1 important formulas/ - answer key results...

1

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Average acceleration, aav1 1 2

1 2

a t + a t=

t + t2

The area under the velocity-time curveis equal to the displacement and slope givesacceleration.

If a body falls freely, the distancecovered by it in each subsequent second starting from firstsecond will be in the ratio 1 : 3 : 5 : 7 etc.If a body is thrown vertically up with an initial velocity u, ittakes u/g second to reach maximum height and u/g secondto return, if air resistance is negligible.If air resistance acting on a body is considered, the timetaken by the body to reach maximum height is less than thetime to fall back the same height.

For a particle having zero initial velocity if s taµ , where2a > , then particle’s acceleration increases with time.

For a particle having zero initial velocity if s taµ , where0a < , then particle’s acceleration decreases with time.

Kinematic equations :v = u + at (t) ; v

2 = u2 + 2at (s)

S = ut + 12

at (t)2; Sn u (2n 1)

2a

= + -

applicable only when t t| a | a=r is constant.

at = magnitude of tangential acceleration, S = distanceIf acceleration is variable use calculus approach.Relative velocity : BA B Av v v= -

r r r

If T is the time of flight, h maximumheight, R horizontal range of a projectile, a

its angle of projection, then therelations among these quantities.

2gTh

8= . ...... (1);

gT2 = 2R tan a ....... (2);R tan a = 4h ....... (3)

2 22u sin u sin;2

T hg g

q q= =

2u sin 2Rg

q=

For a given initial velocity, to get the same horizontal range,there are two angles of projection a and 90° – a.

The SI system : It is the internationalsystem of units. In this system there areseven fundamental and two supplementaryquantities and their corresponding units.These are:

Quantity Unit Symbol1. Length metre m2. Mass kilogram kg3. Time second s4. Electric current ampere A5. Temperature kelvin K6. Luminous intensity candela cd7. Amount of substance mole molSupplementary1. Plane angle radian rad2. Solid angle steradian sr

Dimensions : These are the powers to which the fundamentalunits are raised to get the unit of a derived quantity.Uses of dimensions(i) To check the correctness of a physical relation.(ii) To derive relationship between different physical quantities.(iii) To convert one system of unit into another.

n1u1 = n2u2

1 1 1 1[M L T ]a b cn = 2 2 2 2[M L T ]a b cn

Significant figures : In any measurement, the reliable digitsplus the first uncertain digit are known as significant figures.Error : It is the difference between the measured value andtrue value of a physical quantity.Absolute error : The magnitude of the difference between thetrue value and the measured value is called absolute error.

1 1a aD = -a , 2 2a aD = -a , n na aD = -a

Mean absolute error

1 2| | | | ..... | |na a an

D + D + + DD =a =

1

1 | |n

ii

an =

DåRelative error : It is the ratio of the mean absolute errorto its true value

or relative error = Daa

and percentage error = 100D´

aa

Percentage error : It is the relative error in percent.

Percentage error 100%mean

aa

æ öD= ´ç ÷è ø

UNITS ANDMEASURE-

MENTSMOTION IN A

STRAIGHTLINE

MOTION IN APLANE

IMPORTANT FORMULAS/TERMS for JEE MAIN

PHYSICS

2


The equation to the parabola traced by a body projectedhorizontally from the top of a tower of height y, with a velocityu is y = gx2/2u2, where x is the horizontal distance coveredby it from the foot of the tower.

Equation of trajectory is 2

2 2gxy x tan

2u cos= q -

q, which is

parabola.Equation of trajectory of an oblique projectile in terms of

range (R) is xy x tan 1

Ræ ö= q -ç ÷è ø

Maximum height is equal to n times the range when theprojectile is launched at an angle q = tan–1(4n).In a uniform circular motion, velocity and acceleration areconstants only in magnitude. Their directions change.In a uniform circular motion, the kinetic energy of the body

is a constant. W = 0, a 0, P constant, L=constant¹ ¹r rr

Centripetal acceleration, 2

2r

va r v

r= w = = w (Always

applicable)2

2 2r 2

4a 4 n r rTp

= p = (Applicable in uniform circular motion)

n = frequency of rotation, T = time period of rotation.

ra v= w ´rr r

Newton’s second law :

F ma, F dp / dt= =r rr r

Impulse : 2

2 11

p F t, p p F dtD = D - = òrr

Newton’s third law : 12 21F F= -r r

Frictional force s s max sf (f ) R£ = m ; k kf R= m

Circular motion with variable speed. For complete circles,

the string must be taut in the highest position, 2u 5ga³ .Circular motion ceases at the instant when the stringbecomes slack, i.e., when T = 0, range of values of u for

which the string does go slack is 2ga u 5ga< < .

Conical pendulum : g / hw = where h is height of a pointof suspension from the centre of circular motion.The acceleration of a lift

a =actual weight apparent weight

mass-

If ‘a’ is positive lift is moving down, and if it is negative thelift is moving up.

Work done W = FS cosqRelation between kinetic energy E and

momentum, P 2m= EIf a body moves with constant power,

its velocity (v) is related to distancetravelled (x) by the formula v µ x3/2.

Work due to kinetic force of friction between two contactsurfaces is always negative. It depends on relativedisplacement between contact surfaces. FK K relW F (S )= - .

W KS = S D , WS Þ total work due to all kinds of forces,

KS D Þ total change in kinetic energy..

conservativeW US = -S D ; conservativeWS Þ Total work dueto all kinds of conservative forces.S D Þ Total change in all kinds of potential energy..

Coefficient of restitution velocity of separationevelocity of approach

=

The total momentum of a system of particles is a constant inthe absence of external forces.

The centre of mass of a system ofparticles is defined as the point whose

position vector is i im rR

M= å The centre

of gravity of an extended body is that pointwhere the total gravitational torque on the body is zero.The angular momentum of a system of n particles about the

origin is n

i ii 1

L r p=

= ´å ; L = mvr = Iw

The torque or moment of force on a system of n particles

about the origin is i ii

r Ft = ´å

The moment of inertia of a rigid body about an axis is defined

by the formula 2i iI m r= å

The kinetic energy of rotation is 21K I2

= w

The theorem of parallel axes : 'zI = Iz + Ma2

Theorem of perpendicular axes : Iz = Ix + IyFor rolling motion without slipping vcm = Rw, where vcm isthe velocity of translation (i.e. of the centre of mass), R isthe radius and m is the mass of the body. The kinetic energyof such a rolling body is the sum of kinetic energies of

translation and rotation : 2 2cm

1 1K mv I2 2

= + w

LAWS OFMOTION

WORK,ENERGY AND

POWER

SYSTEM OFPARTICLES AND

ROTATIONALMOTION

3


A rigid body is in mechanical equilibrium if(a) It is translational equilibrium i.e., the total external forceon it is zero : S Fi = 0.(b) It is rotational equilibrium i.e., the total external torqueon it is zero : S ti = Sri × Fi = 0.If a body is released from rest on rough inclined plane, then

for pure rolling rn tan

n 1m ³ q

+ (Ic = nmr2)

Rolling with sliding sn

0 tann 1

æ ö< m < qç ÷è ø+ ;

g sina g sin

n 1q

< < q+

Gravitational force F 1 22

Gm m=

r

G = 6.67 × 10 –11 2

2Nmkg

The acceleration due to gravity.(a) at a height h above the Earth’s surface

E E2 2

EE E

GM GM 2hg(h) 1R(R h) R

æ ö= = -ç ÷è ø+ for h << RE

E

2hg(h) g(0) 1R

æ ö= -ç ÷è ø where E

2E

GMg(0)

R=

(b) at depth d below the Earth’s surface is

E2

E EE

GM d dg(d) 1 g (0) 1R RR

æ ö æ ö= - = -ç ÷ ç ÷è ø è ø

(c) with latitude l g1 = g – Rw2 cos2l

Gravitational potential VgGM

= –r

Intensity of gravitational field 2GM

I =r

The gravitational potential energy

1 2Gm mV constant

r= - +

The escape speed from the surface of the Earth is

Ee E

E

2GMv 2gR

R= = and has a value of 11.2 km s–1.

Orbital velocity, vorbi E

E

GM=

R EgR=

A geostationary (geosynchronous communication) satellitemoves in a circular orbit in the equatorial plane at aapproximate distance of 4.22 × 104 km from the Earth’s centre.

smax

GM 1 ev

a 1 e+æ ö= ç ÷è ø- ; s

minGM 1 e

va 1 e

-æ ö= ç ÷è ø+

Whenever force responsible for orbital motion obeys inversesquare law, then only square of time period is directlyproportional to cube of average distance between planetand sun. This is Kepler’s 3rd law of planetary motion.

2 3T aµ ;2 3

1 12 32 2

T aT a

=

Hooke’s law : stress µ strainYoung’s modulus of elasticity

FY

AD

=l

l

Compressibility 1=Bulk modulus

Y = 3k (1 – 2s)Y = 2n (1 + s)If S is the stress and Y is Young’s modulus, the energydensity of the wire E is equal to S2/2Y.If a is the longitudinal strain and E is the energy density ofa stretched wire, Y Young’s modulus of wire, then E is equal

to 21Y

2a

Pascal’s law : A change in pressureapplied to an enclosed fluid is transmittedundiminished to every point of the fluid andthe walls of the containing vessel.

Bernoulli’s principleP + rv2/2 + rgh = constant

Surface tension is a force per unitlength (or surface energy per unit area) acting in the planeof interface.Stokes’ law states that the viscous drag force F on a sphereof radius a moving with velocity v through a fluid of viscosity hF = – 6phav.

Terminal velocity VT = 22 r ( – σ)g

9r

h

The surface tension of a liquid is zero at boiling point. Thesurface tension is zero at critical temperature.If a drop of water of radius R is broken into n identical drops,the work done in the process is 4pR2S(n1/3 – 1).Two capillary tubes each of radius r are joined in parallel.The rate of flow is Q. If they are replaced by single capillarytube of radius R for the same rate of flow, thenR = 21/4 r.

Coefficient of viscosity n = FdvAdx

-æ öç ÷è ø

Velocity of efflux 2V gh=

GRAVITATION

MECHANICALPROPERTIESOF SOLIDS

MECHANICALPROPERTIESOF FLUIDS

4


The coefficient of linear expansion(al), superficial (b) and volume expansion

(av) are defined by the relations :

TD

= a Dll

l ;

AT

AD

= bD ; VV

TV

D= a D

av = 3al ; 2b = a

l

In conduction, heat is transferredbetween neighbouring parts of a body through molecularcollisions, without any flow of matter. The rate of flow of

heat C DT TH KA

L-

= , where K is the thermal conductivity of

the material of the bar.Convection involves flow of matter within a fluid due tounequal temperatures of its parts.Radiation is the transmission of heat as electromagneticwaves.Stefan’s law of radiation : E = sT4, where the constant s isknown as Stefan’s constant = 5.67 × 10–8 wm–2 k–4.Wein’s displacement law : lmT = constant, where constantis known as Wein’s constant = 2.898 × 10–3 mk.

Newton’s law of cooling: 2 1dQ

k (T T )dt

= - - ; where TT1 is

the temperature of the surrounding medium and T2 is thetemperature of the body.Heat required to change the temperature of the substance,Q = mcDqc = specific heat of the substanceHeat absorbed or released during state change Q = mLL = latent heat of the substance

Mayer’s formula cp – cv = R

First law of thermodynamics: DQ = DU+ DW, where DQ is the heat supplied to thesystem, DW is the work done by the systemand DU is the change in internal energy ofthe system.

In an isothermal expansion of an idealgas from volume V1 to V2 at temperature T

the heat absorbed (Q) equals the work done (W) by the gas,

each given by 2

1

VQ W nRT ln

Væ ö

= = ç ÷è øIn an adiabatic process of an ideal gas PVg = TVg–1

1T

P

g

g -= = constant, where

p

v

CC

g = .

