PHYSICS
FORMULA BOOKLET - GYAAN SUTRA
INDEX
S.No.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
Topic
Page No.
Unit and Dimension
Rectilinear Motion
Projectile Motion & Vector
Relavitve Motion
Newton’s Laws of Motion
Friction
Work, Power & Energy
Circular Motion
Centre of Mass
Rigid Body Dynamics
Simple Harmonic Motion
Sting Wave
Heat & Thermodynamics
Electrostatics
Current Electricity
Capacitance
Alternating Current
Magnetic Effect of Current & Magnetic
force on charge
2
3–4
5–5
5–7
7–9
9–9
10 – 11
11 – 14
14 – 18
18 – 25
26 – 28
29 – 31
31 – 37
37 – 40
41 – 47
47 – 51
52 – 54
Electromagnetic Induction
Geometrical Optics
Modern Physics
Wave Optics
Gravitation
Fluid Mechanics & Properties of Matter
Sound Wave
Electro Magnetic Waves
Error and Measurement
Principle of Communication
Semiconductor
56 – 59
59 – 66
67 – 70
70 – 73
73 – 75
75 – 77
77 – 79
79 – 80
80 – 81
82 – 83
84 – 85
54 – 56
Page # 1
PHYSICS
FORMULA BOOKLET - GYAAN SUTRAA
UNIT AND DIMENSIONS
Unit :
Measurement of any physical quantity is expressed in terms of an
internationally accepted certain basic standard called unit.
*
Fundamental Units.
S.No.
Physical Quantity
SI Unit
Symbol
1
Length
Metre
m
2
Mass
Kilogram
Kg
3
Time
Second
S
4
Electric Current
Ampere
A
5
Temperature
Kelvin
K
6
Luminous Intensity
Candela
Cd
7
Amount of Substance
Mole
mol
*
Supplementary Units :
S.No.
Physical Quantity
SI Unit
Symbol
1
Plane Angle
radian
r
2
Solid Angle
Steradian
Sr
*
Metric Prefixes :
S .N o .
P re fix
S ym b o l
V a lu e
1
C e n ti
c
1 0 –2
2
M ili
m
1 0 –3
3
M icro
µ
1 0 –6
4
Nano
n
1 0 –9
5
P ico
p
10
6
K ilo
K
10
3
7
M ega
M
10
6
–12
Page # 2
RECTILINEAR MOTION
Average Velocity (in an interval) :
r  ri
Total displacement
= f
t
Total time taken
Average Speed (in an interval)
v av = v = <v> =
Average Speed =
Total distance travelled
Total time taken
Instantaneous Velocity (at an instant) :

 r 

v inst = lim


t  0  t 
Average acceleration (in an interval):




v
v f  vi
a av =
=
t
t
Instantaneous Acceleration (at an instant):

 v 

dv


a =
= lim
t 0  t 
dt


Graphs in Uniformly Accelerated Motion along a straight line
(a  0)

x is a quadratic polynomial in terms of t. Hence x  t graph is a
parabola.
x
x
a<0
xi
xi
a>0
t
0
t
0
x-t graph
 v is a linear polynomial in terms of t. Hence vt graph is a straight line of
slope a.
v
e
op
sl
=
v
a
u
slo
pe
=
a
u
a is positive
0
t
a is negative
0
t
Page # 3
v-t graph
 at graph is a horizontal line because a is constant.
a
a
a
positive
acceleration
0
a
t
0
negative
acceleration
a-t graph
Maxima & Minima
dy
d
=0 &
dx
dx
and
 dy 

 < 0 at maximum
 dx 
dy
d
=0&
dx
dx
 dy 

 > 0 at minima.
 dx 
Equations of Motion (for constant acceleration)
(a)
v = u + at
(b)
(c)
1 2
1 2
at s = vt 
at
2
2
v 2 = u2 + 2as
s = ut +
(u  v )
t
(e)
2
For freely falling bodies : (u = 0)
(taking upward direction as positive)
(a)
v = – gt
(d)
s=
(b)
s=–
(c)
1 2
gt
2
v 2 = – 2gs
(d)
sn = –
s = vt 
1 2
gt
2
xf = xi + ut +
sn = u +
1 2
at
2
a
(2n  1)
2
hf = hi –
1 2
gt
2
g
(2n  1)
2
Page # 4
PROJECTILE MOTION & VECTORS
2u sin 
g
Time of flight :
T=
Horizontal range :
R=
u 2 sin 2
g
Maximum height :
H=
u 2 sin 2 
2g
Trajectory equation (equation of path) :
y = x tan  –
gx 2
2
2
2u cos 
= x tan  (1 –
x
)
R
Projection on an inclined plane
y


x
Up the Incline
Down the Incline
2u 2 sin  cos(    )
g cos 2 
2u 2 sin  cos(   )
Time of flight
2u sin 
g cos 
2u sin 
g cos 
Angle of projection with
incline plane for maximum
range
 

4 2
 

4 2
Maximum Range
u2
g(1  sin  )
u2
g(1  sin  )
Range
g cos 2 
RELATIVE MOTION



v AB ( velocity of A with respect to B )  v A  v B



a AB (acceleration of A with respect to B )  a A  aB
Relative motion along straight line -



x BA  x B  x A
Page # 5
CROSSING RIVER
A boat or man in a river always moves in the direction of resultant velocity
of velocity of boat (or man) and velocity of river flow.
1.
Shortest Time :
Velocity along the river, v x = v R.
Velocity perpendicular to the river, v f = v mR
The net speed is given by v m =
2.
2
v mR
 v R2
Shortest Path :
velocity along the river, v x = 0
and velocity perpendicular to river v y =
The net speed is given by v m =
2
v mR
 v R2
2
v mR
 v R2
at an angle of 90º with the river direction.
velocity v y is used only to cross the river,
Page # 6
d
therefore time to cross the river, t = v =
y
d
2
v mR
 v R2
and velocity v x is zero, therefore, in this case the drift should be zero.

v R – v mR sin  = 0
or
v R = v mR sin 
 vR
 = sin–1 
 v mR
RAIN PROBLEMS



v Rm = v R – v m
or




or
v Rm =
2
v R2  v m
NEWTON'S LAWS OF MOTION
1.
From third law of motion
F AB  FBA
F AB = Force on A due to B
FBA = Force on B due to A
2.
From second law of motion
Fx =
5.
6.
7.
8.
dPx
= max
dt
Fy =
dPy
dt
= may
Fz =
dPz
= maz
dt
WEIGHING MACHINE :
A weighing machine does not measure the weight but measures the
force exerted by object on its upper surface.


SPRING FORCE
F  kx
x is displacement of the free end from its natural length or deformation
of the spring where K = spring constant.
SPRING PROPERTY K ×  = constant
= Natural length of spring.
If spring is cut into two in the ratio m : n then spring constant is given
by
1 =
m
;
mn
2 =
n.
mn
k  = k 11 = k 22
Page # 7
9.
11.
12.
For series combination of springs
1
1
1


 .......
k eq k 1 k 2
For parallel combination of spring
keq = k1 + k2 + k3 ............
SPRING BALANCE:
It does not measure the weight. t measures the force exerted by the
object at the hook.
Remember :
a
(m 2  m1 )g
m1  m 2
T
2m1m2g
m1  m2
Vp =
V1  V2
2
aP =
a1  a2
2
WEDGE CONSTRAINT:
Components of velocity along perpendicular direction to the contact
plane of the two objects is always equal if there is no deformations
and they remain in contact.
Page # 8
13.
NEWTON’S LAW FOR A SYSTEM




Fext  m1a1  m2 a2  m3 a3  ......

Fext  Net external force on the system.
m 1, m 2, m 3 are the masses of the objects of the system and
  
a1,a2 ,a3 are the acceleration of the objects respectively..
NEWTON’S LAW FOR NON INERTIAL FRAME :



FRe al  FPseudo  ma
Net sum of real and pseudo force is taken in the resultant force.

a = Acceleration of the particle in the non inertial frame

FPseudo =  m a
Frame
(a) Inertial reference frame: Frame of reference moving with constant velocity.
(b) Non-inertial reference frame: A frame of reference moving with
non-zero acceleration.
FRICTION
Friction force is of two types.
(a)
Kinetic
(b)
Static
KINETIC FRICTION : f k = k N
The proportionality constant k is called the coefficient of kinetic friction
and its value depends on the nature of the two surfaces in contact.
STATIC FRICTION :
It exists between the two surfaces when there is tendency of relative motion but no relative motion along the two contact surfaces.
This means static friction is a variable and self adjusting force.
However it has a maximum value called limiting friction.
f max = sN
0  f s  f smax
f ri
ct
io
n
fstatic
st
at
ic
Friction
14.
maximum
s N
 kN
Applied Force
Page # 9
WORK, POWER & ENERGY
WORK DONE BY CONSTANT FORCE :
 
W= F . S
WORK DONE BY MULTIPLE FORCES




 F = F 1 + F 2 + F 3 + .....


