NANDAKA

PHYSICSFORMULAE BOOKLET

FORJEE Mains

&JEE Advanced

"Physics decides whether you get a rank in JEE;Chemistry decides what rank you get"

CLASS XI : TOPICS

Description

1. Units, Dimensions & Measurements

2. Motion in one Dimension & Newton�s Laws of Motion

3. Vectors

4. Circular Motion, Relative Motion, and Projectile Motion

5. Friction & Dynamics of Rigid Body

6. Conservation Laws & Collisions

7. Simple Harmonic Motion & Lissajous Figures

8. Gravitation

9. Properties of Matter

10. Heat & Thermodynamics

11. Waves

CLASS XII : TOPICS

Description

1. Elecrotstatics and Capacitance

2. Current Electricity

3. Magnetic Effects of Current

4. Magnetic Properties of Current

5. Electromagnetic Induction

6. Alternating Current

7. Light

8. Modern Physics

CONTENTS

UNITS, DIMENSIONS AND MEASUREMENTS (i) SI Units:

(a) Time-second(s); (b) Length-metre (m); (c) Mass-kilogram (kg); (d) Amount of substance�mole (mol); (e) Temperature-Kelvin (K); (f) Electric Current � ampere (A); (g) Luminous Intensity � Candela (Cd)

(ii) Uses of dimensional analysis

(a) To check the accuracy of a given relation (b) To derive a relative between different physical quantities (c) To convert a physical quantity from one system to another system

c

2

1b

2

1a

2

1122221 T

T x

LL

x MM

nnor unun

==

(iii) Mean or average value: N

X...XXX N21 ++−

=

(iv) Absolute error in each measurement: |∆Xi| = | X �Xi|

(v) Mean absolute error: ∆Xm=N

|X| i∆Σ

(vi) Fractional error = XX∆

(vii) Percentage error = 100 x XX∆

(viii) Combination of error: If ƒ = c

ba

ZYX , then maximum fractional error in ƒ is:

Ζ

∆Ζ+∆+∆=ƒƒ∆ |c|

YY |b|

XX |a|

MOTION IN ONE DIMENSION & NEWTON�S LAWS OF MOTION (i) Displacement: | displacement | ≤ distance covered

(ii) Average speed:

2

2

1

1

21

21

21

vs

vs

ssttss

v+

+=

++

=

(a) If s1 = s2 = d, then 21

21vvv v2

v+

= = Harmonic mean

(b) If t1 = t2, then 2

vvv 21 += =arithmetic mean

(iii) Average velocity: (a) 12

12avttrr

v−−

=→→

→; (b) v|v| av ≤

→

(iv) Instantaneous velocity: |v| and dt

rdv→

→→

= = v = instantaneous speed

(v) Average acceleration: 12

12avttvv

a−−

=

→→→

(vi) Instantaneous acceleration: dt/vda→→

=

In one � dimension, a = (dv/dt) = vdxdv

(vii) Equations of motion in one dimension: (a) v = u + at;

(b) x = ut + 21 at2 ;

(c) v2 u2 + 2ax;

(d) x = vt � 21 at2;

(e)

+=

2uvx t;

(f) ;at21 ut xxs 2

0 +=−=

(g) v2 = u2 + 2a (x�x0)

(viii) Distance travelled in nth second: dn = u + 2a (2n�1)

(ix) Motion of a ball: (a) when thrown up: h = (u2/2g) and t = (u/g)

(b) when dropped: v = √(2gh) and t = √(2h/g) (x) Resultant force: F = √(F1

2 + F22 + 2F1F2 cos θ)

(xi) Condition for equilibrium: (a) ; )FF( F 213→→→

+−= (b) F1 + F2 ≥ F3 ≥ |F1 � F2|

(xii) Lami�s Theorem: ( ) ( ) ( )γ−π=

β−π=

α−π sinR

sinQ

sinP

(xiii) Newton�s second law:

==

→→→→dt/pdF ;amF

(xiv) Impulse: =−∆=∆→→

12 pp and tFp Fdt (xv) Newton�s third law:

∫21

(a) →→

−= 1212 FF

(b) Contact force: 2112 FFmM

mF =+

=

(c) Acceleration: a = mM

F+

(xvi) Inertial mass: mI = F/a

(xvii) Gravitational mass: mG = GI

2mm ;

GMFR

gF ==

(xviii) Non inertial frame: If →

0a be the acceleration of frame, then pseudo force →→

−= 0amF

Example: Centrifugal force = rmr

mv22ω=

(xix) Lift problems: Apparent weight = M(g ± a0) (+ sign is used when lift is moving up while � sign when lift is moving down) (xx) Pulley Problems: (a) For figure (2):

Tension in the string, T = gmm

mm

21

21+

Acceleration of the system, a = gmm

m

21

2+

The force on the pulley, F = gmmmm2

21

21+

(b) For figure (3):

Tension in the string, gmmmm2

T21

21+

=

Acceleration of the system, gmmmm

a12

12+−

=

The force on the pulley, gmmmm4

F21

21+

=

• a

T T T

T a

m2

m1

Fig. 3

m2

m1 T

Frictionless surface

m2g Fig. 2

T

F

(1)

(2)

M

F21

m F12

Fig. 1

VECTORS (i) Vector addition: )B(A BA and A B B AR

→→→→→→→→→−+=−+=+=

(ii) Unit vector: )A/A( A^ →

= (iii) Magnitude: )AA(A A 2

z2y

2x ++√=

(iv) Direction cosines: cos α = (Ax/A), cos β = (Ay/A), cos γ = (Az/A) (v) Projection:

(a) Component of →A along

→B =

^B . A

→

(b) Component of →B along

→A =

→B . A

^

(c) If →A = ,jA iA

^y

^x + then its angle with the x�axis is θ = tan�1 (Ay/Ax)

(vi) Dot product:

(a) →→B . A = AB cos θ, (b) zzyyxx BABABA B . A ++=

→→

(vii) Cross product:

(a) =→→B x A AB sin θ

^n ;

(b) ;0Ax A =→→

(c) →→B x A =

zyx

zyx

^^^

BBBAAAkji

(viii) Examples:

(a) ;r . F W→→

= (b) ; v . FP→→

= (c) ;A . E→→

Ε =φ (d) ;A . B →→

Β =φ

(e) ; r x wv→→→

= (f) ;F x →→→

τ=τ (g)

=

→→→B x v qF m

(ix) Area of a parallelogram: Area = |B x A|→→

(x) Area of a triangle: Area = |B x A | 21 →→

(xi) Gradient operator: z

k y

j x

i V^^^

∂∂+

∂∂+

∂∂=

(xii) Volume of a parallelopiped:

=

→→→C x B .AV

CIRCULAR MOTION, RELATIVE MOTION & PROJECTILE MOTION

(i) Uniform Circular Motion: (a) v = ωr; (b) a = (v2/r) = ω2r ; (c) F = (mv2/r);

(d) ;0v.r =→→

(e) 0a.v =→→

(ii) Cyclist taking a turn: tan θ = (v2/rg) (iii) Car taking a turn on level road: v = √(µsrg) (iv) Banking of Roads: tan θ = v2/rg (v) Air plane taking a turn: tan θ = v2/r g (vi) Overloaded truck: (a) Rinner wheel < Router wheel (b) maximum safe velocity on turn, v = √(gdr/2h) (vii) Non�uniform Circular Motion: (a) Centripetal acceleration ar = (v2/r); (b) Tangential acceleration at = (dv/dt);

(c) Resultant acceleration a=√ )aa( 2t

2r +

(viii) Motion in a vertical Circle: (a) For lowest point A and highest point B, TA � TB = 6 mg; v2

A = v2B + 4gl ; vA ≥ √(5gl); and vB≥

√ (gl) (b) Condition for Oscillation: vA < √(2gl) (c) Condition for leaving Circular path: √(2gl) < vA < √(5gl)

(ix) Relative velocity: ABBA vvv→→→

−=

(x) Condition for Collision of ships: 0)vv( x )vr( BABA =−−→→→→

(xi) Crossing a River:

(a) Beat Keeps its direction perpendicular to water current (1) vR =√( ; )vv( 2

b2w + (2) θ = tan�1 );v/v( bw

(3) t=(x/vb) (it is minimum) (4) Drift on opposite bank = (vw/vb)x (b) Boat to reach directly opposite to starting point:

(1) sin θ = (vw/vb); (2) vresultant = vb cos θ ; (3) t=θ cos v

x

b

(xii) Projectile thrown from the ground:

(a) equation of trajectory: y = x tan θ � θ 22

2

cos u2xg

(b) time of flight: gsin u 2T θ=

(c) Horizontal range, R = (u2 sin 2θ/g) (d) Maximum height attained, H = (u2 sin2 θ/2g)

(e) Range is maximum when θ = 450 (f) Ranges are same for projection angles θ and (900�θ) (g) Velocity at the top most point is = u cos θ (h) tan θ = gT2/2R (i) (H/T2) = (g/8)

(xiii) Projectile thrown from a height h in horizontal direction: (a) T = √(2h/g); (b) R = v√(2h/g); (c) y = h � (gx2/2u2) (d) Magnitude of velocity at the ground = √(u2 + 2gh)

(e) Angle at which projectiles strikes the ground, θ = tan�1 ugh2

(xiv) Projectile on an inclined plane:

(a) Time of flight, T = ( )0

0θ

θ−θ cos g

sin u2

(b) Horizontal range, ( )0

0

θθθ−θ

= cos g

cos sin u2R 2

2

FRICTION (i) Force of friction: (a) ƒs ≤ µsN (self adjusting); (ƒs)max = µsN (b) µk = µkN (µk = coefficient of kinetic friction) (c) µk < µs

(ii) Acceleration on a horizontal plane: a = (F � µkN)/M (iii) Acceleration of a body sliding on an inclined plane: a = g sin θ (1� µk cot t2) (iv) Force required to balance an object against wall: F = (Mg/µs) (v) Angle of friction: tan θ = µs (µs = coefficient of static friction)

DYNAMICS OF RIGID BODIES

(i) Average angular velocity: ttt

1 ∆

θ∆=−

θ−θ=ω

2

12

(ii) Instantaneous angular velocity: ω = (dθ/dt)

(iii) Relation between v, ω and r : v=ωr; In vector form →→→

ω= rxv ; In general form, v = ωr sin θ

(iv) Average angular acceleration: ttt 12 ∆ω∆=

−ω−ω

=α 12

(v) Instantaneous angular acceleration: α = (dω/dt) = (d2θ/dt2) (vi) Relation between linear and angular acceleration:

(a) aT = αr and aR = (v2/r) = ω2R (b) Resultant acceleration, a = √ )aa( 2

R2T +

(c) In vector form,

ωω=ω=α=+=

→→→→→→→→→→→→r x x u x a and r x a where,aaa RTRT

(vii) Equations for rotational motion: (a) ω = ω0 + αt;

(b) θ = ω0t + 21 αt2;

(c) ω2 � ω02 = 2αθ

(viii) Centre of mass: For two particle system: (a) ;

mmxmxmx

21

2211CM +

+=

(b) 21

2211CM mm

vmvmv

++

=

(c) 21

2211CM mm

amama

++

=

Also 2CM

2CM

CMCM

CMdt xd

dtdv

a and dt

dxv ===

(ix) Centre of mass: For many particle system:

(a) ;M

xmX iiCM

Σ=

(b) ;M

rmr iiCM

→→ Σ=

(c) ;dtrdv CM

CM

→→

=

(d) ;dtvda CM

CM

→→

=

(e) →→→

Σ== iiCMCM vm vMP ;

(f) .FamaMF iiiCMext→→→→

Σ=Σ== If ;constant V ,0a ,0F CMCMext ===→→→

(g) Also, moment of masses about CM is zero, i.e., 2211ii rmrmor 0rm ==Σ→

(x) Moment of Inertia: (a) I = Σ mi ri

2 (b) I = µr2, where µ = m1m2/(m1 + m2) (xi) Radius of gyration: (a) K = √(I/M) ; (b) K = √[(r1

2 + r22 + � + rn

2)/n] = root mean square distance.