Work done by an ideal gas in an adiabatic change of state

from (P1, V1, T1) to (P2, V2, T2) is 1 2nR (T T )W

1-

=g -

The efficiency of a Carnot engine is given by2

1

T1

Th = -

Second law of thermodynamics: No engine operatingbetween two temperatures can have efficiency greater thanthat of the Carnot engine.

Entropy or disorder Q

S =T

d

Kinetic theory of an ideal gas gives

the relation 21P nmv3

= , Combined with

the ideal gas equation it yields a kineticinterpretation of temperature.

2B

1 3nmv k T

2 2= ,

2 1/2 Brms

3k Tv (v )

m= =

The law of equipartition of energy is stated thus: the energyfor each degree of freedom in thermal equilibrium is 1/2 (kBT)

The translational kinetic energy B3

E k NT2

= . This leads

to a relation 2

PV E3

= .

Root mean square (rms) velocity of the gas

33RT PC =M

=r

Most probable speed Vmp2RT 2KT= =M m

Mean free path l2

KT=2 d Pp

The particle velocity and accelerationduring SHM as functions of time are givenby,v (t) = – wA sin (w + f ) (velocity),a (t) = – w2A cos (wt + f) = – w2x (t)

(acceleration)

Velocity amplitude vm=w A and acceleration amplitude

am =w2A.

A particle of mass m oscillating under the influence of aHooke’s law restoring force given by F = – k x exhibits simple

harmonic motion with km

w = (angular frequency),

mT 2k

= p (period)

Such a system is also called a linear oscillator.A body of mass M is suspended from a spring whose forceconstant is K and mass is m. The time period of this system

will be 2p (M m / 3)

k+

Time period for conical pendulum cosT 2g

æ öq= p ç ÷è ø

l where

q angle between string & vertical.

Energy of the particle 2 21E = mω A2

THERMALPROPERTIESOF MATTER

THERMODY-NAMICS

KINETICTHEORY

OSCILLATIONS

5


The displacement in a sinusoidal wavey (x, t) = a sin (kx – wt + f) where f is thephase constant or phase angle.

The speed of a transverse wave on a

stretched string v T /= m .Sound waves are longitudinal

mechanical waves that can travel throughsolids, liquids, or gases. The speed v of sound wave in a

fluid having bulk modulus B and density µ is v B /= r .The speed of longitudinal waves in a metallic bar is

v Y /= r

For gases, since B = g P, the speed of sound is v P /= g rThe interference of two identical waves moving in oppositedirections produces standing waves. For a string with fixedends, standing wave y (x, t) = [2a sin kx ] cos wtThe separation between two consecutive nodes or antinodesis l/2.A stretched string of length L fixed at both the ends vibrates

with frequencies 1 v

f2 2L

= .

The oscillation mode with lowest frequency is called thefundamental mode or the first harmonic. The secondharmonic is the oscillation mode with n = 2 and so on.A pipe of length L with one end closed and other end open(such as air columns) vibrates with frequencies given by

1 vf n

2 2Læ ö= +ç ÷è ø , n = 0, 1, 2, 3, ....

The lowest frequency given by v/4L is the fundamental modeor the first harmonic.Beats arise when two waves having slightly differentfrequencies, f1 and f2 and comparable amplitudes, aresuperposed. The beat frequency fbeat = f1 – f2The Doppler effect is a change in the observed frequency ofa wave when the source S and the observer O moves relative

to the medium. 00

s

v vf f

v væ ö±

= ç ÷±è ø

Doppler effect formula in light : d

cl n

=l

, where dl is change

in wavelength of a spectral line of original wave length land n, the speed of the source and c is the speed of light.

Coulomb’s Law : 21Fr

= force on q2

due to q1 = 1 2212

21

k (q q )r̂

r where

0

1k4

=pe

= 9 × 109 Nm2 C–2

Electric field due to a point charge qhas a magnitude | q |/4pe0r2

Field of an electric dipole in its equatorial plane

2 2 3/20

p 1E4 (a r )-

=pe +

r

30

p4 r

-@

pe

r

, for r >> a

Dipole electric field on the axis at a distance r from the centre:

2 2 2 30 0

2pr 2pE4 (r a ) 4 r

= @pe - pe

r rr

for r >> a

Dipole moment 2p q a=r

In a uniform electric field Er , a dipole experiences a torque

tr given by p Et = ´

rr r but experiences no net force.

The flux Df of electric field Er through a small area element

SDr

is given by E. SDf = Drr

Gauss’s law: The flux of electric field through any closedsurface S is 1/e0 times the total charge enclosed i.e., QThin infinitely long straight wire of uniform linear charge

density l :0

Ê n2 r

l=

pe

r

Infinite thin plane sheet of uniform surface charge density s

0Ê n

2s

=e

r

Thin spherical shell of uniform surface charge density s :

20

Ê r4 r

s=

pe

r

(r R)³ ; E 0=r

(r < R)

Electric Potential :0

1 QV(r)4 r

=pe

r .

For a charge configuration q1, q2, ..., qn with position vectorsr1, r2, ... rn, the potential at a point P is given by the

superposition principle 1 2 n

0 1P 2P nP

q q q1V .....4 r r r

æ ö= + + +ç ÷pe è ø

,

An equipotential surface is a surface over which potentialhas a constant value. Potential energy of two charges q1, q2 at 1 2r , rr r is given by

1 2

0 12

q q1U4 r

=pe

, where r12 is distance between q1 and q2.

Capacitance C = Q/V, where Q = charge and V = potentialdifferenceFor a parallel plate capacitor (with vacuum between theplates),

0ACd

= e , where A is the area of each plate and d the

separation between them.The energy U stored in a capacitor of capacitance C, withcharge Q and voltage V is

221 1 1 Q

U QV CV2 2 2 C

= = =

For capacitors in the series combination,

eq 1 2 3

1 1 1 1.........

C C C C= + + +

In the parallel combination, Ceq = C1 + C2 + C3 + ...

where C1, C2, C3... are individual capacitances.

WAVES

ELECTRO-STATICS

6


Current density j gives the amount ofcharge flowing per second per unit areanormal to the flow, dJ nqv=

r

Resistance R =A

rl

r = resistivity of the material

Equation E J= rr r

another statement ofOhm’s law, i.e., a conducting material obeys Ohm’s law whenthe resistivity of the material does not depend on themagnitude and direction of applied electric field.r= resistivity of the material.(a) Total resistance R of n resistors connected in seriesR = R1 + R2 +..... + Rn(b) Total resistance R of n resistors connected in parallel

1 2 n

1 1 1 1......R R R R

= + + + .

Kirchhoff’s Rules – (a) Junction rule: At any junction ofcircuit elements, the sum of currents entering the junctionmust equal the sum of currents leaving it.(b) Loop rule: The algebraic sum of changes in potentialaround any closed loop must be zero.The Wheatstone bridge is an arrangement of four resistancesR1, R2, R3, R4. The null-point condition is given by

31

2 4

RRR R

=

The potentiometer is a device to compare potentialdifferences. The device can be used to measure potentialdifference; internal resistance of a cell and compare emf’s

of two sources. Internal resistance 1

2r R 1

æ ö= -ç ÷è ø

l

l

RC circuit : During charging : q = CE (1 – e–t/RC)During discharging : q = q0e–t/RC

The total force on a charge q movingwith velocity v i.e., Lorentz force.

F q (v B E)= ´ +r r rr .

A straight conductor of length l andcarrying a steady current I experiences a

force Fr

in a uniform external magnetic field Br

, F I B= ´rr rl ,

the direction of l is given by the direction of the current.

Biot-Savart law 03

d rdB I4 rm ´

=p

r rr l

.

The magnitude of the magnetic field due to a circular coil ofradius R carrying a current I at an axial distance x from the

centre is 2

02 2 3/2

IRB

2 (x R )m

=+

.

The magnitude of the field B inside a long solenoid carryinga current I is : B = µ0nI, where n is the number of turns per

unit length. For a toroid one obtains, 0NIB

2 rm

=p

.

Ampere’s Circuital Law: 0C

B.d I= mòrrlÑ , where I refers to the

current passing through S.

Force between two long parallel wires 10 1 2I IF Nm

2 a-m

=p

.

The force is attractive if currents are in the same directionand repulsive currents are in the opposite direction.

For current carrying coil M NIA=rr

; torque = M Bt = ´r rr

The magnetic intensity, 0

0

BH =

m

rr

.

The magnetisation Mr of the material is its dipole moment

per unit volume. The magnetic field B in the material is,

0B (H M)= m +r r r

For a linear material M H= cr r

. So that B H= mr r

and c iscalled the magnetic susceptibility of the material.

0 rm = m m ; r 1m = + c .

The magnetic flux

B B.A BA cosf = = qrr

, where q is the angle

between Br

& Ar

.Faraday’s laws of induction :

BdN

dtf

e = -

Lenz’s law states that the polarity of the induced emf is suchthat it tends to produce a current which opposes the changein magnetic flux that produces it.The induced emf (motional emf) across ends of a rod B ve = l

The self-induced emf is given by, dILdt

e = -

L is the self-inductance of the coil.2

0 N AL =

ml

A changing current in a coil (coil 2) can induce an emf in anearby coil (coil 1).

21 12

dIM

dte = - , M12 = mutual inductance of coil 1 w.r.t coil 2.

0 1 2N NM =

Aml

LR circuit : For growth current, Rt/L0i i [1 e ]-= -

For decay of current, Rt/L0i i e-=

CURRENTELECTRICITY

MAGNETISM

ELECTRO-MAGNETICINDUCTION

7


For an alternating current i = im sin wtpassing through a resistor R, the averagepower loss P (averaged over a cycle) due tojoule heating is (1/2)i2mR.

Root mean square (rms) current

mm

iI 0.707 i

2= = .

The average power loss over a complete cycleP = V I cosf. The term cos f is called the power factor.An ac voltage v = vm sinwt applied to a pure inductor L,drives a current in the inductor i = im sin (wt – p/2), whereim = vm/XL. XL = wL is called inductive reactance.An ac voltage v = vm sinwt applied to a capacitor drives acurrent in the capacitor: i = im sin (wt + p/2). Here,

mm C

C

v 1i , XX C

= =w

is called capacitive reactance.

An interesting characteristic of a series RLC circuit is thephenomenon of resonance. The circuit exhibits resonance,i.e., the amplitude of the current is maximum at the resonant

frequency, 0 L C1

(X X )LC

w = = .

Impedance 2 2L Cz = R + (x – x )

Transformation ratio, K S S P

P P S

N E I= = =

N E I

The quality factor Q defined by 0

0

L 1QR CR

w= =

w is an

indicator of the sharpness of the resonance, the higher valueof Q indicating sharper peak in the current.

Reflection is governed by the equationÐ i = Ð r' and refraction by the Snell’s law,sini/sinr = n, where the incident ray, reflectedray, refracted ray and normal lie in the sameplane.

Mirror equation: 1 1 1v u f+ =

Prism Formula 2 m21

1

n sin [(A D ) / 2)]n

n sin (A / 2)+

= = , where Dm

is the angle of minimum deviation.Dispersion is the splitting of light into its constituent colours.The deviation is maximum for violet and minimum for red.

Dispersive power v rd - dw =

d, where dv, dr are deviation

of violet and red and d the deviation of mean ray (usuallyyellow).

For refraction through a spherical interface (from medium 1to 2 of refractive index n1 and n2, respectively)

2 1 2 1n n n nv v R

-- = .