W = [ F ] . S
...(i)






W = F 1 . S + F 2 . S + F 3 . S + .....
or
W = W 1 + W 2 + W 3 + ..........
WORK DONE BY A VARIABLE FORCE
dW =
 
F.ds
RELATION BETWEEN MOMENTUM AND KINETIC ENERGY
p2
and P =
2m
K=
2 m K ; P = linear momentum
POTENTIAL ENERGY

U2
U1
dU  
U

r

r2
r1
 
F dr
i.e.,
U 2  U1  

r2
r1
 
F  dr  W
 
F  d r  W

CONSERVATIVE FORCES
F= –
U
r
WORK-ENERGY THEOREM
W C + W NC + W PS = K
Modified Form of Work-Energy Theorem
W C = U
W NC + W PS = K + U
W NC + W PS = E
Page # 10
POWER
The average power ( P or pav) delivered by an agent is given by P or
pav =
W
t
 

 dS
 
F  dS
P
= F
= F . v
dt
dt
CIRCULAR MOTION
1.
Average angular velocity

av =
 2  1

=
t 2  t1
t

2.
Instantaneous angular velocity
3.
Average angular acceleration
4.
Instantaneous angular acceleration
5.
7.

=
d
dt


 2  1

av = t  t =
t
2
1
d
d
=
dt
d
  
Relation between speed and angular velocity  v = r and v    r
Tangential acceleration (rate of change of speed)

at =

=
dV
d
dr
=r
= 
dt
dt
dt
8.
Radial or normal or centripetal acceleration 
9.
Total acceleration
 


a  a t  a r  a = (at2 + ar2)1/2


 
 
Where a t    r and a r    v
ar =
v2
= 2r
r
at
a
O
ar or a c
v

P
Page # 11
10.
Angular acceleration

d

=
(Non-uniform circular motion)

dt


W
ACotation
R
12.
13.
Radius of curvature R =
v2
mv 2
=
a
F
If y is a function of x.
i.e. y = f(x)

R=
  dy  2 
1    
dx
   
3/2
d2 y
dx 2
Normal reaction of road on a concave bridge
N = mg cos  +

mv 2
r
O

N

concave
bridge
V
mgcos
mg
14.
Normal reaction on a convex bridge
N = mg cos  –

N
mv 2
r
V

convex
bridge
mgcos
mg

O
15.
Skidding of vehicle on a level road

v safe 
gr
16.
Skidding of an object on a rotating platform

max =
g / r
Page # 12
17.
Bending of cyclist
18.
Banking of road without friction
19.
Banking of road with friction 
20.
 tan  =
v2
rg
 tan  =
v2
rg
v2
  tan 

rg 1   tan 
Maximum also minimum safe speed on a banked frictional road
 rg (   tan ) 
Vmax  

 (1   tan  ) 
1/ 2
Vmin
 rg (tan   ) 
 

 (1   tan ) 
1/ 2
21.
Centrifugal force (pseudo force)  f = m2 r, acts outwards when the
particle itself is taken as a frame.
22.
Effect of earths rotation on apparent weight  N = mg – mR2 cos2  ;
where  latitude at a place
23.
Various quantities for a critical condition in a vertical loop at different
positions
C
O
D
B
 
A
P
N
×
(1)
Vmin  4gL
(for completing the circle)
(2)
(3)
Vmin  4gL
Vmin  4gL
(for completing the circle) (for completing the circle)
Page # 13
24.
Conical pendulum :
fixed pointor
suspension
O
/////////////

L
T
T cos 
h

r
mg
T cos  = mg
T sin  = m2 r
Time period =

25.
2
L cos 
g
Relations amoung angular variables :
0  Initial ang. velocity
 = 0 + t
d ,  or 
r
O
 a
r
(Perpendicular
to plane of paper
directed outwards
for ACW rotation)
at or V
  Find angular velocity
  Const. angular acceleration
  Angular displacement
1 2
t
2
2 = 02 + 2 
 = 0t +
CENTRE OF MASS


Mass Moment : M = m r
CENTRE OF MASS OF A SYSTEM OF 'N' DISCRETE PARTICLES



m1r1  m 2 r2  ........  m n rn


; rcm
rcm =
m1  m 2  ........  mn
Page # 14
n

m r
i i
=
i 1
n
m
1

rcm =
M
n

m r
i i
i 1
i
i 1
CENTRE OF MASS OF A CONTINUOUS MASS DISTRIBUTION
x cm =
 x dm , y
 dm
cm
=
 y dm , z
 dm
cm
=
 z dm
 dm
 dm = M (mass of the body)

CENTRE OF MASS OF SOME COMMON SYSTEMS
A system of two point masses m1 r1 = m2 r2
The centre of mass lies closer to the heavier mass.

Rectangular plate (By symmetry)
xc =
b
2
yc =
L
2
Page # 15

A triangular plate (By qualitative argument)
at the centroid : yc =
h
3
yc =
2R


A semi-circular ring

A semi-circular disc

A hemispherical shell

A solid hemisphere

A circular cone (solid)
yc =
h
4

A circular cone (hollow)
yc =
h
3
xc = O
4R
3
yc =
yc =
yc =
R
2
xc = O
xc = O
3R
xc = O
8
Page # 16
MOTION OF CENTRE OF MASS AND CONSERVATION OF MOMENTUM:
Velocity of centre of mass of system
dr1
dr
dr
dr
 m2 2  m3 3 .......... ....  mn n
dt
dt
dt
dt
M




m1 v1  m2 v 2  m3 v 3 ..........  mn v n
=
M
m1

v cm =
P System = M v cm
Acceleration of centre of mass of system
dv
dv
dv
dv
m1 1  m 2 2  m 3 3 ..............  mn n

dt
dt
dt
dt
a cm =
M




m1a1  m 2a 2  m3 a 3 ..........  mn an
=
M
=
Net force on system
Net External Force  Net int ernal Force
=
M
M
=
Net External Force
M


Fext = M a cm
IMPULSE
Impulse of a force F action on a body is defined as :-

J=

tf
ti
Fdt


J  ΔP
(impulse - momentum theorem)
Important points :
1.
Gravitational force and spring force are always non-impulsive.
2.
An impulsive force can only be balanced by another impulsive force.
COEFFICIENT OF RESTITUTION (e)
Im pulse of reformation
e = Im pulse of deformation =
 F dt
 F dt
r
d
Velocity of separation along line of impact
= s Velocity of approach along line of impact
Page # 17
(a)
e=1
(b)
e=0
(c)
0<e<1
 Impulse of Reformation = Impulse of Deformation
 Velocity of separation = Velocity of approach
 Kinetic Energy may be conserved
 Elastic collision.
 Impulse of Reformation = 0
 Velocity of separation = 0
 Kinetic Energy is not conserved
 Perfectly Inelastic collision.
 Impulse of Reformation < Impulse of Deformation
 Velocity of separation < Velocity of approach
 Kinetic Energy is not conserved
 Inelastic collision.
VARIABLE MASS SYSTEM :
If a mass is added or ejected from a system, at rate  kg/s and relative

velocity v rel (w.r.t. the system), then the force exerted by this mass

on the system has magnitude  v rel .

Thrust Force ( Ft )

  dm 
Ft  v rel 

 dt 
Rocket propulsion :
If gravity is ignored and initial velocity of the rocket u = 0;
 m0 
.
 m 
v = v r ln 
RIGID BODY DYNAMICS
1.
RIGID BODY :
VAsin1
A
A
1
VA
VAcos1
B
VB
2
B
VBsin2
VBcos2
Page # 18
If the above body is rigid
VA cos 1 = VB cos 2
VBA = relative velocity of point B with respect to point A.
A
B
VBA
Types of Motion of rigid body
Pure Rotational
Motion
Pure Translational
Motion
2.
Combined Translational and
Rotational Motion
MOMENT OF INERTIA (I) :
Definition : Moment of Inertia is defined as the capability of system
to oppose the change produced in the rotational motion of a body.
Moment of Inertia is a scalar positive quantity.
 = mr12 + m 2 r22 +.........................
=  +  +  +.........................
S units of Moment of Inertia is Kgm 2.
Moment of Inertia of :
2.1
A single particle :  = mr2
where m = mass of the particle
r = perpendicular distance of the particle from the axis about
which moment of Inertia is to be calculated
2.2
For many particles (system of particles) :
n
=
 mr
2
i i
i1
2.3
For a continuous object :
=
 dmr
2
where dm = mass of a small element
r = perpendicular distance of the particle from the axis
Page # 19
2.4
For a larger object :
=
 d
element
where d = moment of inertia of a small element
3.
TWO IMPORTANT THEOREMS ON MOMENT OF INERTIA :
3.1
Perpendicular Axis Theorem
[Only applicable to plane lamina (that means for 2-D objects only)].
z = x + y
3.2
(when object is in x-y plane).
Parallel Axis Theorem
(Applicable to any type of object):
 = cm + Md2
List of some useful formula :
Object
Moment of Inertia
2
MR 2 (Uniform)
5
Solid Sphere
2
MR 2 (Uniform)
3
Hollow Sphere
MR2 (Uniform or Non Uniform)
Page # 20
Ring.
MR 2
(Uniform)
2
Disc
MR 2 (Uniform or Non Uniform)
Hollow cylinder
MR 2
(Uniform)
2
Solid cylinder
ML2
(Uniform)
3
ML2
(Uniform)
12
Page # 21
2m  2
(Uniform)
3
AB = CD = EF =
Ma2
(Uniform)
12
Square Plate
Ma2
(Uniform)
6
Square Plate
=
M(a 2  b 2 )
(Uniform)
12
Rectangular Plate
M(a 2  b 2 )
(Uniform)
12
Cuboid
Page # 22
4.
5.
RADIUS OF GYRATION :
 = MK2
TORQUE :

 
  rF
5.5 Relation between '' & '' (for hinged object or pure rotation)


 ext Hinge = Hinge 

Where  ext Hinge = net external torque acting on the body about Hinge
point
Hinge = moment of Inertia of body about Hinge point
F1t
F1c r1
F
x 2c
r2 F2t
F1t = M1a1t = M1r1
F2t = M2a2t = M2r2
resultant = F 1t r1 + F 2t r2 + ........
= M1  r12 + M2  r22 + ............
resultant ) external = 
Rotational Kinetic Energy =
1
.  . 2
2





P  M v CM
Fexternal  M a CM
Net external force acting on the body has two parts tangential and
centripetal.

FC = maC = m
v2
= m2 rCM
rCM

Ft = mat = m rCM
Page # 23
6.
ROTATIONAL EQUILIBRIUM :
For translational equilibrium.
and
Fx  0
............. (i)
Fy  0
............. (ii)
The condition of rotational equilibrium is
z  0
7.

ANGULAR MOMENTUM ( L )
7.1
Angular momentum of a particle about a point.
  