(xii) Kinetic energy of rotation: K = 21 Iω2 or I = (2K/ω2)

(xiii) Angular momentum: ( ) ( ) d vm csin rp L (b) ; p xr La ; θ==→→→

(xiv) Torque: ( ) ( ) θ=τ=τ→→→

sin Fr b;F x ra

(xv) Relation between τ and L:

=τ

→→dt/dL ;

(xvi) Relation between L and I: (a) L = Iω; (b) K = 21 Iω2 = L2/2I

(xvii) Relation between τ and α: (a) τ = Iα,

(b) If τ = 0, then (dL/dt)=0 or L=constant or, Iω=constant i.e., I1ω1= I2ω2

(Laws of conservation of angular momentum)

(xviii) Angular impulse: tL ∆τ=∆→→

(xix) Rotational work done: W = θτ=θτ av d

(xx) Rotational Power: →→ωτ= .P

(xxi) (a) Perpendicular axes theorem: Iz = Ix + Iy (b) Parallel axes theorem: I = Ic + Md2

(xxii) Moment of Inertia of some objects (a) Ring: I = MR2 (axis); I =

21 MR2 (Diameter);

I = 2 MR2 (tangential to rim, perpendicular to plane); I = (3/2) MR2 (tangential to rim and parallel to diameter)

∫

(b) Disc: I = 21 MR2 (axis); I =

41 MR2 (diameter)

(c) Cylinder: I = ( )axis MR21 2

(d) Thin rod: I = (ML2/12) (about centre); I = (ML2/3) (about one end) (e) Hollow sphere : Idia = (2/3) MR2; Itangential = (5/3) MR2 (f) Solid sphere: Idia = (2/5) MR2 ; Itangential = (7/5) MR2

(g) Rectangular: ( )12

bMI22

C+= l (centre)

(h) Cube: I = (1/6) Ma2 (i) Annular disc: I = (1/2) M ( 2

221 RR + )

(j) Right circular cone: I = (3/10) MR2 (k) Triangular lamina: I = (1/6) Mh2 (about base axis) (l) Elliptical lamina: I = (1/4) Ma2 (about minor axis) and I = (1/4) Mb2 (about major axis) (xxiii) Rolling without slipping on a horizontal surface:

+=ω+= 2

2

222

RK1 MV

21 I

21MV

21K (Q V = Rω and I = MK2)

For inclined plane

(a) Velocity at the bottom, v =

+ 2

2

RK1gh2

(b) Acceleration, a = g sin θ

+ 2

2

rK1

(c) Time taken to reach the bottom, t = θ

+ sin g

RK1s2 2

2

(xxiv) Simple pendulum: = T = 2π√ (L/g) (xxv) Compound Pendulum: T = 2π√ (I/Mg l), where l = M (K2 + l2) Minimum time period, T0 = 2π√ (2K/g) (xxvi) Time period for disc: T = 2π √(3R/2g) Minimum time period for disc, T = 2π√ (1.414R/g) (xxvii) Time period for a rod of length L pivoted at one end: T = 2π√(2L/3g

CONSERVATION LAWS AND COLLISIONS

(i) Work done: (a) ;d.FW→→

= (b) ;θ= cos FdW (c) =W

(ii) Conservation forces: 0rd . F (c) ;rd .F (b) rd . F )a( =→→→→→→

For conservative forces, one must have: →F x V = 0

(iii) Potential energy: (a) ( ) ( ) UVFc ; dU/dX F (b) W; UV −=−=−=→

(iv) Gravitational potential energy: (a) U = mgh ; (b) ( )hRGMmU

+−=

(v) Spring potential energy: ( ) ( ) ( )2

12

22 xx K U b ; Kx Ua −=∆=

(vi) Kinetic energy: (a) ∆K = W = ( ) 22

i2f mv K b ; mv mv =−

(vii) Total mechanical energy: = E = K + U (viii) Conservation of energy: ∆K = � ∆U or, Kƒ + Uƒ = Ki + Ui

In an isolated system, Etotal = constant

(ix) Power: (a) P = (dw/dt) ; (b) P = (dw/dt) ; (c) P = →→v.F

(x) Tractive force: F = (P/v) (xi) Equilibrium Conditions:

(a) For equilibrium, (dU/dx) = 0 (b) For stable equilibrium: U(x) = minimum, (dU/dx) = 0 and (d2U/dx2) is positive (c) For unstable equilibrium: U(x) = maximum, (dU/dx) = 0 and (d2U/dx2) is negative (d) For neutral equilibrium: U(x) = constant, (dU/dx) = 0 and (d2U/dx2) is zero

(xii) Velocity of a particle in terms of U(x): v = ± ( )[ ]xUEm2 −

(xiii) Momentum:

(a) ( )

==

→→→→dt/pdF b ;vmp ,

(b) Conservation of momentum: If ,pp then ,0F ifnet→→→

==

(c) Recoil speed of gun, BG

BG x v

mm

v =

(xiv) Impulse: t F p av ∆=∆→→

(xv) Collision in one dimension:

(a) Momentum conservation : m1u1 + m2u2 = m1v1 + m2v2 (b) For elastic collision, e = 1 = coefficient of restitution (c) Energy conservation: m1u1

2 + m2u22 = m1v1

2 + m2v22

∫ 21

xx

Path 1 Path 2 closed path

∫ba ∫ba ∫

1 2

1 2

1 2

1 2

1 2

1 2

1 2

1 2

1 2

rd.F→→

212

121

21

122

12

21

21

211 u

mmmmu

mmm2 v;u

mmm2u

mmmmv

+−

+

+

=

+

+

+−

=

(e) If m1 = m2 = m, then v1 = u2 and v2 = u1

(f) Coefficient of restitution, e = (v2�v1/u1 = u2) (g) e = 1 for perfectly elastic collision and e=0 for perfectly inelastic collision. For inelastic

collision 0 < e < 1

(xvi) Inelastic collision of a ball dropped from height h0

(a) Height attained after nth impact, hn = e2nh0

(b) Total distance traveled when the ball finally comes to rest, s = h0 (1+e2)/(1�e2)

(c) Total time taken, t =

−+

e1e1

gh2 0

(xvii) Loss of KE in elastic collision: For the first incident particle

( )%100

KK

,mm

If

;mm

m4m

KK

and

mmmm

KK

i

lost212

21

21

i

lost2

21

21

i=∆=

+=∆

+−=ƒ

(xviii) Loss of KE in inelastic collision: ∆ Klost = Ki � Kƒ=21

21mm

mm21

+(u1 � u2)2 (1�e2)

Velocity after inelastic collision (with target at rest)

( )1

21

121

21

211 u

mme1m

vand

u

mmemm

v+

+=

+−

=

(xix) Oblique Collision (target at rest): m1u1 = m1v1 cos θ1 + m2v2 cos θ2 and m1v1 sin θ1 = m2v2 sin θ2

Solving, we get: m1u12 = m1v1

2 + m2v22

(xx) Rocket equation: (a) dt

dM

vdtdVM el−=

(b) V = � vrel loge

−

0

b0M

mM [M0 = original mass of rocket plus fuel and mb = mass of fuel burnt]

(c) If we write M = M0 � mb = mass of the rocket and full at any time, than velocity of rocks at that time is:

V = vrel loge (M0/M)

(xxi) Conservation of angular momentum: (a) If τext = 0, then Lƒ = Li

(b) For planets, min

max

min

maxrr

vv

=

(c) Spinning skater, I1ω1 = I2W2 or ωƒ = ωi

ƒIIi

(d) Velocities of 1st and 2nd body after collision are:

SIMPLE HARMONIC MOTION AND LISSAJOUS FIGURES (i) Simple Harmonic Motion:

(a) F = � Kx ;

(b) a = � mK x or a = � ω2x, where ω = √(K/m);

(c) Fmax = ± KA and amax = ±ω2A

(ii) Equation of motion: 0xdt

xd2

2=ω+ 2

(iii) Displacement: x = A sin (ωt + φ)

(a) If φ = 0, x = A sin ωt ; (b) If φ = π/2, x = A cos ωt

(c) If x = C sin ωt + D cos ωt, then x = A sin (ωt + φ) with A= √(C2+D2) and φ = tan�1 (D/C) (iv) Velocity:

(a) v = A ω cos (ω+ φ); (b) If φ=0, v = A ω cos ωt; (c) vmax =±ωA (d) v = ± ω√(A2 � x2);

(e) 1A

vAx

22

2

2

2=

ω+

(v) Acceleration:

(a) a = �ω2 x = � ω2A sin (ωt+φ) ; (b) If φ=0, a=� ω2A sin ωt (c) |amax| = ω2A; (d) Fmax = ± m ω2A

(vi) Frequency and Time period:

(a) ω = √(K/m) ;

(b) ƒ= ( );m/K21π

(c) T = 2πKm

(vii) Energy in SHM: Potential Energy: (a) U = Kx2 ;

(b) F = � dxdU ;

(c) Umax = mω2A2;

(d) U = mω2 A2 sin2 ωt (viii) Energy in SHM: Kinetic energy: (a) K = mv2; (b) K= mω2 (A2�x2);

(c) K = mω2A2 cos2 ωt ; (d) Kmax = mω2A2

21

21

21

21

21

21

21

(ix) Total energy: (a) E = K + U = conserved; (b) E = (1/2) mω2 A2 ; (c) E = Kmax = Umax

(x) Average PE and KE:

(a) < U > = (1/4) mω2A2 ; (b) < K > = (1/4) mω2A2; (c) (E/2) = < U > = < K >

(xi) Some relations:

(a) ω = ; xxvv

21

22

22

21

−− (b) T = 2π 2

221

21

22

vvxx

−− ; (c) A = ( ) ( )

22

21

122

21

vvxvxv

−− 2

(xii) Spring� mass system: (a) mg = Kx0;

(b) T = 2π g

x 2

Km 0π=

(xiii) Massive spring: T = 2π ( )K

3/mm s+

(xiv) Cutting a spring: (a) K� = nK ; (b) T� = T0/√(n) ; (c) ƒ� = √(n) ƒ0

(d) If spring is cut into two pieces of lengths l1 and l2 such that l1 = nl2, then K1 = K, n

1n

+ K2 =

(n +1) K and K1l1 = K2l2 (xv) Springs in parallel: (a) K = K1 + K2 ; (b) T = 2π √[m/(K1 + K2)] (c) If T1 = 2π√ (m/K1) and T2 = 2π√(m/K2), then for the parallel combination:

22

21

222

21

2122

21

2 and TT

TT Tor T1

T1

T1 ω+ω=ω

+=+=

(xvi) Springs in series: (a) K1x1 = K2x2 = Kx = F applied

(b) 21

21

21 KKKK

Kor K1

K1

K1

+=+=

(c) 22

21

221

TTTor 111 +−ω

+ω

=ω 2

22

(d) T = 2π ( )( )21

21

21

21KK m

KK 21or

KKKKm

+π=ƒ+

(xvii) Torsional pendulum:

(a) Iα=τ�Cθ or 0 IC

dtd

2

2=θ+θ ;

(b) θ=θ0 sin (ωt+φ); (c) ω = √(C/I) ;

(d) ƒ = IC

21π

;

(e) T = 2π√(I/C), where C = πηr4/2l

(xviii) Simple pendulum:

(a) Iα = τ =� mgl sin θ or

+θ

2 l

gdtd2

sin θ = 0 or 0 gdtd

2

2 = θ+θ

l;

(b) ω = √(g/l) ;

(c) ƒ = ( ) ; g/ 21

lπ

(d) T = 2π √(l/g) (xix) Second pendulum: (a) T = 2 sec ; (b) l = 99.3 cm (xx) Infinite length pendulum: (a) ;

R11 g

1 2T

e

+

π=

l

(b) T=2πg

R e (when l→∞)

(xxi) Anharmonic pendulum: T ≅ T0

+≅

θ+ 2

2

0

20

16A1 T

161

l

(xxii) Tension in string of a simple pendulum: T = (3 mg cos θ � 2 mg cos θ0) (xxiii) Conical Pendulum: (a) v = √(gR tan θ) ; (b) T = 2π√ (L cos θ/g)

(xxiv) Compound pendulum: T = 2π ( )2

/K2 ll +

(a) For a bar: T = 2π√(2L/3g) ; (b) For a disc : T = 2π√ (3R/2g) (xxv) Floating cylinder: (a) K = Aρg ; (b) T = 2π√(m/Aρg) = 2π√(Ld/ρg) (xxvi) Liquid in U�tube: (a) K = 2A ρg and m = ALρ ; (b) T = 2π√(L/2g) = 2π√(h/g) (xxvii) Ball in bowl: T = 2π√[(R � r)/g] (xxviii) Piston in a gas cylinder:

(a) ; V

EAK2

=

(b) ; EA

mV 2 T 2π=

(c) PA

V 2T 2

mπ= (E�P for Isothermal process);

(d) P

V2T m

γΑπ= 2 (E = γ P for adiabatic process)

(xxix) Elastic wire:

(a) K = ;AYl

(b) T = AY

m 2 lπ

(xxx) Tunnel across earth: T = 2π√(Re/g) (xxxi) Magnetic dipole in magnetic field: T = 2π√(I/MB)

(xxxii) Electrical LC circuit: T = 2π LC or LC 2

1π

=ƒ

(xxxiii) Lissajous figures � Case (a): ω1 = ω2 = ω or ω1 : ω2 = 1 : 1

General equation: φ = φ−+ sin cos abxy2

by

ax 2

2

2

2

2

For φ = 0 : y = (b/a) x ; straight line with positive slope

For φ = π/4 : ellipse oblique ; 21

abxy2

by

ax

2

2

2

2=−+

For φ = π/2 : ellipse lsymmetrica ; 1by

ax

2

2

2

2=+

For φ = π : y = �(b/a) x ; straight line with negative slope. Case (b): For ω1 : ω2 = 2:1 with x = a sin (2ωt + φ) and y = b sin ωt

For φ = 0, π: Figure of eight

For φ = :3 ,4π

4π Double parabola

For φ = :3 ,2π

2π Single parabola

GRAVITATION (i) Newton�s law of gravitation:

(a) F = G m1m2/r2 ; (b) a = 6.67 x 10�11 K.m2/(kg)2 ; (c) rdr 2

FdF −=

(ii) Acceleration due to gravity (a) g = GM/R2 ; (b) Weight W = mg (iii) Variation of g: (a) due to shape ; gequator < gpole (b) due to rotation of earth: (i) gpole = GM/R2 (No effect)

(ii) gequator = RRGM 2

2 ω−

(iii) gequator < gpole (iv) ω2R = 0.034 m/s2 (v) If ω ≅ 17 ω0 or T = (T0/17) = (24/17)h = 1.4 h, then object would

float on equator

(c) At a height h above earth�s surface g� = g R h if ,gh21 <<

−

(d) At a depth of below earth�s surface: g� = g

−

Rd1

(iv) Acceleration on moon: gm = earth2m

m g61

RGM

≅

(v) Gravitational field: (a) ( ) ( ) ( )inside rrRGMg b ; outside r

rGMg

^

3

^

2 −=−=→→

(vi) Gravitational potential energy of mass m: (a) At a distance r : U(r) = � GMm/r (b) At the surface of the earth: U0 = � GMm/R (c) At any height h above earth�s surface: U � U0 = mgh (for h < < R)

or U = mgh (if origin of potential energy is shifted to the surface of earth)

(vii) Potential energy and gravitational force: F = � (dU/dR) (viii) Gravitational potential: V(r) = �GM/r (ix) Gravitational potential energy of system of masses: (a) Two particles: U = � Gm1m2/r

(b) Three particles: U = � 23

32

13

31

12

21r

mGmr

mGmr

mGm−−

(x) Escape velocity:

(a) ve = RGM2 or ve = √(2gR) = √(gD)

(b) ve = 3G8R ρπ

(xi) Maximum height attained by a projectile:

( ) R

h vhR

h vor v 1v/v

Rh ee2e

≅+

=−

= (if h < < R)

(xii) Orbital velocity of satellite:

( ) ( ) ( ) ;hR 2

R v vb ;r

GMva e00 +== (c) v0 ≅ ve/√2 (if h<<R)

(xiii) Time period of satellite: (a) ( ) ( ) ( )Rh if gR 2 T b ;

GMhR 2T

3<<π=+π=

(xiv) Energy of satellite: (a) Kinetic energy K = r

GMm 21 mv

21 2

0 =

(b) Potential energy U =� r

GMm =� 2K ;

(c) Total energy E=K + U=�21 ;

rGMm

(d) E = U/2 = � K ; (e) BE = �E = 21

rGMm

(xv) Geosynchronous satellite: (a) T = 24 hours ; (b) T2 = ( ) ;hR GM4 3+π2

(c) 4

GMTh3/1

2

2

π= �R ; (d) h ≅ 36,000 km.

