a current loop

qBv

m=

v B

F qv B= ×

F qvBsin= θ

qBv

m>

Electricity and Magnetism

Biot-Savart Law: This law is used to findthe magnetic field at a point whoseposition vector with respect to a currentelement is . According to this law:

The total field where is unit vector in thedirection of vector .Magnetic field due to current carryingconductor:

The direction of magnetic fieldcan be obtained from the righthand thumb rule.Magnetic field of amoving charge . Biot-Savart's law is the magnetic analogue ofcoulomb's law in Electrostatics. The chargeelement dq appearing in Coulomb's law is ascalar but the current element idl appearing inBiot-Savart's law is a vector .E.g. A steady current I goes through a wireloop PQR having shape of a right angletriangle with PQ = 3x, PR = 4x and QR = 5x.If the magnitude of the magnetic field at Pdue to this loop is , find the value of k.

Analysis:

K = 7.

Magnetic field along the axis of a circularloop:

When x >> r, A current carrying loop acts as a magneticdipole whose magnetic moment is ,where A is area of the coil.Magnetic field along the axis of solenoid

where n is number ofturns per unit length. If

solenoid is very long ¸ .

Force on a moving charge in a uniformmagnetic field :The force on a charge q moving with avelocity in a magnetic field is given by

. i.e. where θ is anglebetween and .Nature of path followed by a chargedparticle in magnetic field : Ifθ = 0° or 180°, F = 0, the path is straight

line. Ifθ = 90°, F = Bqv, the path is circle of

radius , and frequency (cyclotronfrequency).

Ifθ lies between 0° and 90°, the chargedparticle describes a helical path of radiusmv sin θ/qB; and pitch x = .

After time t, the deviation will be interms of length of the magnetic field x.

e.g. A magnetic field exists in theregion a < x < 2a and , in the region2a < x < 3a, where B0 is a positive constant.A positive point charge moving with avelocity , where v0 is a positive const-ant, enters the magnetic field at x = a. Thetrajectory of the charge inthis region can be like,

a) b)

c) d)

Analysis:for a < x < 2a path will be concave upward2a < x < 3a path will be concave downward;So (a) is correct.e.g. A particle of mass m and charge qmoving with velocity v enters Region IInormal to the boundary as shown in thefigure. Region II has a uniform magneticfield B perpendicular to the plane of thepaper. The length of the Region II is l.Choose the correct choice(s). a) The particle enters Region III only if itsvelocity

b) Theparticle enters Region IIIonly if its velocity

c) Path length of the particle in Region II ismaximum when velocity

d) Time spent in Region II is same for anyvelocity v as long as the particle returns toRegion I.

Analysis:In region II, the particle follows a circularpath of radius.

Therefore, the particle can enter region III if

r = , i.e. if .

In region II, the maximum path length isr = , which gives .

The time period of the circular motion is

The particle will return to region I if the time

spent by it in region II is , which

is independent of the velocity. Hence thecorrect choices are (a), (c) and (d).Lorentz Force : The force experienced by a moving

charge with velocity in electric fieldand a magnetic field ,

When is parallel to and particlevelocity is perpendicular to both of thesefields, the path of a particle is a helix withincreasing pitch.

When electric field is perpendicular toand the particle is released at rest from

origin, the path of a particle is cycloid. Force on a current carrying conductor in

a magnetic field . Force between two parallel wires carryi-

ng current is given by

Torque on a current loop:The torque experienced by a current carryingloop in a magnetic field .

Where is magnetic dipole moment = niA.Potential energy of a magnetic dipole in amagnetic field is .Force on magnetic dipole is non-uniform

magnetic field.

Model Questions

1. The magnetic field at the centre of acircular coil of radius r and carrying acurrent I is B. What is the magnetic fieldat a distance fromthe centre, on the axis ofthe coil?a) B b) 2B c) 4B d) 8B

2. If an electron and a protonare projected at rightangles to a uniformmagnetic field with thesame linear momentum,a) the electron trajectorywill be less curved thanthe proton trajectoryb) the proton trajectorywill be less curved thanthe electron trajectoryc) both trajectories will beequally curvedd) both particles move instraight lines

3. In a hydrogen atom, anelectron of mass m andcharge e is in an orbit of

radius r making n revolutions per second.If the mass of the hydrogen nucleus is M,the magnetic moment associated with theorbital motion of the electron is

a) b)

c) d) 4. A rectangular loop carrying a

current i is situated near a longstraight wire such that the wireis parallel to one of the sides of the loop.If a steady current I is established in thewire, as shown in figure, the loop willa) rotate about an axis parallel to the wireb) move away from the wirec) move towards the wired) remain stationary

5. A straight section PQ of a circuit liesalong the x-axis from x = -a/2 to x = a/2and carries a current I. The magnetic fielddue to the section PQ at point x = +a willbea) proportional to ab) proportional to a2

c) proportional to 1/a d) equal to zero6. Two long parallel wires are at a distance

2d apart. They carry steady equal currentsflowing out of the plane of the paper asshown. The variation of the magneticfield B along the line 'XX' is given by

a) b) c) d)

7. A coil having N turns is wound tightly inthe form of a spiral with inner and outerradii 'a' and 'b' respectively. When acurrent I passes through the coil, themagnetic field at the centre is

d

X

B

d

X′

d

X

B

d

X ′

d

X

B

d

X ′

d

X

B

d

X′

2nerπ2ner m

(M m)

π+

2ner M

m

π2ner m

M

π

x 3 r=

dF (M B)

dx= − ⋅

− ⋅U = M B

τ = ×

B

0 1 2F i i

2 r

µ=π

F i( B)= ×

B

E

B

E

F qE v B= + ×

Bv

T m

2 qB

π=

2 r 2 mv 2 mT

v v qB qB

π π π= = × =

M niA=

r

r

B dB= ∫

r

i d

dB

02

î d rdB

4 r

µ ×=π

dB

P r

i

θ d

B

v B

mvr

Bq=

vcos 2 m

Bq

θ π

Bqt

mθ =

0ˆB B j=

0ˆB B j= −

0ˆv v i=

a 0

B0

x 3a 2a

–B0

a

z

x 3a 2a

a

z

x 3a 2a

a

z

x 3a 2a

a

z

x 3a 2a

Region I

v

× × × × × × × × × × × ×

Region II Region III

q Bv

m>

q Bv

m<

q Bv

m=

i1 i2

r

mvr

qB=

1 xsin

r− θ =

qBf

2 m=

π

90α = β = ° 0B ni= µ

0I

k48 x

µ π

i d

0iB sin sin 4 r

µ= α + βπ

α β

i

0niB sin sin

2

µ= α + β

20

3

ni rB

4 x

µ π=π

20

2 2 3/ 2

ni rB

2 (r x )

µ=

+

P

x

r i

37°

53°

3x 12x/5

4x

5x

P

Q

R

0 0I IB [cos53 cos 37 ] 7

12x 48 x45

µ µ = ° + ° = π π

02

ˆqv rB

4 r

µ ×=

π

P

i

r αβ

A

I

i


4

Motional emf

Electro Magnetic Induction

a) b)

c) d)

8. A particle of mass m and charge q moveswith a constant velocity v along thepositive x direction. It enters a regioncontaining a uniform magnetic field Bdirected along the negative z direction,extending from x = a to x = b. Theminimum value of v required so that theparticle can just enter the region x > b isa) qbB/m b) q(b - a)B/mc) qaB/m d) q(b + a)B/2m

9. An electron is movingin the xy-plane alongthe positive x-axis.There is a suddenchange in its path due to the presence ofelectric and/or magnetic fields beyond Pas shown in figure. The curved path lies inthe xy-plane and is found to be non-circular. Which of the followingcombinations is possible?

a) b)

c) d)

Key: (1) d ; (2) c ; (3) d ; (4) c ; (5) d ; (6) b ;(7) c ; (8) b ; (9) b

ELECTRO MAGNETIC INDUCTION

1. The Magnetic flux (φ) through a givensurface A is given by

2. Faraday's laws of electromagnetic

induction says

Where e = induced emf and φ= magneticflux.According to Lenz law, negative signshows that emf induced will oppose thechange in magnetic flux (causing theemf).

3. Motional emf: The emf induced due tomotion of conductor in magnetic field isgiven by , where is themagnetic field in the region. Is thelength of conductor and is the velocityof conductor.The formula given above is very useful tofind the induced emf.The direction of induced current can begiven by Fleming Right Hand Rule. If thethumb and first two fingers of the righthand are spread out so that they point inthree directions at right angles to oneanother, the first finger giving thedirection of magnetic field, the thumbindicating the direction of the motion ofthe conductor, then second fingerindicates the direction of induced emf orcurrent.

4. Motional emf induced in a rotatingconductor: A conducting rod of length lrotates with constantangular speed ω about oneend P in uniform magneticfield B.Consider a segment of rod of length dr ata distance r from P.Induced emf in this segment

.Summing the emf induced across allsegments we get, total emf across the rod.

From Fleming Right Hand Rule we cansee that Q is at higher potential and P is atlower potential.

5. Self induction: Whenever the electriccurrent passes through a coil changes, themagnetic flux linked with it also changes.As a result, an emf is induced in the coil.Be careful in checking the polarity of thisinduced emf (or called back emf).a) The magnetic flux produced in a coil isdirectly proportional to the currentflowing in it, i.e., . Theconstant of proportionality L is defined asthe coefficient of self induction.b) The induced e.m.f. generated in the coilis given by

c) The inductance of long solenoid or atoroid is give by L= . Where N istotal number of turns, l is length and A isarea of cross- section.