L = r P

L = r  × P

L = rpsin

L = P× r
7.3 Angular momentum of a rigid body rotating about fixed axis :


L H =  H
LH = angular momentum of object about axis H.
IH = Moment of Inertia of rigid object about axis H.
 = angular velocity of the object.
7.4 Conservation of Angular Momentum
Angular momentum of a particle or a system remains constant if
 ext = 0 about that point or axis of rotation.
7.5
Relation between Torque and Angular Momentum

dL

 =
dt
Torque is change in angular momentum
Page # 24
7.6
Impulse of Torque :
 dt  J
J  Change in angular momentum.
For a rigid body, the distance between the particles remain unchanged
during its motion i.e. rP/Q = constant
For velocities
with respect to Q
with respect to ground
P
r
Q
P
r
r

Q


VQ
wr
VQ
VP  VQ 2  r 2  2 VQ r cos 
For acceleration :
, ,  are same about every point of the body (or any other point
outside which is rigidly attached to the body).
Dynamics :

 

 cm   cm  , Fext  Ma cm


Psystem  Mv cm ,
1
1
Mv cm2 +
 cm  2
2
2


Angular momentum axis AB = L about C.M. + L of C.M. about AB


 
L AB   cm   rcm  Mv cm
Total K.E.
=
Page # 25
SIMPLE HARMONIC MOTION
S.H.M.
F = – kx
General equation of S.H.M. is x = A sin (t + ); (t + ) is phase of the
motion and  is initial phase of the motion.
Angular Frequency () :
=
2
= 2f
T
Time period (T) :
T=
2
m
= 2

k
k
m
Speed :
v   A 2  x2
2
a =  x
Acceleration :
Kinetic Energy (KE) :
1
1
1
mv 2 =
m2 (A2 – x2) = k (A2 – x2)
2
2
2
1
Kx2
2
Potential Energy (PE) :
Total Mechanical Energy (TME)
= K.E. + P.E. =
1
1
1
k (A2 – x2) +
Kx2 =
KA2 (which is constant)
2
2
2
SPRING-MASS SYSTEM
k
(1)

smooth surface

T = 2
m
k
m
(2)
T = 2
m1m 2

where  =
known as reduced mass
(m1  m 2 )
K
Page # 26
COMBINATION OF SPRINGS
Series Combination :
Parallel combination :
SIMPLE PENDULUM
T = 2
1/keq = 1/k1 + 1/k2
keq = k1 + k2


=
2
g
geff . (in accelerating Refer-
ence Frame); geff is net acceleration due to pseudo force and gravitational
force.
COMPOUND PENDULUM / PHYSICAL PENDULUM
Time period (T) :

T = 2 mg
where,  = CM + m2 ;  is distance between point of suspension and
centre of mass.
TORSIONAL PENDULUM
Time period (T) :
T = 2

C
where, C = Torsional constant
Superposition of SHM’s along the same direction
x1 = A1 sin t &
x2 = A2 sin (t + )
A2
A
A1
If equation of resultant SHM is taken as x = A sin (t + )
A=
1.
A12  A 22  2A1A 2 cos 
&
tan  =
A 2 sin 
A1  A 2 cos 
Damped Oscillation
 Damping force


F  – bv
 equation of motion is
mdv
= –kx – bv
dt
 b2 - 4mK > 0 over damping
Page # 27
 b2 - 4mK = 0 critical damping
 b2 - 4mK < 0 under damping
 For small damping the solution is of the form.
k  b 
1

x = A 0 e – bt / 2m sin [ t +  ], where '    – 
 m   2m 


2
For small b
 angular frequency '  k / m,  0
– bt
 Amplitude A  A 0 e 2m
l
 Energy E (t) =
1
KA2 e – bt / m
2
 Quality factor or Q value , Q = 2
where , ' 
2.
k b2
.
m 4m 2
, Y 
'
E
= 2
| E |
Y
b
2m
Forced Oscillations And Resonance
External Force F(t) = F0 cos d t
x(t) = A cos (dt + )
F0
A
 2 2
2
 m   d



2
 and
 d2 b2 

(a) Small Damping A 
tan  
v0
d x0
F0

m 2  2d

(b) Driving Frequency Close to Natural Frequency A 
F0
db
Page # 28
STRING WAVES
GENERAL EQUATION OF WAVE MOTION :
2y
t
2
= v2
2y
x 2
x
)
v
where, y (x, t) should be finite everywhere.
y(x,t) = f (t ±


x

f  t   represents wave travelling in – ve x-axis.
 v
x

f  t   represents wave travelling in + ve x-axis.
v

y = A sin (t ± kx + )
TERMS RELATED TO WAVE MOTION ( FOR 1-D PROGRESSIVE
SINE WAVE )
(e) Wave number (or propagation constant) (k) :

k = 2/ =
(rad m –1)
v
(f) Phase of wave : The argument of harmonic function (t ± kx + )
is called phase of the wave.
Phase difference () : difference in phases of two particles at any
time t.
2
2
 =
x
Also. 
t

T
SPEED OF TRANSVERSE WAVE ALONG A STRING/WIRE.
v=
T

where
T  Tension
  mass per unit length
POWER TRANSMITTED ALONG THE STRING BY A SINE WAVE
Average Power P = 22 f 2 A2 v
Intensity
I=
P
s
= 22 f 2 A2 v
REFLECTION AND REFRACTION OF WAVES
yi = Ai sin (t – k1x)
y t  A t sin ( t  k 2 x) 
 if incident from rarer to denser medium (v2 < v1)
y r   A r sin ( t  k 1x) 
Page # 29
y t  A t sin (t – k 2 x)
 if incident from denser to rarer medium. (v 2 > v 1)
y r  A r sin (t  k1x) 
(d)
Amplitude of reflected & transmitted waves.
2k 1
k1  k 2
Ai
A i & At =
Ar =
k1  k 2
k1  k 2
STANDING/STATIONARY WAVES :(b)
y1 = A sin (t – kx + 1)
y2 = A sin (t + kx + 2)

2  1 

  

 sin  t  1 2 
y1 + y2 = 2 A cos  kx 
2 
2 



 2  1 

 represents resultant amplitude at
The quantity 2A cos  kx 
2 

x. At some position resultant amplitude is zero these are called nodes.
At some positions resultant amplitude is 2A, these are called antinodes.

(c) Distance between successive nodes or antinodes = .
2
(d) Distance between successive nodes and antinodes = /4.
(e) All the particles in same segment (portion between two successive
nodes) vibrate in same phase.
(f) The particles in two consecutive segments vibrate in opposite phase.
(g) Since nodes are permanently at rest so energy can not be transmitted across these.
VIBRATIONS OF STRINGS ( STANDING WAVE)
(a) Fixed at both ends :
1. Fixed ends will be nodes. So waves for which
2

3
L=
L=
L=
2
2
2
are possible giving
L=
as
n
2
or  =
v=
T

fn =
2L
where n = 1, 2, 3, ....
n
n
2L
T
, n = no. of loops

Page # 30
(b) String free at one end :
1. for fundamental mode L =

= or  = 4L
4
fundamental mode
First overtone L =
4L
3
Hence  =
3
4
first overtone

so f 1 =
3
4L
T
(First overtone)

Second overtone f 2 =
1

n  
2

so f n = 
2L
5
4L
T

T (2n  1) T


4L

HEAT & THERMODYNAMICS
Total translational K.E. of gas =
< V2 > =
3P

Vrms =
1
3
3
M < V2 > =
PV =
nRT
T
2
2
2
3P
=

3 RT
=
Mmol
3 KT
m
Important Points :
– Vrms 
T
V
8 KT
 m = 1.59
Most probable speed Vp =
KT
m
2 KT
= 1.41
m
Vrms = 1.73
KT
m
KT
 Vrms > V > Vmp
m
Degree of freedom :
Mono atomic f = 3
Diatomic f = 5
polyatomic f = 6
Page # 31
Maxwell’s law of equipartition of energy :
Total K.E. of the molecule = 1/2 f KT
For an ideal gas :
f
nRT
2
Internal energy U =
Workdone in isothermal process :
W = [2.303 nRT log10
Internal energy in isothermal process :
Work done in isochoric process :
Change in int. energy in isochoric process :
Vf
]
Vi
U = 0
dW = 0
f
R T = heat given
2
Isobaric process :
Work done W = nR(Tf – Ti)
change in int. energy U = nCv T
heat given Q = U + W
U = n
f

f
R
Cp =   1 R
2
2 
Molar heat capacity of ideal gas in terms of R :
Specific heat :
Cv =
(i) for monoatomic gas :
Cv
(iii) for triatomic gas :
Cp

= 1 

Mayer’s eq.  Cp – Cv = R
=
Cv
Cp
(ii) for diatomic gas :
In general :
Cp
Cv
Cp
Cv
= 1.67
= 1.4
= 1.33
2
f 
for ideal gas only
Adiabatic process :
Work done W =
nR (Ti  Tf )
 1
Page # 32
In cyclic process :
Q = W
In a mixture of non-reacting gases :
Mol. wt. =
n1M1  n2M2
n1  n 2
Cv =
n1C v 1  n 2C v 2
n1  n 2
n1Cp  n2 Cp  .....
Cp (mix )
1
2
= C
= n C  n C  ....
v ( mix )
1 v1
2 v2
Heat Engines
Efficiency ,  
work done by the engine
heat sup plied to it
Q
W QH – Q L
 1– L
= Q  Q
Q
H
H
H
Second law of Thermodynamics
 Kelvin- Planck Statement
It is impossible to construct an engine, operating in a cycle, which will
produce no effect other than extracting heat from a reservoir and performing an equivalent amount of work.
Page # 33
 Rudlope Classius Statement
It is impossible to make heat f low f rom a body at a lower
temperature to a body at a higher temperature without doing external work
on the working substance
Entropy
f
Q
 change in entropy of the system is S =
 S f – Si 
T

i
Q
T
 In an adiabatic reversible process, entropy of the system remains constant.
Efficiency of Carnot Engine
(1) Operation I (Isothermal Expansion)
(2) Operation II (Adiabatic Expansion)
(3) Operation III (Isothermal Compression)
(4) Operation IV (Adiabatic Compression)
Thermal Efficiency of a Carnot engine
V2 V3
Q 2 T2
T


  1– 2
V1 V4  Q1 T1 
T1
Page # 34
Refrigerator (Heat Pump)
Refrigerator
Hot (T2)
Hot (T1)
Q2
Q1
W
 Coefficient of performance,  
Q2
1
1
= 
=
T
T
W
1
1
–1
–1
T2
T2
Calorimetry and thermal expansion
Types of thermometers :
(a) Liquid Thermometer :
   0 
 × 100
T = 
 100   0 
(b) Gas Thermometer :
Constant volume :
 P  P0 
T =  P  P  × 100
0
 100
; P = P0 + g h
 V 
T= 
 T
 V  V  0
(c) Electrical Resistance Thermometer :
Constant Pressure :
 R t  R0 
T =  R  R  × 100
0
 100
Thermal Expansion :
(a) Linear :
L
 = L T
0
or
L = L0 (1 + T)
Page # 35
(b) Area/superficial :
A
 = A T
0
or
A = A0 (1 + T)
or
V = V0 (1 +  T)
(c) volume/ cubical :
r=
V
V0 T
 

2 3
Thermal stress of a material :