(xvi) Kepler�s law: (a) Law of orbits: Orbits are elliptical (b) Law of areas: Equal area is swept in equal time (c) Law of period: T2 ∝ r3 ; T2 = (4π2/GM)r3

SURFACE TENSION

(i) (a) l

FLengthForceT == ; (b)

AW

area Surfaceenergy SurfaceT ==

(ii) Combination of n drops into one big drop: (a) R = n1/3r

(b) Ei = n(4πr2T), Eƒ = 4πR2T, (Eƒ/Ei) = n�1/3,

−=∆1/3

i n11

EE

(c) ∆E = 4πR2T (n1/3 �1) = 4πR3T

−

R1

r1

(iii) Increase in temperature: ∆θ = sT3

ρ

−

R1

r1 or

−

ρ R1

r1

sJT3

(iv) Shape of liquid surface:

(a) Plane surface (as for water � silver) if Fadhesive > 2

Fcohesive

(b) Concave surface (as for water � glass) if Fadhesive > 2

Fcohesive

(c) Convex surface (as for mercury�glass) if Fadhesive < 2

Fcohesive

(v) Angle of contact: (a) Acute: If Fa> Fc/√2 ; (b) obtuse: if Fa<Fc/√2 ; (c) θc=900 : if Fa=Fc√/2

(d) cos θc = a

ssaT

TT

l

l− , (where Tsa, Tsl and Tla represent solid-air, solid- liquid and liquid-air

surface tensions respectively). Here θc is acute if Tsl < Tsa while θc is obtuse if Tsl > Tsa (vi) Excess pressure:

(a) General formula: Pexcess =

+

21 R1

R1T

(b) For a liquid drop: Pexcess = 2T/R (c) For an air bubble in liquid: Pexcess = 2T/R (d) For a soap bubble: Pexcess = 4T/R (e) Pressure inside an air bubble at a depth h in a liquid: Pin = Patm + hdg + (2T/R)

(vii) Forces between two plates with thin water film separating them:

(a) ∆P = T ; R1

r1

−

(b) ; R1

r1 AT F

−=

(c) If separation between plates is d, then ∆P = 2T/d and F = 2AT/d

(viii) Double bubble: Radius of Curvature of common film Rcommon = rR

rR−

(ix) Capillary rise:

(a) ;rdg

cos T2h θ=

(b) rdg2Th = (For water θ = 00)

(c) If weight of water in meniseus is taken into account then T = θ

+

cos 23rh rdg

(d) Capillary depression, ( )rdg

cos T2h θ−π−

(x) Combination of two soap bubbles:

(a) If ∆V is the increase in volume and ∆S is the increase in surface area, then 3P0∆V + 4T∆S = 0 where P0 is the atmospheric pressure

(b) If the bubbles combine in environment of zero outside pressure isothermally, then ∆S = 0 or R3 = √ ( )2

221 RR +

ELASTICITY

(i) Stress: (a) Stress = [Deforming force/cross�sectional area]; (b) Tensile or longitudinal stress = (F/π r2); (c) Tangential or shearing stress = (F/A); (d) Hydrostatic stress = P (ii) Strain: (a) Tensile or longitudinal strain = (∆L/L); (b) Shearing strain = φ;

(c) Volume strain = (∆V/V)

(iii) Hook�s law:

(a) For stretching: Stress = Y x Strain or ( )LAFLY∆

=

(b) For shear: Stress = η x Strain or η = F/Aφ

(c) For volume elasticity: Stress = B x Strain or B = � ( )V/VP

∆

(iv) Compressibility: K = (1/B)

(v) Elongation of a wire due to its own weight: ∆L = Y

gL 21

YAMgL

21 2ρ=

(vi) Bulk modulus of an idea gas: Bisothermal = P and Badiabatic = γP (where γ = Cp/Cv) (vii) Stress due to heating or cooling of a clamped rod Thermal stress = Yα (∆t) and force = YA α (∆t) (viii) Torsion of a cylinder:

(a) r θ = lφ (where θ = angle of twist and φ = angle of shear); (b) restoring torque τ = cθ (c) restoring Couple per unit twist, c = πηr4/2l (for solid cylinder) and C = πη (r2

4 � r14)/2l (for hollow cylinder)

(ix) Work done in stretching:

(a) W = 21 x stress x strain x volume = Y

21 (strain)2 x volume = ( ) volumex

Ystress

21 2

(b) Potential energy stored, U = W = 21 x stress x strain x volume

(c) Potential energy stored per unit volume, u = 21 x stress x strain

(a) depression, δ = ( )rrectangula Ybd4W

3

3l

(b) Depression, δ = ( )lcylindrica rY12

W2

3

πl

(xi) Position�s ratio:

(a) Lateral strain = r

rDD ∆−=∆−

(b) Longitudinal strain = (∆L/L)

(c) Poisson�s ratio σ = L/Lrr

strain allongitudinstrain lateral

∆/∆−=

(d) Theoretically, �1 < σ < 0.5 but experimentally σ ≅ 0.2 � 0.4 (xii) Relations between Y, η, B and σ: (a) Y = 3B (1�2σ) ;

(b) Y = 2η (1+ σ);

(c) η

+=31

B91

Y1

(xiii) Interatomic force constant: k = Yr0 (r0 = equilibrium inter atomic separation)

KINETIC THEORY OF GASES

(i) Boyle�s law: PV = constant or P1V1 = P2V2 (i) Chare�s law: (V/T) = constant or (V1/T1) = (V2/T2) (ii) Pressure � temperature law: (P1/T1) = (P2/T2) (iii) Avogadro�s principle: At constant temperature and pressure, Volume of gas, V ∝ number of moles, µ Where µ = N/Na [N = number of molecules in the sample and NA = Avogadro�s number = 6.02 x 1023/mole]

M

Msample= [Msample = mass of gas sample and M = molecular weight]

(iv) Kinetic Theory:

(a) Momentum delivered to the wall perpendicular to the x�axis, ∆P = 2m vx (b) Time taken between two successive collisions on the same wall by the same molecule: ∆t =

(2L/vx) (c) The frequency of collision: νcoll. = (νx/2L) (d) Total force exerted on the wall by collision of various molecules: F = (MN/L) <vx

2>

(e) The pressure on the wall : P = 2rms

2rms

22x v

31v

VmN

31 v

V3mNv

VmN ρ==><=><

(v) RMS speed: (a) νrms = √(v1

2 + v22 + � + v 2

N /N); (b) νrms = √(3P/ρ) ; (c) νrms = √(3KT/m);

(d) νrms = √(3RT/M) ; (e) ( )( ) 1

2

1

2

2rms

1rms

MM

mm

==νν

(x) Loaded beam:

(a) (1/2) Mv2 rms = (3/2) RT ;

(b) (1/2) mv2 rms = (3/2) KT (c) Kinetic energy of one molecule = (3/2) KT ; (d) kinetic energy of one mole of gas = (3/2) RT (e) Kinetic energy of one gram of gas (3/2) (RT/M)

(ix) Maxwell molecular speed distribution:

(a) n (v) = 4πN KT2/mv-23/2

2e v

KT 2m

π

(b) The average speed: MRT60.1

MRT 8

mKT8v =

π=

π=

(c) The rms speed: vrms = MRT 73.1

MRT3

mkT3 ==

(d) The most probable speed: νp = MRT 41.1

MRT2

mKT2 ==

(e) Speed relations: (I) vp < v < vrms (II) vp : v : vrms = √(2) : √(8/π) : √(3) = 1.41 : 1.60 : 1.73 (x) Internal energy: (a) Einternal = (3/2)RT (for one mole) (b) Einternal = (3/2 µRT (for µ mole)

(c) Pressure exerted by a gas P = E32

VE

32 =

(xi) Degrees of freedom: (a) Ideal gas: 3 (all translational) (b) Monoatomic gas : 3 (all translational) (c) Diatomic gas: 5 (three translational plus two rotational) (d) Polyatomic gas (linear molecule e.g. CO2) : 7 (three translational plus two rotational plus two

vibrational) (e) Polyatomic gas (non�linear molecule, e.g., NH3, H2O etc): 6 (three translational plus three

rotational) (f) Internal energy of a gas: Einternal = (f/2) µRT. (where f = number of degrees of freedom)

(xii) Dalton�s law: The pressure exerted by a mixture of perfect gases is the sum of the pressures

exerted by the individual gases occupying the same volume alone i.e., P = P1 + P2 + �. (xiii) Van der Wall�s gas equation:

(a) ( ) Τµ=µ

µ+ R b-V V

aP 2

2

(b) ( ) RT bV V

aP m2m

2=−

µ+ (where Vm = V/µ = volume per mole);

(c) b = 30 cm3/mole

(d) Critical values: Pc = ;Rb 27a8T ,b3V ,

b27a

CC2 ==

(e) 375.083

RTVP

C

CC ==

(xiv) Mean free path: λ = nd2

12ρπ

,

Where ρn = (N/V) = number of gas molecules per unit volume and d = diameter of molecules of the gas

(vi) Kinetic interpretation of temperature:

FLUID MECHANICS

(i) The viscous force between two layers of area A having velocity gradient (dv/dx) is given by: F = � ηA (dv/dx), where η is called coefficient of viscosity

(i) In SI system, η is measured I Poiseiulle (Pl) 1Pl = 1Nsm�2 = 1 decapoise. In egs system, the unit of η is g/cm/sec and is called POISE

(ii) When a spherical body is allowed to fall through viscous medium, its velocity increases, till the sum of viscous drag and upthrust becomes equal to the weight of the body. After that the body moves with a constant velocity called terminal velocity.

(iii) According to STOKE�s Law, the viscous drag on a spherical body moving in a fluid is given by: F = 6πηr v, where r is the radius and v is the velocity of the body.

(iv) The terminal velocity is given by: vT = ( )η

σ−ρ g r 92 2

where ρ is the density of the material of the body and σ is the density of liquid (v) Rate of flow of liquid through a capillary tube of radius r and length l

Rp

r/8p

8pr V 4

4=

πη=

ηπ=

ll

where p is the pressure difference between two ends of the capillary and R is the fluid resistance (=8 ηl/πr4)

(vi) The matter which possess the property of flowing is called as FLUID (For example, gases and liquids)

(vii) Pressure exerted by a column of liquid of height h is : P = hρg (ρ = density of the liquid) (viii) Pressure at a point within the liquid, P = P0 + hρg, where P0 is atmospheric pressure and h is the

depth of point w.r.t. free surface of liquid (ix) Apparent weight of the body immersed in a liquid Mg� = Mg � Vρg (x) If W be the weight of a body and U be the upthrust force of the liquid on the body then

(a) the body sinks in the liquid of W > U (b) the body floats just completely immersed if W = U (c) the body floats with a part immersed in the liquid if W < U

(xi) solid of densitysolid of density

solid of volumetotalsolid a ofpart immersed of Volume =

(xii) Equation of Continuity: a1v1 = a2v2

(xiii) Bernouilli�s theorem: (P/ρ) + gh + 21 v2 = constant

(xiv) Accelerated fluid containers : tan θ = g

a x

(xv) Volume of liquid flowing per second through a tube: R=a1v1 = a2v2 ( )22

21 aa

gh2−

(xvi) Velocity of efflux of liquid from a hole: v = √(2gh), where h is the depth of a hole from the free surface of liquid

ax

ρ

θ

Fig. 4

HEAT AND THERMODYNAMICS (i) L2 � L1 = L1α(T2 � T1); A2 � A1 = A2 β(T2 � T1); V2 � V1 = V1γ(T2 � T1) where, L1, A1, V1 are the length, area and volume at temperature T1; and L2, A2, V2 are that at

temperature T2.α represents the coefficient of linear expansion, β the coefficient of superficial expansion and γ the coefficient of cubical expansion.

(ii) If dt be the density at t0C and d0 be that at 00C, then: dt = d0 (1�γ∆T) (iii) α : β: γ = 1 : 2 : 3 (iv) If γr, γa be the coefficients of real and apparent expansions of a liquid and γg be the coefficient of the

cubical expansion for the containing vessel (say glass), then γr = γa + γg (v) The pressure of the gases varies with temperature as : Pt = P0 (1+ γ∆T), where γ = (1/273) per 0C (vi) If temperature on Celsius scale is C, that on Fahrenheit scale is F, on Kelvin scale is K, and on

Reaumer scale is R, then

(a) 4R

5273K

932F

5C =−=−= (b) 32 C

59F +=

(c) ( )32F 95C −=

(d) K = C + 273 (e) ( )459.4 F 95K +=

(vii) (a) Triple point of water = 273.16 K

(b) Absolute zero = 0 K = �273.150C

(c) For a gas thermometer, T = (273.15) ( )Kelvin P

P

triple

(d) For a resistance thermometer, Rθ = R0 [1+ αθ] (viii) If mechanical work W produces the same temperature change as heat H, then we can write: W = JH, where J is called mechanical equivalent of heat (ix) The heat absorbed or given out by a body of mass m, when the temperature changes by ∆T is: ∆Q

= mc∆T, where c is a constant for a substance, called as SPECIFIC HEAT. (x) HEAT CAPACITY of a body of mass m is defined as : ∆Q = mc (xi) WATER EQUIVALENT of a body is numerically equal to the product of its mass and specific heat

i.e., W = mc (xii) When the state of matter changes, the heat absorbed or evolved is given by: Q = mL, where L is

called LATENT HEAT (xiii) In case of gases, there are two types of specific heats i.e., cp and cv [cp = specific heat at constant

pressure and Cv = specific heat at constant volume]. Molar specific heats of a gas are: Cp = Mcp and Cv = Mcv, where M = molecular weight of the gas.