Thin lens formula 1 1 1v u f- =

Lens maker’s formula : 2 1

1 1 2

(n n )1 1 1f n R R

æ ö-= -ç ÷è ø

The power of a lens P = 1/f. The SI unit for power of a lens isdioptre (D): 1 D = 1 m–1.If several thin lenses of focal length f1, f2, f3,.. are in contact,

the effective focal 1 2 3

1 1 1 1= + + +.....f f f f

The total power of a combination of several lensesP = P1 + P2 + P3 +.......Chromatic aberration is the colouring of image produced bylenses. This can be avoided by combining a convex and aconcave lens of focal lengths f1 and f2 and dispersive powersw1, w2 respectively satisfying the equation

1 2

1 20

f fw w

+ = or in terms of powers 1 1 2 2P P 0w +w = .

Wave front : It is the locus of all theparticles vibrating in the same phase.

The resultant intensity of two wavesof intensity I0/4 of phase difference f atany points

20I I cos

2fé ù= ê úë û , where I0 is the maximum

density.

Condition for dark band : (2n 1)2l

d = - , for bright band :

= md l

Fringe width Ddl

b =

A thin film of thickness t and refractive index µ appears darkby reflection when viewed at an angle of refraction r if

2µt cos r = nl (n = 1, 2, 3, etc.)A single slit of width a gives a diffraction pattern with acentral maximum. The intensity falls to zero at angles of

2,

a al l

± ± , etc.

Amplitude of resultant wave R 2 2= 2 cosa b ab+ + f

Intensity of wave I = I1 + I2 + 2 1 2I I cosf

ALTERNATINGCURRENT

RAY OPTICS

WAVE OPTICS

8


Energy of a photon hcE = h =nl

Momentum of a photon hP =l

Einstein’s photoelectric equation

2max 0 0 0

1 mv V e h h ( )2

= = n - f = n- n

Mass defect,DM = (Z mp + (A – Z )mn) – M ; D Eb = DM c2.

1 amu = 931 MeV2

n 2ZE 13.6eVn

= - ´ (For hydrogen like atom)

Radius of the orbit of electron 2 2

2 2n hr =

4π mkzeBragg’s law : 2d sin q = nl.

Radius of the nucleus 1/3oR = R A

Law of radioactive decay : N = N0e–lt.

Activity = dN

NN

= -l (unit is Becquerel)

Half life period, 1/20.693T =

l

X-rays : min12400 Å

Vl =

Characteristics X-rays : K La al < l

Moseley law : n = a (Z – b)2

Pure semiconductors are called ‘intrinsic semiconductors’.The presence of charge carriers (electrons and holes) numberof electrons (ne ) is equal to the number of holes (nh).The number of charge carriers can be changed by ‘doping’of a suitable impurity in pure semiconductors known asextrinsic semiconductors (n-type and p-type).In n-type semiconductors, ne >> nh while in p-typesemiconductors nh >> ne.n-type semiconducting Si or Ge is obtained by doping withpentavalent atoms (donors) like As, Sb, P, etc., while p-typeSi or Ge can be obtained by doping with trivalent atom(acceptors) like B, Al, In etc.p-n junction is the ‘key’ to all semiconductor devices. Whensuch a junction is made, a ‘depletion layer’ is formed consistingof immobile ion-cores devoid of their electrons or holes. Thisis responsible for a junction potential barrier.In forward bias (n-side is connected to negative terminal ofthe battery and p-side is connected to the positive), the barrieris decreased while the barrier increases in reverse bias.Diodes can be used for rectifying an ac voltage (restrictingthe ac voltage to one direction).

Zener diode is one such special purpose diode. In reversebias, after a certain voltage, the current suddenly increases(breakdown voltage) in a Zener diode. This property hasbeen used to obtain voltage regulation.The important transistor parameters for CE-configuration are:

Input resistance, BEi

B VCE

Vr

Iæ öD

= ç ÷Dè ø

Output resistance, CE0

C IB

Vr

Iæ öD

= ç ÷Dè ø

Current amplification factor, C

B VCE

II

æ öDb = ç ÷Dè ø

The voltage gain of a transistor amplifier in common emitterconfiguration is:

0 Cv

i B

v RA

v Ræ ö

= = bç ÷è ø , where RC and RB are respectively the

resistances in collector and base sides of the circuit.The important digital circuits performing special logicoperations are called logic gates. These are: OR, AND, NOT,NAND, and NOR gates. NAND gate is the combination ofNOT and AND gate. NOR gate is the combination of NOTand OR gate.

Transmitter, transmission channel andreceiver are three basic units of acommunication system.

Two important forms of communicationsystem are: Analog and Digital. The informationto be transmitted is generally in continuouswaveform for the former while for the latter it

has only discrete or quantised levels.Low frequencies cannot be transmitted to long distances.Therefore, they are superimposed on a high frequency carriersignal by a process known as modulation.In the process of modulation, new frequencies called sidebandsare generated on either side.If an antenna radiates electromagnetic waves from a height

hT, then the range dT is given by T2Rh where R is theradius of the earth.

Effective range, d T R= 2Rh + 2RhhT = height of transmitting antenna; hR = height of receivingantennaCritical frequency Vc = 9(Nmax)1/2

where Nmax = no. density of electrons/m3

Skip distance, Dskip 2

max2 1c

Vh

Væ ö

= -ç ÷è ø

h= height of reflecting layer of atmosphere.

Power radiated by an antenna 21

µl

MODERNPHYSICS

COMMUNI-CATION

SYSTEMS

9


Energy of electron in species with oneelectron.

En = 2 4 2

2 22 me Z

n h- p

For energy in SI system,

En = 2 4 2

2 2 20

2 me Zn h (4 )- p

pe

En = 2

12

1312Z kJ moln

--

where e0 is permitivity constant and its value is 8.8542 ×10–12 coulomb2 newton–1 metre–2

r = 2 2

2 2n h

4 mZep = 0.529

2nZ

æ öç ÷ç ÷è ø

Å

Total energy of electron in the nth shell

= K.E. + P.E. 2 2 2

n n n

e kZe kZekZ2r r 2r

æ ö= + - = -ç ÷è ø

1u =

l = RZ2

2 21 2

1 1n n

é ù-ê ú

ê úë û, [R = 1.0968 × 107 m–1]

No. of spectral lines produced when an electron drops from

nth level to ground level = n(n 1)

2-

Heisenberg Uncertainty Principle (Dx) (Dp) ³ h/4pNodes (n – 1) = total nodes, l = angular nodes,(n – l – 1) = Radial nodes

Orbital angular momentum : h( 1) ( 1)h2

+ = +p

l l l l

Radial probability density curves:

®r

1s

n=1

R.4

r.dr

22

p

R.4

r.d

r2

2p

® r

2s

n=2

R.4

r.d

r2

2p

® r

3sn=3

R.4

r.d

r2

2p

®r

2p

n =2

R.4

r.d

r2

2p

® r

3p

n=3

R.4

r.d

r2

2p

® r

3dn=3

(a) % ionic character =

Actual dipole moment100

Calculated dipole moment´

(b) Pauling equation % ioniccharacter

A B1 (X X )4100 1 exp

- -é ùê ú= -ë û

Fajan’s Factors : Following factors are helpful in bringingcovalent character in Ionic compounds(a) Small cation (b) Big anion(c) High charge on cation/anion(d) Cation having pseudo inert gas configuration (ns2p6d10)e.g. Cu+, Ag+, Zn+2, Cd+2

M.O. theory :(a) Bond order = ½(Nb–Na)(b) Higher the bond order, higher is the bond dissociation

energy, greater is the stability, shorter is the bond length.(c) Species Bond order Magnetic properties

H2 1 DiamagneticH2

+ 0.5 ParamagneticLi2 1 Diamagnetic

Relative bond strength : sp3d2 >dsp2 >sp3 >sp2 >sp >p-p(Co-axial) > s - p > s - s > p - p (Co-lateral)VSEPR theory(a) (LP-LP) repulsion > (LP-BP) > (BP-BP)(b) NH3 ® Bond Angle 106° 45’ because (LP-BP) repulsion> (BP-BP) H2O ® 104° 27’because (LP-LP) repulsion > (LP-LB) > (BP-BP)Bond angle :(a) NH3 > PH3 > AsH3 (b) H2O > H2S > H2Se

(c) NH3 > NF3(d) Cl2O > OF2

gΔnp cK = K (RT) where Dng' nP – nR

Free Energy change (DG)(a) If DG = 0 then reversible reaction

would be in equilibrium, Kc = 0(b) If DG = (+) ve then equilibrium will

be displace in backward direction;Kc < 1

(c) If DG = (–) ve then equilibrium will displace in forwarddirection; Kc > 1

(a) Kc unit ® (moles/lit)Dn,(b) Kp unit ® (atm)Dn

ATOMICSTRUCTURE

CHEMICALEQUILIBRIUM

CHEMICALBONDING

CHEMISTRY

10


Reaction Quotient and Equilibrium ConstantConsider the following reversible reaction

A + B � C + D

\ Qc = [C][D][A][B]

Case I : If Qc < Kc then : [Reactants] > [Products]then the system is not at equilibriumCase II : If Qc = Kc then : The system is at equilibrium andthe concentration of the species C, D, B,A are at equilibrium.Case III : If Qc > Kc then : [Products] > [Reactants]The system is not at equilibrium.A relationship between the equilibrium constant KC, reactionquotient and Gibbs energy.DG = DG° + RT ln QAt equilibrium DG = 0 and Q = K then

DG° = –RT ln Kc\ DG° = –RT ln KpLe chatelier’s principal(i) Increase of reactant conc. (Shift forward)(ii) Decrease of reactant conc. (Shift backward)(iii) Increase of pressure (from more moles to less moles)(iv) Decrease of pressure (from less moles to more moles)(v) For exothermic reaction decrease in temp. (Shift

forward)(vi) For endothermic increase in temp. (Shift backward)

(a) Lewis Acid (e– pair acceptor) ®CO2, BF3, AlCl3, ZnCl2, normalcation

(b) Lewis Base (e– pair donor) NH3,ROH, ROR, H2O, RNH2, normalanion

Dissociation of weak Acid & WeakBase ®

(a) Weak Acid ® Ka = Cx2/(1 – x) orKa = Cx2 ; x << 1

(b) Weak Base ® Kb = Cx2/(1 – x) or Kb = Cx2 ; x << 1Buffer solution {Henderson equation} :(a) Acidic ® pH = pKa + log {Salt/Acid}.

For Maximum buffer action pH = pKaRange of Buffer pH = pKa ± 1

(b) Alkaline ® pOH = pKb + log {Salt/Base} for max. bufferaction pH = 14 – pKbRange pH = 14 – pKb ± 1

(c) Buffer Capacity = Moles / lit of Acid or Base Mixed

change in pHNecessary condition for showing neutral colour of IndicatorpH = pKln or[HIn] = [In–] or [InOH] = [In+]

Relation between ionisation constant(Ki) & degree of ionisation(a):-

Ki = 2

(1 )Va- a

= 2C

(1 )a- a

(Ostwald’ss

dilution law)It is applicable to weak electrolytesfor which a <<1 then

a = iK V = iKC

or V C ¯ a

Common ion effect : By addition of X mole/L of a commonion, to a weak acid (or weak base) a becomes equal to

a bK Kor

X Xæ öç ÷è ø [where a = degree of dissociation]

(A) If solubility product > ionic product then the solution isunsaturated and more of the substance can be dissolved in it.(B) If ionic product > solubility product the solution is supersaturated (principle of precipitation).Salt of weak acid and strong base :

pH = 0.5 (pKw + pKa + log c); h = hKc

; Kh = w

a

KK

(h = degree of hydrolysis)Salt of weak base and strong acid :

pH = 0.5 (pKw – pKb – log c); h = w

b

KK c´

Salt of weak acid and weak base :

pH = 0.5 (pKw + pKa – pKb ); h = w

a b

KK K´

Differences between order andmolecularity of reaction:

Order of reaction Molecularity1. It is experimentally It is a theoretical concept.

determined quantity2. It can have integral, Always integral values only,

fractional or negative never zero or negativevalues

3. It cannot be obtained It can be obtained.from balanced orstoichiometric equation.

4. It tells about the slowest It does not tell anythingstep in the mechanism about mechanism

5. It is sum of the powers It is the number of reactingof the concentration species undergoing terms in the rate law simultaneous collision in theequation. reaction.