F

Y
A

Energy stored per unit volume :
1
1 AY
E
( L )2
K(L)2
or
2
2 L
Variation of time period of pendulum clocks :
E=
T =
1
T
2
T’ < T - clock-fast : time-gain
T’ > T - clock slow : time-loss
CALORIMETRY :
Specific heat S =
Q
m.T
Q
n.T
Water equivalent = mWSW
Molar specific heat C =
HEAT TRANSFER
Thermal Conduction :
dQ
dT
= – KA
dt
dx
Thermal Resistance :
R=

KA
Page # 36
Series and parallel combination of rod :
 eq
=
1  2

 ....... (when A = A = A = .........)
1
2
3
K1 K 2
(i) Series :
K eq
(ii) Parallel :
Keq Aeq = K1 A1 + K2 A2 + ...... (when  1 = 2 = 3 = .........)
for absorption, reflection and transmission
r+t+a=1
U
Emissive power :
E=
A t
dE
Spectral emissive power :
E =
d
E of a body at T temp.
Emissivity :
e=
E of a black body at T temp.
E (body )
= E (black body)
a (body )
Wein’s Displacement law :
m . T = b.
b = 0.282 cm-k
Stefan Boltzmann law :
u =  T4
s = 5.67 × 10–8 W/m2 k4
u = u – u0 = e A (T4 – T04)
Kirchoff’s law :
Newton’s law of cooling :
d
= k ( – 0) ;  = 0 + (i – 0) e–k t
dt
ELECTROSTATICS
Coulomb force between two point charges
1 q1q2
1 q1q2 

 r̂
 3 r =
4 0r | r |2
4 0  r | r |
The electric field intensity at any point is the force experienced


F
by unit positive charge, given by E 
q0

F


Electric force on a charge 'q' at the position of electric field



intensity E produced by some source charges is F  qE

Electric Potential
Page # 37
If (W  P)ext is the work required in moving a point charge q from infinity
to a point P, the electric potential of the point P is
Vp 


( Wp )ext 

q
 acc 0
Potential Difference between two points A and B is
VA – VB

Formulae of E and potential V
Kq
Kq
Kq 
(i)
Point charge E=  2  r̂ = 3 r , V =
r
|r|
r
(ii)
(iii)

Infinitely long line charge 2 r r̂ = 2Kr̂
0
r
V = not defined, v B – v A = –2K ln (rB / rA)

Infinite nonconducting thin sheet 2 n̂ ,
0
V = not defined, v B  v A  
(iv)

rB  rA 
2 0
Uniformly charged ring
KQx
Eaxis =
R
2
 x2

3/2
,
Ecentre = 0
KQ
KQ
,
Vcentre =
R
R2  x2
x is the distance from centre along axis.
Vaxis =
(v)
Infinitely large charged conducting sheet

n̂
0

rB  rA 
0
Uniformly charged hollow conducting/ nonconducting /solid
conducting sphere
V = not defined, v B  v A  
(vi)

kQ
(a)
for E  
(b)
KQ

E  0 for r < R, V = R
|r|
2
r̂ , r  R, V =
KQ
r
Page # 38
(vii) Uniformly charged solid nonconducting sphere (insulating material)
(a)

kQ
KQ
E   2 r̂ for r
 R,V=
|r|
r
(b)

 KQ r
 r for r
E 3 
 R,
3 0
R
(viii)
V=

(3R2–r2)
6 0
thin uniformly charged disc (surface charge density is )

Eaxis = 2
0

x
1 
2

R  x2



Vaxis =
  2
R  x 2  x

2 0 

Work done by external agent in taking a charge q from A to B is
(W ext)AB= q (VB – VA) or (W el ) AB = q (VA – VB) .

The electrostatic potential energy of a point charge
U = qV

U = PE of the system =
U1  U2  ...
= (U12 + U13 + ..... + U1n ) + (U23 + U24 + ...... + U2n )
2
+ (U34 + U35 + ..... + U3n ) ....

Energy Density =

Self Energy of a uniformly charged shell = Uself 

1 2
E
2
KQ 2
2R
Self Energy of a uniformly charged solid non-conducting sphere
3KQ 2
5R
Electric Field Intensity Due to Dipole

2KP

(i) on the axis E = 3
r


KP
(ii) on the equatorial position : E = – 3
r
= Uself 

(iii) Total electric field at general point O (r,) is Eres =
KP
r
3
1  3 cos 2 
Page # 39



Potential Energy of an Electric Dipole in External Electric Field:
 
U = - p. E
Electric Dipole in Uniform Electric Field :

  
torque   p x E ; F = 0
Electric Dipole in Nonuniform Electric Field:
 
  
E
torque   p x E ; U =  p  E , Net force |F| = p
r


Electric Potential Due to Dipole at General Point (r, ) :
 
P cos
p.r

V=
4 0 r 2
4 0 r 3
The electric flux over the whole area is given by

 = E.dS = En dS

E


S
S
Flux using Gauss's law, Flux through a closed surface

qin
E = E  dS =
.
0


Electric field intensity near the conducting surface
=


n̂
0
Electric pressure : Electric pressure at the surface of a conductor is
given by formula
P=

2
where  is the local surface charge density..
2 0
Potential difference between points A and B
B
VB – VA = –


 E .d r
A


 

 
 
 
E =   î x V  ĵ x V  k̂ z V  = –   î x  ĵ x  k̂ z  V




= – V = –grad V
Page # 40
CURRENT ELECTRICITY
1.
ELECTRIC CURRENT
Iav =
q
and instantaneous current
t
i =. Lim
t  0
2.
q dq

t
dt
ELECTRIC CURRENT IN A CONDUCTOR
I = nAeV.

vd  ,

3.
4.
1  eE  2


2 m 
1 eE
vd 
=
,

2 m
I = neAVd
CURRENT DENSITY
 dI 
J
n
ds
ELECTRICAL RESISTANCE
 eE 
I = neAVd = neA 
 =
 2m 
 ne 2  


 2m  AE


 ne 2    A 
A
so
I =  2m     V =   V = V/R  V = IR
  

 
 is called resistivity (it is also called specific resistance) and
V
E=

2m
1
,  is called conductivity. Therefore current in conductors
ne  
is proportional to potential difference applied across its ends. This is
Ohm's Law.
Units:
=
2
=
R  ohm(),   ohm  meter(  m)
also called siemens,    1m 1 .
Page # 41
Dependence of Resistance on Temperature :
R = Ro (1 + ).
Electric current in resistance
V2  V1
R
ELECTRICAL POWER
P = V
I=
5.
Energy =
 pdt
P = I2R = V =
V2
.
R
H = Vt = 2 Rt =
V2
t
R
H = 2 RT Joule =
 2RT
Calorie
4.2
9.
KIRCHHOFF'S LAWS
9.1
Kirchhoff’s Current Law (Junction law)
 in = out
9.2
Kirchhoff’s Voltage Law (Loop law)
IR + EMF =0”.
10.
COMBINATION OF RESISTANCES :
Resistances in Series:
R = R1 + R2 + R3 +................ + Rn (this means Req is greater then any
resistor) ) and
V = V1 + V2 + V3 +................ + Vn .
V1 =
2.
R2
R1
V ;
V ; V2 =
R1  R 2  .........  R n
R1  R 2  .........  R n
Resistances in Parallel :
Page # 42
11.
WHEATSTONE NETWORK : (4 TERMINAL NETWORK)
When current through the galvanometer is zero (null point or balance
point)
13.
P R
= , then PS = QR
Q S
GROUPING OF CELLS
13.1
Cells in Series :

Equivalent EMFEeq = E1  E 2  .......  En [write EMF's with polarity]
Equivalent internal resistance
13.2
req = r1  r2  r3  r4  ....  rn
Cells in Parallel:
1
E eq
  2  ....   n
r1
r2
rn

[Use emf with polarity]
1  1  .....  1
r1
r2
rn
1
1 1
1
 
 .... 
req
r1 r2
rn
15.
AMMETER
A shunt (small resistance) is connected in parallel with galvanometer
to convert it into ammeter. An ideal ammeter has zero resistance
Page # 43
Ammeter is represented as follows -
If maximum value of current to be measured by ammeter is  then
I G . RG = (I – I G )S
 G .R G
  RG
S= G
when  >> G.
  G

where  = Maximum current that can be measured using the given
ammeter.
S=
16.
VOLTMETER
A high resistance is put in series with galvanometer. It is used to
measure potential difference across a resistor in a circuit.
For maximum potential difference
V = G . RS + G RG
17.
V
RS =  – RG
G
POTENTIOMETER
=
If
RG << RS  RS 
V
G

r R
VA – VB =

.R
Rr
Potential gradient (x)  Potential difference per unit length of wire
x=
VA  VB

R
=
.
L
R r L
Page # 44
Application of potentiometer
(a) To find emf of unknown cell and compare emf of two cells.
In case ,
In figure (1) is joint to (2) then balance length = 1
1 = x1
....(1)
in case ,
In figure (3) is joint to (2) then balance length = 2
2 = x2
....(2)
1  1

2  2
If any one of 1 or 2 is known the other can be found. If x is known then
both 1 and 2 can be found
(b)
To find current if resistance is known
VA – V C = x 1
IR1 = x1
x 1
= R
1
Similarly, we can find the value of R2 also.
Potentiometer is ideal voltmeter because it does not draw any current
from circuit, at the balance point.
(c)
To find the internal resistance of cell.
I st arrangement
2nd arrangement
Page # 45
by first arrangement
’ = x1
by second arrangement IR = x2
 =

x 2
,
R
also  =
'
x 2
=
r 'R
R
...(1)
'
r 'R

x 1
x 2
=
r 'R
R
 1   2 
r’ =    R
2 

(d)Ammeter and voltmeter can be graduated by potentiometer.
(e)Ammeter and voltmeter can be calibrated by potentiometer.
18.
METRE BRIDGE (USE TO MEASURE UNKNOWN RESISTANCE)
If AB =  cm, then BC = (100 – ) cm.
Resistance of the wire between A and B , R  
[  Specific resistance  and cross-sectional area A are same for whole
of the wire ]
or
R = 
...(1)
where  is resistance per cm of wire.
If P is the resistance of wire between A and B then
P 
P = ()
Similarly, if Q is resistance of the wire between B and C, then
Q  100 – 

Q = (100 – )
....(2)
Dividing (1) by (2),
P

=
Q
100  
Page # 46
Applying the condition for balanced Wheatstone bridge, we get R Q = P X
Q
100  
or
X=
R
P

Since R and  are known, therefore, the value of X can be calculated.