(xiv) Cp > Cv and according to Mayer�s formula Cp � Cv = R (xv) For all thermodynamic processes, equation of state for an ideal gas: PV = µRT

(a) For ISOBARIC process: P = Constant ; TV =Constant

(b) For ISOCHORIC (Isometric) process: V = Constant; TP =Constant

(c) For ISOTHERMAL process T = Constant ; PV= Constant (d) For ADIABATIC process: PVγ = Constant ; TVγ�1=Constant and P(1�γ) Tγ = Constant

(a) For isobaric process: zero (b) For isochoric process: infinite (c) For isothermal process: slope = �(P/V) (d) For adiabatic process: slope = �γ(P/V) (e) Slope of adiabatic curve > slope of isothermal curve.

(xvii) Work done (a) For isobaric process: W = P (V2 � V1) (b) For isochoric process: W = 0 (c) For isothermal process: W=µRT loge (V2/V1)

µRT x 2.303 x log10 (V2/V1) P1V1 x 2.303 x log10 (V2/V1) µRT x 2.303 x log10 (P1/P2)

(d) For adiabatic process: ( )( )

( )( )1−γ

−=1−γ−µ= 221121 VPVP TT RW

(e) In expansion from same initial state to same final volume

Wadiabatic < Wisothermal < Wisobaric (f) In compression from same initial state to same final volume:

Wadiabatic < Wisothermal < Wisobaric (xviii) Heat added or removed:

(a) For isobaric process: Q = µCp∆T (b) For isochoric process = Q = µCv∆T (c) For isothermal process = Q = W = µRt loge (V2/V1) (d) For adiabatic process: Q = 0

(xix) Change in internal energy

(a) For isobaric process = ∆U = µCv∆T (b) For isochoric process = ∆U = µCv∆T (c) For isothermal process = ∆U = 0

(d) For adiabatic process: ∆U = �W = ( )( )1

TTR 12

−γ−µ

(xx) Elasticities of gases (a) Isothermal bulk modulus = BI = P (b) Adiabatic bulk modulus BA = γP

(xxi) For a CYCLIC process, work done ∆W = area enclosed in the cycle on PV diagram.

Further, ∆U = 0 (as state of the system remains unchanged) So, ∆Q = ∆W

(xxii) Internal energy and specific heats of an ideal gas (Monoatomic gas)

(a) U = 23 RT (for one mole);

(b) 23U = µRT (for µ moles)

(c) ∆U = 23 µR∆T (for µ moles);

(d) Cv= R23

ΤU 1 =

∆∆

µ

(xvi) Slope on PV diagram

(e) Cp = Cv + R = 23 R + R =

25 R

(f) γ = 67.135 R

23R

25

C

C

v

p ==

=

(xxiii) Internal energy and specific heats of a diatomic gas

(a) 25U = µRT (for µ moles);

(b) ∆U = 25 µR∆T (for µ moles)

(c) Cv = ;R25

TU1 =

∆∆

µ

(d) Cp = Cv + R = 25 R + R =

27 R

(e) 4.157

2R5

2R7

CC

v

p ==

=

=γ

(xxiv) Mixture of gases: µ = µ1 + µ2

21

221121NN

mNmNMMM

++

=µ+µ

+µ+µ=

21

21

21

21

21

21

µ+µµ+µ

=µ+µµ+µ

= 2121 ppp

vvv

CCC and

CCC

(xxv) First law of thermodynamics

(a) ∆Q = ∆U + ∆W or ∆U = ∆Q � ∆W (b) Both ∆Q, ∆W depends on path, but ∆U does not depend on the path (c) For isothermal process: ∆Q = ∆W = µRT log | V2/V1|, ∆U = 0, T = Constant, PV = Constant

and Ciso = ± ∞

(d) For adiabatic process: ∆W = ( )( ) ,1

TTR 12γ−−µ ∆Q = 0, ∆U = µCv (T2�T1), Q = 0,

PVγ = constant, Cad = 0 and ƒ

+==γ 21CC

v

p

(where ƒ is the degree of freedom)

(e) For isochoric process: ∆W = 0, ∆Q = ∆U = µCv∆T, V = constant, and Cv = (R/γ�1) (f) For isobaric process: ∆Q = µCp∆T, ∆U = µCv∆T., ∆W = µR∆T, P = constant and Cp = (γR/γ�1)

(g) For cyclic process: ∆U = 0, ∆Q = ∆W (h) For free expansion: ∆U = 0, ∆Q = 0, ∆W = 0 (i) For polytropic process: ∆W = [µR(T2�T1)/1�n], ∆Q = µ C (T2�T1), PVn = constant and

n1RRC−

+1−γ

=

(xxvi) Second law of thermodynamics

(a) There are no perfect engines (b) There are no perfect refrigerators

(c) Efficiency of carnot engine: η = 1�1

2

1

2

1

2TT

QQ

,QQ

=

(d) Coefficient of performance of a refrigerator:

β=21

2

21

22TT

TQQ

QWQ

orrefrigeraton done Workreservoir cold from absorbed Heat

−=

−==

For a perfect refrigerator, W = 0 or Q1 = Q2 or β = ∞

(xxvii) The amount of heat transmitted is given by: Q = �KA tx∆θ∆ , where K is coefficient of thermal

conductivity, A is the area of cross section, ∆θ is the difference in temperature, t is the time of heat flow and ∆x is separation between two ends

(xxviii) Thermal resistance of a conductor of length d = RTh = A K

d

(xxix) Flow of heat through a composite conductor:

(a) Temperature of interface, θ = ( ) ( )( ) ( )2211

2211d/K d/K

d/ K d/K+

θ+θ 21

(b) Rate of flow of heat through the composite conductor: H = ( )( ) ( )2211 K/d K/d

AtQ

+θ−θ= 21

(c) Thermal resistance of the composite conductor

( ) ( )2Th1Th2

2

1

1TH RR

AKd

AKd

R +=+=

(d) Equivalent thermal conductivity, K = ( ) ( )2211

21K/d K/d

dd++

(xxx) (a) Radiation absorption coefficient: a = Q0/Q0 (b) Reflection coefficient: r = Qr/Q0 (c) Transmission coefficient: t = Qt/Q0 (d) Emissive power: e or E = Q/A .t [t = time]

(e) Spectral emissive power: eλ = ( )λd AtQ and e = eλeλ

(f) Emissivity: ε = e/E ; 0 ≤ ε ≤ 1 (g) Absorptive power: a = Qa/Q0 (h) Kirchhoffs law: (eλ/aλ)1 = (eλ/aλ)B = ���= Eλ (i) Stefan�s law: (a) E=σT4 (where σ=5.67x10�8 Wm�2 K�4) For a black body: E = σ (T4�T0

4) For a body: e = εσ (T4�T0

4)

(j) Rate of loss of heat: � )( AdtdQ 4

04 θ−θσε=

For spherical objects: ( )( ) 2

2

21

2

1

rr

dt/dQdt/dQ

=

(k) Rate of fall of temperature: ( ) ( )40

440

4 θ−θρσε=θ−θσε=θ s V

A msA

dtd

∴ ( )( ) 1

2

1

2

2

1

2

1rr

VV x

AA

dtddtd

==/θ/θ

∫∞0

(For spherical bodies)

(l) Newton�s law of cooling: dtdθ = �K (θ�θ0) or (θ�θ0) α e�KT

(m) Wein�s displacement law: λmT = b (where b = 2.9 x 10�3 m � K)

(n) Wein�s radiation law: Eλdλ= A

λ5 ƒ (λT) dλ=

λ5A e�a/λTdλ

(o) Solar Constant: S =2

ES

SRR

σT4 or T =2/1

S

ES4/1

RR

S

σ

WAVES 1. Velocity: v = nλ and n = (1/T)

2. Velocity of transverse waves in a string: v = dr

TmT

2π=

3. Velocity of longitudinal waves: (a) In rods: v = √(Y/ρ) (Y � Young�s modulus, ρ = density) (b) In liquids: v = √(B/ρ) (B = Bulk modulus) (c) In gases: v = √(γP/ρ) (Laplace formula)

4. Effect of temperature: (a) v = v0√ (T/273) or v = v0 + 0.61t

(b) (vsound/vrms) = √(γ/3)

5. Wave equation: (a) y = a sin λπ2 (vt�x)

(b) y = a sin 2π

λ− x

Tt

(c) y = a sin (ωt � kx), where wave velocity v = λ=ω nk

6. Particle velocity: (a) vparticle = (∂y/∂t)

(b) maximum particle velocity, (vparticle)max = ω a

7. Strain in medium (a) strain = � (∂y/∂x) = ka cos (ωt � kx) (b) Maximum strain = (∂y/dx)max = ka (c) (vparticle/strain) = (ω/k) = wave velocity

i.e., vparticle = wave velocity x strain in the medium

8. Wave equation:

∂∂=

∂∂

2

22

2

2

xy v

ty

9. Intensity of sound waves: (a) I = (E/At) (b) If ρ is the density of the medium; v the velocity of the wave; n the frequency and a the

amplitude then I = 2π2 ρ v n2 a2 i.e. I ∝ n2a2 (c) Intensity level is decibel: β 10 log (I/I0). Where, I0 =Threshold of hearing = 10�12 Watt/m2

10. Principle of superposition: y = y1 + y2 11. Resultant amplitude: a = √(a1

2 + a22 + 2a1a2 cos φ)

12. Resultant intensity: I = I1 + I2 + 2√(I1I2 cos φ)

(a) For constructive interference: φ = 2nπ, amax = a1 + a2 and Imax = (√I1 + √I2)2 (b) For destructive interference: φ = (2n�1) π, amin = a2 �a2 and Imin = (√I1=√I 2)2

13. (a) Beat frequency = n1 � n2 and beat period T = (T1T2/T2�T1)

(b) If there are N forks in successive order each giving x beat/sec with nearest neighbour, then nlast = nfirst + (N�1)x

14. Stationary waves: The equation of stationary wave,

(a) When the wave is reflected from a free boundary, is:

y = + 2a cos tsin kx cos a2T

t2sin x2 ω=πλπ

(b) When the wave is reflected from a rigid boundary, is:

Y + �2a sin T

t2 cos x2 πλπ =�2a sin kx cos ωt

15. Vibrations of a stretched string:

(a) For fundamental tone: n1 = mT1

λ

(b) For p th harmonic : np = mTp

λ

(c) The ratio of successive harmonic frequencies: n1 : n2 : n3 :��.. = 1 : 2 : 3 : ��

(d) Sonometer: mT

2lnl

= (m = π r2 d)

(e) Melde�s experiment: (i) Transverse mode: n = mT

2pl

(ii) Longitudinal mode: n = mT

2p2l

16. Vibrations of closed organ pipe

(a) For fundamental tone: n1 =

L4v

(b) For first overtone (third harmonic): n2 = 3n1 (c) Only odd harmonics are found in the vibrations of a closed organ pipe and n1 : n2 : n3 : �..=1 : 3 : 5 : ��

17. Vibrations of open organ pipe:

(a) For fundamental tone: n1 = (v/2L) (b) For first overtone (second harmonic) : n2 = 2n1 (c) Both even and odd harmonics are found in the vibrations of an open organ pipe and

n1 : n2 : n3 : ��=1 : 2 : 3 : ��. 18. End correction: (a) Closed pipe : L = Lpipe + 0.3d

(b) Open pipe: L = Lpipe + 0.6 d where d = diameter = 2r

19. Resonance column: (a) l1 +e = 4λ ; (b) l2 + e =

43λ

(c) e = 2

3 12 ll − ; (d) n = ( ) ( )1212

2 or 2

vll

ll−=λ

−

20. Kundt�s tube: rod

air

rod

airvv

λλ

=

21. Longitudinal vibration of rods

(a) Both ends open and clamped in middle: (i) Fundamental frequency, n1 = (v/2l) (ii) Frequency of first overtone, n2 = 3n1 (iii)Ratio of frequencies, n1 : n2 : n3 : �� = 1 3: 5 : �..