Unit of Rate constant :k = mol1–n litn–1 sec–1

Order of reaction It can be fraction, zero or any whole number.Molecularity of reaction is always a whole number. It is nevermore than three. It cannot be zero.First Order reaction :

k = 2.303

t log10 a

(a x)- & t1/12 = 0.693

k[A]t = [A]0e–kt

Second Order Reaction :When concentration of A and B taking same.

CHEMICALKINETICS

ACIDS ANDBASES

IONICEQUILIBRIUM

11


k2 = 1t

xa(a x)

æ öç ÷-è ø

When concentration of A and B are taking different -

k2 = 2.303

t(a b)- log b(a x)a(b x)

--

Zero Order Reaction :

x = kt & t1/2 = a

2kThe rate of reaction is independent of the concentration ofthe reacting substance.Time of nth fraction of first order process,

t1/n =12 303 log 11

æ ö× ç ÷-ç ÷è øk

nArrhenius equation :

k = Ae–Ea/RT & slope = aE2.303R

- & Temperature Coefficient

log 2

1

kk

æ öç ÷è ø

= aE2.303

2 1

1 2

T TT T

æ ö-ç ÷è ø

It has been found that for a chemical reaction with rise intemperature by 10 °C, the rate constant gets nearly doubled.

Oxidant itself is reduced (gives O2)Or Oxidant ¾¾® e– (s) AcceptorReductant itself is oxidised (gives H2)Or reductant ¾¾® e– (s) Donor(i) Strongness of acid µ O.N(ii) Strongness of base µ 1/ O.N(a) Electro Chemical Series:- Li, K, Ba,

Sr, Ca, Na, Mg, Al, Mn, Zn, Cr, Fe, Cd, Co,Ni, Sn, Pb, H2, Cu, Ag, Pt, Au.(b) As we move from top to bottom in this series(1) Standard Reduction Potential (2) Standard Oxidation Potential ̄(3) Reducing Capacity ̄(4) IP (5) Reactivity ̄

Equivalent weight of element

= Atomic wt of the element

Valency of elementThe law of Dulong and PetitAtomic wt.×specific heat » 6.4Normality (N)

= number of equivalents

volume of the solution in litres

Molarity (M) = number of moles

volume of the solution in litresCommon acid-base indicators

Indicator Acid colour Alkaline colour pH range of change

Methyl orange Red Yellow 3.2–4.4Methyl red Red Yellow 4.2–6.2 Litmus Red Blue 4.5–8.3 Phenolphthalein Colourless red 8.3–10

First Law : DE = Q + WExpression for pressure volume workW = –PDVMaximum work in a reversibleexpansion :

W = –2.303n RT log 2

1

VV = –2.303 nRTT

log 1

2

PP

Wrev ³ Wirr

qv = cvDT = DU, qp = cpDT = DHEnthapy changes during phase transformation(i) Enthalpy of Fusion(ii) Heat of Vapourisation(iii) Heat of SublimationEnthalpy : DH = DE + PDV = DE + DngRTKirchoff’s equation :

2 1T T VE E CD = D + D (T2 – T1) [constant V]

2 1T T PH H CD = D + D (T2 – T1) [constant P]Entropy(s) : Measure of disorder or randomnessDS = SSp–SSR

DS = revqT

= 2.303 nR log 2

1

VV = 2.303 n R log

1

2

PP

Free energy change : DG = DH – TDS–DG = W(maximum) – PDV

DH DS DG Reaction characteristics– + Always negative Reaction is spontaneous at

all temperature.+ – Always positive Reaction is nonspontaneous

at all temperature– – Negative at low Spontaneous at low temp. &

temperature but non spontaneous at highpositive at high temperaturetemperature

+ + Positive at low Non spontaneous at lowtemp. but temp. & spontaneous at highnegative at high temp.temperature

m = Z.I.t

Degree of dissociation : a = eq0eq

l

l =

Equivalent conductance at given concentration

equivalent conductance at infinite dilutionSpecific conductance

spL = 1 1

/= =

l

lS RA RA ; m sp1000 .L = ´ L

CKohlrausch’s law : 0 0 0

m A Bx yL = l + lNernst Equation

E = Eº – 0.0591

n log10

[Products][Reactants]

OXIDATION -REDUCTION

VOLUMETRICANALYSIS

ELECTRO-CHEMISTRY

THERMO-DYNAMICS

12


& EºCell = Eºright + Eºleft & Keq. = antilog nEº

0.0591é ùê úë û

DG = – nFEcell & DGº = –nFEº cell

& Wmax= +nFEº & DG = DH + T P

GT

¶Dæ öç ÷¶è ø

Calculation of pH of an electrolyte by using a calomel

electrode : cellE 0.2415pH

0.0591-

=

Thermodynamic efficiency of fuel cells :ocellnFEG

H--D

h= =D DH

For H2–O2 fuel cells it is 95%.

Raoult’s lawP = pA + pB = p°AXA + p°BXBCharacteristics of an ideal solution:(i) DsolV = 0(ii) DsolH = 0Relative lowering of vapour pressure

=–o

A AoA

P PP

–o

A A BBo

A A B

P P nX

P n n= =

+Colligative µ Number of particlesproperties µ Number of ions (in case of electrolytes)

µ Number of moles of soluteDepression of freezing point, DTf = KfmElevation in boiling point with relative lowering of vapourpressure

ob

b o1

1000K p pTM p

æ ö-D = ç ÷ç ÷

è ø (M1 = mol. wt. of solvent)

Osmotic pressure (P) with depression in freezing point DTf

ff

dRTP T1000K

= D ´

Relation between Osmotic pressure and other colligativeproperties:

(i)B

oA

AoA

MdRT

ppp

´÷÷ø

öççè

æ -=p Relative lowering of vapour pressure

(ii)b

b K1000dRTT ´D=p Elevation in boiling point

(iii)f

f K1000dRTT ´D=p Depression in freezing point

i = Normal molar mass

Observed molar mass = Observed colligative propertyNormal colligative property

i = Observed osmotic pressureNormal osmotic pressure

Degree of association a = (1 – i) n

n 1-

& degree of dissociation (a) = i 1n 1

--

Ideal gas equation : PV = nRT(i) R = 0.0821 liter atm. deg–1 mole–1

(ii) R = 2 cals. deg.–1 mole–

(iii) R = 8.314 JK–1 mole–1

Velocities related to gaseous state

RMS velocity C = 3PVM

= 3RTM

=

3Pd

Average speed = 8RTM

& Most probable speed = 2RTM

Average speed = 0.9213 × RMS speedRMS speed = 1.085 × Average speedMPS = .816 × RMS; RMS = 1.224 MPSMPS : A.V. speed : RMS = 1 : 1.128 : 1.224

Rate of diffusion 1density of gas

µ

van der Waal’s equation2

2n aP (V nb) nRTV

æ ö+ - =ç ÷

è ø for n moles

Z (compressibility factor) = PVnRT

; Z = 1 for ideal gas

Available space filled up by hardspheres (packing fraction):

Simple cubic = 6p

= 0.52

bcc = 3

8p

= 0.68

fcc = 2

6p

= 0.74

hcp = 2

6p

= 0.74 diamond = 3

6p

= 0.34

Radius ratio and co-ordination number (CN)Limiting radius ratio CN Geometry[0.155– 0.225] 3 [plane triangle][0.255–0.414] 4 [tetrahedral][0.414–0.732] 6 [octahedral][0.732–1] 8 [bcc]

Atomic radius r and the edge of the unit cell:Pure elements :

Simple cubic = r = a2

; bcc r = 3a4

; fcc = 2a4

Relationship between radius of void (r) and the radius of thesphere (R) : r (tetrahedral) = 0.225 R ; r (octahedral) = 0.414 RParamagnetic : Presence of unpaired electrons [attracted bymagnetic field]Ferromagnetic : Permanent magnetism [ ]Antiferromagnetic : net magnetic moment is zero [ ̄ ̄ ]Ferrimagnetic : net magnetic moment is three [ ̄ ̄ ]

SOLID ANDLIQUID STATE

SOLUTION ANDCOLLIGATIVEPROPERTIES

GASEOUSSTATE

13


Emulsion : Colloidal soln. of twoimmiscible liquids [O/W emulsion,W/O emulsion]Emulsifier : Long chainhydrocarbons are added tostabilize emulsion.Lyophilic colloid : Starchy gum, gelatinhave greater affinity for solvent.

Lyophobic colloid : No affinity for solvent, special methodsare used to prepare sol. [e.g. As2S3, Fe(OH)3 sol]Preparation of colloidal solution :(i) Dispersion methods (ii) Condensation method.Properties of colloidal solution :(i) Tyndall effect (ii) Brownian movement (iii) Coagulation(iv) Filtrability.

INORGANIC CHEMISTRYGeneral electronic configuration(of outer orbits)s-block ns1–2

p-block ns2np1–6

d-block (n–1)d1–10 ns1–2

f-block (n–2) f1–14s2p6d10(n–1)s2p6d0 or 1 ns2

Property Pr (L To R) Gr (T to B)(a) atomic radius ¯ (b) ionisation potential ¯(c) electron affinity ¯(d) electro negativity ¯(e) metallic character or ¯

electropositive character(f) alkaline character ¯

of hydroxides(g) acidic character ¯(i) reducing property ¯ (j) oxidising property ¯(k) non metallic character ¯

IP µ 1

Metallic character µ 1

Reducing character

EA µ 1

size µ nuclear charge.

Second electron affinity is always negative.Electron affinity of chlorine is greater than fluorine (smallatomic size).The first element of a group has similar properties with thesecond element of the next group. This is called diagonalrelationship. The diagonal relationship disappears after IVgroup.

Atomic radii : Li < Na < K < Rb < CsElectronegativity : Li > Na > K >Rb > CsFirst ionization potential : Li > Na> K > Rb > CsMelting point Li > Na > K > Rb >CsColour of the flame Li - Red, Na -Golden, K - Violet, Rb - Red, Cs -Blue, Ca - Brick red, Sr - Blood red,Ba-Apple green

Rb and Cs show photoelectric effect.Stability of hydrides : LiH > NaH > KH > RbH > CsHBasic nature of hydroxides : LiOH < NaOH < KOH < RbOH <CsOH

Hydration energy : Li > Na > K > Rb > CsReducing character : Li > Cs > Rb > K > Na

Stability of +3 oxidation state : B >Al > Ga > In > TlStability of +1 oxidation state : Ga< In < TlBasic nature of the oxides andhydroxides :B < Al < Ga < In < TlRelative strength of Lewis acid :BF3 < BCl3 < BBr3 < BI3

Ionisation energy : B > Al < Ga > In < TlElectronegativity : Electronegativity first decreases from B toAl and then increases marginally.