x=R
CAPACITANCE
1.
(i)
q V

q = CV
q : Charge on positive plate of the capacitor
C : Capacitance of capacitor.
V : Potential difference between positive and negative plates.
(ii)
Representation of capacitor :
(iii)
Energy stored in the capacitor : U =
(iv)
Energy density =
(
1
QV
Q2
CV2 =
=
2
2C
2
1
1
r E2 = K E2
2
2
r = Relative permittivity of the medium.
K= r : Dielectric Constant
For vacuum, energy density =
(v)
(a)
,
1
 E2
2 
Types of Capacitors :
Parallel plate capacitor
 0r A
0 A
= K
d
d
A : Area of plates
d : distance between the plates( << size of plate )
Spherical Capacitor :
Capacitance of an isolated spherical Conductor (hollow or solid )
C= 4 r R
R = Radius of the spherical conductor
Capacitance of spherical capacitor
C=
(b)


b

4 0K 2ab
C=
(b  a)
1 2
a
ab
C= 4
(b  a )
b
K1 K2 K3
a
Page # 47
(c)
Cylindrical Capacitor :  >> {a,b}
Capacitance per unit length =
(vi)
(vii)
2 
F/m
n(b / a) 

b
Capacitance of capacitor depends on
(a)
Area of plates
(b)
Distance between the plates
(c)
Dielectric medium between the plates.
Electric field intensity between the plates of capacitor

V
E= 
0
d
Surface change density
(viii)
2.
Force experienced by any plate of capacitor :
F=
q2
2A 0
DISTRIBUTION OF CHARGES ON CONNECTING TWO CHARGED
CAPACITORS:
When two capacitors are C1 and C2 are connected as shown in figure
(a)
Common potential :

V=
(b)
Q 1' = C1V =
Total ch arg e
C1V1  C 2 V2
= Total capaci tan ce
C1  C 2
C1
(Q 1 + Q 2)
C1  C 2
C2
Q 2' = C2 V = C  C (Q 1 +Q 2)
1
2
Page # 48
(c)
Heat loss during redistribution :
1 C1C 2
(V1 – V2)2
2 C1  C 2
The loss of energy is in the form of Joule heating in the wire.
H = Ui – Uf =
3.
Combination of capacitor :
(i)
Series Combination
1
1
1
1



Ceq C1 C2 C3
+Q
C1
–Q +Q
C2
V1
(ii)
V1 : V2 : V3 
–Q +Q
C3
1 1 1
:
:
C1 C2 C3
–Q
V3
V2
Parallel Combination :
Q+ –Q
C1
Q+ –Q
C2
Q+ –Q
C3
V
Ceq = C1 + C2 + C3
4.
Q1: Q2 :Q3 = C1 : C2 : C3
Charging and Discharging of a capacitor :
(i)
Charging of Capacitor ( Capacitor initially uncharged ):
q = q0 ( 1 – e– t /)
R
V
C
q0 = Charge on the capacitor at steady state
q0 = CV
Page # 49
Time constant = CReq.
I=
(ii)
q0

e – t / 
V –t/
e
R
Discharging of Capacitor :
q = q0 e – t / 
q0 = Initial charge on the capacitor
I=
q0

e –t/
q
R
q0
C
0.37v0

5.
t
Capacitor with dielectric :
(i)
Capacitance in the presence of dielectric :
C=
+
V
K0 A
= KC0
d
+
– – – – – – – – – – – – b
 0
 b0
+ + + + + + + + + + + + + + + b
–
–
C0 = Capacitance in the absence of dielectric.
Page # 50
b


V
Ein = E – Eind =  –  = K =
d
0
0
0
(ii)
E:

Electric field in the absence of dielectric
0
Eind : Induced (bound) charge density.
(iii)
1
).
K
Force on dielectric
(i)
F
When battery is connected
0b(K  1)V 2
2d

b
b
+
F
x
(ii)

6.
b = (1 –
d
–
When battery is not connected
F=
Q2
2C2
dC
dx
* Force on the dielectric will be zero when the dielectric is fully inside.
Page # 51
ALTERNATING CURRENT
1.
AC AND DC CURRENT :
A current that changes its direction periodically is called alternating current (AC). If a current maintains its direction constant it is called direct
current (DC).
3.
ROOT MEAN SQUARE VALUE:
Root Mean Square Value of a function, from t1 to t2, is defined as
t2
f
dt
t1
f rms =
4.
2
.
t 2  t1
POWER CONSUMED OR SUPPLIED IN AN AC CIRCUIT:
2

 Pdt
Average power consumed in a cycle =
Vm
o
2

=
1
V  cos 
2 m m
m
. cos = Vrms rms cos.
2
2
Here cos is called power factor.
=
.
Page # 52
5.
SOME DEFINITIONS:
The factor cos  is called Power factor.
m sin  is called wattless current.
Vrms
Vm
=
m
rms
L is called inductive reactance and is denoted by XL.
Impedance Z is defined as Z =
6.
1
is called capacitive reactance and is denoted by XC.
C
PURELY RESISTIVE CIRCUIT:
I=
v s Vm sin t
=
= m sin t
R
R
m =
Vm
R
rms =
Vrms
R
2
<P> = Vrmsrmscos 
7.
Vrms
R
PURELY CAPACITIVE CIRCUIT:
I= =
=
Vm
cos t
1
C
Vm
cos t = m cos t.
XC
XC =
1
and is called capacitive reactance.
C
v
V
T
t
i
I
t
Page # 53
IC leads by v C by /2 Diagrammatically
(phasor diagram) it is represented as
m
.
Vm
Since º, <P> = Vrms rmscos 
MAGNETIC EFFECT OF CURRENT & MAGNETIC FORCE ON
CHARGE/CURRENT
1.
Magnetic field due to a moving point charge
  q( v  r )
B 0 
4
r3
v
2.
r
Biot-savart's Law

0I  d  r 
dB 

4  r 3 


3.

Magnetic field due to a straight wire
B=
1
r 2
P
0 I
(sin 1 + sin 2)
4 r

4.
Magnetic field due to infinite straight wire
r
0 I
2 r
Magnetic field due to circular loop
B=
5.
(i)
At centre
(ii)
At Axis

B=
0 
NR2
P
 0NI
2r


B = 2  2 2 3/2 
 (R  x )

Page # 54
6.
Magnetic field on the axis of the solenoid
2
7.
1
B=
0 nI
(cos 1 – cos 2)
2
Ampere's Law
 
 B.d   0I
8.
Magnetic field due to long cylinderical shell
B = 0, r < R
0 I
,r  R
2 r
=
9.
Magnetic force acting on a moving point charge

 
a.
F  q (  B)
×
×
×
×
 
(i)
B
×
×
m
r
qB
×
B×
× r
×
×
×
×
×
×
×
2m
T=
qB

(ii)

r
B
2m
qB




F  q (  B)  E
T=

b.
10.
12.
Pitch =
2m cos 
qB

Magnetic force acting on a current carrying wire

 
F  I  B
Magnetic Moment of a current carrying loop
M=N·I·A
Torque acting on a loop
  
  MB

11.
m sin 
qB

Page # 55
13.
Magnetic field due to a single pole
0 m
·
4 r 2
Magnetic field on the axis of magnet
B=
14.
 0 2M
·
4 r 3
Magnetic field on the equatorial axis of the magnet
B=
15.
0 M
·
4 r 3
Magnetic field at point P due to magnet
B=
16.
B=
0 M
4 r 3
1  3 cos 2 
P
r

S
N
ELECTROMAGNETIC INDUCTION
=

 
B.ds
1.
Magnetic flux is mathematically defined as
2.
Faraday’s laws of electromagnetic induction
3.
d
dt
Lenz’s Law (conservation of energy principle)
According to this law, emf will be induced in such a way that it will oppose
the cause which has produced it.
Motional emf
E=–
4.
Induced emf due to rotation
Emf induced in a conducting rod of length l rotating with angular speed 
about its one end, in a uniform perpendicular magnetic field B is 1/2 B 
2.
Page # 56
1.
EMF Induced in a rotating disc :
Emf between the centre and the edge of disc of radius r rotating in a
Br 2
2
Fixed loop in a varying magnetic field
magnetic field B =
5.
If magnetic field changes with the rate
dB
, electric field is generated
dt
r dB
2 dt
This electric field is non conservative in nature. The lines of force associated with this electric field are closed curves.
whose average tangential value along a circle is given by E=
6.
Self induction
 =   (Nt )    (LIt )   Lt I .
The instantaneous emf is given as
 =  d (dtN)   d(dtLI )   LdI
dt
Self inductance of solenoid = µ0 n2 r2.
6.1
Inductor
It is represent by
electrical equivalence of loop

VA  L
dI
 VB
dt
1 2
L
2
Growth Of Current in Series R–L Circuit
If a circuit consists of a cell, an inductor L and a resistor R and a switch S
,connected in series and the switch is closed at t = 0, the current in the
Energy stored in an inductor =
7.
circuit I will increase as I =  (1  e
R
Rt
L
)
Page # 57
The quantity L/R is called time constant of the circuit and is denoted by .
The variation of current with time is as shown.
1. Final current in the circuit =
8

, which is independent of L.
R
2. After one time constant , current in the circuit =63% of the final current.
3. More time constant in the circuit implies slower rate of change of current.
Decay of current in the circuit containing resistor and inductor:
Let the initial current in a circuit containing inductor and resistor be 0.
Current at a time t is given as I = 0
Rt
e L
1
Current after one time constant : I = 0 e =0.37% of initial current.
9.
10.
Mutual inductance is induction of EMF in a coil (secondary) due to
change in current in another coil (primary). If current in primary coil is I,
total flux in secondary is proportional to I, i.e. N  (in secondary)  I.
or
N  (in secondary) = M I.
The emf generated around the secondary due to the current flowing around
the primary is directly proportional to the rate at which that current changes.
Equivalent self inductance :
L
1.
2.
VA  VB
dI / dt
..(1)
Series combination :
L = L1 + L2
( neglecting mutual inductance)
L = L1 + L2 + 2M (if coils are mutually coupled and they have
winding in same direction)
L = L1 + L2 – 2M (if coils are mutually coupled and they have
winding in opposite direction)
Parallel Combination :
1 1
1