(b) One end clamped (i) Fundamental frequency, n1 = (v/4l) (ii) Frequency of first overtone, n2 = 3n1 (iii) Ratio of frequencies, n1 : n2 : n3 : ��= 1 : 3 : 5 : ���

22. Frequency of a turning fork: n α ρE t

2l

Where t = thickness, l = length of prong, E = Elastic constant and ρ = density 23. Doppler Effect for Sound

(a) Observer stationary and source moving:

(i) Source approaching: n� = vv

v

s−x n and λ� =

vvv s− x λ

(ii) Source receding: n� = svv

v+

x n and λ� = vvv s+ x λ

(b) Source stationary and observer moving:

(i) Observer approaching the source: n� = vvv 0+ x n and λ� = λ

(ii) Observer receding away from source: n� = vvv 0− x n and λ� = λ

(c) Source and observer both moving:

(i) S and O moving towards each other: n� = s

0vvvv

−+ x n

(ii) S and O moving away from each other: n� = s

0vvvv

+− x n

(iii) S and O in same direction, S behind O : n� = s

0vvvv

−− x n

(iv) S and O in same direction, S ahead of O: n�=s

0vvvv

++ x n

(d) Effect of motion of medium: sm

0mv vv

v vv 'n

±±±±

=

(e) Change in frequency: (i) Moving source passes a stationary observer: ∆n = 2s

2s

vvvv2−

x n

For vs <<v, ∆, = vv2 s x n

(ii) Moving observer passes a stationary source: ∆ n= vv2 0 x n

(f) Source moving towards or away from hill or wall

(i) Source moving towards wall (a) Observer between source and wall

n� = n x vv

v

s− (for direct waves)

n� = svv

v−

x n (for reflected waves)

(b) Source between observer and wall

n� = n x vv

v

s+ (for direct waves)

n� = svv

v−

x n (for reflected waves)

(ii) Source moving away from wall

(a) Observer between source and wall

n� = n x vv

v

s+ (for direct waves)

n� = svv

v+

x n (for reflected waves)

(b) Source between observer and wall

n� = n x vv

v

s− (for direct waves)

n� = svv

v+

x n (for reflected waves)

(g) Moving Target: (i) S and O stationary at the same place and target approaching with speed u

n� = n x uvuv

−+ or n� = n x

vu21

+ (for u <<v)

(ii) S and O stationary at the same place and target receding with speed u

n� = n x uvuv

+− or n� = n x

vu21

− (for u <<v)

(h) SONAR: n� = n x v

v2 1 n x

vv v v sub

sub

sub

±≅±±

(upper sign for approaching submarine while lower sign for receding submarine)

(i) Transverse Doppler effect: There is no transverse Doppler effect in sound. For velocity component vs cos θ

n�= n x cos v v

v

s θ± (� sign for approaching and + sign for receding)

24. Doppler Effect for light

(a) Red shift (when light source is moving away):

n� = n x c/v1c/v1

+− or λ� = λ

−+ x

c/v1c/v1

For v << c, ∆ n = � n x cv

or ∆λ� =

cv x λ

(b) Blue shift (when light source is approaching)

n� = n x c/v1c/v1

−+ or λ� = λ

+− x

c/v1c/v1

For v << c, ∆ n = n cv

or ∆λ� =�

cv λ

(c) Doppler Broadening = 2∆λ = 2

cv λ

(d) Transverse Doppler effect:

For light, n� = n x cv

211 n x

cv1 2

2

2

2

−=− (for v << c)

(e) RADAR: ∆n =

cv2 n

ELECTROSTATICS & CAPACITANCE

ELECTROSTATICS 1. Coulomb�s Law

(a) Fm = K ,rqq

K41

221

0πε= Dielectric constant or relative permittivity of the medium

(b) Fm = KF0 [F0 � Force between point charges placed in vacuum]

(c) [ε0] = [M�1L�3T4A2]

(d) g

e

FF

=2.4 x 1039 [For electron�proton pair)

=1.2 x 1036 (For proton�proton pair) 2. Electric field

(a) 0q

FE→

→=

(b) Electric field due to a point charge: (i) ^

2 r.rq

41E

0

→

πε= (if charge q is placed at the origin)

(ii) 3

0

0

|rr|

)rr(q 4

1E →→

→→

0

→

−

−πε

= (if charge q is placed at some point having position vector 0r→

)

(c) [E] = [M1L1T�3A�1] 3. Electric field on the axis of a uniformly charged ring

(a) Eaxial = ( ) 2/322 xR

qx 4

1

+πε0 (R = radius of the ring)

(b) Ecentre = 0 4. Electric dipole (a) dipole moment p = q(2l) (where 2l = length of the dipole)

(b) Eaxial = [ ]222r

2pr 4

1

l−πε0 (r = distance of axial point w.r.t. centre of dipole)

3r2p

41

0πε≅ (if r >>l)

(c) Eequat. = [ ] 32/322 rp

41

r

p 4

1

00 πε≈

+πε l (if r >>l)

(d) (Eaxial/Eequat.) = 2/1 (e) Dipole field at an arbitrary point (r, θ)

(i) Er = ; r

cos 2p 4

13

0

θπε

(ii) Eθ = 3rsin p

41 θπε0

(iii) E = 22r EE θ+ = θ+

πε0 cos 31

rp

41 2

3

(f) Dipole field component at (x, y, z) point

(i) Ex = ;r

p 3xz 4

1s

0πε (ii) Ey = ;

rp 3yz

41

50πε

(iii) Ez = ( )5

22

rr3zp

41 −πε0

(g) Torque on a dipole : (i) ; E x p →→→

=τ (ii) τ = pE sin θ

(h) Potential energy of a dipole: (i) U = θ−=−→→

cos pE E . p (ii) Work done in rotating a dipole from angle θ1 to angle θ2 W = U2 � U1 = pE (cos θ1 � cos θ2) 5. Electric flux

(a) dφ = →→dS . E

(b) φ→→

∫ dS . E = EA cos θ (If electric field is constant over the whole surface) (c) Unit of φ = (Nm2/Coulomb) = J.m/Couplomb (d) [φ]= [M1L3T�3A�1]

6. Gauss�s Law: ∫0

→→

ε= qdS . E

7. Electric field due to various systems of charges

(a) Isolated Charge: 2rq

41E

0επ=

(b) Electric dipole: (i) 3||x

p2 4

1 E→

0

→

επ=

(ii) 3yp

41E

→

0⊥

→

επ−=

(iii)

(c) A ring of charge: E = ( ) 2/322 xR

qx 4

1

+επ 0

Fig. 1

r

q P

• •

•

• • P

+q �q

E⊥ y

x

E||

Fig. 2

2a

•

• • θ �q

→p+q

r

Fig. 3

+ + + + + + + +

+ + +

+ +

+ + + + + + + +

• •

R q

x P

Fig. 4

θ+επ

=0

cos 31 rp

41E 2

3

(d) A disc of charge: E =

+−

εσ

0 22 Rx

x1 2

(e) Infinite sheet of charge: E = 0ε2

σ

(f) Infinitely long line of charge: E = x0επ2

λ

(g) Finite line of charge: E ⊥ = x0επ4

λ (sin α + sin β)

E ||=0επ4

λ (cos α+cos β)

(h) Charged spherical shell: (i) Inside: 0 ≤ r ≤ R E = 0 (ii) Outside: r ≥ R

2r4qE0επ

=

(i) Solid sphere of charge: (i) Inside: 0 ≤ r ≤ r E = ρr/3ε0 (ii) Outside : r ≥ R

2

rR

3RE

ερ=

0

+ + + +

•

+ + + + + + + +

+

+ +

+ + + + + +

•

R q

x P

Fig. 5

+ + +

+ + +

σ

Fig. 6

+ + + +

• P

λ

x

Fig. 7

Pαβ

++++++

x Fig. 8

+ ++

++

+++

+ +

+ + + +

+

R q

Fig. 9

R++++++

+ ++

+ ++ +

Fig. 10

8. Force on a charged conductor: The force per unit area or electric pressure

0

2

εσ==2dA

dFP .elec

9. Charged soap bubble: (a) Pin � Pout = 0ε

σ−2r

T4 2

(b) If air pressure inside and outside are assumed equal then: Pin = Pout and 0

2

εσ=2r

T4

or 20

σε= T8r or

0

2

εσ=8

rT or σ = √(8ε0T/r) or Q = 8πr√(2ε0rT)

or r = [Q2/128π2ε0T]1/3

10. Electric potential: (a) V = (W/q) (b) Unit of V = Volt (c) [V] = [ML2T�3A�1]

(d) VVE→→

−=

(e) Potential due to a point charge, V = rq

41

0επ

(f) Potential due to a group of charges, V =

++

επ 0 3

3

2

2

1

1rq

rq

rq

41

(g) Potential due to a dipole:

(i) Axial point, V = 2rp

41

0επ; (ii) equatorial point, V = 0;

(iii) V (r, θ) = 20 ρ

θεπ

cos p 4

1

(h) Potential due to a charged spherical shell

(i) outside: V = rq

41

0επ (ii) surface: V =

Rq

41

0επ;

(iii) inside : V = Vsurface = Rq

41

0επ

(i) Potential due to a charged spherical conductor is the same as that due to a charged spherical shell. (j) Potential due to a uniformly charged nonconducting sphere

(i) outside: V= ;rq

41

0επ (ii) surface: V =

Rq

41

0πε

(iii) inside: V= ( )3

22

R2rR 3q

41 −πε0

; (iv) centre: V = 5.1Rq

41 x

23 =

επ 0 Vsurface

(k) Common potential (Two spheres joined by thin wire)

(i) common potential V=

++

επ 0 21

21rrQQ

41

(ii) ( )( ) 21

22

21

1

21

2111 rr

Qrq ;rr

QrrrQQ rq

+=

+=

++=

(iii) 1

2

2

1

2

1

rror

rr

qq =

σσ=

2

1

11. Potential energy

(a) U = 1121 Vq

rqq

41 =πε0

(For a system of two charges)

(b) U=23

32

13

31

12

21r

qq 4

1rqq

41

rqq

41

000 επ+

επ+

επ (For a system of three charges)

(c) →→

−= E . pU (For an electric dipole) 12. If n drops coalesce to form one drop, then (a) Q = nq; (b) R=n1/3r ; (c) V = n2/3 Vsmall ; (d) σ = n1/3 σsmall (e) E=n1/3 Esmall

13. Energy density of electrostatic field: 2E21u 0ε=

CAPACITANCE

14. Capacitance: (a) C = (q/V) (b) Unit of C = farad (F) (c) Dimensions of C = [M�1L�2T4A2] 15. Energy stored in a charged capacitor

(a) U=21 CV2 ; (b) U= ;QV

21 (c)

CQ

21U

2=

16. Energy density: (a) u = 21 ε0E2 ; (b) u =

0

2

εσ

21

17. Force of attraction between plates of a charged capacitor

(a) F=21 ε0E2A; (b) F =

0

2

εσ2

A ; (c) F=A2

Q2

0ε

18. Capacitance formulae (a) Sphere: (i) Cair = 4π ε0R; (ii) Cmed = K (4π ε0R)

(b) Spherical capacitor: (i) Cair = ab

barr

rr4−

επ 0 ; (ii) Cmed = ( )ab

barr

rrK4−

επ 0

(c) Parallel plate capacitor: (i) Cair = ;dA0ε (ii) Cmed =

dAK 0ε

(d) Cylindrical capacitor: (i) Cair = ( )abe r/r log2 l0επ ; (ii) Cmed = ( )abe r/rlog

K2 l0επ

(e) Two long parallel wires: C= ( )d/a loge

l0ε where d is the separation between wires and a radius of

each wire (d>>a) 19. Series Combination of Capacitors (a) q1 = q2 = q3 = q (Charge remains same)

(b) V1 = 3

32

21 C

qV,CqV,

Cq == (Potential difference is different)

(c) 321 C

1C1

C1

C1 ++=

(d) For two capacitors in series: C = C1C2/(C1 + C2) (e) Energy stored: U = U1 + U2 + U3 20. Parallel Combination of Capacitors (a) V1 = V2 = V3 = V (Potential difference remains same) (b) q1 = C1V, q2 = C2V, q3 = C3V (Charges are different) (c) C = C1 + C2 + C3 (d) U = U1 + U2 + U3 21. Effect of dielectric (a) Field inside dielectric, Ed = E0/K (b) Polarization charges on surface of dielectric:

(i) Qp = Q

−

K11 ; (ii) σp =

AQp =σ

−

K11

(c) Polarization vector: (i) |P|→

=Qp/A ; (ii) |P|→

=ε0χEd 22. Capacitance formulae with dielectric

(a) 1)(Kt d K

A K

K11 td

AC−−

ε=

−−

ε− 00 (For a dielectric slab of thickness t)

(b) td

AC−

ε= 0 (For a metallic slab of thickness t)

(c)

+ε= 0

2KK

dAC 21

(d)

+

ε= 0

21

21

KKKK

dA2C

(e)

+

+ε= 0

32

321 KK

K K2K d 4AC

K2K1 d

A/2A/1

Fig. 11

K2

K1

A/2A/2

Fig. 12

d/2

d/2

K2

Fig. 13

d

d K3

K1

(f) For n plates with alternate plates connected: C = (n�1) ε0A/d

(g)

++

ε= 0

3

3

2

2

1

1

Kt

Kt

Kt

AC

23. Spherical capacitor with inner sphere grounded

(a) ( ) 212

21 r 4rr

rr4C 0

0 επ+−

επ=

(b) Charge on inner sphere = �q1, while charge on outer sphere = +q2

(c) Magnitude of charge on inner sphere: q1 =

2

1

rr

q2

24. Insertion of dielectric slab (a) Battery remains connected when slab is introduced (i) V� = V; (ii) C� = KC ; (iii) Q� = KQ ; (iv) E� = E; (v) U� = KU (b) Battery is disconnected after charging the capacitor and slab is introduced (i) Q� = Q; (ii) C� = KC ; (iii) E� = E/K; (iv) V� = V/K; (v) U� = U/K

25. Charge transfer, Common potential and energy loss when two capacitors are connected

(a) Common potential: 21

21

21

2211

CCqq

CCVCVCV

++=

++=

(b) Charge transfer: ∆q = ( )2121

21 VV CC

CC −+

(c) Energy loss: ∆U = ( )221

21

21 VV CC

CC 21 −

+

26. Charging and discharging of a capacitor (a) Charging: (i) q = q0 (1�e�t/RC) ; (ii) V = V0 (1�e�t/RC); (iii) I = I0e�t/RC; (iv) I0 = V0/R (b) Discharge: (i) q = q0e�t/RC ; (ii) V = V0e�t/RC ; (iii) I = � I0e�t/RC

(b) Time constant: τ = RC

K1 K2 K3

t1 t2 t3Fig. 14

CURRENT ELECTRICITY

1. Electric Current (a) I = (q/t); (b) I = (dq/dt); (c) I = (ne/t); (d) q = IdT 2. Ohm�s law, Resistivity and Conductivity (a) V = IR ; (b) R= ρ(l/A) ; (c) σ = (1/ρ) (d) vd = (eEτ/m); (e) I = neAvd;

(f) ; A

ne

m R 2

τ= l (g) ρ =

τ2nem ; (h) σ =

mne2τ

3. Current density (a) J = (I/A); (b) J = nevd; (c) J = σE; (d) µ = (vd/E); (e) σ =neµ 4. Temperature Coefficient of Resistance

(a) R = R0[1 + α(T�T0)] ; (b) α = ( )00

0TT R

RR−

− ; (c) ρ = ρ0 [1+α(T�T0)] ;

(d) α = ( )0TT −ρρ−ρ

0

0

5. Cell: (a) ;QWE = (b) I =

RrE+

; (c) V = E � Ir (where V = IR)

6. Series Combination of Resistances (a) R = R1 + R2 + R3; (b) V = V1 + V2 + V3; (c) I = constant = I1 = I2 = I3; (d) V1 = IR1, V2 = IR2, V3 = IR3

7. Parallel Combination of resistances

(a) ; R1

R1

R1

R1

321++=

(b) I= I1 + I2 + I3; (c) V = constant = V1 = V2 = V3;