Reactivity : C < Si < Ge < Sn < PbMetallic character : C < Si < Ge <Sn < PbAcidic character of the oxides :CO2 > SiO2 > GeO2 > SnO2 > PbO2Weaker acidic (amphoteric)Reducing nature of hydrides

CH4 < SiH4 < GeH4 < SnH4 < PbH4Thermal stability of tetrahalidesCCl4 > SiCl4 > GeCl4 > SnCl4 > PbCl4Oxidising character of M+4 speciesGeCl4 < SnCl4 < PbCl4Ease of hydrolysis of tetrahalidesSiCl4 < GeCl4 < SnCl4 < PbCl4

Acidic strength of trioxides : N2O3> P2O3 > As2O3Acidic strength of pentoxidesN2O5 > P2O5 > As2O5 > Sb2O5 >Bi2O5Acidic strength of oxides ofnitrogenN2O < NO < N2O3 < N2O4 < N2O5Basic nature, bond angle, thermalstability and dipole

moment of hydridesNH3 > PH3 > AsH3 > SbH3 > BiH3Stability of trihalides of nitrogen : NF3 > NCl3 > NBr3Lewis base strength : NF3 < NCl3> NBr3 < NI3Ease of hydrolysis of trichloridesNCl3 > PCl3 > AsCl3 > SbCl3 > BiCl3Lewis acid strength of trihalides of P, As and SbPCl3 > AsCl3 > SbCl3Lewis acid strength among phosphorus trihalidesPF3 > PCl3 > PBr3 > PI3Nitrogen displays a great tendency to form pp – pp multiplebonds with itself as well as with carbon and oxygen.The basic strength of the hydrides

NH3 > PH3 > AsH3 > SbH3

SURFACECHEMISTRY &COLLOIDAL

STATE

PERIODICTABLE

s-BLOCKELEMENTS

BORONFAMILY

CARBONFAMILY

NITROGENFAMILY

14


The thermal stability of the hydrides decreases as the atomicsize increases.

Melting and boiling point of hydridesH2O > H2Te > H2Se > H2SVolatility of hydridesH2O < H2Te < H2Se < H2SReducing nature of hydridesH2S < H2Se < H2TeCovalent character of hydridesH2O < H2S < H2Se < H2TeThe acidic character of oxides

(elements in the same oxidation state)SO2 > SeO2 > TeO2 > PoO2 ; SO3 > SeO3 > TeO3Acidic character of oxide of a particular element (e.g. S)SO < SO2 < SO3 ; SO2 > TeO2 > SeO2 > PoO2

Bond energy of halogens : Cl2 >Br2 > F2 > I2Solubility of halogen in water : F2> Cl2 > Br2 > I2Oxidising power : F2 > Cl2 > Br2 >I2Enthalpy of hydration of X– ion :F– > Cl– > Br– > I–

Reactivity of halogens : F > Cl > Br > IIonic character of M - X bond in halides

M – F > M – Cl > M – Br > M – IReducing character of X– ion : I– > Br– > Cl– > F–

Acidic strength of halogen acids : HI > HBr > HCl > HFConjugate base strength of halogen acidsI– < Br– < Cl– < F–

Reducing property of hydrogen halidesHF < HCl < HBr < HIOxidising power of oxides of chlorineCl2O > ClO2 > Cl2O6 > Cl2O7Acidic character of oxyacids of chlorineHClO < HClO2 < HClO3 < HClO4Oxidising power of oxyacids of chlorineHClO > HClO2 > HClO3 > HClO4

The element with exceptionalconfiguration areCr24[Ar] 3d54s1, Cu29[Ar] 3d104s1

Mo42[Kr] 4d55s1, Pd46[Kr]4d105s0

Ag47[Kr] 4d105s1, Pt78[Xe]4f145d106s0

Ferromagnetic substances arethose in which there are largenumber of electrons with unpairedspins and whose magneticmoments are aligned in the samedirection.

Inner Transition Elements(i) Electronic Configuration - The general electronic

configuration of these elements is[ ]

0 10 14 2Xe 4f 5d 6s--

(iii) Magnetic properties - Magnetic properties have spin andorbit contributions (Contrast “spin only”of transitionmetals). Hence magnetic momentums are given by theformula.

( ) ( )114 +++= LLSSm

where L = Orbital quantum number, S = Spin quantum number

Coordination number is the number ofthe nearest atoms or groups in thecoordination sphere.

Ligand is a Lewis base donor ofelectrons that bonds to a central metalatom in a coordination compound.

Paramagnetic substance is one that isattracted to the magnetic field, this resultson account of unpaired electrons present

in the atom/molecule/ion.Effective atomic number EAN

= (Z – Oxidation number) + (2 × Coordination number)Factors affecting stability of complex(i) Greater the charge on the central metal ion, greater is thestability.(ii) Greater the ability of the ligand to donate electron pair(basic strength) greater is the stability.(iii) Formation of chelate rings increases the stability.Isomerism in coordination compounds :(i) Structural Isomers (ii) Ionization Isomers(iii) Hydration Isomers (iv) Linkage Isomers(v) Coordination Isomerism (vi) Ligand isomerism(vii) Polymerisation Isomerism (viii) Valence Isomerism(ix) Coordination position isomerism(x) Stereo isomerism

(a) Geometrical(I) Square planar complexes of the type

MA2X2 ; MABX2 ; MABXY(II) Octahedral of the type : MA4XY, MA4X2 MA3X3

MA2X2Y2. M(AA)2X2 and M(ABCDEF).(b) Optical isomerism

The order of decreasingelectronegativity of hybrid orbitals is sp> sp2 > sp3.

Conformational isomers are thoseisomers which arise due to rotationaround a single bond.

A meso compound is opticallyinactive, even though it has asymmetriccentres (due to internal compensation ofrotation of plane polarised light)

An equimolar mixture of enantiomersis called racemic mixture, which is optically inactive.Reaction intermediates and reagents :Homolytic fission ® Free radicalsHeterolytic fission ® ions (Carbonium, ion carbonium, etc.)Nucleophiles – electron richTwo types : 1. Anions 2. Neutral moleculeswith lone pair of electrons (Lewis bases)Electrophiles : electron deficient.Two types : 1. Cations 2. Neutral molecules with vacantorbitals (Lewis acids).Inductive effect is due to s electron displacement along achain and is permanent effect.+I (inductive effect) increases basicity, – I effect increasesacidity of compounds.Resonance is a phenomenon in which two or more structurescan be written for the same compound but none of themactually exists.

OXYGENFAMILY

HALOGENFAMILY

TRANSITIONELEMENTS

(d- and f-BLOCKELEMENTS

COORDINATIONCOMPOUNDS

GOC

15


ORGANIC CHEMISTRY Pyrolytic cracking is a process in

which alkane decomposes to a mixtureof smaller hydrocarbons, when it isheated strongly, in the absence ofoxygen.

Ethane can exist in an infinite numberof conformations. They are

HH

HH HHH H

H

H HHH HHH HH

Eclipsed q < 60° > 0 Skewq = 60° StaggeredConformations of Cyclohexane : It exists in two nonplanar,strainless forms, the boat and the chair form

Chair formMost Stable

Half Chair

Most StableTwist Boat Boat form

(Least Stable)

In dehydration anddehydrohalogenation the preferentialorder for removal of hydrogen is 3° > 2° >1° (Saytzeff’s rule).

The lower the DHh (heat ofhydrogenation) the more stable thealkene is.

Alkenes undergo anti-Markonikovaddition only with HBr in the presenceof peroxides.

Alkynes add water molecule inpresence of mercuric sulphate and dil.H2SO4 and form carbonyl compounds.

Terminal alkynes have acidic H-atoms,so they form metal alkynides with Na,ammonical cuprous chloride solution andammoniacal silver nitrate solution.

Alkynes are acidic because of H-atomswhich are attached to sp ‘C’ atom which (a) has moreelectronegativity (b) has more ‘s’ character than sp2 andsp3 ‘C’ atoms.

All o and p-directing groups are ringactivating groups

(except – X)They are : – OH, – NH2, – X, – R, –

OR, etc.All m-directing groups are ring

deactivating groups.They are : – CHO, – COOH, – NO2, –

CN, – 3NR+

, etc.

The order of reactivity is(a) RI > RBr > RCl > RF(b) Allyl halide > Alkyl halide > Vinyl

halide(c) Alkyl halide > Aryl halide SN1 reaction : Mainly 3° alkyl halides

undergo this reaction and form racemicmixture. SN1 is favoured by polar solventand low concentration of nucleophile.

SN2 reaction : Mainly 1° alkyl halides undergo thissubstitution. Walden inversion takes place. SN

2 reaction ispreferred by non-polar solvents and high concentration ofnucleophile.Reaction with metals:

(i) Dry ether

Alkyl Grignard reagenthalides

R – X + Mg R – Mg – X¾¾¾¾®

(ii) Wurtz reaction:R – X + 2 Na + X – R Dry ether –

AlkaneR R 2Na X+¾¾ ¾ ¾® - +

Alkenes are converted to alcohol indifferent ways as follows

Reagent Types of additiondil H2SO4

– MarkovnikovB2H6 and H2O2, OH– –

Anti-MarkovnikovOxymercuration demercuration –

MarkovnikovOxidation of1° alcohol ¾¾® aldehyde ¾¾® carboxylic acid

(with same no. (with same no. of of C atom) C atom)

2° alcohol ¾¾® ketone ¾¾® carboxylic acid (with same no. (with less no. of of C atom) C atom)

3° alcohol ¾¾® ketone ¾¾® carboxylic acid (with less no. (with less no. of of C atom) C atom)

Phenol CHCl /OH3Q

¾¾¾¾¾¾® Phenolicaldehyde

(Reimer-Tieman reaction)

Phenol CO2D

¾¾¾® Phenolic carboxylic

acid (Kolbe reaction)Acidity of phenols(a) Increase by electron withdrawing substituents like

– NO2, – CN, – CHO, – COOH, –X, 3N R+

-(b) decrease by electron releasing substituents like

– R, – OH, – NH2, – NR2, – OR

ALKANES

ALKENES

ALKYNES

ARENES

HALOGENCOMPOUNDS

ALCOHOLS

PHENOLS

16


2 3Al O2250ºC

2ROH R O R H O¾¾¾¾® - - +

RONa X R' ROR ' NaX+ - ¾¾® +(Williamson's synthesis)

2 4dil. H SO2ROR H O 2ROH

D+ ¾¾¾¾¾®

Formation of alcohols using RMgX(a) Formaldehyde + RMgX

Hydrolysis¾¾¾¾¾® 1° alcohol

(b) Aldehyde + RMgX Hydrolysis¾¾¾¾¾®2° alcohol

(other than HCHO)

(c) Ketone + RMgX Hydrolysis¾¾¾¾¾®3° alcohol

Cannizzaro reaction (Disproportionation)

Aldehyde Hot conc.

alkali¾¾¾¾® Alcohol + Salt of acid

(no a H-atom)Aldol condensation :Carbonyl compound + dil. alkali ––® b-hydroxy carbonyl(with a H-atom) compoundBenzoin condensation

Benzaldehyde ethanolic

NaCNBenzoin¾¾¾¾®

The relative reactivities of different acid derivatives towardsnucleophilic acyl substitution reactions follow the order:

2Acid chloride Anhydride Ester Amide

O O O O O|| || || || ||

R – C – Cl R – C – O – C – R ' R – C – OR ' R – C – NH> > >

The rate of esterfication decreaseswhen alcohol, acid or both havebranched substituents.

Ortho effect : All orthosubstituted benzoic acids (irrespectiveof type of substituent) are stronger thanbenzoic acid.

Order of basicity : (R = – CH3 or– C2H5)

2° > 1° > 3° > NH3Hofmann degradation

Amides Br /KOH2¾¾¾¾® 1° amineThe basicity of amines is

(a) decreased by electron withdrawinggroups

(b) increased by electron releasing groupsReduction of nitrobenzene in different media gives differentproductsMedium ProductAcidic AnilineBasic Azoxy, Azo and finally hydrazobenzeneNeutral Phenyl hydroxylamine

Carbohydrates are polyhydroxyaldehydes or ketones.

Monosaccharides are simple sugars,containing three to nine carbon atoms.