6. Combination of inductors.i) Coils in series

ii) Coils in parallel

7. Coefficient of mutual induction (M):a) The coefficient of mutual inductionbetween two coils is equal to thatmagnetic flux linked with the secondarycoil which is produced as a result of unitcurrent flow in the primary coil.

When I1=1 amp, then

b) The coefficient of mutual induction isnumerically equal to that induced e.m.f.in the secondary coil, which is producedas a result of unit rate of change of currentin the primary coil,

When =1 amp/s, then

M21=e2

c) Mutual inductance of two solenoids orcoils is M = . N1, N2 are number ofturns in primary coil, secondary coil, λ islength and A is area of cross section of thecoils.d) If L1 and L2 are self inductances of twocoils, the mutual inductance between thecoils M= .

8. Oscillation in L-Ccircuit: When a chargedcapacitor C having aninitial charge q0 isdischarged through an inductor L then

Or or.

This equation is similar to SimpleHarmonic Motion. So, charge oscillates inthe circuit with a natural frequency

, q = q0 cos ωt,.

Comprehension Passage:A moving coil galvanometer consists of a coilof N turns and area A suspended by a thinphosphor bronze strip in radial magnetic fieldB. The moment of inertia of the coil about theaxis of rotation is I and C is the torsionalconstant of the phosphor bronze strip. When acurrent i is passed through the coil, it deflectsthrough an angle θ (in radian). Choose thecorrect statement from the following.1. The magnitude of the torque experienced

by the coil is independent ofa) N b) B c) i d) I

2. The current sensitivity of thegalvanometer is increased ifa) N, A and B are increased and C is

decreasedb) N and A are increased and B and C are

decreasedc) N, B and C are increased and A is

decreasedd) N, A, B and C are all increased

3. When a charge Q is passed almostinstantly through the coil, the angularspeed ω acquired by the coil is

a) b) c) d)

4. In the above question, the maximumangular deflection (in radian) of the coil is

a) b)

c) d)

Analysis:The above problem is best example ofcombination of different concepts i.e.Electromagnetism, Rotatory Motion,Elasticity, Conservation of energy.1. The magnitude of torque experienced by

the coil is given by where αis the angle which is the normal to the pla-ne of the coil makes with the direction ofthe magnetic field. If the magnetic field isradial, the plane of the coil is always para-llel to the direction of the magnetic field,i.e. = 90°. Hence = Ki, where K= NAB. So, the correct choice is (d).

2. Let θ be the angular deflection (in radian)when a current i is passed through thecoil. Then, restoring torque = Cθ. Whenthe coil is in equilibrium, deflectingtorque = restoring torque, i.e.

.

∴ Current sensitivity is.

Hence the correct choice is (a).3. If ω is the angular speed acquired by the

coil when a charge Q is passed through itfor very short time ∆ t, then

Or

Or or,

which is choice (c).4. From the principle of conservation of

energy, we have ,

which gives , which is

choice (a).

Model Questions

1. A thin semicircularconducting ring ofradius R is fallingwith its plane verticalin a horizontalmagnetic field B(figure). At the position MNQ the speedof the ring is v and the potentialdifference across the ring isa) zero

b) and M is at higher potential

c) πRBv and Q is at higher potentiald) 2 RBv and Q is at higher potential

2. A small square loop of wire of side 1 is pl-aced inside a large square loop of wire ofside L (L>>1). The loops are coplanar andtheir centres coincide. The mutual induc-tance of the system is proportional to

a) b)

c) d)

Key: (1) d ; (2) b

2L

L

2

L

L

21Bv R

2π

maxI

Cθ = ω

2 2max

1 1I C

2 2ω = θ

NABQω =I NABQω =

QI t Ki t KQ i

t ω = τ∆ = ∆ = ∴ = ∆

angular momentum I

time interval t

ωτ = =∆

NABθ =

iNAB C= θ

iNABτ =

iNABsinτ = α

max ICθ = ωmax IC

ωθ =

maxI

IC

θ = ωmaxI

Cθ = ω

NAQI

BNABQ

I

BAQ

NI

NAB

QI

0i q cos t2

π = ω ω + 1

LCω =

2

2

d q 1q 0

LCdt+ =

2

2

q d qL 0

C dt

− − =

q Ldi0

C dt− =

1 2L L

0 1 2N N Aµλ

1Id

dt−

221

1Ie

Md

dt

=

21 2M ϕ=221

1IM

ϕ=

1 2P

1 2

L LL

L L=

+

S 1 2L L L= +

Circuit is made ON or current is increasing

i i di

e Ldt

=

i i di

e Ldt

=

Circuit is made OFF or current is decreasing

20 N Aµ

d dle L

dt dt

ϕ= − = −

I Ior Lϕ α ϕ =

2

0

Be dr B r dr e

2

ω= = ω ⇒ =∫ ∫

e Bv dr B r dr= = ω

v

B

e B ( v)= ⋅ ×

de

dt

φ= −

NB A NBA cosφ = ⋅ = θ

ˆ ˆ Ê ai; B ck bj= = + ˆ Ê 0; B cj bk= = +

ˆ ˆ Ê ai; B ck bi= = +

ˆ Ê 0; B bi ck= = +

0NI aln

2(b a) b

µ−

0NI bln

2(b a) a

µ−

02 NI

a

µ0NI

b

µ

Q dr ω

B

r

× × × × ×

× × × × ×

× × × × ×

L

C

v

N

M Q

B × × × × × × × × × × × × × × × ×

P x

y


5

Application and Analysis

Gauss's law.. Amperes law

GAUSS'S LAW AND AMPERE'S LAW

Gauss's Law:1. This law is the mathematical analysis of

the relation between the electric flux froma closed surface and its enclosed charge.

2. This law states that the total fluxemerging out from a closed surface isequal to the product of sum of enclosedcharge by the surface and the constant1/ε0.

3. Mathematically Gauss's Law is written as

here the sign φ repre-

sents the integration over a closed surfacewhich enclosed a total charge Σqencl..

4. Gauss law is a very helpful tool in findingthe electric field strength due to variousdistribution of changes.

5. Application of Gauss's Law in theRegion of varying electric field:Consider the situationas shown in figure. In aregion electric fielddepends on x directionas E = E0 x2. Find the net charge enclosedin the cube.In the cube of edge as shown in figurefrom front face electric flux goes in whichcan be given as From the other surface flux coming outcan be given as

Here for the cubical surface

hence net charge enclosed in the cube ispositive, we can apply Gauss Law to findthe net charge enclosed in the cube as

Ampere's Law:Similar to the Gauss's law of electrostatics,this law provides us shortcut methods offinding magnetic field in cases of symmetry.According to this law, the line integral ofmagnetic field over the closed path

is equal to µ0 times the net currentcrossing the area enclosed by the path.

Example:Consider a coaxial cable which consists of aninner wire of radius 'a' surrounded by an outershell of inner and outer radii b and c,respectively. The inner wire carries an electriccurrent i0 and the outer shell carries an equalcurrent in opposite direction. Find themagnetic field at a distance x from the axiswhere (a) x < a, (b) a < x < b, (c) b < x < c and(d) x > c. Assume that the current density is

uniform in the inner wire and also uniform inthe outer shell.Analysis:

A cross section of the cable is shown infigure. Draw a circle of radius x with thecentre at the axis of the cable. The parts (a),(b), (c) and (d) of the figure correspond to thefour parts of the problem. By symmetry, themagnetic field at each point of a circle willhave the same magnitude and will betangential to it. The circulation of B along this circle is, therefore, in each of thefour parts of the figure.

a) The current enclosed within the circle inpart (a) is i0 so that

Ampere's law

gives

The direction will be along the tangent tothe circle.

b) The current enclosed within the circle inpart (b) is i0 so that

c) Current density of outer shell:

So current from x = 0 to x = x :I = i0 – J(πr2 – πb2)

d) For x > c, magnetic field will be zero,because net current is zero.

Gauss Law and Ampere's Law:A ball of radius R carries a positivecharge whose volume density depends ona separation r from the ball's centre as

where ρ0 is a constant.

Assuming the permittivities of the balland environment is equal to unity.

1. For r < R, electric field is given by E =

a) b)

c) d)

2. For r >> R, electric field is given by E =

a) b)

c) d)

3. The maximum electric field is

a) b) c) d)

Model Questions

Comprehension Passage:A long, straight, solidcylinder, oriented with its axisin the z-direction carries acurrent whose current densityis . The current density,although symmetrical aboutthe cylinder axis, is notconstant but varies according to relation

Where R is the radius of the cylinder, r isthe radial distance from the cylinder axisand i0 is constant having units of amperes.

1. What is the total current passing throughthe entire cross section of the wire?

2. Find the magnetic field in the region r ≥R.

3. Derive the expression for the magneticfield in the region r ≤ R.

4. Obtain the expression for the current icontained in a circular cross section ofradius r ≤ R and centred at the cylinderaxis.