L L1 L 2
( neglecting mutual inductance)
Page # 58
For two coils which are mutually coupled it has been found that M  L1L 2
or M =k L1L 2 where k is called coupling constant and its value is less
than or equal to 1.
E s Ns p


where denotaEp Np  s ,
tions have their usual meanings.
NS > NP
ES > EP 

for step up transformer.
12.
LC Oscillations
2 
Magnetic
Core
S
EP
Primary
coil
ES
Secondary
coil
1
LC
GEOMETRICAL OPTICS
1.
Reflection of Light
(b)
  i = r
1.3
Characteristics of image due to Reflection by a Plane
Mirror:
(a) Distance of object from mirror = Distance of image from the mirror.
(b) The line joining a point object and its image is normal to the reflecting
surface.
(c) The size of the image is the same as that of the object.
(d) For a real object the image is virtual and for a virtual object the image
is real
2.
Relation between velocity of object and image :
From mirror property : xim = - xom , yim = yom and zim = zom
Here xim means ‘x’ coordinate of image with respect to mirror.
Similarly others have meaning.
Page # 59
Differentiating w.r.t time , we get
v (im)x = -v(om)x ;
v (im)y = v (om)y ;
3.
v (im)z = v (om)z ,
Spherical Mirror
1 1 2 1
+ = =
v u R f
(b)
.....
Mirror formula
x co–ordinate of centre of Curvature and focus of Concave
mirror are negative and those for Convex mirror are positive.
In case of mirrors since light rays reflect back in - X direction,
therefore -ve sign of v indicates
real image and +ve
sign of v indicates virtual image
Lateral magnification (or transverse magnification)
m=
h2
h1
m =
v
.
u
dv
v2 .
=
du
u2
(d)
On differentiating (a) we get
(e)
On dif f erentiating (a) with respect to time we get
dv
dv
v 2 du
 2
,where
is the velocity of image along Principal
dt
dt
u dt
axis and
(f)
du
is the velocity of object along Principal axis. Negative
dt
sign implies that the image , in case of mirror, always moves
in the direction opposite to that of object.This discussion is
for velocity with respect to mirror and along the x axis.
Newton's Formula:
XY = f 2
X and Y are the distances ( along the principal axis ) of the object
and image respectively from the principal focus. This formula can
be used when the distances are mentioned or asked from the
focus.
1
f
(g)
Optical power of a mirror (in Diopters) =
(h)
f = focal length with sign and in meters.
If object lying along the principal axis is not of very small size, the
v 2  v1
longitudinal magnification = u  u
2
1
(it will always be inverted)
Page # 60
4.
Refraction of Light
speed of light in vacuum c
 .
speed of light in medium v
vacuum.  
4.1
Laws of Refraction (at any Refracting Surface)
Sin i
(b)
= Constant for any pair of media and for light of a given
Sin r
wave length. This is known as Snell's Law. More precisely,
Sin i
n
v

= 2 = 1 = 1
Sin r
n1
v2
2
4.2
Deviation of a Ray Due to Refraction
Deviation () of ray incident at i and refracted at r is given by  = |i  r|.
5.
7.
Principle of Reversibility of Light Rays
A ray travelling along the path of the reflected ray is reflected along the
path of the incident ray. A refracted ray reversed to travel back along its
path will get refracted along the path of the incident ray. Thus the incident
and refracted rays are mutually reversible.
Apparent Depth and shift of Submerged Object
At near normal incidence (small angle of incidence i) apparent depth (d)
is given by:
d=
d
n relative
nrelative =

n i (R .I.of medium of incidence )
n r (R.I.of medium of refraction )

1 
Apparent shift = d 1  n 
rel 

Refraction through a Composite Slab (or Refraction through a
number of parallel media, as seen from a medium of R. I. n0)
Apparent depth (distance of final image from final surface)
=
t1
n1 rel
+
t2
n2 rel
+
t3
n3 rel
+......... +
tn
nn
rel
Page # 61

1

n 1rel
Apparent shift = t1  1
8.


1
 + t2  1
 n 2 rel



n 
 +........+ 1  n 

n rel 


Critical Angle and Total Internal Reflection ( T. I. R.)
C = sin 1
nr
nd
(i)
(a)
(b)
9.
Conditions of T. I. R.
light is incident on the interface from denser medium.
Angle of incidence should be greater than the critical
angle (i > c).
Refraction Through Prism
9.1
Characteristics of a prism
9.2
 = (i + e)  (r1 + r2) and r1 + r2 = A
 = i + e  A.
Variation of  versus i
Page # 62
(1)
(2)
There is one and only one angle of incidence for which the angle
of deviation is minimum.
When  = min , the angle of minimum deviation, then i = e and
r1 = r2, the ray passes symmetrically w.r.t. the refracting surfaces.
We can show by simple calculation that min = 2imin – A
where imin = angle of incidence for minimum deviation and r = A/2.
 nrel =
(3)

A  m
2
sin A2
sin
 
 , where n
=
rel
nsurroundings
Alsomin = (n  1) A (for small values of  A)
For a thin prism ( A 10o) and for small value of i, all values of
 = ( nrel  1 ) A
10.
n prism
where nrel =
nprism
nsurrounding
Dispersion Of Light
The angular splitting of a ray of white light into a number of components
and spreading in different directions is called Dispersion of Light. This
phenomenon is because waves of different wavelength move with same
speed in vacuum but with different speeds in a medium.
The refractive index of a medium depends slightly on wavelength also.
This variation of refractive index with wavelength is given by Cauchy’s
formula.
Cauchy's formula n () = a 
b
2
where a and b are positive constants
of a medium.
Angle between the rays of the extreme colours in the refracted (dispersed) light is
called angle of dispersion.
For prism of small ‘A’ and with small ‘i’ :
 = (nv – nr)A
Deviation of beam(also called mean deviation)
 = y = (ny – 1)A
Dispersive power () of the medium of the material of prism is given by:
n v  nr
 = n 1
y
For small angled prism ( A 10o ) with light incident at small angle i :
n v  nr

 v  r
=
=
ny  1
y

y
=
angular dispersion
deviation of mean ray (yellow)
Page # 63
[ ny =
=
n v  nr
if ny is not given in the problem ]
2
nv  nr
n v  nr
 v  r
= n  1 [take ny =
if value of ny is not given in
y
y
2
the problem]
nv, nr and ny are R. I. of material for violet, red and yellow colours respectively.
11.
Combination of Two Prisms
Two or more prisms can be combined in various ways to get different
combination of angular dispersion and deviation.
(a)
Direct Vision Combination (dispersion without deviation)
The condition for direct vision combination is :
 n v nr 
 n v  n r 
 1 A  
 1 A  n y 1 A = ny 1 A

 2

 2

(b)
12.
Achromatic Combination (deviation without dispersion.)
Condition for achromatic combination is: (nv  nr) A = (nv  nr) A
Refraction at Spherical Surfaces
For paraxial rays incident on a spherical surface separating two media:
n
n2
n n1
 1 = 2
u
v
R
where light moves from the medium of refractive index n1 to the medium
of refractive index n2.
Transverse magnification (m) (of dimension perpendicular to principal axis)
due to refraction at spherical surface is given by m =
13.
v  R  v / n 2 
=

u  R  u / n1 
Refraction at Spherical Thin Lens
A thin lens is called convex if it is thicker at the middle and it is
called concave if it is thicker at the ends.
For a spherical, thin lens having the same medium on both sides:
1 1

= (nrel  1)
v
u
 1
1



 R1 R2 
where nrel =
nlens
nmedium
Page # 64
1
= (nrel  1)
f
 1
1



 R1 R2 
1 1
1

=
f
v
u
m=
 Lens Maker's Formula
v
u
Combination Of Lenses:
1 1 1 1
   ...
F f1 f2 f3
OPTICAL INSTRUMENT
SIMPLE MICROSCOPE
D
 Magnifying power : U
0
 when image is formed at infinity M 
D
f
 When change is formed at near print D. MD  1 
COMPOUND MICROSCOPE
Magnifying power
M
V0D 0
U0U e
M 
V0D
U0 f e
MD 
V0
U0

D
1  


f
e 

D
f
Length of Microscope
L = V0 + Ue
L = V0 + f e
D.f e
LD = V0  D  f
e
Page # 65
Astronomical Telescope
Magnifying power
Length of Microscope
f0
M= 
e
L = f + ue.
M 
f0
fe
L = f0 + fe
MD 
f0
fe
Df e
LD= f 0 + D  f
e
f 

1  e 
D

Terrestrial Telescope
Magnifying power
M
Length of Microscope
f0
Ue
M 
f0
fe
MD 
f0
fe
L= f 0 + 4f + Ue.
L = f 0 + 4f + fe.
Df e
LD = f 0 + 4f + D  f
e
 fe 
1  
D

Galilean Telescope
Magnifying power
M
Length of Microscope
f0
Ue
M 
f0
fe
MD 
f0
fe
L = f 0 - Ue.
L = f 0 - fe.
fe D
LD = f 0 – D – f
f 

1 – e 
d

e
Resolving Power
Microscope R 
Telescope. R 
1
2 sin 

d

1
a

 1.22
Page # 66
MODERN PHYSICS

Work function is minimum for cesium (1.9 eV)

work function W = h0 =

Photoelectric current is directly proportional to intensity of incident radiation.
( – constant)

Photoelectrons ejected from metal have kinetic energies ranging from 0 to
KEmax
Here
KEmax = eVs
Vs - stopping potential
Stopping potential is independent of intensity of light used (-constant)
Intensity in the terms of electric field is


I=


hc
0
1
 E2.c
2 0
h
.

Einstein equation for photoelectric effect is
Momentum of one photon is
h = w0 + kmax 


Energy E =
hc
hc
=  + eVs

0
12400
eV
( A 0 )
Force due to radiation (Photon) (no transmission)
When light is incident perpendicularly
(a)
a=1 r=0
A

, Pressure =
c
c
r = 1, a = 0
F=
(b)
2A
2
,
P=
c
c
when 0 < r < 1 and a + r = 1
F=
(c)
F=
A

(1 + r), P =
(1 + r)
c
c
Page # 67
When light is incident at an angle  with vertical.
(a)
a = 1, r = 0
A cos 
,
c
r = 1, a = 0
F=
(b)
P=

F cos 
=
cos2 
c
A
2A cos 2 
2 cos2 
,
P=
c
c
0 < r < 1,
a+r=1
F=
(c)

 cos 2 
(1 + r)
c
De Broglie wavelength

h
h
h
=
=
2mKE
mv P
Radius and speed of electron in hydrogen like atoms.
P=
=
rn =
n2
a0
Z
a0 = 0.529 Å
Z
v
n 0
Energy in nth orbit
v 0 = 2.19 x 106 m/s
vn =

En = E1 .