Characteristic reactions :Homologous series

Type of reactions(a) Alkanes

Substitution(Mostly free radical

(b) Alkenes and alkynes Electrophillic addition(c) Arenes Electrophillic substitution(d) Alkyl halides Nucleophillic substitution(e) Aldehyde and ketones Nucleophillic additionTests to differentiate :1°, 2° and 3° alcohols (1) Lucas test

(2) Victormeyer’s test1°, 2° and 3° amines Hinsberg test1°, 2° and 3° nitro compounds Test with HNO2 and KOHAryl halides and alkyl halides Test with AgNO3

solutionAldehydes and ketones Tollen’s test/Fehling’s

testAromatic aldehydes and Fehling’s testAliphatic aldehydes

Dil H2SO4 [or Conc. H2SO4 + H2O]Use ® Hydrating agent (+HOH)Alc. KOH or NaNH2(Use ® -HX)CH3CH2Cl alc.KOH¾¾¾¾® CH2=CH2Lucas reagent ZnCl2 + Conc. HClUse ® for distinction between 1º, 2º

& 3º alc.Tilden Reagent NOCl (Nitrosyl

chloride)C2H5NH2 NOCl¾¾¾® C2H5ClAlkaline KMnO4(Strong oxidant)

Toluene ® Benzoic acidBayer’s Regent : 1% alkaline KMnO4(Weak oxidant)Use: ® For test of > C = C < or –C = C –CH2=CH2+H2O+[O] BR¾¾® CH2OH–CH2OH

Acidic K2Cr2O7 (Strong oxidant) : RCH2OH [O]¾¾® RCHOSnCl2/HCl or Sn/HCl use ® for redn of nitrobenzene in acidicmedium.

C6H5NO2 SnCl / HCl2

6H¾¾¾¾¾® C6H5NH2

Lindlar’s Catalyst = Pd/CaCO3+ in small quantity (CH3COO)2Pb2 – butye + H2

"¾¾® Cis-2-butene (main product)

Ziegler –Natta Catalyst (C2H5)3Al + TiCl4Use ®In Addition polymerisation

IDENTIFICATION TESTS :(a) Unsaturated compound (Bayer’s reagent)Decolourising the reagent(b) Alcohols (Ceric ammonium nitrate solution)Red colouration(c) Phenols (Neutral FeCl3 solution)Violet/deep blue colouration(d) Aldehydes and ketones (2, 4-D.N.P.)Orange precipitate(e) Acids (NaHCO3 solution)Brisk effervescence (CO2 is evolved)(f) 1° amine (CHCl3 + KOH)Foul smell (isocyanide)(g) 2° amine (NaNO2 + HCl)Yellow oily liquid (Nitrosoamine)

ETHERS

CARBONYLCOMPOUNDS

CARBOXYLICACIDS

NITROGENCOMPOUNDS

IMPORTANTREAGENT

CARBOHYDRATES,AMINO ACIDS AND

POLYMERS

17


A relation R from a set A to a set B isa subset of the cartesian product A × Bobtained by describing a relationshipbetween the first element x and the secondelement y of the ordered pairs in A × B.

Function : A function f from a set A toa set B is a specific type of relation forwhich every element x of set A has oneand only one image y in set B. We write

f : A ® B, where f (x) = y.A function f : X ® Y is one-one (or injective) iff (x1) = f (x2) Þ x1 = x2 " x1, x2 Î X.A function f : X ® Y is onto (or surjective) if given anyy Î Y, $ x Î X such that f (x) = y.Many-One Function :A function f : A ® B is called many- one, if two or moredifferent elements of A have the same f- image in B.Into function :A function f : A ® B is into if there exist at least one elementin B which is not the f - image of any element in A.Many One -Onto function :A function f : A ® R is said to be many one- onto if f is ontobut not one-one.Many One -Into function :A function is said to be many one-into if it is neither one-onenor onto.A function f : X ® Y is invertible if and only if f is one-oneand onto.

General Solution of the equationsinq = 0:when sinq = 0q = np : n Î I i.e. n = 0, ± 1, ±2...........General solution of the equationcosq = 0 :when cosq = 0q = (2n + 1)p/2, n Î I. i.e. n = 0, ±1, +2.......

General solution of the equation tanq = 0:General solution of tanq = 0 is q = np; n Î IGeneral solution of the equation(a) sinq = sina : q = np + (–1)na ; n Î I(b) sinq = k, where –1 < k < 1.

q = np + (–1)na, where n Î I and a = sin–1k(c) cosq = cosa : q = 2np ± a, n Î I(d) cosq = k, where –1 < k < 1.

q = 2np ± a, where n Î I and a = cos–1 k(e) tanq = tana : q = np + a ; n Î I(f) tanq = k, q = np + a, where n Î I and a = tan–1k(g) sin2q = sin2a : q = np ± a; n Î I(h) cos2q = cos2a : q = np ± a ; n Î I(i) tan2q = tan2a : q = np ± a ; n Î I

sina + sin (a + b) + sin (a +2b) +........ to n terms

=

n 1 nsin sin2 2

sin ( / 2)

é ù é ù- bæ ö æ öa + bç ÷ ç ÷ê ú ê úè ø è øë û ë ûb ; b¹2np

cosa + cos (a + b) + cos (a +2b) +........ to n terms

=

n 1 ncos sin2 2

sin2

é ù é ù- bæ ö æ öa + bç ÷ ç ÷ê ú ê úè ø è øë û ë ûbæ öç ÷è ø

; b ¹ 2np

B C b c Atan cot2 b c 2- -æ ö æ ö æ ö=ç ÷ ç ÷ ç ÷è ø è ø è ø+

sinA2

æ öç ÷è ø =

(s b)(s c)bc

- -

tanA2

æ öç ÷è ø =

(s b)(s c)s (s a)- -

-

R = a b c

2sin A 2sin B 2sin C= =

R =abc4D

r = 4R sinA2

æ öç ÷è ø . sin

B2

æ öç ÷è ø . sin

C2

æ öç ÷è ø

a = c cos B + b cos C

Maximum value of a sin q + b cos q 2 2a b= + and minimum

value of a sin q + b cos q 2 2a b= - +

Properties of inverse trigonometricfunction• tan–1 x + tan–1 y =

1

1

1

x ytan , if x,y 11 xy

if x 0, y 0x ytan ,and xy 11 xy

if x 0, y 0x ytan ,and xy 11 xy

-

-

-

ì æ ö+<ï ç ÷-è øï

ï > >æ ö+ï p +í ç ÷ >-è øïï < <æ ö+ï -p + ç ÷ï >-è øî

• tan–1 x – tan–1y =

1

1

1

x ytan , if xy 1

1 xyx ytan , if x 0, y 0and xy 1

1 xyx ytan , if x 0, y 0 and xy 1

1 xy

-

-

-

ì æ ö-> -ï ç ÷è + øï

æ ö-ïp + > < < -í ç ÷è + øïæ ö-ï-p + < > < -ç ÷ï è + øî

RELATIONSAND

FUNCTIONS

TRIGONOMET-RIC FUNCTIONS

ANDEQUATIONS INVERSE

TRIGONOMETRICFUNCTIONS

MATHEMATICS

18


• sin–1 x + sin–1 y =

2 21 2 22 2

1 2 22 2

2 21 2 2

if 1 x, y 1 and x y 1sin {x 1 y y 1 x },or if xy 0 and x y 1

if 0 x, y 1sin {x 1 y y 1 x }, and x y 1

if 1 x, y 0 and x y 1sin {x 1 y y 1 x },

-

-

-

ì - £ £ + £- + -ï< + >ï

ï< £í - - + -ï + >

ïï - £ < + >- - - + -î

p

p

• cos–1 x + cos–1 y =

1 2 2

1 2 2cos {xy 1 x 1 y } , if 1 x, y 1 and x y 02 cos {xy 1 x 1 y }, if 1 x, y 1and x y 0

-

-

ìï - - - - £ £ + ³í

p - - - - - £ £ + £ïî

2 sin–1x =

1 2

1 2

1 2

1 1sin (2x 1 x ) , if x2 2

1sin (2x 1 x ) , if x 12

1sin (2x 1 x ) , if 1 x

2

-

-

-

ì- - £ £ï

ïï p - - £ £íïï-p - - - £ £ -ïî

2 tan–1 x =

12

12

12

2xtan , if 1 x 11 x

2xtan , if x 11 x

2xtan , if x 1

1 x

-

-

-

ì æ ö - < <ï ç ÷è ø-ïï æ öp + >í ç ÷è ø-ïï æ ö-p + < -ï ç ÷è ø-î

Roots of a Quadratic Equation : Theroots of the quadratic equation are given by

2b b 4acx2a

- ± -=

Nature of roots : In Quadratic equationax2 + bx + c = 0. The term b2 – 4ac is calleddiscriminant of the equation. It is denoted

by D or D.(A) Suppose a, b, c Î R and a ¹ 0(i) If D > 0 Þ Roots are Real and unequal(ii) If D = 0 Þ Roots are Real and equal and each equal to

– b/2a(iii) If D < 0 Þ Roots are imaginary and unequal or

complex conjugate.(B) Suppose a, b, c Î Q and a ¹ 0(i) If D > 0 and D is perfect square Þ Roots are unequal and

Rational(ii) If D > 0 and D is not perfect square Þ Roots are irrational

and unequal.Condition for Common Root(s)Let ax2 + bx + c = 0 and dx2 + ex + f = 0 have a common roota (say).

Condition for both the roots to be common is fc

eb

da

==

If p + iq (p and q being real) is a root of the quadratic equation,

where 1i -= , then p –iq is also a root of the quadraticequation.Every equation of nth degree (n ³ 1) has exactly n roots andif the equation has more than n roots, it is an identity.

Exponential Form: If z = x + iy is acomplex number then its exponential formis z = reiq where r is modulus and q isamplitude of complex number.

(i) 1 2 1 2| z | | z | | z z |+ ³ + ; hereequality holds when arg(z1/z2) = 0 i.e. z1 andz2 are parallel.

(ii) 1 2 1 2|| z | | z || | z z |- £ - ; here equality holds whenarg(z1/z2) = 0 i.e. z1 and z2 are parallel.(iii) | z1 + z2 |

2 + | z1 – z2 |2 = 2( | z1 |

2 + | z2 | 2 )

1 2 1 2 1 2arg(z z ) arg(z ) arg(z )= q + q = +

11 2 1 2

2

zarg arg(z ) arg(z )

zæ ö

= q - q = -ç ÷è ø

For any integer k, i4k = 1, i4k + 1 = i, i4k + 2 = – 1, i4k + 3 = – i

1 2| z z | | z z |- + - = l , represents an ellipse if| z1 – z2 | < l, having the points z1 and z2 as its foci. And if

1 2| z z |- = l , then z lies on a line segment connecting z1and z2.Properties of Cube Roots of Unity

(i) 21 0+w + w = (ii) w3 = 1

(iii) n 2n1 3+ w + w = (if n is multiple of 3)

(iv) n 2n1 0+w + w = (if n is not a multiple of 3).

The number of permutations of ndifferent things, taken r at a time, whererepetition is allowed, is nr.

Selection of Objects with Repetition :The total number of selections of r

things from n different things when eachthing may be repeated any number of timesis n+r+1Cr

Selection from distinct objects :The number of ways (or combinations) of n different thingsselecting at least one of them isnC1 + nC2 + nC3 + .....+ nCn = 2n – 1. This can also be stated asthe total number of combination of n different things.Selection from identical objects :The number of ways to select some or all outof (p + q + r) things where p are alike of first kind, q are alikeof second kind and r are alike of third kind is(p + 1) (q + 1) (r + 1) – 1Selection when both identical and distinct objects arepresent:If out of (p + q + r + t) things, p are alike one kind, q are alikeof second kind, r are alike of third kind and t are different,then the total number of combinations is(p + 1)(q + 1)(r + 1) 2t – 1Circular permutations:(a) Arrangements round a circular table :The number of circular permutations of n different things

taken all at a time is n

nPn

= (n – 1) !, if clockwise and

anticlockwise orders are taken as different.