5. The magnetic field withincylindrical region whosecross-section is indicatedstarts increasing at aconstant rate α tesla/sec.The graph showing the variation ofinduced electric field with distance r fromthe axis of cylinder isa) b)

c) d)

6. A square non-conductingloop, 20 cm on a side isplaced in a magnetic field.The centre of side ABcoincides with the centreof magnetic field. The magnetic field isincreasing at the rate of 2 T/s. Thepotential difference between B and C isa) 10 mV b) 20 mVc) 30 mV d) zero

7. In the above problem, the potentialdifference between C and D isa) 40 mV b) 60 mVc) 80 mV d) zero

8. A charge q is placed at adistance from disc ofradius R as shown infigure. The electric fluxpassing through disc is

a) b)

c)

d)

9. The electric field in a region is given by

with

. Find the flux of

this field through a rectangular surface ofarea 0.2 m2 parallel to the Y - Z plane.

a) b)

c) d)

10. A charge Q is placed at adistance a/2 above the centreof a horizontal, squaresurface of edge a as shown infigure. Find the flux of the electric fieldthrough the square surface.

a) b) c) d)

Key: (1) i0; (2) ;

(3);

(4) (5) a; (6) b; (7) a;

(8) c; (9) a; (10) b

FLUID STATICS AND DYNAMICS

1. Hydrostatic Pressure: Pressure existseverywhere within a fluid. The hydrosta-tic pressure at a depth h below the surfaceof a fluid is given by , where ρ isthe density of the fluid and g, theacceleration due to gravity. The pressureis the same at all points at same horizontallevel. The pressure at any point in a fluidcontained in a vessel is independent of theshape or size of the vessel.

P h g= ρ

2 20

2 2

i r ri 2

R R

= −

20 0

2 2

i r rB 2

2 R R

µ= − π

0 0i

2 r

µπ

0

Q

ε0

Q

(2 )ε0

Q

(6 )ε0

Q

(4 )ε

2N m280

C

−2N m480

C

−

2N m120

C

−2N m240

C

−

30E 2.0 10 N / C= ×

o 0

3 4E E i E j

5 5= +

2 20

q1

R

− ε +

2 20

q1

2 R

− ε +

2 20

q

2 Rε +

0

q

ε

B

For R≥J 0=

2

02

2i r ˆJ 1 k For r RR R

= − ≤ π

J

0

0

Rρε

0

0

R

2

ρε

0

0

R

3

ρε

0

0

R

9

ρε

30

20

R

4 r

ρε

30

20

R

6 r

ρε

30

20

R

12 r

ρε

20

30

R

12 r

ρε

30

20

R

12 r

ρε 0

0

r 3r1

3 4R

ρ −ε

0

0

r 3r1

3 7R

ρ −ε 0

0

r 2r1

3 3R

ρ −ε

0

r1

R ρ = ρ −

2 20 0 0

2 2

I i (c x )B .

2 x 2 x(c b )

µ µ −= = =π π −

2 2 2 20

0 0 2 2 2 2

x b i (c x )i i

c b c b

− −= − = − −

02 2

iJ

c b=

π − π

0 00 0

iB2 x i or B .

2 x

µπ = µ =π

20 0 0 0

2 2

i x i xB2 x or B

a 2 a

µ µπ = =π

0B d i⋅ = µ∫

2 20 02 2

i ix x .

a aπ =

π

B d B2 x⋅ = π∫

c b

a x

(a) (b) (c) (d)

0 1 3 2B d (I I I )⋅ = µ + −∫

0 enclosed

closed lopp

B d I⋅ = µ∫

( B d )⋅∫

4enclout in encl 0 0

0

qor q 5 E a .φ − φ = = ε

ε

out inφ > φ

2 2 4out 0 0E (3a) a 9E a .φ = ⋅ =

2 2 4E (2a) a 4E aφ = ⋅ =

encl.

0

qdS ,

Σε ⋅ =ε∫

a

φout

x

y

O 2a

a

a

φin

20E E x=

dl

I5

B

⊗

⊗ ⊗

I3

I2

I4

I1

× × × × × × × × × × ×

× × ×

O

D C

B A

× × × × × × × × × × ×

× × ×

q

2 2R +

disc

R φ

r

E

r

E

r

E

r

E

z i y

x

i R

a

a

a/2


6

Surface Tension

Fluid Statics and Dynamics

2. Gauge Pressure: The pressure at anypoint in a fluid is equal to the sum of theatmospheric pressure P0 acting on itssurface and the hydrostatic pressure due to the weight of the fluid above thatpoint which is at a depth h below thesurface of the fluid. This is called thegauge pressure which is given by

.3. Archimedes' Principle: 'When a solid

body is wholly or partly immersed in afluid, it experiences an upward thrust orbuoyant force equal to the weight of thefluid displaced by it.' The word 'fluid'includes both liquids and gases.

, where V is volume of theliquid displaced, is density of liquid,g is acceleration due to gravity.

4. Surface Tension: Surface tension is theforce acting per unit length of animaginary line drawn on the surface of aliquid at rest. The work done to increasethe surface area of a liquid is calledsurface energy. Surface tension is equalto surface energy.The SI unit of surface tension is Nm-1 orJm-2.

i) Rise of a liquid in capillary tube: Aliquid of density ρ, surface tension σ,angle of contact θ, rises in a capillarytube of radius r to a height h given by

.ii) Excess pressure: The excess pressure

inside a liquid drop of radius r is given

by .

The excess pressure inside a liquid

bubble of radius r is .

The excess pressure inside an air bubbleof radius r in a liquid of surface tension σ

is .

iii) Capillary tube of varying radius:Suppose radius of tube varies from r1 andr2 in its total length l. The radius at the

position of meniscus.

The rise of the liquid in the tube can be

obtained by the formula , by

placing the value of r in terms of r1 and r2.5. Bernoulli's Principle: During stream-

line flow of a liquid, the sum of the kinet-ic energy of the fluid, and the gravitatio-

nal potential energy of the fluid, and thepotential energy due to the pressure of thefluid, is a constant. Mathematically

constant.

6. Continuity of Flow: If A1 and A2 are theareas of cross-section of a tube of avariable cross-section and v1 and v2 arethe velocities of flow of a fluid crossingthese areas, then . This is theequation of continuity of flow.

7. Torricelli's Theorem: Velocity of Efflux The velocity of efflux of a liquid throughan orifice at a depth 'h' below the surfaceof the liquid is the same as that of a freelyfalling body from that height, i.e.

.8. Viscosity: The viscous force between

two layers of a fluid of area A having avelocity gradient dv/dx is given by

, where η is called the

coefficient of viscosity. In SI units, η isexpressed in a unit called poiseuille (Pl).

9. Stroke's Law: The viscous force on aspherical body of radius r movingthrough a fluid of viscosity η is given by

, where v is the velocity ofthe sphere.

10. Terminal Velocity: When a body isallowed to fall in a viscous fluid, itaccelerates first, but soon begins toexperience a viscous (retarding) force. Ifthe weight of the body is sufficientlysmall, it is found that after attaining acertain velocity, the retarding force justbecomes equal to the effective weight ofthe body in the fluid. The body then doesnot experience any net force (and henceany acceleration) and, therefore, fallswith a constant velocity known as theterminal velocity (vt).The viscous force on a spherical body ofradius r moving with a terminal velocityvt in a fluid is given by .

.Example:

Through a very narrow gap of height 'h',a thin plate of large extension is pulled ata velocity 'v' on one side of the plate is oilof viscosity η1 and on the other side oil ofviscosity η2. Calculate the position of theplate so that

i) the shear force on the two sides of theplate is equal

ii) the pull required to drag the plate isminimum.

Analysis:Let y is the distance ofthe plate from one of thesurfaces.i) Froce per unit area of the upper surface of

the plate

and force per unit are on the bottomsurface of the plate

Equating the two, we get

.ii) Let F is the force required to pull the

plate. Suppose A is the area of the plate,then

For F to be minimum,.

or

.Ex. A container of width '2a' is

filled with a liquid. A thinwire of weight per unit lengthλ is gently placed over theliquid surface in the middle of the surfaceas shown in the figure. As a result, theliquid surface is depressed by a distance y(y << a). Find the surface tension of theliquid.

Analysis:Let ' ' be the length of the wire. As 'λ' isthe mass per unit length of the wire, soweight of the wire .If σ is the surface tension, thensurface tension force = σThe weight of the wire isbalanced by the vertical component of thesurface tension force σ , so

.

From the figure :

(As y << a)

MODEL QUESTIONS

1. Figure shows a container having liquid of

variable density. The density of liquid

varies as . Here h0 and ρ0

are constants and h is measured formbottom of the container. A solid block ofsmall dimensions whose

density is and mass m is

released from bottom of thetank. If it executes simple harmonicmotion, the frequency of oscillation is

a) b)

c) d)

Passage 1:A container of large uniform cross-sectionalarea 'A' resting on a horizontal surface, holdstwo immiscible, non-viscous andincompressible liquids ofdensities 'd' and '2d' each of heightH/2 as shown in figure. The lowerdensity liquid is open to the atmospherehaving pressure 'P0'.