(d) 3

32

21

1 RVI ,

RV I,

RVI ===

(e) For a parallel combination of two resistances:

(i) ;RR

RRR21

21

+= (ii) I1 = ;I

RRR

21

2

+ (iii) I2 = I

RRR

21

1

+

8. Heating effect of current (a) W = VI t; (b) P = VI ; (c) P = I2R = V2/R;

(d) H = I2 Rt Joule = JRtI2

Calorie

9. Electric bulb: (a) Resistance of filament R = V2/P; (b) Maximum current that can be allowed to pass through

bulb, Imax = (P/V) 10. Total Power Consumed (a) Parallel combination: P = P1 + P2 + P3

(b) Series combination: 321 P

1P1

P1

P1 ++=

∫

11. Effect of stretching a resistance wire

4

2

12

2

12

1

2

2

1

2

1

1

2

rr

AA

AA x

RR

=

=

==

l

l

l

l [Q l1A1 = l2A2]

12. Cells in series: I = RE

RnrnE =+

(if n r <<R)

13. Cells in parallel: ( ) RE

R n/rEI =

+= (if r << R)

r

nE= (if r >>R)

14. Mixed Combination (m rows with each containing n cells in series)

(a) ( ) ;R mnr

En mRm/nr

nEI+

=+

=

(b) I is maximum when n r = m R ;

(c) Imax = Rr n m2

En m

15. Chemical effect of current: (a) Faraday�s first law of electrolysis: m = Zq = ZIt (b) Faraday�s second law of electrolysis: (i) m ∝ W (W = ECE) or m/W = constant (where W = atomic weight/valency)

(ii) As 2

1

2

1

2

1

2

1

2

1

2

1

WW

ZZ so ;

WW

mm and

ZZ

mm ===

(c) Faraday : 1 Faraday = 96,500 Coulomb

(d) == FZW Faraday�s constant

16. Thermo e.m.f. : e = αθ + 2

2βθ (where θ = θH = θC)

17. Neutral temperature: θN = �

βα

=

θΝθ

0dde

18. Temperature of inversion:

2Cθ+θ=θ 1

Ν [QθI � θN = θN � θC]

19. Thermoelectric power or Seebeck Coefficient: S = θd

de =α + βθ

20. Peltier effect: (i) Heat absorbed per second at a junction when a current I flows = πI (where π = Peltier coefficient) (ii) Peltier coefficient, π = SθH

21. Thomson Coefficient:

(i) Heat absorbed/ sec = I θσd

∫ Η

Χ

θθ

(ii) Thomson coefficient, σ = θ∆

∆I

Q/time

MAGNETIC EFFECTS OF CURRENT

22. Biot�Savart law : dB = 20 θπ4

µrsin d I l

23. Field due to a long straight wire: r2IB

πµ= 0

24. Field due to a circular coil:

(a) at centre: B = a2

IN0µ ;

(b) at an axial point: B = ( ) 2/322

2

xa 2

a I N

+

µ0

(c) on axis when x >> a : B = 3

20

x2a I N µ

(d) point of inflexion: It occurs at x = a/2

Field at the point of inflexion: B =

µ0

a 55I N4

= 0.716 Bcentre

25. Magnetic moment of circular coil: (a) M = NIA ; (b) Field: B= 3x2M

π4µ0

26. Field due to an arc of current: (a) B = (µ0Il/4πR2) ; (b) B = (µ0Iθ/4πR) (c) At the centre of a semicircular coil: B = (µ0I/4R)

27. Field due to finite length of wire: B = a4I0

πµ (sin φ1 + sin φ2)

28. Field at the centre of a square loop: B =

πµ0

l I 22

29. Ampere�s law: 1 d .H (b) ; I d . B )a( =µ=→→

0→→

ll 30. Field due to a current in cylindrical rod:

(a) outside: B = (µ0 I/2πr) ; (b) surface: B = (µ0 I/2πR); (c) inside: B= 2R2r I

πµ0

31. Field due to a current carrying solenoid: (a) inside: B = µ0n I ; (b) at one end : B = (µ0n I/2)

(c) at an axial point: B = 2

In0µ (cos α2 � cos α1)

32. Field due to a toroid: (a) inside: B=µ0nI � µ0NI/2πR ; (b) outside: B = 0

33. Force on electric current: →→→

= Bx HF

34. Force between two parallel conductors: d2IIF 21

πµ= 0

l

35. Comparison of magnetic and electric forces between two moving charges: (Fmagnetic/Felectric) =

(v2/c2)

∫ ∫

36. Force on a current loop in a magnetic field: 0F =→

(any shape)

37. Torque on a current loop in a magnetic field: →→→

=τ B x M or τ M B sin θ 38. Moving coil galvanometer: (a) τ � N I A B ; (b) τ =Kθ ;

(c) I= ; AB NK θ

(d) Current sensitivity = (θ/I) = (NaB/K) ; (e) Voltage sensitivity = (θ/V) = (θ/IR) = (NAB/KR) 39. Ammeter: (a) Shunt resistance S = (IgG/ I� Ig); (b) Length of shunt wire, l = S πr2/ρ; (c) Effective resistance of ammeter, RA = GS/(G+S); (d) For an ideal ammeter, RA = 0 40. Voltmeter:

(a) High resistance in series, R = ; GIV

g

−

(b) For converted Voltmeter, RV = R + G; (c) For an ideal Voltmeter, RV = ∞ 41. Force on a moving charge:

(a)

=→→→B x vq F ; (b) F = q v B sin θ

42. Path of a moving charge in a magnetic field

(a) When →v is ⊥ to

→B :

(i) path = circular; (ii) r = (mv/qB) ; (iii) ν = (qB/2πm); (iv) T = (2πm/qB) ; (v) ω = qB/m)

(b) When angle between →v and

→B is θ:

(i) path=helical ; (ii) r = (mv⊥ /qB) = (mv sin θ/qB);

(iii) ν = (qB/2πm); (iv) T = ; B qm2π (v) ω = (qB/m);

(vi) pitch p = 2πr/tan θ (where tan θ = (v⊥ /v||) 43. Cyclotron: (i) T = (2πm/qB) ; (ii) ν = (qB/2πm) ; (iii) ω = (θB/m) ; (iv) radius of particle acquiring energy E, r = [√(2mE)/qB]; (v) velocity of particle at radius r, v = qBr/m; (vi) the maximum kinetic energy (with upper limit of radius = R)

m

RBq 21K

222

max =

44. Magnetic field produced by a moving charge:

(a) ;r

)r x v( q 4

B 3

→→0

→

πµ=

(b) 20 θπ4

µ=rsin vq B

MAGNETIC PROPERTIES OF CURRENT

45. Magnetic field:

(a) ; vq

FB max= (b) maxd

dF I1B

=

l

46. Atomic magnetic moments:

(a) µL = � ;m2

eL (b) µS = � ;meS

(c) µJ = � g ;m2

eJ (d) µB = m 4

h e π

= 0.927 x 10�23 J/T

47. Intensity of magnetization: I = (M/V) 48. Magnetizing field:

(a) H = IB −µ0

;

(b) For vacuum, H = 0µ

B ;

(c) For medium, h = B/µ; (d) H =nI (solenoid); (e) H = I/2 πr (straight wire);

(f) 2θ=

rsin d IH l (Biot-Savart law) ;

(g) freeI d . H =→→l

49. Magnetic susceptibility: χ = (I/H) 50. Magnetic permeability: (a) µ = (B/H) ; (b) µr = (µ/µ0) ; (c) µr = (B/B0) 51. Other relations: (a) µ = µ0 (1+χ) ; (b) µr = 1 + χ or χ = µr � 1; (c) B = B0 (1+ χ) : (d) B = µ0 (H + I) 52. Pole strength: m = F/B 53. Magnetic moment of dipole : M = m x 2l

54. Field due to a pole:

π4µ= 0

2rm B

55. Field due to a bar magnet:

(a) Axial point: ( )

π4

µ=−π4

µ= 003222 r

2M r

Mr 2 Bl

(if r > > l)

(b) Equatorial point: ( )

π4

µ=+π4

µ= 0032/322 r

M r

M Bl

(c) At arbitrary point: θ+π4

µ= 0 cos 31 rM B 2

3

∫

56. Force and torque on a dipole in uniform magnetic field

(a) ; 0F =→

(b) →→→

=τ B x M ; (c) τ MB sin θ

57. Potential energy of a dipole in magnetic field: →→

−= B.MU = � MB cos θ 58. Tangent galvanometer: (a) B = BH tan θ;

(b) I = K tan θ, where K = n

B r2

0

H

µ

59. Vibration magnetometer: T = 2πHB M

I

ELECTROMAGNETIC INDUCTION 60. Magnetic flux:

(a) dφ = ;θ=→→

cosBdA Ad .B (b) φ= ; Ad .B→→

(c) φ = BA cos θ; (d) ; 0Sd .B =→→

(e) 0 B.V =→

61. Faraday�s laws of e.m. induction: (a) Induced e.m.f., e = � (dφ/dt);

(b) Induced current, I = ; dtd

R1

Re φ−=

(c) Induced charge, q = (φ1 � φ2)/R 62. Motion of a conducting rod:

(a) ; )B x v( e F→→→

−= (b) Induced e.m.f., e = B/v (c) For a rod rotating with angular frequency ω or rotating disc, induced e.m.f.,

f B 21e 22 =πΒ=ω= ll Baƒ

63. Motion of conducting loop in a magnetic field: (a) Induced e.m.f. e = Blv ; (b) Induced current, I = (e/R) = (Blv/R) (c) F = IlB = B2l2v/R ; (d) P = Fv = IlBv = B2l2v2R; (e) H = I2R = (B2l2v2/R); (f) In non uniform magnetic field, e = (B1�B2) lv and I = (B1�B2)lv/R 64. Rotating loop: (a) φ = NAB cos ωt = φ0 cos ωt, with φ0 = NAB; (b) e = e0 sin ωt, where e0 = NaBω; (c) I = (e0 sin ωt/R) = I0 sin ωt, with I 0 = e0/R

65. Induced electric field: Induced e.m.f. = ∫→→ld.E

66. Self Inductance: (a) L = φ/ I ; (b) e = � (LdI/dt); (c) L = µ0N2A/l = µ0n2Al (For a solenoid with air core); (d) L = µrµ0N2A/l (For a solenoid with a material core); (e) L = µ0N2πR/2 (For a plane circular coil) 67. Mutual inductance: (a) M = (φ2/ I1) ; (b) e2 = � M(dI 1/dt); (c) M = µ0NsNp A/lp 68. Series and parallel combination (a) L = L1 + L2 (if inductors are kept far apart and joined in series) (b) L = L1 + L2 ±2M (if inductors are connected in series and they have mutual inductance M)

(c) 2121

21

L1

L1

L1or

LLLLL +=+

=

(if two inductors are connected in parallel and are kept for apart)

∫

∫

(d) M = K√(L1L2) (if two coils of self inductances, L1 and L2 are over each other)

69. Energy stored in an inductor:

(a) 2LI21U = ; (b) uB = (B2/2µ0)

70. Growth and decay of current in LR circuit (a) I = I0 (1�e�t/τ) (for growth), where τ = L/R (b) I = I0 e�t/τ (for decay), where τ = L/R

ALTERNATING CURRENT 71. A.C. Currrent and e.m.f. : (a) I = I0 sin (ωt ± φ) ; (b) e= e0 sin (ωt ± φ);

(c) < I > = 0, < I >1/2 = π

0I2 =0.637 I0 ;

(d) ;2/I I 20

2 >< (e) Irms = (I0/√2) = 0.707 I0 ; (f) form factor = π/2√2 72. A.C. response of R, L, C and their series combinations (a) Resistance only: (i) e = e0 sin ωt; (ii) I = I0 sin ωt ; (iii) phase difference φ = 0; (iv) e0 = I0R; (v) erms = Irms R (b) Inductance only: (i) e = e0 sin ωt; (ii) I = I0 sin (ωt�π/2) ; (iii) current lags the voltage or voltage leads the current by a phase π/2; (iv) e0 = I0XL; (iv) erms = Irms XL ; (vi) XL = ωL (c) Capacitance only:

(i) e = e0 sin ωt ; (ii) I = I0 sin (ωt + π/2); (iii) current leads the voltage or voltage lags the current by a phase π/2 ; (iv) e0 = I0XC; (v) erms = Irms XC ; (vi) XC = (1/ωC)

(d) Series LR circuit: (i) e = e0 sin ωt ; (ii) I = I0 sin (ωt + φ); (iii) the current lags the voltage or voltage leads the current by a phase φ = tan�1 (XL/R); (iv) cos φ = (R/Z) and sin θ = (XL/Z); (v) Impedance, Z = √ [R2 +(ωL)2)] ; (vi) e0 = I0Z; (vii) erms = Irms Z (e) Series RC circuit: (i) e = e0 sin ωt ; (ii) I = I0 sin (ωt + φ); (iii) The current leads the voltage or voltage lags behind the current by a phase φ = tan�1 (XC/R) (iv) cos φ = (R/Z); (v) Impedance, Z = √[R2 + (1+ωC)2)]; (vi) e0 = I0Z ; (vii) erms = Irms Z (f) Series LCR circuit:

(i) e = e0 sin ωt ; (ii) I = I0 sin (ωt � φ);

(iii) φ=tan�1

−R

XX CL , φ is positive for XL > XC, φ is negative for XL<XC;

(iv) current lags and circuit is inductive if XL < XC ; (v) current leads and circuit is capacitive if XL < XC ; (vi) e0 = I0Z; (vi) Impedance, Z = √[R2 + (XL � XC)2];

(viii) cos φ = (R/Z) and sin φ =

−Z

XX CL

73. Resonance

(a) Resonance frequency, ƒr =

π LC 21

(b) At resonance, XL = XC, φ = 0, Z = R (minimum), cos φ = 1, sin φ = 0 and current is maximum (=E0/R)