QUADRATICEQUATIONS

ANDINEQUALITIES

COMPLEXNUMBERS

PERMUTA-TIONS ANDCOMBINA-

TIONS

19


(b) Arrangements of beads or flowers (all different) arounda circular necklace or garland:The number of circular permutations of ‘n’ different things

taken all at a time is 12

(n – 1)!, if clockwise and anticlockwise

orders are taken to be some.Sum of numbers :(a) For given n different digits a1, a2, a3 ......an the sum of thedigits in the unit place of all numbers formed (if numbers arenot repeated) is (a1 + a2 + a3 + .....+ an) (n – 1) !(b) Sum of the total numbers which can be formed with givenn different digits a1, a2, .........an is(a1 + a2 + a3 + .........+ an) (n – 1)! . (111 ........n times)

Greatest binomial coefficients : In abinomial expansion binomial coefficients ofthe middle terms are called as greatestbinomial coefficients.

(a) If n is even : When r =n2

i.e. nCn/2 takes

maximum value.

(b) If n is odd : r =n 1

2-

or n 1

2+

i.e. n n

n 1 n 12 2

C C- += and take maximum value.

Important Expansions :If | x | < 1 and n Î Q but n Ï N, then

(a) (1+ x)n = 1 + nx + n(n 1)

2!-

x2

+ ......+ n(n 1).....(n r 1)

r!- - +

xr + .......

(b) (1 – x)n = 1 – nx + n(n 1)

2!-

x2 – n(n 1)(n 2)

3!- -

x3

+ ......+n(n 1).....(n r 1)

r!- - +

(–x)r + .......

Properties related to A.P. :(i) Common difference of AP is given by

d = S2 – 2S1 where S2 is sum of firsttwo terms and S1 is sum of first term.

(ii) If for an AP sum of p terms is q, sum ofq terms is p, then sum of (p + q) term is(p + q).(iii) In an A.P. the sum of terms

equidistant from the beginning and end is constant andequal to sum of first and last terms.

(iv) If terms a1, a2, ..., an, an+1, ..., a2n+1 are in A.P., then sum ofthese terms will be equal to (2n + 1)an+1.

(v) If for an A.P. sum of p terms is equal to sum of q termsthen sum of (p + q) terms is zero

(vi) Sum of n AM's inserted between a and b is equal to n

times the single AM between a and b i.e. n

rr 1

A=å = nA

where A =a b

2+

The geometric mean (G.M.) of any two positive numbers aand b is given by ab i.e., the sequence a, G, b is G.P..n GM's between two given numbers: If in between twonumbers 'a' and 'b', we have to insert n GM G1,G2,.......... Gnthen a1, G1,G2,........Gn , b will be in G.P.The series consist of (n +2) terms and the last term is b andfirst term is a.

Þ arn + 2 –1 = b Þ r = 1

n 1ba

+æ öç ÷è ø

G1 = ar, G2 = ar2 ........Gn = arn or Gn = b/rUse of inequalities in progression :(a) Arithmetic Mean ³ Geometric Mean(b) Geometric Mean ³ Harmonic Mean :

A G H³ ³

An acute angle (say q) between linesL1 and L2 with slopes m1 and m2 is given by

2 21 2

1 2

m mtan , 1 m m 0

1 m m-

q = + ¹+

Three points A, B and C are collinear,if and only if slope of AB = slope of BC.

The equation of the line having normal distance from originis p and angle between normal and the positive x-axis is w, isgiven by x cos w + y sin w = p.Co-ordinate of some particular points :Let A(x1,y1), B(x2,y2) and C(x3,y3) are vertices of anytriangle ABC, thenIncentre : Co-ordinates of incentre

1 2 3 1 2 3ax bx cx ay by cy,

a b c a b c+ + + +æ ö

ç ÷è ø+ + + +

where a, b, c are the sides of triangle ABCArea of a triangle : Let (x1, y1), (x2, y2) and (x3, y3)respectively be the coordinates of the vertices A, B, C of atriangle ABC. Then the area of triangle ABC, is

12

[x1 (y2 – y3)+ x2 (y3 – y1) + x3 (y1 – y2)]

Or

= 12

1 1

2 2

3 3

x y 1x y 1x y 1

BINOMIALTHEOREM

SEQUENCEAND SERIES

STRAIGHTLINES

20


Condition of Tangency : Circlex2 + y2 =a2 will touch the line.

y = mx + c if c = ± 2a 1 m+Pair of Tangents : From a given point

P( x1,y1) two tangents PQ and PR can bedrawn to the circle S = x2 + y2 + 2gx + 2fy +c = 0. Their combined equation is SS1 = T2.

Condition of Orthogonality : If the angle of intersection ofthe two circle is a right angle (q = 90°) then such circle arecalled Orthogonal circle and conditions for their orthogonalityis 2g1g2 + 2f1f2 =c1 +c2Tangent to the parabola :Condition of Tangency : If the line y = mx + c touches aparabola y2 = 4ax then c = a/mTangent to the Ellipse:Condition of tangency and point of contact :The condition for the line y = mx + c to be a tangent to the

ellipse 2 2

2 2x ya b

+ = 1 is that c2 = a2m2 + b2 and the coordinates

of the points of contact are 2 2

2 2 2 2 2 2

a m b,

a m b a m b

æ ö±ç ÷

è ø+ +m

Normal to the ellipse(i) Point Form : The equation of the normal to the ellipse

2 2

2 2x ya b

+ = 1 at the point (x1, y1) is 2 2

1 1

a x b yx y

- = a2 – b2

(ii) Parametric Form : The equation of the normal to the

ellipse 2 2

2 2x ya b

+ = 1 at the point (a cosq, b sinq) is

ax secq – by cosecq = a2 – b2

Tangent to the hyperbola :Condition for tangency and points of contact : The conditionfor the line y = mx + c to be a tangent to the hyperbola

2 2

2 2x ya b

- = 1 is that c2 = a2m2 – b2 and the coordinates of the

points of contact are 2 2

2 2 2 2 2 2

a m b,

a m b a m b

æ ö± ±ç ÷

è ø- -Chord of contact :The equation of chord of contact of tangent drawn from a

point P (x1, y1) to the hyperbola 2 2

2 2x ya b

- = 1 is T = 0

where T º 1 12 2

xx yya b

- – 1

Equation of normal in different forms :Point Form : The equation of the normal to the hyperbola

2 2

2 2x ya b

- = 1 at the point (x1, y1) is 2 2

1 1

a x b yx y

+ = a2 + b2

Slope Form : The equation of normal to

the hyperbola 2 2

2 2x ya b

- = 1 in terms of slope

'm' is

y = mx ± 2 2

2 2 2

m(a b )

a b m

+

-Conditions of Parallelism and Perpendicularity of TwoLines:Case-I : When dc's of two lines AB and CD, say l1 , m1, n1and l2 , m2, n2 are known.

AB | | CD Û l1 = l2 , m1 = m2, n1 = n2AB ^ CD Û l1l2 + m1m2 + n1n2 = 0

Case-II : When dr's of two lines AB and CD, say a1, b1 c1 anda2, b2, c2 are known

AB | | CD Û 1 1 1

2 2 2

a b ca b c

= =

AB ^ CD Û a1a2 + b1b2 + c1c2 = 0If l1, m1, n1 and l2, m2, n2 are the direction cosines of twolines; and q is the acute angle between the two lines; then

cos q = | l1l2 + m1m2 + n1n2 |.Equation of a line through a point (x1, y1, z1) and having

direction cosines l, m, n is 1 1 1x x y y z zm n

- - -= =

l

Shortest distance between 1 1r a b= + lrr r and 2 2r a b= + m

rr r

is 1 2 2 1

1 2

(b b ).(a a )| b b |

´ -´

r r r r

r r

Let the two lines be

1 1 1

1 1 1

x y zm n

- a - b - g= =

l..........(1)

and2 2 2

2 2 2

x y zm n

- a -b - g= =

l..........(2)

These lines will coplanar if

2 1 2 1 2 1

1 1 1

2 2 2

m nm n

a - a b - b g - gl

l

= 0

The plane containing the two lines is

1 1 1

1 1 1

2 2 2

x y zm nm n

- a - b - gl

l

= 0

The equation of a plane through a point whose positionvector is ar and perpendicular to the vector N

r is

(r a).N 0- =rr r

THREEDIMENSIONAL

GEOMETRY

CONICSECTIONS

21


Vector equation of a plane that passes through theintersection of planes 1 1r.n d=

r r and 2 2r.n d=r r is

1 2 1 2r.(n n ) d d+ l = + lr r r , where l is any nonzero constant.

Two planes 1 1r a b= + lrr r and 2 2r a b= + m

rr r are coplanar if

2 1 1 2(a a ) (b b ) 0- + ´ =r rr r

LIMIT

Existence Of Limit :

x alim® f (x) exists Þ

x alim

-® f (x) =

x alim

+® f (x) = l

Where l is called the limit of the function

(i) If f(x) £ g(x) for every x in the deleted

nbd of a, then x alim®

f(x) £ x alim®

g(x)

(ii) If f(x) £ g(x) £ h (x) for every x in the deleted nbd of a

and x alim® f(x) = l = x a

lim® h (x) then x a

lim® g(x) = l

(iii) x alim® fog (x) = f

x alim g (x)®

æ öè ø = f (m) where x a

lim® g (x) = m

(iv) If x alim® f(x) = + ¥ or – ¥ , then x a

lim®

1f (x)

= 0

CONTINUITY AND DIFFERENTIABILITY OF FUNCTIONS

A function f(x) is said to be continuous at a point x = a if

x alim

+® f (x) =

x alim

-®f (x) = f(a)

Discontinuous Functions :(a) Removable Discontinuity:A function f is said to have removable discontinuity at x = a

if x alim

-®f (x) =

x alim

+® f (x) but their common value is not

equal to f (a).(b) Discontinuity of the first kind: A function f is said to

have a discontinuity of the first kind at x = a if x alim

-® f(x) and

x alim

+®f (x) both exist but are not equal.

(c) Discontinuity of second kind: A function f is said tohave a discontinuity of the second kind at x = a if neither

x alim

-® f (x) nor

x alim

+®f (x) exists.

Similarly, if +®axlim f (x) does not exist, then f is said to have

discontinuity of the second kind from the right at x = a.For a function f :Differentiability Þ Continuity;Continuity Þ/ derivabilityNot derivibaility Þ/ discontinuous ;But discontinuity Þ Non derivability

Differentiation of infinite series:

(i) If y f (x) f (x) f (x) ........= + + + ¥

Þ y = f (x) y+ Þ y2 = f (x) + y

2y dydx

= f ' (x) + dydx

\ dydx

= f '(x)2y 1-

(ii) If y = ....f (x)f (x)f (x)

¥ then y = f (x)y.

\ log y = y log [f(x)]

1 dy y '.f '(x) dylog f (x).

y dx f (x) dxæ ö= + ç ÷è ø

\2dy y f '(x)

dx f (x)[1 y log f (x)]=

-

(iii) If y = 1 1

f (x) f (x)

1f (x) ........f (x)+ +

+ then

dy yf '(x)dx 2y f (x)

=-

Interpretation of the Derivative : Ify = f (x) then, m= f ¢ (a) is the slope of thetangent line to y = f (x) at x = a

Increasing/Decreasing :(i) If f ¢ (x) > 0 for all x in an interval I thenf (x) is increasing on the interval I.

(ii) If f ¢ (x) < 0 for all x in aninterval I then f (x) is decreasing on the interval I.

(iii) If f ¢ (x) = 0 for all x in an interval I then f (x) is constant onthe interval I.

Test of Local Maxima and Minima –First Derivative Test – Let f be a differentiable function definedon an open interval I and c Î I be any point. f has a local maximaor a local minima at x = c, f ¢ (c) = 0.

Put dy 0dx

= and solve this equation for x. Let c1, c2........cn

be the roots of this.