Answer the following questions.2. A homogeneous solid cylinder of length

'L' (L<H/2), cross sectional area 'A/5' isimmersed such that it floats with its axisvertical at the liquid-liquid interface withlength 'L/4' in the denser liquid. Thedensity of the solid cylinder is

a) b) c) d)

3. The total pressure at the bottom of thecontainer is

a) b)

c) P0 + dgL d)

4. The cylinder is removed and the originalarrangement is restored. A tiny hole ofarea 'a' (a<<A) is punched on the verticalside of the container at a height 'h' (h<H/2). The initial speed of efflux of theliquid at the hole is

a) b)

c) d)

5. The horizontal distance 'x' traveled by theliquid initially is

a) b)

c) d)

6. The height 'hm' at which the hole shouldbe punched so that the liquid travels themaximum distance is

h(H 4h)−h(3H h)−

h(3H 2h)−h(3H 4h)−

(3H 2h)g

2

−(3H 4h)g

4

−

(3H 4h)g

2

−(3H 4h)g−

0

dg(H L)P

++

0

dg(6H L)P

++0

dg(H L)P

4

++

5d

2

5d

6

4d

5

5d

4

0

1 6g

2 hπ0

1 6g

2 5hπ

0

1 g

2 2hπ0

1 g

2 hπ

0

5

2ρ

00

3h4

h

ρ = ρ −

g ga

2(y / a) 2y

λ λσ = =

2 2

y ycos

aa yθ = =

+

W g g

2 cos 2 cos 2cos

λ λ∴σ = = =θ θ θ

2 cos Wσ θ =

W g= λ

1 22 2

1

2

A v v h0 y

(h y) y1

η η− = ⇒ =− η+

η

dF0

dy=

1 2

1 2

F f A f A

v vF A A

h y y

= +

= η + η −

21 2

v v vy

(h y) y ( )

ηη = η ⇒ =− η + η

2 2 2

dv vf

dy y= −η = −η

1 1 1

dv vf

dy (h y)= η = η

−

2

t

2( )grv

9

ρ − σ=η

tF 6 r v= πη

F 6 r v= πη

21Pl 1 Ns m −=

dvF A

dx= −η

v 2gh=

1 1 2 2A v A v=

21p u gh

2+ ρ + ρ =

2 cosh

r g

τ θ=ρ

1 21

r rr r h

− = −

2p

r

σ=

4p

r

σ=

2p

r

σ=

2 cosh

r g

σ θ=ρ

ρ

upF V g= ρ

0P P h g= + ρ

h gρy

h-y h

v

y

2a

h0

d

2d H/2

H/2

W

σ

a

θ θ

a

σ


7

Radiation.. Heat conduction

Thermal Physics

a) H/2 b) 3H/2 c) 3H/3 d) 3H/4Passage 2:A cylindrical tank of base area 'A' has a smallhole of area 'a' at the bottom. At time t = 0, atap starts to supply water into the tank at aconstant rate Q m3/s. (Answer the following 2questions) 7. The maximum level of water in the tank

is

a) b) c) d)

8. The time when level of water becomes h(<hmax) is

a)

b)

c)

d)

Key:(1) c; (2) a; (3) b; (4) b; (5) a;(6) c; (7) b; (8) b

THERMAL PHYSICS

1. Expansion of solids:a) Coefficient of linear expansion:

b) Coefficient of superficial expansion:

c) Coefficient of volume expansion:

d) :2. Density of material at any temperature:

3. Expansion of liquids: If γa and γg are theapparent coefficient of expansion ofliquid and volume coefficient ofexpansion of container, then

.4. Expansion of gases:

.5. Law of mixture : Heat lost = Heat gained.6. First law of thermodynamics:

.Here heat supplied to the system,

= change in internal energy, =work done by gas.

Some important points: Work is path dependent in

thermodynamics.

Work is taken as positive when systemexpands against external force.

The area of P-V diagram gives workdone.

In cyclic process, work done is area of P-V diagram cycle. It is positive whenprocess is clockwise. It is negative whenprocess is anticlockwise.

The change in internal energy isindependent of path. It depends only oninitial and final states.

For ideal gas

If PVn = constant

7. γ of the mixture.

8. Molar heat capacity 'C' is the heatrequired to raise the temperature of 1mole of a gas by 1°C or 1 K.

The most general expression for C in theprocess constant is

For isobaric process P = constant or x = 0

For isothermal process x = 1

For adiabatic process x = rC = 0.

9. Heat conduction:

Here, = temperature gradient, K =

thermal conductivity.Unit of 'K' is watt per metre per Kelvin.

10. Radiation:

a) Stefan's law:

Here, = rate of radiated energy per

unit areaσ = Stefan's constant = 5.67x10-8 Wm-2

K-4 ; e = emissivity, T = temperature ofbody in Kelvinb) Kirchhoff's law:If a = absorptive power, e0 = emissivepower then e/a = emissive power of blockbody.

11. Wein's displacement law:

12. Newton's law of cooling:

, Here θ= temperature of

body, θ0 = temperature of surroundings.

Model Questions:

1. A vertical cylinder closed from both endsis equipped with an easily moving piston,dividing the volume in two parts, eachcontaining one mole of air. In equilibriumat T0=300 K the volume of the upper partis 4 times that of lower part. At whattemperature, will the ratio of thesevolumes be equal to 3 ?a) 222 K b) 322 Kc) 422 K d) 522 K

2. An ideal gas is initially at temperature 'T'and volume 'V'.Its volume is increased by∆V due to an increase in temperature ∆T,pressure remaining constant. The quantityδ=∆V/V∆T varies with temperature as

a) b)

c) d)

3. An ideal gas is taken thro-ugh the cycle A→ B→C→ A, as shown in thefigure. If the net heat supplied to the gasin the cycle is 5 J, the work done by thegas in the process C→ A isa) -5J b) -10J c) -15J d) -20J

4. The molar heat capacity of a certainsubstance varies with temperature variesaccording to the empirical equation C =27.2 J/mol-K+(4.0x10-3J/mol-K2)T. Howmuch heat is necessary to change thetemperature of 2.0 mol of this substancefrom 27°C to 427°C?a) 23360 J b) 23630 Jc) 32360 J d) 26330 J

5. A point source of heat of power 'P' isplaced at the centre of a spherical shell ofmean radius 'R'. The material of the shellhas thermal conductivity 'k'. Calculate thethickness of the shell if temperaturedifference between the outer and innersurfaces of the shell in steady state is 'T'.

a) b)

c) d)

6. P-V diagram of 'n' moles of an ideal gas isas shown in figure. Find the maximumtemperature between A and B.

a) b)

c) d)

Passage:0.01 mole of an ideal diatomic gas is enclosedin an adiabatic cylinder of cross-sectionalarea A = 10-4m2. In the arrangement shown in

figure, a block of massM=0.8 kg is placed on ahorizontal support andanother block of massm=1kg is suspended from a spring of stiffnessconstant k=16 N/m. Initially, the spring isrelaxed and the volume of the gas isV=1.4x10-4m3 (Atomic pressure =1x105N/m2)7. The initial pressure of the gas is

a) 1x105 N/m2 b) 2x105 N/m2

c) 3x105 N/m2 d) 4x105 N/m2

8. If the block 'm' is slightly pushed downand released it oscillates harmonically.The angular frequency isa) 4 rad/s b) 5 rad/sc) 6 rad/s d) 7 rad/s

9. When the gas in the cylinder is heated bya heating coil arranged inside thecylinder, the piston starts moving up andthe spring gets compressed so that, theblock 'M' is just lifted up. The heatsupplied to the gas is (atmosphericpressure = 105N/m2, g = 10 m/s2)a) 25 J b) 50 J c) 75 J d) 100 J

Key :(1) c; (2) c; (3) a; (4) a; (5) c; (6) a;(7) b; (8) c; (9) c

MECHANICSSIGNIFICANT FIGURES

1. The number significant figure in anymeasurement indicates the degree ofprecision of that measurement. Thedegree of precision is determined by theleast count of the measuring instrument.

2. Estimation of appropriate significantfigures in calculations: The importanceof significant figures lies in calculation tofind the result of addition ormultiplication of measured quantitieshaving a different number of significantfigures. The least accurate quantitydetermines the accuracy of the sum orproduct. The result must be rounded off tothe appropriate digit.

3. Significant figures in Addition andSubtraction: For addition andsubtraction rule is as follows : Round offthe final result such that it has the samenumber of digits after the decimal placeas in the least accurate measurement.

4. Significant figures in Multiplicationand Division: We use the following ruleto determine the number of significantfigures in the result of multiplication anddivision of various physical quantities.Do not worry about the number of digitsafter the decimal place. Round off theresult so that, it has the same number ofsignificant figures as in the least accuratequantity.

0 03P V

2nR0 09P V

2nR

0 04P V

9nR0 09P V

4nR

4 kRT

P

π24 kR T

P

π

22 kR T

P

π2kR T

4P

π

0

dK( )

dt

θ = θ − θ

3mT constant 2.892 10 mK−λ = = ×

H

A t

∆∆

4He T

A t

∆ = σ∆

d

dx

θ

dH dKA

dt dx

θ= −

C = ∞

VC C R= +

R RC

1 1 x= +

γ − −

xPV =

QC

n T

∆=∆

1 2 1 2

1 2

n n n n

1 1 1

+ = +γ − γ − γ −

W PdV= ∫

W∆U∆H∆ =

H U W∆ = ∆ + ∆

v p

1 1/ C, / C

273 273γ = ° γ = °

t 0r a g a

0

V V

V t

−γ = γ + γ ⇒ γ =

0t (1 t)

ρρ =+ γ

2 ; 3β α γ α

0

V

V t

∆γ =∆

0

A

A t

∆β =∆

0

L

L t

∆α =∆

Q a 2ghA Qln 2gh

2ag a Q

−−

Q a ghA Qln gh

ag a Q

−−

Q a 2ghA Qln 2gh

ag a Q

−−

Q a 2ghA Qln gh

ag a Q

−−

2

2

Q

4ga

2

2

Q

ga

2

2

Q

2ga

2Q

2ga

P(Pressure)

U (Internal Energy)

Isobaric Isochoric

Adiabatic Isothermal

V

P

n=0, (Isobaric)

n=1(Isothermal)

n=2(Adiabatic) n=∞(Isochoric)

T T

δ

T+∆T

T T

δ

T+∆T

T T

δ

T+∆T

T T

δ

T+∆T

V(m3) 2 C

A

B

P(N/m2) 10

1

P

P0

2P0 A

B

V0 V

2V0

M

m

K


8

Motion along a Curve

Kinematics... Collision

e.g. A man runs 100.5 m in 10.3 s. Findhis average speed up to appropriatesignificant figure.Analysis: Average speed (v) =

.The distance 100.5 m has four significantfigures but the time 10.3 s has only three.Hence the value of the result must beround off to three significant figures. Thecorrect result is v = 9.71 ms-1.