Z2
E1 = – 13.6 eV
n2
Wavelength corresponding to spectral lines
 1
1
1
=R  2  2

 n1 n2 
for Lyman series
n1 = 1
n2 = 2, 3, 4...........
Balmer
n1 = 2
n2 = 3, 4, 5...........
Paschen
n1 = 3
n2 = 4, 5, 6...........
The lyman series is an ultraviolet and Paschen, Brackett and Pfund series
are in the infrared region.
n(n  1)
Total number of possible transitions, is
, (from nth state)
2
If effect of nucleus motion is considered,
n2 m
rn = (0.529 Å)
.

Z
En = (–13.6 eV)
Z2
n
2
.

m
Page # 68





Here µ - reduced mass
Mm
µ=
, M - mass of nucleus
(M  m)
Minimum wavelength for x-rays
hc
12400
min = eV = V (volt) Å
0
0
Moseley’s Law
v = a(z – b)
a and b are positive constants for one type of x-rays (independent of Z)
Average radius of nucleus may be written as
R = R0A1/3,
R0 = 1.1 x 10–15 M
A - mass number
Binding energy of nucleus of mass M, is given by B = (ZMp + NMN – M)C2
Alpha - decay process
A
Z
X  Az 24 Y  24 He
Q-value is
  


Q = m AZ X  m
Beta- minus decay
A
Z
A4
z 2 Y
  m  He C
4
2
2
A
X  z 1 Y      
A

Q- value = [ m ( zA X)  m ( Z1 Y )] c 2
Beta plus-decay
A
z X

A
Z 1 Y
+ + + 
A

Q- value = [ m ( zA X)  m ( Z1 Y )  2 me] c 2
Electron capture : when atomic electron is captured, X-rays are emitted.
A
z X
+ e 
A
Z 1 Y
+
A

Q - value = [ m ( zA X)  m ( Z1 Y )] c 2
In radioactive decay, number of nuclei at instant t is given by N = N0 e–t ,
-decay constant.

Activity of sample :

Activity per unit mass is called specific activity.
0.693
Half life : T1/2 =

T1/ 2
Average life : Tav =
0.693


A = A0 e–t
Page # 69

A radioactive nucleus can decay by two different processes having half
lives t1 and t2 respectively. Effective half-life of nucleus is given by
1 1 1
  .
t t1 t 2
WAVE OPTICS
Interference of waves of intensity 1 and 2 :
resultant intensity,  = 1 + 2 + 2 1 2 cos () where,  = phase
difference.

=  
For Constructive Interference :
max =
1
For Destructive interference :
min
1
If sources are incoherent
YDSE :
Path difference, p = S2P – S1P = d sin 
if
d<<D
if
y << D
for maxima,
p = n

for minima
p =

=


2
2
 2
 = 1 + 2 , at each point.
dy
D
y = n


(2n  1) 2
p = 
(2n  1) 

2


(2n  1) 2
y= 
(2n  1) 
2

 2
n = 0, ±1, ±2 .......
n  1, 2, 3.............
n  -1, - 2, - 3........
n  1, 2, 3.............
n  -1, - 2, - 3.......
D
d
Here,  = wavelength in medium.
where, fringe width  =
d
nmax =  

total number of maxima = 2nmax + 1
 d 1
Highest order minima :
nmax =   
 2
total number of minima = 2nmax.
Highest order maxima :
Page # 70
Intensity on screen :
 = 1 + 2 + 2  1 2 cos () where,  =
2
p

  

1 = 2,  = 41 cos2 
 2 
YDSE with two wavelengths 1 & 2 :
The nearest point to central maxima where the bright fringes coincide:
y = n11 = n22 = Lcm of 1 and 2
If
The nearest point to central maxima where the two dark fringes
coincide,
1
1
y = (n1 – ) 1 = n2 – ) 2
2
2
Optical path difference
popt = p
 =
2
2
p =
 vacuum

 = ( – 1) t.
popt.
D
B
= ( – 1)t .
d

YDSE WITH OBLIQUE INCIDENCE
In YDSE, ray is incident on the slit at an inclination of 0 to
the axis of symmetry of the experimental set-up
S1
P1
1
0
dsin0
2
O
P2
S2
O'
B0
We obtain central maxima at a point where, p = 0.
or
2 = 0.
This corresponds to the point O’ in the diagram.
Hence we have path difference.
d(sin 0  sin )  for points above O

d(sin 0  sin )  for points between O & O'
p = 
d(sin   sin  )  for points below O'
0

... (8.1)
Page # 71
THIN-FILM INTERFERENCE
for interference in reflected light
2d
for destructiv e interference
n

= 
1
for constructi ve interference
(n  2 )
for interference in transmitted light
2d
n

= 
1
(n  2 )

for constructive interference
for destructive interference
Polarisation

 = tan .(brewster's angle)
 +  r = 90°(reflected and refracted rays are mutually
perpendicular.)

Law of Malus.
I = I 0 cos2
I = KA2 cos2

Optical activity
t
C


LC
 = rotation in length L at concentration C.
Diffraction

a sin  = (2m + 1) /2 for maxima.

sin  =

Linear width of central maxima =

Angular width of central maxima =
where m = 1, 2, 3 ......
m
, m =  1,  2,  3......... for minima.
a
2d
a
2
a
Page # 72
2

 sin  / 2 
  0 

 /2 

Resolving power .
where  =
a sin 



R =  –   
2
1
where ,  
1   2
,  = 2 - 1
2
GRAVITATION
GRAVITATION :
Universal Law of Gravitation
m1 m 2
m m
F
or F = G 1 2 2
r2
r
where G = 6.67 × 10–11 Nm2 kg–2 is the universal gravitational constant.
Newton's Law of Gravitation in vector form :
Gm1m2

r̂12
F12 =
r2

Gm1m2
& F2 1 =
r2

 G m1 m2
Now r̂12   r̂21 , Thus F21 
r̂12 .
r2

Comparing above, we get F12   F21
GM
F
= 2
m
r
Gravitational potential : gravitational potential,
Gravitational Field
V=–
GM
.
r
Ring.
V=
E=
E=–
dV
.
dr
 GM
1.
or E = –
2
2 1/ 2
x or (a  r )
&
E=
 GM r
2
(a  r 2 )3 / 2
r̂
GM cos 
x2
Page # 73
Gravitational field is maximum at a distance,
2 and it is – 2GM 3 3 a 2
r=± a
2.
Thin Circular Disc.


2GM
2GM 
r

1  cos 
1
& E=–
2 
1 = –
a2
a
 r 2  a 2 2 
Non conducting solid sphere
(a) Point P inside the sphere. r < a, then
 2GM  2 2
 a r
V=
a 2 

3.
V= 
GM
2a
3
1
2


 r



GM r
(3a 2  r 2 ) & E = –
, and at the centre V = –
a3
(b) Point P outside the sphere .
3GM
and E = 0
2a
GM
GM
& E=– 2
r
r
4.
Uniform Thin Spherical Shell / Conducting solid sphere
(a)
Point P Inside the shell.
 GM
r < a , then V =
&
E=0
a
(b)
Point P outside shell.
GM
 GM
r > a, then V =
&
E=– 2
r
r
VARIATION OF ACCELERATION DUE TO GRAVITY :
1.
Effect of Altitude
r > a, then V = 
gh =
2.
3.
GMe

h
= g 1 
 Re
R e  h2




2
~



Re 
2h
 when h << R.
g 1 


d 

gd = g 1 

 Re 
Effect of the surface of Earth
The equatorial radius is about 21 km longer than its polar radius.
Effect of depth
We know, g =
GMe
R 2e
Hence gpole > gequator.
SATELLITE VELOCITY (OR ORBITAL VELOCITY)
1
1
 g R 2e  2
 GM e  2
v0 = 
=



 R e  h 
 R e  h 
Page # 74
When h << Re then v 0 =
gRe
3
–1
1
9.8  6.4  10 6 = 7.92 × 10 ms = 7.92 km s
Time period of Satellite

v0 =
1
T=
2 R e  h
1
3
2  R e  h
=

R e 
g
 g R 2e  2


 R e  h 
Energy of a Satellite
2


GMem
GMem
GM e m
K.E. =
; then total energy E = –
r
2r
2R e
Kepler's Laws
Law of area :
The line joining the sun and a planet sweeps out equal areas in equal
intervals of time.
1
r (rd)
area swept
1 2 d
2
Areal velocity =
=
=7
r
= constant .
dt
time
2
dt
U=
Hence
1 2
r  = constant. Law of periods :
2
T2
R3
= constant
FLUID MECHANICS & PROPERTIES OF MATTER
FLUIDS, SURFACE TENSION, VISCOSITY & ELASTICITY :
1.
f
F
A

or F   f .
a A
a
Hydrostatic Paradox
PA = PB = PC
(i) Liquid placed in elevator : When elevator accelerates upward with
acceleration a0 then pressure in the fluid, at depth ‘h’ may be given by,
p = h [g + a0]
Hydraulic press.
p=
and force of buoyancy, B = m (g + a0)
(ii) Free surface of liquid in horizontal acceleration :
tan  =
a0
g
Page # 75
p1 – p2 =  a0
where p1 and p2 are pressures at points 1 & 2.
a 0
Then h1 – h2 =
g
(iii) Free surface of liquid in case of rotating cylinder.
h=
v2
 2r 2
=
2g
2g
Equation of Continuity
a1v1 = a2v2
In general av = constant .
Bernoulli’s Theorem
i.e.
P
1 2
+
v + gh = constant.