If dydx

changes sign from +ve to –ve as x increases

through c1 then the function attains a local max at x = c1

If dydx

changes its sign from –ve to +ve as x increases

through c1 then the function attains a local minimum at x = c1

If dydx does not changes sign as increases through c1

then x = c1 is neither a point of local maxm nor a point of localminm. In this case x is a point of inflexion.

DIFFERENTIALCALCULUS

DIFFERENTIA-TION AND

APPLICATION

22


Rate of change of variable :

The value of dydx

at x = x0 i.e. x x0

dydx =

æ öç ÷è ø represents the rate

of change of y with respect to x at x = x0

If x = f (t) and y = y (t), then dy dy / dtdx dx / dt

= , provided that dxdt

¹ 0

Thus, the rate of change of y with respect to x can becalculated by using the rate of change of y and that of x eachwith respect to t.

Length of Sub–tangent = dxydy ; Sub–normal =

dyy

dx ;

Length of tangent =

2dxy 1dy

ì üæ öï ï+í ýç ÷è øï ïî þ

Length of normal = 2dyy 1

dx

ì üï ïæ ö+í ýç ÷è øï ïî þEquations of tangent and normal : The equation of thetangent at P (x1, y1) to the curve y = f(x) is

y – y1 = P

dydx

æ öç ÷è ø

(x – x1)

The equation of the normal atP (x1, y1) to the curve y = f (x) is

y – y1 =

P

1dydx

-æ öç ÷è ø

(x – x1)

Two standard forms of integral :xeò [f (x) + f ' (x) dx = ex f (x) + c

Þ xeò [f (x) + f '(x)] dx = xeò f (x)dx +

xeò f ' (x) dx

= ex f (x) – xeò f '(x) dx + xeò f' (x)(on integrating by parts) = ex f (x) + cTable shows the partial fractions corresponding to differenttype of rational functions :

S. Form of rational Form of partialNo. function fraction

1.px q

(x a) (x b)+

- -A B

(x a) (x b)+

- -

2.2

2px qx r

(x a) (x b)+ +

- - 2A B C

(x a) (x b)(x a)+ +

- --

3.2

2px qx r

(x a) (x bx c)+ +

- + + 2A Bx C

(x a) x bx C+

+- + +

Leibnitz rule :

g(x)

f (x)

d F(t) dt g (x)F(g(x)) f (x)F(f (x))dx

= -¢ ¢ò

If a series can be put in the formr n 1

r 0

1 rfn n

= -

=

æ öç ÷è øå or

r n

r 1

1 rfn n

=

=

æ öç ÷è øå , then its limit as n ® ¥

is 1

0

f (x) dxòArea between curves :

b

a

y f (x) A [upper function] [lower function] dx= Þ = -ò

and d

c

x f (y) A [right function] [left function] dy= Þ = -ò

If the curves intersect then the area of each portion must befound individually.Symmetrical area : If the curve is symmetrical about acoordinate axis (or a line or origin), then we find the area ofone symmetrical portion and multiply it by the number of

symmetrical portion to get the required area.Probability of an event: For a finite

sample space with equally likely outcomesProbability of an event is

n(A)P(A)

n(S)= , where n (A) = number of

elements in the set A, n (S) = number ofelements in the set S.Theorem of total probability : Let {E1, E2, ...,En} be a partitionof a sample space and suppose that each of E1, E2, ..., En hasnonzero probability. Let A be any event associated with S,then P(A) = P(E1) P (A | E1) + P (E2) P (A | E2) + ... + P (En)P(A | En)Bayes' theorem: If E1, E2, ..., En are events which constitutea partition of sample space S, i.e. E1, E2, ..., En are pairwisedisjoint and E1 È E2 È... È En = S and A be any event withnonzero probability, then

i ii n

j jj 1

P(E ) P (A | E )P (E | A)

P(E ) P(A | E )=

=

å

Let X be a random variable whose possible values x1, x2, x3,..., xn occur with probabilities p1, p2, p3, ... pn respectively.

The mean of X, denoted by µ, is the number n

i ii 1

x p=å

The mean of a random variable X is also called theexpectation of X, denoted by E (X).

INTEGRALCALCULUS

PROBABILITY

23


Trials of a random experiment are called Bernoulli trials, ifthey satisfy the following conditions :(a) There should be a finite number of trials. (b) The trialsshould be independent. (c) Each trial has exactly twooutcomes : success or failure. (d) The probability of successremains the same in each trial.For Binomial distribution B (n, p),P (X = x) = nCx q

n–x px, x = 0, 1,..., n (q = 1 – p)Properties of adjoint matrix : If A, B

are square matrices of order n and In iscorresponding unit matrix, then(i) A (adj. A) = | A | In = (adj A) A(ii) | adj A | = | A |n–1 (Thus A (adj A) is

always a scalar matrix)(iii) adj (adj A) = | A |n–2 A

(iv) | adj (adj A) | = 2(n 1)| A | -

(v) adj (AT) = (adj A)T

(vi) adj (AB) = (adj B) (adj A)(vii) adj (Am) = (adj A)m, m Î N(viii) adj (kA) = kn–1 (adj. A), k Î R(ix) adj (In) = InProperties of Inverse Matrix : Let A and B are two invertiblematrices of the same order, then(i) (AT)–1 = (A–1)T

(ii) (AB)–1 = B–1A–1

(iii) (Ak)–1 = (A–1)k, k Î N(iv) adj (A–1) = (adj A)–1

(v) (A–1)–1 = A

(vi) | A–1 | = 1

| A | = | A |–1

(vii) If A = diag (a1,a2.....,an), thenA–1 = diag (a1

–1, a2–1, .........an

–1)(viii) A is symmetric matrix Þ A–1 is symmetric matrix.

Differentiation of a matrix : If A = f (x) g(x)h(x) (x)é ùê úë ûl

then

f '(x) g '(x)dAh '(x) '(x)dxé ù

= ê úë ûl

is a differentiation of Matrix A.

Properties of Transpose(i) (AT)T = A (ii) (A ± B)T = AT ± BT

(iii)(AB)T = BT AT (iv) (kA)T = k(A)T

(v) IT = I (vi) tr (A) = tr (A)T

(vii) (A1A2A3.....An–1 An)T = An

T An–1T.....A3

TA2TA1

T

Rank of a Matrix : A number r is said to be the rank of am × n matrix A if(a) Every square sub matrix of order (r + 1) or more is singularand (b) There exists at least one square submatrix of order rwhich is non-singular.Thus, the rank of matrix is the order of the highest ordernon-singular sub matrix.

Using Crammer's rule of determinantwe get

1 2 3

x y z 1= = =D D D D i. e. x = 1D

D, y = 2D

D,

z = 3DD

Case-I : If D ¹ 0

Then x = 1DD

, y = 2DD

, z = 3DD

\ The system is consistent and has unique solutions.Case-II if D = 0 and(i) If at least one of D1, D2, D3 is not zero then the system

of equations a inconsistent i.e. has no solution.(ii) If d1 = d2 = d3 = 0 or D1, D2, D3 are all zero then the

system of equations has infinitely many solutions.

Given vectors 1 1 1x a y b z c+ +r r r

,

2 2 2x a y b z c+ +r r r

, 3 3 3x a y b z c+ +r r r

, where

a, b, crr r are non-coplanar vectors, will be

coplanar if and only if 1 1 1

2 2 2

3 3 3

x y zx y zx y z

= 0

Scalar triple product :

(a) If 1 2 3ˆ ˆ ˆa a i a j a k= + +

r, 1 2 3

ˆ ˆ ˆb b i b j b k= + +r

and

1 2 3ˆ ˆ ˆc c i c j c k= + +

r then

1 2 3

1 2 3

1 2 3

a a a(a b).c [a b c] b b b

c c c´ = =

r r r r r r

(b) [a b c] = volume of the parallelopiped whosecoterminous edges are formed by a, b, c

rr r

(c) a, b, crr r are coplanar if and only if [a b c]

r r r = 0

(d) Four points A, B, C, D with position vectors a, b, c,dr r r r

respectively are coplanar if and only if

[AB AC AD]uuur uuur uuur

= 0 i.e. if and only if

[b a c a d a]- - -r r r r r r

= 0(e) Volume of a tetrahedron with three coterminous edges

1a ,b,c [a b c]

6=

r r r r r r

(f) Volume of prism on a triangular base with three

coterminous edges 1a ,b,c [a b c]2

=r r r r r r

Lagrange's identity :

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

b.c b.d´ ´ = = -

r r urrr r r r r r r r r r r r

r r r r

MATRICES

DETERMI-NANTS

VECTORALGEBRA

24


Reciprocal system of vectors : If a, b, crr r be any three non

coplanar vectors so that

[a b c] 0¹rr r then the three vectors a 'b 'c '

r r r

defined by the

equations b c

a '[a b c]

´=

r rr

rr r , c ab '[a b c]

´=

r rr

rr r , a bc '

[a b c]´

=

rrr

rr r are called

the reciprocal system of vectors to the given vectors a, b, crr r

.Relation between A.M., G.M. and H.M.A.M. ³ G.M. ³ H.M.

Equality sign holds only when all theobservations in the series are same.

Relationship between mean, mode andmedian :(i) In symmetrical distribution

Mean = Mode = Median(ii) In skew (moderately symmetrical) distribution

Mode = 3 median – 2 meanMean deviation for ungrouped data

i| x x |M.D.(x)

n-

= å , i| x M |M.D.(M)

n-

= å

Mean deviation for grouped data

i if | x x |M.D.(x)

N-

= å , i if | x M |M.D.(M)

N-

= å ,

where iN f= åVariance and standard deviation for ungrouped data

2 2 2i i

1 1(x x) , (x x)

n ns = - s = -å å

Variance and standard deviation of a discrete frequencydistribution

2 2 2i i i i

1 1f (x x) , f (x x)

n Ns = - s = -å å

Variance and standard deviation of a continuous frequencydistribution

2 2 2 2i i i i i i

1 1f (x x) , f x ( f x )

n Ns = - s = -å å å

Coefficient of variation (C.V.) = 100, x 0xs

´ ¹

For series with equal means, the series with lesser standarddeviation is more consistent or less scattered.

Methods of solving a first order firstdegree differential equation :(a) Differential equation of the formdy f (x)dx

=

dy f (x)dx

= Þ dy = f (x) dx

Integrating both sides we obtain

dy f (x)dx c= +ò ò or y = f (x)dx c+ò

(b) Differential equation of the form dy f (x) g(y)dx

=

dy f (x) g(y)dx

= dy f (x) dx cg (y)

Þ = +ò ò(c) Differential equation of the form of

dy f (ax by c)dx

= + + :

To solve this type of differential equations, we put

ax + by + c = v and dy 1 dv

adx b dx

æ ö= -ç ÷è ø

\ dv

dxa b f (v)

=+

So solution is by integrating dv dxa bf (v)

=+ò ò

(d) Differential Equation of homogeneous type :An equation in x and y is said to be homogeneous if it

can be put in the form dy f (x, y)dx g (x, y)

= where f (x ,y) and g

(x ,y) are both homogeneous functions of the samedegree in x & y .So to solve the homogeneous differential equation

dy f (x, y)dx g(x, y)

= , substitute y = vx and so dy dVv xdx dx

= +

Thus dv dx dvv x f (v)dx x f (v) v

+ = Þ =-

Therefore solution is dx dv cx f (v) v

= +-ò ò

Linear differential equations :dy Py Qdx

+ = ....... (1)

Where P and Q are either constants or functions of x.

Multiplying both sides of (1) by P dxeò , we get

P dx P dxdye Py Q edx

æ öò ò+ =ç ÷è ø

On integrating both sides with respect to x we getP dx P dxy e Q e cò ò= +ò

which is the required solution, where c is the constant andP dxeò is called the integration factor..

P dx Pdxy e Q e cò ò= +ò

DIFFERENTIALEQUATIONS

STATISTICS

1 important formulas/ - answer key results...

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