5. Least Counts of Some MeasuringInstruments:a) Least count of metre scale = 1 mm =0.1 cmb) Vernier constant (or least count) ofvernier calipers = value of 1 main scaledivision - value of 1 vernier scaledivision = 1 M.S.D. - 1 V.S.D.Let the value of 1 M.S.D. = a unitIf n vernier scale divisions coincide withm main scale divisions, then value of 1 V.S.D. = m/n of 1 M.S.D = ma/n unit.

∴ Least count = unit.

c) Least count of a micrometer screw isfound by the formula:

where pitch = lateral distance moved inone complete rotation of the screw.

6. Error in product and division: Supposewe determine the value of a physicalquantity u by measuring three quantitiesx, y and z whose true values are related tou by the equation .

Then,.

KINEMATICS

1. Projectile:a) The equation of trajectory of projectile

is.

b) The equation oftrajectory of projectile is aparabola with

i) Latus rectum =

ii) The coordinates of the focus =

iii) The equation of directrix,.

c) Time of flight = ,

Range R =

d) Velocity at any time t,.

e) Position at any time t,

f) Range of projectile on inclined plane

.g) Time of flight

h) Maximum Range =

.2. Motion along a curve: Radial and

Transverse components:

Position vector

;Where = unit vector along radial

direction, = unit vectoralong transverse direction.

Acceleration

where .

DYNAMICS

1. Force=.

2. A jet of water of density 'd' from a tube ofarea of cross section 'A' comes out with avelocity 'v'. Average force exerted by tubeon water is 'dAv2'.

3. The impulse of a force is defined as theproduct of the average force and the timeinterval for which it acts.Impulse J = FAV∆t = Impulse momentum theorem =

dt =

4. A rope of length 'L' is pulled by aconstant force 'F'. The tension in the ropeat a distance 'x' from the end where it isapplied is F(1-x/L).

5. The amount of work done is given by thedot product of force and displacement.

6. If the force is varying non-uniformly,then the work done = .

7. The work done on a spring in stretchingor compressing it through a distance x isgiven by W=1/2kx2 where k is the forceconstant or spring constant.

8. The work done in rotating a rodor bar of mass 'm' through anangle 'θ' about a point ofsuspension is W= mgL/2(1-cosθ) = mgL sin2(θ/2) where L is thedistance of the centre of gravity from thepoint of suspension.

9. The work done in lifting a body of mass'm' and density 'ds' in a liquid of density'dl' through a height 'h' under gravity is W

= m g h

10. Rate of doing work is called power.

Power = = Force x velocity.

11. If a body is rotated in circular path, the

power exerted is given by P =

12. When water is coming out from a hosepipe of area of cross section 'A' with avelocity 'v' and hits a wall normally andi) stops dead, then force exerted by thewater on the wall is Av2ρ. And the powerexerted by water is P = A v3ρ(ρ= densityof water)ii) If water rebounds with same velocity'v' after striking the wall, P = 2Av3ρ.

COLLISION

1. Collision is an interaction between two ormore bodies in which sudden changes ofmomenta take place. e.g.: Striking a ballwith a bat.

2. Coefficient of restitution (e): Thecoefficient of restitution between twobodies in a collision is defined as the ratioof the relative velocity of separation aftercollision to the relative velocity of theirapproach before their collision.

(i) e = ;

(ii) e =

3.

a) V1 =

V2 =

V1 & V2 are velocities of m1 and m2 afterthe collision

b) Loss of Kinetic energy from the system is

∆K = Elost =

CENTRE OF MASS

1. Every particle is attracted towards thecentre of the earth by the force of gravityand the centre of gravity of a body is thepoint where the resultant force ofattraction of the weight of the body acts.

2. Centre of mass: If a system of parallelforces proportional to the masses of thevarious particles of a body are assumed toact on it, their resultant passes through afixed point, irrespective of the directionof the parallel forces and that point iscalled centre of mass.

3. In vector notation, each particle of thesystem can be described by a positionvector and the centre of mass can belocated by the position vector .

4. A circular portion of radius r2 is removedfrom a circular disc of radius r1 from oneedge. Then the shift in the center of massof the disc is x =

.

FRICTION

1. When a body is in motion over anothersurface or when an object moves througha viscous medium like air or water orwhen a body rolls over another, there is aresistance to the motion because of theinteraction of the object with itssurroundings. Such a resistance force iscalled force of friction.

2. Motion on a rough horizontal plane:(a) Frictional force = µ x normal reaction.body moving with uniform velocity F=µkmg.(b) Pulled with a force F inclined at anangle θ with the horizontal and the bodymoving with uniformvelocity.

(c) Pushed with a force F inclined at an angle'θ' with the horizontal and the bodymoving with uniformvelocity:

k

k

mgF

cos sin

µ=θ − µ θ

k

k

mgF

cos sin

µ=θ + µ θ

22

1 2

r

r r+

i icm i i i i

i

m r ˆ ˆ ˆr where r ix jy kzm

= = + +∑∑

cmr

r

2 21 21 2

1 2

m m1[1 e ] [u u ]

2 m m

− − +

( ) ( )1 2 11 2

1 2 1 2

m 1 e m emu u

m m m m

+ −+

+ +

( ) ( )1 2 21 2

1 2 1 2

m em 1 e mu u

m m m m

− ++

+ +

2 1

1 2

v v

u u

−−

relative velocity of separation

relative velocity of approach

d

dt

θτ = τω

work

time

l

s

d1

d

−

F.ds F.ds.cos= θ∫ ∫

F.s Fs cos= θ

mv mu− 2

1

t

t

F∫

mv mu−

change in momentum m(v u), i.e.F

time t

−=

2

2

dr d rr , r

dt dt= =

2r r

2r

2 2r

dˆ ˆ ˆ ˆ â (r e r e ) r e re r e

dtˆ ˆ(r r )e (2r rd)e

1 dˆ â (r r )e (r ) e

r dt

θ θ θ

θ

θ

ω= + ω + ω + − ω

= − ω + ω +

= − ω + ω

dva

dt=

rr r

r r

ˆded drˆ ˆv (re ) e r

dt dt dtdr

ˆ ˆ ˆ ˆv e r e r e r edt θ θ

= = +

= + ω = + ω

eθ

rer

ˆ ˆ ˆ ˆˆ ê cos i sin j;e sin i cos jθ= θ + θ = − θ + θ

rˆr re=

2u

g(1 sin )+ β

2u sin( )T

g cos

θ − β=β

2

2

uR [sin(2 ) sin ]

g cos= θ − β − β

β

21ˆ ˆr u cos t i u sin t gt j2

= θ + θ −

ˆ ˆv u cos i (u sin gt) j= θ + θ −

2u sin 2

g

θ

2u sin

g

θ

2uy

2g=

2 2 2u sin cos u sin,

g 2g

θ θ θ−

2 22u cos

g

θ

2

2 2

gxy x tan

2u cos= θ −

θ

max

u x y z

u x y z

δ δ δ δ = α + β + γ

u x y zα β −γ=

Pitch of screweast count =

Total number of divisions on circular scale

ma ma 1 a

n n − = −

1100.2m9.708737ms

10.3s−=

Motion on a Straight Line (one dimensional motion)

Uniform velocity i) s = vt ii) a = 0

Motion with constant acceleration

) u v

s t2

+ =

i) 21s ut at

2= +

ii) 2 2v u 2as= + v) v u at= +

Motion with variable acceleration

i) dv

If a f (t),adt

= =

ii) dv

If a f (s),a vds

= =

iii) dv

If a f (v), adt

= =

iv) ds

vdt

=

v) s vdt= ∫

vi) v adt= ∫

y

O x

u

θ

u

θ β

m1

C1

m2

C2C

x

F

s

m1 m2 u1 u2

A B

m2 v1 v2 A B

m1

L.

.

Radial

i

y

x

j

θ

r

tv

v

re

θ


9

Radius of Gyration

Rotatory Motion..

3. Block on a rough inclined plane a) Angle of repose αα : It is

the angle of inclinationof the inclined plane withthe horizontal for which block justbegins to slide down.

b) If αα is the angle of repose µs=tanαc) The angle of repose is the angle of

static frictiond) The angle of inclination is(θ) less than

(α), the block does not slide down, itis at rest. The force of friction f<fs andis equal to f = mg sinθ [mg sinθ< fs]

e) If the angle of inclination is [θ] equalto [α]. Then the block is in limitingequilibrium. The force of friction is f= fs = µs mg cos α

f) If the block slides down the inclinedplane with uniform velocity µk=tan θwhere θ is the angle of inclination ofthe inclined plane.

ROTATORY MOTION

1. Equations of motion of rotating bodywith constant angular acceleration:

2. Moment of inertia or rotational inertia:M.I. of a particle of mass m at a distancer from axis of rotation is given by I=mr2.

M.I. of a body can be defined

For a system of discrete particles.

3. Radius of gyration: Radius of gyration ofsystem of particles about an axis is given

by , where and

.Radius of gyration of a body of mass Mand moment of inertia about an axis is

given by .