2
2gh
2
(vi) Torricelli’s theorem – (speed of efflux) v= 1  A 22 ,A2 = area of hole
A1
A1 = area of vessel.
restoring force
F
ELASTICITY & VISCOSITY : stress = area of the body  A
change in configuration
original configuration
L
Longitudinal strain =
L
V
v = volume strain =
V
Strain,  =
(i)
(ii)
(iii)
1.
Shear Strain : tan  or  =
Young's modulus of elasticity Y =
x

F/ A
FL

L / L AL
1
1
(stress × strain) =
(Y × strain2 )
2
2
k = Yr0.
Potential Energy per unit volume =
Inter-Atomic Force-Constant
Page # 76
dv
dv
or F = – A
dx
dx
Newton’s Law of viscosity,
F A
Stoke’s Law
Terminal velocity =
F = 6 r v.
2 r 2 (  )g
9

SURFACE TENSION
Surface tension(T) =
Total force on either of the imaginary line (F )
;
Length of the line ( )
W
A
Thus, surface tension is numerically equal to surface energy or work
done per unit increase surface area.
T=S=
4T
= pexcess ;
r
2T
Inside the drop :
(p – pa) =
= pexcess
r
2T
Inside air bubble in a liquid :(p – pa) =
= pexcess
r
Inside a bubble :
(p – pa) =
Capillary Rise
h=
2T cos 
rg
SOUND WAVES
(i)
(ii)
(iii)
Longitudinal displacement of sound wave
 = A sin (t – kx)
Pressure excess during travelling sound wave

Pex =  B
(it is true for travelling
x
= (BAk) cos(t – kx)
wave as well as standing waves)
Amplitude of pressure excess = BAk
E

Where E = Ellastic modulus for the medium
 = density of medium
Speed of sound C =
–
for solid
C=
Y

Page # 77
where Y = young's modulus for the solid
B

where B = Bulk modulus for the liquid
–
for liquid
C=
B
P
 RT




M0
where M0 is molecular wt. of the gas in (kg/mole)
Intensity of sound wave :
–
for gases
C=
Pm2
<> = 2 f A v =
2v
2 2
(iv)
2
<>
 Pm2
  
L = 10 log10    dB
 0
W/m 2 (This the minimum intensity human ears can
Loudness of sound :
where I 0 = 10–12
listen)
Intensity at a distance r from a point source =  
P
4r 2
Interference of Sound Wave
if
P1 = pm1 sin (t – kx 1 + 1)
P2 = pm2 sin (t – kx 2 + 2)
resultant excess pressure at point O is
p = P1 + P2
p = p0 sin (t – kx + )
p0 =
2
2
pm
 pm
 2pm1 p m2 cos 
1
2
where  = [k (x 2 – x 1) + (1 – 2)]
(i)
(ii)
and
I = I 1 + I 2 + 2 1  2
For constructive interference
 = 2n and  p0 = pm1 + pm2 (constructive interference)
For destructive interfrence
 = (2n+ 1) and  p0 = | pm1 – pm2 | (destructive interference)
2
x.

Condition for constructive interference : x = n
If  is due to path difference only then  =
Condition for destructive interference : x = (2n + 1)


2
Page # 78
(a)
If
pm1 = pm2 and 
resultant p = 0 i.e. no sound
(b)
If
pm1 = pm2 and  = 0 , 2, 4, ...
p0 = 2pm & I 0 = 4I 1
p0 = 2pm1
Close organ pipe :
v 3v 5v
(2n  1)v
, , ,..........
4 4 4
4
Open organ pipe :
v 2v 3v
nV
, , ,..........
f=
2 2 2
2
Beats : Beatsfrequency = |f 1 – f 2|.
Doppler’s Effect
f=
n = overtone
vv 
The observed frequency,
0
f = f  v  v 
s 

and
 v  vs 

 =  
 v 
Apparent wavelength
ELECTRO MAGNETIC WAVES
Maxwell's equations
 E  dA  Q / 
 B  dA  0
 E  d 
(Gauss's Law for electricity)
0
(Gauss's Law for magnetism)
– d B
dt
 B  d   i
0 c
(Faraday's Law)
 0 0
d E
dt
(Ampere-Maxwell Law)
Oscillating electric and magnetic fields
E= Ex(t) = E0 sin (kz - t)
 z
  z t 

= E0 sin 2  – vt  = E0 sin 2  – T 


 
 
E0/B0 = c
c = 1/  0  0
c is speed of light in vaccum
v  1/ 
v is speed of light in medium
Page # 79
U energy transferred to a surface in time t is U, the magnitude of
p
c
the total momentum
delivered to this surface (for complete
absorption) is p
Electromagnetic spectrum
Type
Radio
Wavelength
range
> 0.1m
Microwave
0.1m to 1mm
Infra-red
1mm to 700nm
Light
700nm to
400nm
Ultraviolet
400nm to 1nm
X-rays
Gamma
rays
1nm to 10
–3
–3
nm
< 10 nm
Production
Rapid
acceleration
and
decelerations of electrons in
aerials
Klystron value or magnetron
value
Vibration of atoms and
molecules
Electrons in atoms emit light
when they move from one
energy level to a lower
energy
Inner shell electrons in
atoms moving from one
energy level to a lower level
X-ray tubes or inner shell
electrons
Radioactive decay of the
nucleus
Detection
Receiver's aerials
Point contact diodes
Thermopiles Bolometer,
Infrared
photographic
film
The eye, photocells,
Photographic film
photocells photographic
film
Photograpic film, Geiger
tubes, lonisation chamber
do
ERROR AND MEASUREMENT
1.
Least Count
mm.scale
L.C =1mm
2.
Screw gauge Stop Watch Temp thermometer
Vernier
L.C=0.1mm L.C=0.1mm L.C=0.1Sec L.C=0.1°C
Significant Figures
 Non-zero digits are significant
 Zeros occurring between two non-zeros digits are significant.
 Change of units cannot change S.F.
 In the number less than one, all zeros after decimal point and to
the left of first non-zero digit are insignificant
 The terminal or trailing zeros in a number without a decimal
point are not significant.
Page # 80
3.
Permissible Error
 Max permissible error in a measured quantity = least count of
the measuring instrument and if nothing is given about least count
then Max permissible error = place value of the last number
 f (x,y) = x + y then (f)max = max of (  X  Y)
 f 
f (x,y,z) = (constant) xa yb zc then  
 f  max

x
y
z 
= max of   a x  b y  c z 


4.
Errors in averaging
 Absolute Error an = |amean -an|


 Mean Absolute Error amean = 

 Relative error =
n

| a | 
i
i1

n
a mean
a mean
a mean
 Percentage error = a
×100
mean
5.
Experiments
 Reading of screw gauge
Thicknes of object  Re ading of screw gauge
 main   circular 

 
 Least 

  scale    scale 
 reading   reading  count 

 

least count of screw gauge =
pitch
No. of circular scale division
 Vernier callipers
Thicknes of object  Re ading of vernier calliper
 main   vernier 

 
 Least 

  scale    scale 
 reading   reading  count 

 

Least count of vernier calliper = 1 MSD –1 VSD
Page # 81
PRINCIPLE OF COMMUNICATION
Transmission from tower of height h
 the distance to the horizon dT =
 dM =
2RhT
2RhT  2RhR
Amplitude Modulation
 The modulated signal cm (t) can be written as
cm(t) = Ac sin ct +
A c
A c
cos (C - m) t –
cos (C + m)
2
2
 Modulation index ma 
Change in amplitude of carrier wave kA m

Amplitude of original carrier wave
Ac
where k = A factor which determines the maximum change in the
amplitude for a given amplitude Em of the modulating. If k = 1 then
ma =
A m A max – A min

Ac
A max – A min
 If a carrier wave is modulated by several sine waves the total modulated
index mt is given by mt =
m12  m22  m23  .........
 Side band frequencies
(f c + f m) = Upper side band (USB) frequency
(fc - f m) = Lower side band (LBS) frequency
 Band width = (fc + fm) - (f c - f m) = 2fm
 Power in AM waves : P 
2
Vrms
R
2
 Ac 


A2
2
(i) carrier power
Pc  
 c
R
2R
Page # 82
2
 ma A c 
 ma A c 
2 2




2 2  ma A c
(ii) Total power of side bands Psb =  2 2 


R
2R
4R
A c2  ma2
(iii) Total power of AM wave PTotal = Pc + Pab = 2R  1  2

Pt  ma2
(iv) P   1  2
c






P
ma2 / 2
 and sb 

Pt
 ma2 

1


2 

(v) Maximum power in the AM (without distortion) will occur when
ma = 1 i.e., Pt = 1.5 P = 3Pab
(vi) If Ic = Unmodulated current and It = total or modulated current

t
ma2
Pt
2t

1



 P
 

2
2c
c
c





Frequency Modulation
 Frequency deviation  = = (fmax - fc) = f c - fmin = kf .
Em
2
 Carrier swing (CS) = CS = 2 × f
 Frequency modulation index (mf)
fmax – fc fc – fmin k f Em



=. mf = f 
fm
fm
fm
m
 Frequency spectrum = FM side band modulated signal consist of infinite number of side bands whose frequencies are (fc ± f m), (f c ± 2f m),
(f c ± 3fm).........
(f )max
 Deviation ratio = (f )
m max
( f )actual
 Percent modulation , m = (f )
max
Page # 83
SEMICONDUCTOR
Conductivity and resistivity

P (– m)
Metals
10–2 -10–6
(–1m–1)
102 – 108
semiconductors 10–5 -10–6
105 – 10–6
Insulators
1011 –1019
10–11 – 10–19
Charge concentration and current
[ n = e]
In case of intrinsic semiconductors
 P type
n >> e
 i = ie + ih
 e n = i2
 Number of electrons reaching from valence bond to conduction bond.
= A T 3 / 2 e – Eg / 2kT (A is positive constant)
  = e ( e me + n n)
for  hype
n = Na >> e.
for  – type
e = Na >> h
 Dynamic Resistance of P-N junction in forward biasing =
V

Transistor
 CB amplifier
Samll change in collector current (ic )
(i) ac current gain c = Samll change in collector current ( i )
e
Collector current (ic )
(ii) dc current gain dc = Emitter current (i ) value of dc lies
e
between 0.95 to 0.99
Change in output voltage ( V0 )
(iii) Voltage gain AV = Change in input voltage ( V )
f
AV = aac × Resistance gain
Change in output power ( P0 )
(iv) Power gain = Change in input voltage ( P )
C
 Power gain = a2ac × Resistance gain
(v) Phase difference (between output and input) : same phase
(vi) Application : For High frequency
Page # 84
CE Amplifier
 i c 
(i) ac current gain ac =  i  VCE = constant
 b
ic
(ii) dc current gain dc = i
b
V0
(iii) Voltage gain : AV = V = ac × Resistance gain
i
P0
(iv) Power gain = P = 2ac × Resistance
i
(v) Transconductance (gm) : The ratio of the change in collector in
collector current to the change in emitter base voltage is called trans
i c
AV
conductance i.e. gm = V
. Also gm = R RL = Load resistance.
EB
L


or  =
1


1– 
(v) Transconductance (gm) : The ratio of the change in collector in collector current to the change in
emitter base voltage is called trans conductance
i.e. gm = . Also gm = RL = Load resistance.
 Relation between  and  :  
Page # 85
ROUGH WORK
Page # 86