74. Half power frequencies

(a) lower, ƒ1 = ƒr � L4

Rπ

or ω1 = ωr � L2

R

(b) upper, ƒ2 = ƒr + L4

Rπ

or ω2 = ωr + L2

R

75. Band width: ∆ƒ = L2

Rπ

or ∆ω = LR

76. Quality factor

(a) ;R

LQ rr ω=ω∆

ω=

(b) As ωr = LC1 , hence Qα√L, Qα

R1 and Q α ;

C1

(c) ;CR1Q

rω=

(d) ( )

RX

Q resL= or ( )

RX resC ;

(e)

ƒ∆

ƒ= rQ or ∆f = Q

rƒ

77. At resonance, peak voltages are (a) (VL)res = e0Q; (b) (VC)res = e0Q ; (c) (VR)res = e0 78. Conductance, susceptance and admittance (a) Conductance, G= (1/R); (b) Susceptance, S = (1/X); (c) SL = (1/XL) and SC=(1/XC) =ωC; (d) admittance Y = (1/Z); (e) Impedance add in series while admittance add in parallel 79. Power in AC circuits

(a) Pav = 21 E0I0 cos φ = Erms Irms cos φ;

(b) Power factor, cos φ = rmsrms

av

I EP

power Virtualpower alRe =

(c) Cos φ = (R/Z)

(d) (i) R only : φ = 0, cos φ = 1, Pav = I2rms R =

Re2

rms

(ii) C only : φ = � 900 = �π/2, cos φ = 0, Pav = 0 (iii) L only : φ = 900 = π/2, cos φ = 0, Pav = 0

(iv) Series RL or RC: φ = tan�1

RXL or φ = tan�1

RXC

Pav = Erms Irms cos φ = R IZ

R E 2rms2

2rms =

(iv) Series LCR: φ = tan�1 φ = tan�1

−R

XX CL , PAV = R, IZ

R Erms

22

2rms =

At resonance, φ = 0, cos φ = 1 and Pav = I2rms R = E2

rms/R

80. Parallel LCR circuit

(a) ; X1

X1

R1

Z1

2

CL2

−+=

(b) ( ) ; SSGY 2CL

2 −+= (c) I0 = E0Y;

(d) tan φ = ;G

SS CL −

(e) ωr = LC1 or ωr = ;

LR

LC1

2

2

−

(f) in parallel resonance circuit, impedance is maximum, admittance is minimum and current is minimum.

81. Transformer:

(a) Cp = Np

φ

dtd and es = Ns

φ

dtd

(b)

=

s

p

s

p

NN

ee

(c) Q ep Ip = es Is, so

=

=

s

p

s

p

p

s

NN

ee

II

(d) Step down: es < ep, Ns < Np and Is > Ip (e) Step up : es > ep, Ns > Np and Is < Ip

(f) Efficiency, η =

pp

ss

I eI e

82. AC generator: e = e0 sin (2πƒt), (where e0 = NBAω) 83. DC motor:

(a)

−=

ReEI

(b) IE = Ie = I2R

(c) efficiency, η emf Applied

emfBack Ee =

LIGHT 1. Intensity of light

(a) Spherical wave front: (i) I = ,r4

P2π

(ii) amplitude ∝ r1

(b) Cylindrical wave front: (i) I ∝ r1 , (ii) amplitude ∝

r1

(c) Plane wave front: (i) I ∝ r0, (i) A ∝ r0 (i.e. I and A are both constants)

2. Law of reflection: Angle of incidence (i) = Angle of reflection (r)

3. Law of reflection: Snell�s law: η = rsin i sin

4. Other relations

(a) 2η1 = vc and

vv

2

1 =η

(b) λmedium = η

λ air or vmedium = ηairv (Qνmedium = νair)

(c) η1 sin i = η2 sin r 5. Electromagnetic nature of light

(a) The magnitude of →→B and E are related in vacuum by: B=

CE

(b) →→B and E are such that

→→B x E is always in the direction of propagation of wave

(c) c=0εµ0

1 and v=µε1

(d) Refractive index, η = √(µr εr) (µr = µ/µ0 and εr = ε/ε0) For non�magnetic material, µr ≈ 1 and η = √(εr) (e) The EM wave propagating in the positive x�direction may be represented by: Ey = E0 sin (kx � ωt) and Bz = B0 sin (kx � ωt) 6. Energy transmitted by an electromagnetic wave

(a) Energy density of electromagnetic wave is: u = ue + um = 21 ε0 E2 +

21

0µ

2B

(b) As for EM wave, B = CE and

c1 = √ (µ0 ε0), hence

2222

2

0

2 E E 21 E

21

cE

21E

21u 0000 ε=ε+ε=

µ+ε=

(c) Time averaged value of energy density is: 20E

21u 0ε=

7. Intensity of an electromagnetic wave

(a) In a medium: I = vE21 2

0

ε0

(b) In free space: I = c E21 2

0

ε0

(a) )B x E( c)B x E( 1 H x ES 2 →→0

→→

0

→→→ε=

µ==

(b) S = cε0E2 = √(ε0/µ0)E2 (c) SI = and uc S = (d) Impedance of free space, Z = √ (µ0/ε0) ≅ 377 ohm 9. Pressure of EM Radiation (a) Change in momentum (normal incidence)

c

tA S cUp ∆==∆ (absorber)

c

tA S cU2p ∆==∆ (reflector)

(b) Pressure (normal incidence)

ucSP == (absorber)

u2cS2P == (reflector)

(c) Pressure for diffused radiation

u31

cS

31P == (absorber)

u 32

cS

32P == (reflector)

10. Quantum theory of light: (a) Energy of photon, E = hν = hc/λ

(b) Momentum, p = λ

= hcE

(c) Rest mass of photon = 0 (d) Mass equivalent of energy, m = (E/c2) 11. Inclined mirrors: number of images (a) When 3600 is exactly divisible by θ0 and 3600/θ0 is an even integer then the number of images

formed is

1360n −θ

= (whatever may be location of the object)

(b) When 3600 is exactly divisible by θ0 and 360/θ) is an odd integer, then the number of images

formed is

1360n −θ

= (for symmetrical placement)

θ

= 360 (for unsymmetrical placement)

8. Pointing vector

(c) When 3600 is not exactly divisible by θ, then the number of images formed is = integer value of n (where n = 360/θ) 12. Reflection amplitude and intensity (a) When a ray of light is incident (with angle of incidence i ≈ 0) from a medium 1 of refractive index η1

to the plane surface of medium 2 of refractive index η2, then reflection amplitude is

21

21

η+ηη−η

=R

(b) The ratio of the reflected intensity and the incident intensity is: 2

i

r

II

η+ηη−η

=21

21 .

13. Refraction of light

(a) ; rsin iin s=η (b) 1η2 = ;

sin sin

2

1

θθ

(c) 1η2 = 1η2

1 ; (d) Cauchy�s relation: η = A + 2λB

14. Parallel slab (a) Angle of incidence, i = Angle of emergence, e (b) Lateral shift = [t sin (i � r)/cos r] 15. Composite block: η1 sin θ1 = η2 sin θ2 = η3 sin θ3 = constant 16. Apparent depth

(a) a = η

=η

tR (where R = Real depth)

(b) If there is an ink spot at the bottom of a glass slab, if appears to be raised by a distance

x = t � a = t �

η

−=η

11 tt , where t is the thickness of the glass slab

(c) If a beaker is filled with immissible transparent liquids of refractive indices η1, η2, η3 and individual depths t1, t2, t3 respectively, then the apparent depth of the beaker is:

321 η

+η

+η

= 321 ttta

17. Total internal reflection: Critical angle iC is given by: sin iC =η1

18. For a luminous body at a depth d inside a liquid: Radius of bright circular patch at the surface

r = d tan iC = 1

d

−η2

19. For optical fibre: sin i ≤ ( )[ ]1n/n 212 −

20. Prism: (a) i + e = A + δ (b) r1 + r2 = A;

(c) At minimum deviation: i = e and r1 = r2. Hence, η =

δ+

2A sin

2A sin m

(d) For small angle prism: δ = (η�1) A

21. Dispersion: (a) δred < δviolet because ηred < ηviolet (b) Angular dispersion: θ = δV � δR = (ηV�ηR)A

(c) Dispersive power: ω = 11 Y

RB

Y

RV

Y

RV−ηη−η=

−ηη−η=

δδ−δ (In practice)

(d) Dispersion without deviation: (i) δC + δF = 0 or ( )( )1

1AA

F

C

C

F

−η−η

−=

(ii) Also, angular dispersion, θ=AC (ηC�1) (ωC�ωF)

(e) Deviation without dispersion: (i) θC + θF = 0 or, FRFV

CRCV

C

F

AA

η−ηη−η

−=

(ii) Also, FY

CY

C

F

δδ

−=ωω

22. Principle of superposition: y = y1 + y2 23. Superposition of waves of equal frequency and constant phase difference (a) Resultant wave amplitude, a = √(a1

2 + a22 + 2a1a2 cos φ)

(b) Resultant wave intensity, I = I1 + I2 + 2√(I1I2) cos φ (c) If a1 = a2 = a0, and I1 = I2 = I0, then a = 2a0 cos (φ/2) and I = 4I0 cos2 (φ/2) 24. Constructive interference (a) conditions: φ = 2nπ ≡ 0, 2π, 4π, 6π,�.. or, ∆ = nλ ≡ 0, λ, 2λ, 3λ, �� (b) amax = a1 + a2 (c) Imax ∝ (a1 + a2)2 (d) Imax = I1 + I2 + 2 √ (I1I2) = (√I1 + √I2)2 (e) Imax = 4I0 ; If I1 = I2 = I0 25. Destructive interference

(a) conditions: φ = (2n �1) π ≡ π, 3π, 5π,�� or, ∆ = (2n�1) ,.........5,2

3 , 2λλ

2λ=

2λ

(b) amin = a1 � a2 (c) Imin ∝ (a1 � a2)2 (d) Imin = I1 + I2 � 2√(I1I2) = (√I1 � √I2)2 (e) Imin = 0 if I1 � I2 = I0 26. Young�s double slit experiment

(a) Phase difference, φ = λπ2 (S2P � S1P) =

λπ2 x path difference

(b) A = 2a0 cos (φ/2) and I 4I0 cos2 (φ/2) (c) Position of nth fringe on the screen:

(i) for bright fringe, d

nDx nλ=

(ii) for dark fringe, ( )d2

D 1n2x nλ−=

27. Fringe width:

(a) Linear fringe width, β = d

Dλ

(b) Angular fringe width, α = dλ

(c) βliquid = liquid

airair

ηβ or λ liquid =

liquidair

air

ηλ

(d) βwater = 43 βair

28. When a thin shit is introduced in the path of one of the interfering waves: (a) (η�1) t = nλ

(b) Shift of the central fringe = ( )λ

β1−η t

29. Fringe visibility: V =minmax

minmax

IIII

−−

30. Frensel�s biprism: (a) d = 2a (η�1) α ; (b) d = √(d1d2) (c) β = (Dλ/d); (d) dliquid < dair, for example, dwater = dair/4

(e) βliquid > βair ; βliquid = βair

η−η−η

t1

g

g

31. Newton�s rings: (a) Diameter of nth dark fringe, Dn = √(4nλR)

(b) λ = pR 4

DD 2n

2pn −+ and 2

n2

pn

2n

2pn

'D'D

DD

−

−=η

+

+

32. Thin films: For reflected light 2ηt cos r = nλ (Dark fringe)

2ηt cos r = (n�21 )λ (Bright fringe)

33. Diffraction: (a) a sin θ = nλ (a = width of slit) (b) Half angular width of central maxima, θ = sin�1 (λ/a)

(c) Intensity distribution of the screen I = I0 2

sin

φ

φ

where, φ = Dy a

λπ and I0 = Intensity at central point of screen

(d) Limit of resolution of telescope: θ = a

22.1 λ

(e) Resolving power of telescope = λ

=θ 22.1

a1

34. Spherical mirrors: (a) Focal length: ƒ = (R/2)

(b) Mirror formula: u1

v11 +=

ƒ

(c) Newton�s formula: ƒ2 = xy (x and y are the distances of the object and image from the principal focus respectively)

(d) Linear magnification: ƒ−ƒ=

−ƒƒ=−== v

uuv

oIm

(e) Longitudinal magnification: 2

2

uvm −=

35. Spherical lenses: (c) A single spherical surface:

(i) ( )Ruv

1212 η−η=η−η [For an object placed in a medium of refractive index η1]

(ii) ( )R

uv

2121 η−η=η−η [For an object placed in a medium of refractive index η2]

(iii) First principal focus: ƒ1 = ( )1−ηR where η = η2/η1

(iv) Second principal focus: ƒ2 = ( )1−ηηR

(v) Magnification: m =1η

ηu//v 2

(d) Lens Maker�s formula:

(i) ( )

−η−η=η−η

−

−

ηη=

ƒ 121

1 21

1

21

2R1

R1

uv or,

R1

R1 1 1

[When medium is same on both sides of the lens]

(ii)

η−η+

η−η=η−η 23

1

121

2

3

R

Ruv

[When different medium exist on two sides of the lens]

(e) Biconvex or biconcave lens of the same radii for two surfaces: ( )R

2 1 1−η=ƒ

(f) Linear magnification: u

vuv

OIm

+ƒƒ=

ƒ−ƒ===

(g) Power of lens: P= ƒ1

(h) Lenses in contact:

(i) ; 111

21 ƒ+

ƒ=

ƒ (ii) P = P1 + P2

(iii) For lenses separated by a distance d = 2121

d111ƒƒ

−ƒ

+ƒ

=ƒ

(i) Achromatic lens combinations: Condition of achromatism, yy ''

ƒω−=

ƒω

36. Silvering at one surface:

(a) ( )R

1- 2 2111F1

m

η=ƒ

=ƒ

+ƒ

+ƒ

=lll

η R

Fig. 1

(b) ( )Rn2

R2

R 2 12

F1

m=+

1−η=

ƒ+

ƒ=

l

(c) ( )221m R

2R1

R1 2 12

F1 +

+1−η=

ƒ+

ƒ=

l

37. Optical Instruments (a) Astronomical Telescope:

(i) For normal adjustment: m = e

0ƒƒ

(ii) For near�point adjustment: m =

ƒ+

ƒƒ

D1 e

e

0

(b) Simple Microscope:

(i) For normal adjustment: m = ƒD

(ii) For near�point adjustment = m = 1+ƒD

(c) Compound Microscope:

(i) For normal adjustment: m =

ƒe0

0 D uv

(ii) For near�point adjustment: m =

ƒ+

e0

0 D1 uv

η R

Fig. 2

η R2

Fig. 3

R1

MODERN PHYSICS CATHODE RAYS AND POSITIVE RAYS

1. Cathode rays (a) Thomson identified cathode rays as an electron beam. (b) Specific charge q/m as measured by Thomson is: (q/m) = 1.759 x 1011 Coulomb/Kg 2. Positive rays (a) Positive rays were discovered by Goldstein. (b) (q/m) for positive rays is much less than that of electrons. 3. Motion of charge particle through electric field (Field ⊥ to initial velocity) (a) The path is parabolic: y = (qE/2mu2)x2 (b) The time spent in the electric field: t = (L/u) (c) The y�component of velocity acquired: vy = (qEL/mu) (d) The angle at which particle emerges out tan θ = qEL/mu2 (e) The displacement in y-direction, when the particle emerges out of the field: y1=(qEL2/2mu2) (f) The displacement on the screen = Y = (qELD/mu2) 4. Motion of charged particle through magnetic field (Field ⊥ to initial velocity) (a) The path is circular with radius: r = (mu/qB) (b) Momentum of the particle: p = qBr (c) The deflection on the screen: X = (qBLD/mu) 5. Mass spectrographs (a) Thomson�s mass spectrograph

(i) Traces on the screen are parabolic in nature (ii) Inner parabola corresponding to heavy M white outer parabola to light M. (iii) The upper portion of parabola is due to small v ions, while lower portion is due to high v ions. (iv) Only v = ∞ ions can reach vertex of parabola. (v) Equation of parabola: X2 = (B2LD/E) (q/M) Y = K (q/M) Y

(b) Brain bridge mass spectrograph (i) Velocity selector: v = (E/B)

(ii) Other relations: r = (Mv/qB�) = (ME/qBB�) (whre B� is the magnetic field in dome); d =2r; (d2 � d1) ∝ (M2�M1) ; M1 : M2 = d1 : d2 [where d1 and d2 are the

distances of traces 1 and 2 from the slit S2 of velocity selector].

PHOTOELECTRIC EFFECT

6. Threshold frequency: v0 = hW

hfunction Work =

7. Threshold wavelength: λ0 = Whc

hvhc

vc

00==

(To calculate λ0, use hc = 1240 (eV) (nm) = 1.24 x 10�6 eV) (m) 8. Maximum kinetic energy of emitted photoelectrons

(a) Kmax = 21 mv2

max = eV0

(b) Kmax = hv � W = h (v � v0) = hc

λ

−λ 0

11

9. Slope of (V0 v) graph = eh

10. Energy, momentum and mass of a photon

(a) Rest mass of photon = 0 (b) E = hv = λhc

(c) p = λ

= hcE (d) m =

λ=

ch

cE2

11. Number of photons:

(a) number of photons per sec per m2, np = ( )vh

Watt/m Intensity 2

(b) number of photons incident per second, np = ( )hv

Watt Power

(c) number of electrons emitted per second = (efficiency of surface) x number of photons incident per second.

12. Compton wavelength:

(a) λc = Cm

h

0=2.426 pm

(b) Change in wavelenth, (λ� � λ) = λc (1� cos φ)

ATOMIC STRUCTURE 13. Rutherford�s α�particle scattering (a) N(θ) ∝ cosec4 (θ/2)

(b) Impact parameter, b = ( ) ( )( )E 4

2cot Ze2

0επ/θ , (where E =

21 mu2 = KE of the α�particle)

14. Distance of closest approach: r0 ( )E4 Ze2 2

0επ= (where E =

21 mu2 = KE of the

α�particle) 15. Bohr�s atomic model

(a) L = mvr = π2

nh

(b) hv = Ei = Ef = λhc

(c) Radius of nth orbit:

(i) rn ∝ ,Zn2

(ii) rn =

π2 2

22

ke m 4h

Zn

(iii) Bohr�s radius: a0 = (h2/4π2mke2) = 0.529 Å

(iv) Ratio of radii: r1:r2 : r3 = 1 : 4 : 9 ; rN : rHe+ : rLi++ = 1 : 31:

21 = 6 : 3 : 1

(d) Velcotiy of electron in nth orbit:

(i) vn = c nZ

137c

nZ α=

(where α = ==π

1371

hcKe 2 2

fine structures constant)

(ii) v1 : v2 ; v3 = 1 : 31:

21 = 6 : 3 : 2

(iii) v1 = velocity of electron is 1st orbit of H�atom = (c/137) (e) Total energy of electron: (i) Potential energy, U = � (kZe2/r)

(ii) K = 21 mv2 = (kZe2/2r)

(iii) E = K + U = � (kZe2/2r) = (U/2) = �K (iv) K = � (U/2) or U = 2K = 2E

(v) En = � 2

2

n Z6.13 eV = � 2

2

nZ J

n Z10 x 2.18

hek m 2

2

2-18

2

42−=

π2

(f) Ionization energy = � E1 = + (13.6Z2)eV (i) For H�atom, I.E. = 13.6 eV

(ii) For He+ � ion, I.E = 54.4 eV (iii) For Li++�ion, I.E. = 122.4 eV

(g) Ionization potential: (i) For H�atom, I.P. = 13.6 V (ii) For He+ ion, I.P. = 54.42 (h) Series formula (wave number λ1/=v

n1

n1 RZ1

22

21

2

−=

λ where 1-7

3

42m 10 x 097.1

hcek m 2R =π=

2

(i) Series formula for H�atom

(i) Lyman series: ∞=

−=λ

...... 4, 3, 2, n ,n11 R 12

(ii) Balmer series: ∞=

−=

λ5.... 4, 3, n ,

n1

21 R 1

22

(iii) Paschen series: ∞=

−=

λ6.... 5, 4, n ,

n1

31R1

22

(iv) Brackett series: ,n1

41 R 1

22

−=

λn = 5, 6, 7�.∞

(v) P�fund series: 8 7, 6, n ,n1

51 R 1

22 =

−=

λ�.∞

(j) Series limits (λmin) (i) Lyman: λmin = 912 Å (ii) Balmer: λmin = 3645 Å (iii) Paschen: λmin = 8201 Å 16. Number of emission lines from excited state n = n(n�1)/2 17. Time period of revolution (a) Tn ∝ (n3/Z2) ; (b) T1 = 1.5 x 10�16 sec ; (c) T1 : T2 : T3 = 1 : 8 : 27

18. Frequency of revolution

(a) vn ∝ (Z2/n3); (b) v1 = 6.6 x 1015 Hz ; (c) v1 : v2 : v3 = 1 : 271:

81

19. Current due to orbital motion (a) In ∝ (Z2/n3) ; (b) I1 = 1 mA 20. Magnetic field at nucleus due to orbital motion of electron (a) Bn ∝ (Z3/n5) ; (b) B1 = 12.5 Tesla 21. Magnetic moment: (a) Mn = (eL/2m) = (nhe/4πm); (b) M1 = (eh/4πm) = µB = Bohr Magneton = 9.27 x 10�24 Am2 22. Magnitude of angular momentum: L = √[l(l+1)] (h/2π) 23. Angle of angular momentum vector from z�axis (a) cos θ = [ml√{l(l+1)}]; (b) the least angle is for ml = l i.e. cos θmin = [l/√{l(l+1})] 24. Magnitude of spin angular momentum

S = √[s (s+1)] (h/2π) = 23 (h/2π)

X�RAYS 25. Continuous X�rays: (a) vmax = (eV/h) ; (b) λmin = (hc/eV) = (12400/V) Å

26. Characterictic X�rays: (a) λKα < λLα < λMα ; (b) vKα > vLα > vMα

27. Frequency of Kα line : v (Kα) = 4cR3 (Z�1)2 = 2.47 x 1015 (Z�1)2

28. Wavelength of Kα line: λ(Kα) = [4/3R(Z�1)2] = [1216/(Z�1)2]Å 29. Energy of Kα X�ray photon: E(Kα) = 10.2 (Z�1)2 eV 30. Mosley�s law: (a) v = a (Z�b)2, where a = (3cR/4) = 2.47 x 105 Hz (b) For Kα line, b = 1; (c) √v α Z 31. Bragg�s law: 2d sin θ = nλ 32. Absorption formula: I = I0 e�µx

33. Half�value thickness: x1/2 = (0.693/µ)

MATTER WAVES 34. For photons: (a) E = hv = (hc/λ) ; (b) p = (hv/c) = (E/c) = (h/λ) ; (c) m = (E/c2) = (hv/c2) = h/cλ (d) rest mass = 0, charge = 0, spin = 1 (h/2π) 35. Matter waves: (a) de Broglie wavelength,

V q m 2h

E m2h

mvh

ph ====λ ]qVmv E [ 2

2

1 ==Q

(b) (i) For electron λe = ÅV27.12

(ii) For proton, λp = ÅV

286.0

(iii) For alpha particle λα = ÅV

101.0

(c) For particle at temperature T :

=

3=λ kT

23E

KT m h

(i) For neutron or proton: λ = (25.2/√T) Å [if E = (3/2) kT, average energy]

but λ = T8.30 Å [if E = kT, most probable energy]

(d) The wavelength of electron accelerated by potential difference of V volts is: λe =v27.12 Å

Hence, accelerating potential required for obtaining de Broglie wavelength for as electron is:

volt6.150V 2eλ

=

(e) Condition for obtaining stable orbit: 2πrn = nλ (f) The phase velocity of a de Broglie wave of wavelength λ and frequency v is

.c vi.e. vc

vmh x

hmc

mvh x

hvv p

22

p >==Ε=λ=

(g) Group velocity, vg = (dω/dk). It is found that group velocity is equal to particle velocity i.e., vg = v

RADIOACTIVITY 36. Decay law: (a) (dN/dt) = � λN ; (b) N = N0 e�λt; (c) (N/N0) = (1/2)t/T 37. Half life and decay constant:

(a) ( ) ;N

dN/dt−=λ (b) λT = loge 2 or T = (0.693/λ) or λ = (0.693/T)

38. Mean life: (a) τ = (1/λ) or λ=(1/τ); (b) T = 0.693τ or τ = 1.443 T 39. Activity: (a) R = |dN/dt| ; (b) R = λN ; (c) R = R0e�λt; (d) (R/R0) = (1/2)t/T ; (e) 1 Becquerel = 1 dps ; (f) 1 curie = 1 ci = 3.7 x 1010 dps; (g) 1 Rutherford = 1Rd = 106 Rd = 106 dps 40. Decay of active mass:

(a) m = m0 e�λt ; (b) (m/m0) = (1/2)t/T ; (c) N =A

m x 10 x 023.6 23

41. Radioactive equilibrium: NAλA = NBλB

42. Decay constant for two channels: (a) λ = λ1 + λ2 ; (b) T = 21

21

TTT T

+

43. Gamma intensity absorption: (a) I = I0e�µx ; (b) Half value thickness, x1/2 = (0.693/µ)

NUCLEAR PHYSICS 44. Atomic mass unit: (a) 1 amu = 1.66 x 10�27 kg ; (b) 1 amu ≡ 1u ≡ 931.5 MeV 45. Properties of nucleus (a) Radius: R = R0A1/3 where R0 = 1.2 fermi

(b) Volume: V αA

π=π= A R

34R

34V 3

03Q

(c) Density: ρ = 2.4 x 1017 Kg/m3 (ρ is independent of A) 46. Mass defect: ∆M = Zmp + (A�Z) mn � M 47. Packing fraction: ƒ = ∆/A = mass excess per nucleon [∆ = �∆M = mass excess] 48. Binding energy: ∆E = BE = (∆M)c2 49. Binding energy per ncuelon: (a) BEN = (BE/A); (b) BEN for Helium = 7.1 MeV/nucleon (c) BEN for Deuterium = 1.1 MeV/nucleon

ELECTRONICS

50. Richarson equation (a) J = AT2e�W/KT where A = 60 x 104 A/K2m2 (b) J = AT2e11600 W/T [∴ K = Boltzmann�s constant = 1.38 x 10�23 J/K = 8.62 x 10�5 eV/K

Hence, 1/K = 11600 kelvin/eV] (c) I = AT2Se�W/KT

51. Child�s law: Ip = KVp

3/2 [K = constant of proportionality] 52. Diode resistance (a) Static plate resistance: (i) Rp = (Vp/ Ip); (ii) Rp ∝ Vp

�1/2 (iii) Rp ∝ I 3/1p−

(b) Dynamic plate resistance: (i) rp = (∆Vp/∆ Ip); (ii) rp ∝ v 2/1p− ; (iii) rp ∝ I .3/1

p−

53. Triode Constants:

(a) rp = constant Vp

p

gIV

=

∆∆

; (b) gm = ;VI

ttanconsVg

p

p =

∆∆

(c) µ = consantIg

p

pVV

=

∆∆

; (d) µ = rp x gm ; (e) rp ∝ Ip�1/3 ; (f) gm ∝ Ip

1/3

54. Plate current equation: Ip = K 2/3

pg

VV

µ

+

55. Cut off voltage: Vg = � (Vp/µ) 56. Triode as an amplifier: (a) Ip = (µVg/RL + rp); (b) A = (µRL/RL + rp)

(c) Amax = µ; (d) µ = A ; Rr

1L

p

+ (e) A = µ/2 if RL = rp

57. Conductivity of semi conductors (a) Intrinsic: (i) σ = e (neµe + nh µh) ; (ii)

KT2/Ege−0σ=σ

(b) Extrinsic: (i) n�type : σ = eneµe; (ii) p�type : σ = enhµh

58. Transistor: (a) IE = IC + IB (IB << IE, IB << IC) (b) Current gains:

(c) Relation between α and β: α = α−

α=ββ+1

β1 or

( )

( )B

Cac

B

C

Cac

E

C

I I ,

I ii

I , i

∆∆=βΙ=β

∆Ι∆=α

ΙΙ=α

Ε