4. Rolling motion:i) For a wheel of radius R rolling with

constant velocity .ii) For rolling with constant acceleration

.5. Velocity of different points of a rolling

wheel:

6. Angular momentum of a rolling wheel:or

.

7. Kinetic energy of a rolling wheel:

.

8. Rolling on rough inclined plane:Acceleration of C.M.

Velocity of C.M. after falling height h:

Minimum friction required to cause purerolling:

9. Rolling of cylinder on inclined plane:

,

and .

Model Questions

1. A disc of radius R is rolling on astationary horizontal ground withoutslipping. The distance moved by a pointon circumference of the disc when disccompletes one rotation isa) 2πR b) 2R c) 4R d) 8R

2. When a disc rolls down without slippingon an inclined planea) friction force is zerob) friction force non-zero but work done

by friction force is zeroc) friction force and workdone by it are

both non-zerod) friction force is zero but work done by

friction force is non-zero3. A small object of uniform density rolls up

a curved surfacewith an initialvelocity V. Itreaches up to amaximum height

, with respect to initial position.

The object isa) ring b) solid spherec) hollow sphere d) disc

4. A disc of mass m0 rotates freely about afixed smooth horizontal axis through itscentre. A thin cotton pad is fixed to itsrim, which can absorb water. The mass ofwater dripping onto the pad is µ kg persecond. After what timewill the angular velocityof the disc get reduced tohalf of its initial value?

a) b) c) d)

5. A simple pendulum issuspended from a peg on avertical wall. The pendulumis pulled away from the wallto a horizontal position (see figure) andreleased. The ball hits the wall, thecoefficient of restitution being .What is the minimum number ofcollisions after which the amplitude ofoscillations becomes less than 60degrees?a) 1 b) 2 c) 3 d) 4

6. A wedge of mass m and triangular cross-section (AB = BC = CA = 2R) is movingwith a constant velocity towards asphere of radius R fixed on a smoothhorizontal table as shown in the figure.The wedge makes an elastic collisionwith the fixed sphere and returns alongthe same path without any rotation.Neglect all friction and suppose that thewedge remains in contact with the spherefor a very short time ∆t during which thesphere exerts aconstant force on the wedge.The force F is

a) b)

c) d)

Comprehension Passage:A uniform thin rod of mass M and length L ispivoted at its end P as shown in figure. Therod is released from its nearly verticalposition.7. When the rod reaches the

horizontal position, thenormal component of totalreaction at the support P isa) 3Mg/2 b) Mg/4 c) Mg/2 d) Mg

8. When the rod reaches the horizontalposition, the tangential component oftotal reaction at the support P isa) Mg/2 b) Mg/3 c) 2/3Mg d) Mg/4

9. When the rod reaches the horizontalposition (at t = 0) the support at P iswithdrawn. The velocity of the centre ofmass of the rod after time t is

a) gt b) c) d)

Comprehension Passage:A block of mass M witha semicircular track ofradius R, rests on ahorizontal frictionlesssurface. A uniform cylinder of radius r andmass m is released from rest at the top pointA(see in figure). The cylinder slips on thesemicircular frictionless track.

10. How far has the block moved when thecylinder reaches the bottom (point B) ofthe track?

a) b)

c) d)

11. How fast is the block moving when thecylinder reaches the bottom of the track?

a) b)

c) d)

Comprehension Passage:Two blocks of masses m1 and m2 connectedby an ideal spring of spring constant K are atrest on a smooth horizontal table. A constanthorizontal force F acts on m1. During themotion the maximumelongation of the springis x0. Answer thefollowing questions.12. Both m1 and m2 move with same velocity

when the elongation of the spring isa) x0 b) x0/2 c) x0/4 d) x0/8

13. Both m1 and m2 move with sameacceleration when the elongation of thespring isa) x0 b) x0/2 c) x0/4 d) x0/8

14. Select the correct alternativea) velocity of centre of mass is same as

that of m1 and m2 at the instant whenthe elongation is x0

b) velocity of centre of mass is same asthat of m1 and m2 at the instant whenthe elongation is x0/2

c) velocity of centre of mass can neverbe the same as that of m1 and m2 atany instant

d) velocity of centre of mass can becomezero at an instant during the motion ofthe system

Key:(1) d; (2) b; (3) d; (4) d; (5) d; (6) c; (7) a;(8) d; (9) c; (10) a; (11) b; (12) a; (13) b;(14) a.

ELECTROSTATICS

1. Quantization of Charge: Any chargedbody, big or small, has a total charge qwhich is an integral multiple of e, i.e. q =± ne, where n is an integer having values1, 2, 3, … etc.

2. Coulomb's law:

3. Definition of Electric Field:

Field due to a point charge:

4. Electric Potential:Electric potential due to a number of

20

Q 1E

4 R=

πε

FE

Q=

1 22

Q Q1F

4 r= ⋅

πε

2g(R r)m

M(M m)

++

2grm

M(M m)+

2g(R r)m

M(M m)

−+

2gRM

M(M m)+

MR

M m+mR

M m+

M(R r)

M m

−+

m(R r)

M m

−+

3gLgt

2−

3gLgt

2+

3gL

2

2mv ˆ ˆ( 2 i k)3 t

−∆

2mv ˆ ˆ( 3 i k)3 t

−∆

mv ˆ ˆ( 3 i k)2 3 t

−∆

mv ˆ ˆ( 3 i k)3 t

−∆

F

ˆvi

2 / 5

0m

2µ0m

µ03m

µ02m

µ

23Vh

4g=

min

tan

3

θµ =

min

mgsin

3= θcm

2ga sin

3= θ

min 2

2

mg sinf

r1

k

θ=+

cm

2

2ghv

I1

mR

=+

cm

2

g sina

I1

mR

θ = +

22

2

1 kmv 1

2 r

= +

2 2Translation Rotation cm cm

1 1K.E. K K mv I

2 2= + = + ω

cm cmL m(R v ) I= × + ω

Translation RotationL L L= +

cma R= α

cmv R= ω

IK

M=

iM m= ∑

2i iI m r= ∑I

KM

=

2r dm= ∫

2i iI m r= ∑

0

20

2 20

n 0

t

1t t

2

2

(2n 1)2

ω = ω + α

θ = ω + α

ω = ω + αθαθ = ω + −

θ

mg mg cos θ

θ

mg sin θ

N

2vcm

O

cm2 v

ω cm2 v

v

L

y

z

v

A

R

x B C

L P

B M

R m

r

F m2 m1


10

Decay of Current

Kirchhoff's Law.. Junction Rule

point charges

The electric field is along the directionwhere the potential decreases at themaximum rate.

5. Electric Potential Energy:Electric potential energy between two

point charges

Also, .

6. Electric Dipole Moment:Two point charges of equal magnitudebut opposite signs separated by a smalldistance form an electric dipole. Dipole moment is a vector quantity, p =2ql.

Direction of dipole is from -q to +q.Potential at the point P:

This equation gives the electric field atany point P at a distance r from thedipole.Torque on a Dipole in an Electric field:Torque =

In vector form: Torque = Potential Energy of a Dipole in anElectric Field:

No electric field can exist inside aconducting material.

Force on a dipole: is the

derivative of electric field with respectiveto distance along the direction of thedipole.

Gauss' Theorem: , where q0

is equal to the net charge enclosed withthe surface. Work done by the electricfield along the closed path is zero.

Mathematically: .

7. Capacitance:Capacitor is a device for storing charge.Mathematically, Q = CV, the unit of C isfarad.

For parallel plate capacitor: ,

where A = Plate area, d = Distancebetween the plates

8. Energy Stored in a Capacitor:

9. Energy Stored per Unit Volume in an

Electric Field (E): .

10. Electrostatic Pressure: , where

σ is surface charge density.11. Kirchhoff's Law:

In electrical circuits in which no tworesistors are in series or in parallel, it isnecessary to employ methods other thanthe series parallel method, one alternativemethod is → Kirchoff's rules.

Junction Rule: The total current directedinto a junction must equal to the totalcurrent directedout of thejunction.

Loop Rule: Around any closed circuitloop, the sum of the potential drops equalthe sum of the potential rises.

In problems involving Kirchhoff's rules italways helpful to mark resistors with +and - signs to keep the track of potentialrises and drops in the circuit.

12. Sign convention:

13. Potential rise and fall in a circuit

14. Loop Rule:

15. Junction Rule:

16. GROWTH OF CURRENT IN LRCIRCUITWhen switch "S" is

closed at t=0;

At time t, current

17. The constant L/R has dimensions of timeand is called the inductive time constant(τ) of the LR circuit.

18. Decay of current: When switch "S" is

open at t = 0; at t= 0, i = i0 at

time t,

19. Charging of a capacitor: When acapacitor is connectedto a battery, positivecharge appears on oneplate and negative charge on the other.The potential difference between theplates ultimately becomes equal to e.m.fof the battery. The whole process takessome time and during this time there is anelectric current through connecting wiresand the battery.

20. At any time t,

21. The constant RC has dimensions of timeand is called capacitive time constant (τ).

22. Discharging of capacitor:

23. LCR series circuit

;

When , the impedance

becomes minimum and hence current will bemaximum. The circuit is then said to beresonance and the corresponding frequency isknown as resonant frequency. The resonant

frequency = . The peak current in

this case is .

MODEL QUESTIONS

Passage:Two capacitors C1 and C2 bothhaving same capacitance C,are connected to two batteriesA and B each emf V as shownin figure. Initially both switches S1 and S2 areopen and capacitor C1 has charge q0 = CVand C2 has no charge. Now switch S1 isclosed and S2 remain open till the capacitorC2 becomes fully charged. Then S1 is openand S2 is closed till the flow of chargethrough battery is stopped. This processconstitutes one cycle. The cycle is repeatedfor a number of times.Answer the following questions:1. At the end of 1st cycle the charge on C1is

a) 2CV b) 3/2 CV c) CV/2 d) 5/2 CV2. After 2nd cycle the charge on capacitor C2

isa) 7/8 CV b) 7/4 CV c) CV/2 d) 3/2 CV

3. After 3rd cycle, the charge on capacitorC2 isa) 3/4CV b) 1/4CV c) 5/4CV d) 7/4CV

4. Work done by the battery B in 5 cycles isa) 127/64 CV2 b) 31/16 CV2

c) 63/32 CV2 d) 31/32 CV2

5. In the circuit in figure, ifno current flows throughthe galvanometer when thekey k is closed, the bridgeis balanced. The balancing condition forbridge is

a) b)

c) d)

6. Three concentric metallic spherical shellsof radii R, 2R, 3R, are given charges Q1,Q2, Q3 respectively. It is found that thesurface charge densities are same given tothe shells, Q1 : Q2 : Q3 isa) 1 : 2 : 3 b) 1 : 3 : 5c) 1 : 4 : 9 d) 1 : 8 : 18

21 222 1

C R

C R=

2 21 12 22 2

C R

C R=

1 2

2 1

C R

C R=1 1

2 2

C R

C R=

0

R

ε

1 1.

2 LCπ

L C

1X X or L

C= ω =

ω

0

1

LCω =L 1 / C

tanR

ω − ωϕ =

2 2rms R L C

22 2 2

L C

( ~ )

1Z R (X ~ X ) R L

C

ε = ε + ε ε

= + = + ω − ω

t

RCq Qe−

=

( )t

RCq C 1 e−

= ε −

t

0i i e−

τ=

diL Ri

dt− =

RtLi 1 e

R

− ε= −

diL Ri

dtε − =

+ + =

− − + + =

=

= = = −

1 2 3

1 2 3

Let V be potential difference between tw o junctions P and Q.

i i i 0

V 10 V V 200

2 2 2

V 10 Volts

i 0, i 5A, i 5A

+V i3 i1 2Ω 20V

10V i2

2Ω 2Ω

Q

P

1 1 2

1 2 2

1 2

For loop ABEFA

2i 2(i i ) 10 0

For loop BEDCB

2(i i ) 2i 20 0

By solving equation (1) and (2)

we get i 0, i 5A

− − + + =

− + − + =

= =

1 2 3

A O O B O C

1 2 3

i i i

V V V V V V

R R R

= +− − −= +

2

0

P2

σ=ε

20

U 1E

Volume 2= ε

21 1U CV , U QV

2 2= =

2QU

2C=

0AC

d

ε=

E d 0⋅ =∫

0

0

qE ds⋅ =

ε∫

E EF p ,

∂ ∂=∂ ∂

p E= − ⋅

p E×

2q sin Eθ×

20

t 30

30

23

0

1 p cosV

4 r

dV 1 pE 2cos

dr 4 r

1 dV 1 pE sin

r d 4 r

1 pE 1 3cos

4 r

1tan tan

2

θ

θ=πε

= − = ⋅ θπε

= − = ⋅ θθ πε

= + θπε

α = θ

1 22 1

0 2 1

Q Q 1 1U U

4 r r − = − πε

1

0

Q Q 1U

4 r

⋅=πε

2

1

r

2 1

r

dVV V E dr E

dr− = − ⋅ ⇒ = −∫

ni

i 10 i

Q1v

4 r=

=πε ∑

Ee

a

p –q

2

+q

Eb

θ

α

E

+q

–q

Junction Rule

The law of conservation of electric charge

Electric current

R1

i2

i1 O

R3

R2

i3

Junction

A

C

B

– + q0=CV

B V

S2

S1

V

A

C2

C1

R1

G

R2

C1 C2

K

I1

I2

E

R b

S a i

C i = 0

– +

E

R b

S a

••

• i

L

Loop Rule

The principle of conservation of energy

Electric potential

i + –

5Ω

i i=2A

+ –

+

1Ω

10V 2V

ε1

+ – + –

ε2

ε = ε + ε1 2

+ – – +

ε1 ε2

ε = ε − ε1 2

2A + –

12 2Ω

+ – – + 4Ω

8V

8V

12

A i2 i1 2Ω 20V

10V i1+i2

2Ω 2Ω

i2 F E

D

C B

R

~

i

L

RωL

1/ω

C

Zφ

C

i

ω

ωο

1/C

-L

ωω

Moseley's Law

Modern Physics and Optics


11

7. For the circuit shown inthe figurea) the current I through thebattery is 7.5 mAb) the potential difference across RL is 18Vc) ratio of powers dissipated in R1 and R2

is 3d) if R1 and R2 are interchanged,magnitude of the power dissipated in RL

will decrease by a factor of 9.8. A spherical portion has

been removed from a solidsphere having a chargedistributed uniformly in itsvolume as shown in the figure. Theelectric field inside the emptied space isa) Zero everywhereb) non-zero and uniformc) Non-uniformd) zero only at its centre

9. Positive and negative point charges ofequal magnitude are kept at (0, 0, a/2)and (0, 0, -a/2) respectively. The workdone by the electric field when anotherpositive point charge is moved from (-a,0, 0) to (0, a, 0) isa) positive b) negative c) zerod) depends on the path connecting theinitial and final positions

10. Find the potential difference between Aand B. VA - VB = ?

a) 6 V b) 12 V c) 15 V d) 18 VKey:

(1) b; (2) c; (3) a; (4) b; (5) b; (6) b;(7) a,d; (8) b; (9) c; (10) a

MODERN PHYSICS & OPTICS

1. Photo electric effect supports quantumnature of light becausea) there is a minimum frequency below

which no photoelectrons are emittedb) the maximum kinetic energy of

photoelectrons depends on thefrequency of light and not on itsintensity

c) even when the metal surface is finallyilluminated the photoelectrons leavethe surface immediately

d) electric charge of the photoelectrons isquantized

2. If the wavelength of light in anexperiment on photoelectric effect isdoubleda) the photoelectric emission will not

take placeb) the photoelectric emission may or

may not take placec) the stopping potential will increased) the stopping potential will decrease

3. An electron with kinetic energy 5 eV isincident on a hydrogen atom in its groundstate. The collisiona) must be elasticb) may be partially elasticc) must be completely inelasticd) may be completely inelastic

4. Moseley's law for characteristic x-rays isgiven by . Choose the correctstatement/sa) both 'a' and 'b' depend on the target

materialb) both 'a' and 'b' are independent of

target materialc) both 'a' and 'b' depend on the energy of

the electron beamd) both 'a' and 'b' depends on the nature

of the line (i.e. K, L, M, … etc)5. A point source of monochromatic light of

power P is placed at a distance of 'r' mfrom a metal surface of area 'a'. Lightfalls perpendicularly on the surface andwork function of the material is φ. If weassume wave theory of light to hold andalso that all the light falling on thesurface is absorbed by a single electronon the surface, the time required by theelectron to receive sufficient energy tocome out of the metal is

a) b) c) d)

6. A monochromatic light of wavelength λis incident on an isolated metallic sphereof radius r. The threshold wavelength isλ0 which is larger than λ. The number ofphotoelectrons emitted before theemission of photoelectrons will stop is

a) b)

c) d)

7. A small metal plate (work function φ) iskept at distance 'd' from a singly ionized,fixed ion. A monochromatic light beam isincident on the metal plate andphotoelectrons emitted. The maximumwavelength of the light beam so thatsome of the photoelectrons may go roundthe ion along a circle is

a) b)

c) d)

8. In a photoelectric effect set up a pointsource of light of power 3.2x10-3 W emitsmonoenergetic photons of energy 5 eV.The source is located at a distance of 0.8m from the centre of stationary metallicsphere of work function 3 eV and ofradius 8x10-3 m. The efficiency ofphotoelectron emission is one for every106 incident photons. It is observed that

photoelectron emission stops at a certaintime t after the light source is switchedon. The time t =a) 55 s b) 111 s c) 222 s d) 11 s

9. A radio active substance x decays intoanother radioactive substance y. Initiallyonly x was present. λx and λy are thedisintegration constants of x and y, Nx

and Ny are the number of nuclei of x andy at any time t. Number of nuclei Ny willbe maximum when

a) b)

c) d)

PassageThe spectrum of x-rays observed is as shownin figure (I → intensity of x-rays, λ→ wave length of x-rays)10. In figure which of the following represe-

nts the source of emission peaks 1 and 2?a) Bremsstrahlungb) Absorption of x-ray photons resulting

in electronic excitations in atomc) Emission of x-ray photons as a result

of electronic transitions in atomd) both (1) and (3)

11. In general, Bremsstrahlung will not beproduced by collisions between electronsanda) He b) He2+ c) Li1+ d) Protons

12. In figure, peaks P1 and P2 were producedby events that occurred with unequalprobabilities. Which peak was producedby the more probable event?a) P1, because the peak has shorter

wavelengthb) P1, because the peak has lower

intensity c) P2, because the peak has longer

wavelengthd) P2, because the peak has higher

intensityPassageAn experimental set up ofverification of photoelectriceffect is shown in the diagram. The voltageacross the electrodes is measured with thehelp of potentiometer as shown in figure. theresistance of 100 cm long potentiometer wire(PQ) is 8Ω . Both electrodes A and C aremade of photomaterial potassium oxide.13. When radiation falls on the cathode phot-

ometal C a current of 2 µA is recorded inthe ideal ammeter A. If we assume thatthe vacuum tube set up followsapproximately ohm's law, the equivalentresistance of vacuum tube operating inthis case when jockey is at end P isa) 8 x108Ω b) 16 x106Ωc) 8 x106Ω d) 10 x106Ω

14. When other light falls on the anode platethe ammeter reading remains zero till,jockey is moved from the end P to themiddle point of the wire PQ. There afterthe deflection is recorded in the ammeter.The maximum kinetic energy of theemitted electron isa) 16 eV b) 8 eVc) 4 eV d) 10 eV

15. Column-I shows four situations ofstandard Young's double slit arrangementwith the screen placed far away from theslits S1 and S2. In each of these cases S1P0

= S2P0, S1P1 - S2P1 = λ/4 and S1P2 - S2P2

= λ/3, where λ is the wavelength of thelight used. In the cases B, C and D, atransparent sheet of refractive index µand thickness t is pasted on slit S2. Thethicknesses of the sheets are different indifferent cases. The phase differencebetween the light waves reaching a pointP on the screen from the two slits isdenoted by δ(P) and the intensity by I(P).Match the situation given in Column Iwith the statement(s) in Column II validfor that situation.

16. Match the following.

Key: (1) a,b,c; (2) b,d; (3) a; (4) b,d; (5) c; (6)b; (7) a; (8) b; (9) c (10) c; (11) a; (12) d;(13) c; (14) b; (15) A→ (p,s); B→ (q);C→ (t); D→ (r,s,t) (16) a→ p,q,r,s; b→p,q,r,s; c → p,q,r,s; d→ p,q,r,s

y x xN Nλλ = λy x xN Nλλ = λ

y x

x y x y

N

N N

λ=− λ − λ

y y

x y x y

N

N N

λ=

− λ − λ

20

4 dhc

e 4 d0πε

+ πε φ20

8 dhc

e 8 d0πε

− πε φ

20

4 dhc

e 8 d0πε

+ πε φ20

8 dhc

e 8 d0πε

+ πε φ

20

4 rhc 1 1

3e0πε − λ λ

20

8 rhc 1 1

e0πε − λ λ

20

4 rhc 1 1

e0πε − λ λ

20

2 rhc 1 1

e0πε − λ λ

2Er

aP

π24 Er

aP

π22 Er

aP

π24 Er

P

π

a(z b)ν = −

I P1 P2

λ

J Q P

A C

A

ν

2Ω 20V

A 1Ω 2A 9V B 2Ω 3V 3Ω

I

24V RL

2kΩ

6kΩ 1.5kΩ

R1

R2

Column I Column IIA) p) δ(P0) = 0

q) δ(P1) = 0r) I(P1) = 0s) I(P0) > I(P1)t) I(P2) > I(P1)

B) (µ-1)t = λ/4

C) (µ-1)t = λ/2

D) (µ-1)t = 3λ/4

S1

S2 P2 P1 P0

S1

S2 P2 P1 P0

S1

S2 P2 P1 P0

S1

S2 P2 P1 P0

Column Ia) Mass of the products

is less than theoriginal mass of thesystem

b) Binding energy pernucleon increases

c) Mass number isconserved

d) Charge number isconserved

Column IIp) α-decay

q) β-decay

r) Nuclear fission

s) Nuclear fusion


12

Quick Review

Factors to Remember..

EIsolated charge

r q p =

πε 20

1 qE .

4 r=

πε0

qV

4 r

y

E⊥

E||–q +

2a x

⊥

=πε

=πε

|| 30

30

1 2pE .

4 x

1 pE .

4 y ⊥

=πε

=

|| 20

1V

4 x

V 0

r

P

x θ

θ +=πε

2

30

1 p 3cos 1E

4 r

θ=πε 2

0

pcosV

4 r

p x

q R

+ + +

+ +

+

+ + + + + + +

=πε +2 2 3/2

0

1 qxE .

4 (R x )=

πε +2 20

1 qV

4 R x

+ + +

+ +

+

+ + + + + + +

σ p x

q R σ

σ = − ε + 2 20

xE 1

2 x R

2 2

0V [ R x x]

2σ= + −ε

+ + +

σ + + + +

+

0E

2σ=ε

p

+ + + +

r λ λ=

πε0

1 2E

4 r

P

x + + + +

β α

⊥λ= α+ β

πε0E (sin sin )

4 x

λ= α − βπε||

0E (cos cos )

4 x

0

sec tanV ln

4 sec tanλ β + β=πε α − α

R

+ + + +

+ +

+ + + + + + + +

a) Inside, 0 ≤ r ≤ R, E = 0

b) Outside, r ≥ R, E =2

0

q

4 rπε

a) Inside, 0 ≤ r ≤ R,V=0

q4 Rπε

b) Outside, r ≥ R, V=0

q4 rπε

R +

+ + +

+ + + + +

a) Inside, 0 ≤ r ≤ R, E=0

r3ρε

b) Outside, r≥R, E=2

0

R R3 rρ

ε

a) Inside,

0 ≤ r ≤ R, V=2 2

20

R r3

6 R

ρ − ε

b) Outside,

r ≥ R, V =3

0

R 13 rρ

ε

System Electric field intensity Potential

Isolated charge

Diplole

x,y>>a

A ring of

charge

A disc of

charge

Infinite sheet ----

of charge

Infinitely long

line of charge ----

Finite line

of charge

Charged

spherical

shell

Solid shpere

of charge

S.No. Process Law applicable

Quantity that remains constant

U∆ W∆ H U W∆ = ∆ + ∆

1. Isochoric Gay-

Lussac’s law Volume VnC dT 0 VnC dT

2. Isobaric Charles’ law Pressure VnC dT P(V2 – V1) PnC dT

3. Free expansion

– Temperature 0 0 0

4. Isothermal process

Boyle’s law Temperature 0 2

1

VnRTln

V

2

1

VnRT ln

V

5. Adiabatic process

– PV constantγ =

W−∆ 1 2nR (T T )

1

−γ −

0

6. Polytropic process ×

nPV constant= VnC dT 1 1 2 2P V P V

(n 1)

−−

×

7. Cyclic process × × 0

Area of P-V diagram

∆W

Thermodynamic processes

Electric field and potential for typical situations

Finally it is clear that more emphasisshould be given for Mechanics, Electricityand Magnetism.

In Thermodynamics and Modern Physics,some topics being common to bothPhysics and Chemistry, it is better to gothrough these topics keeping both Physicsand Chemistry syllabus in mind.

In Optics, Wave Optics may be coveredfirst as the content is less compared toGeometrical Optics.

Of Waves, transverse waves and soundwaves have almost same weightage. Themathematical part involved in S.H.M.,Physical Optics, LC-oscillations and A.C.Circuits is useful.

Another common area is Gravitation,Electrostatics and Magnetism with sameprinciples and applications by change ofvariables. Coulomb's law becomes New-ton's Gravitational Law. The analogue ofGauss's Law can be used for evaluation ofgravitational field. Electricity and

Magnetism can be studied together. In Physics, usually questions connected to

different chapters will be asked. It issuggested that students take a basicprinciple and cover all topics where it canbe applied. In Mechanics, most of theproblems related to conservation of LinearMomentum, Energy, Angular Momentumand Newton's second law will be asked.Similarly S.H.M. can be clubbed withmany other chapters likei) Oscillation of a floating body is

S.H.M.ii) In Thermodynamics small oscillation

of piston is S.H.M.iii) A small negative charge at the centre

of a positively charged ring will beS.H.M.

iv) In YDSE source undergoing S.H.M.may be linked to fringe width.

v) A small current element placed above/ below a current carrying wire mayunder go S.H.M.

WHAT TO DO IN THE LAST 5 TO 6 DAYS

Already you might have written number ofmodel tests. In the last 5 to 6 days it is betterto revise basic concepts and form-ulae of allthe three subjects. If possible it is better towrite basic formulae on paper. Do not studyanything new, you can also go through last2 to 3 three model examin-ation papers thatyou have done and mark-ing schemeapplicable. But, be careful the pattern of thepaper may be different from the paperwhich you have written in your college/inst-itute. Do not panic, relax and try to underst-and the new pattern and marking scheme.

On The Examination day:Do's

Read the instructions given for every typeof question. First 5 to 6 minutes gothrough complete paper, during this timeyour brain comes to relaxed state.

In the first 1½ hour, it is better to allothalf-an-hour for each subject.

After Paper-1, the only analysis that you

should do is to understand the pattern ofthe question paper-1.

Give more attention to subject in thePaper-2 if you have not attempted thatsubject to your satisfaction in Paper-1.

If you think a problem involves multiplesteps, try to attempt in the end.

Don't's Don't discuss any subject matter with

your friends or classmates, be cool andrelax. You will feel little tension, it isquite natural. It is like tension in vibr-ating string. It vibrates and gives beauti-ful music only when tension exists with-in limits. If you apply tension beyond li-mits it may break.

Don't spend more than 1½ hour on anysingle subject in any paper. That wouldhamper your chances with respect tosubject wise cut offs.

Don't discuss the Paper-1 with yourfriends / classmates or teachers.

Relax during two hours break. Your mindnee-ds to be focused on the next Paper-II.

Preparation Tips

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