physics formulla booklet (1)

PHYSICS
FORMULA BOOKLET

INDEX

S.No. Topic Page No.

1. Unit and Dimension 2

2. Rectilinear Motion 3–4

3. Projectile Motion & Vector 5–5

4. Relavitve Motion 5–7

5. Newton’s Laws of Motion 7–9

6. Friction 9–9

7. Work, Power & Energy 10 – 11

8. Circular Motion 11 – 14

9. Centre of Mass 14 – 18

10. Rigid Body Dynamics 18 – 25

11. Simple Harmonic Motion 26 – 28

12. Sting Wave 29 – 31

13. Heat & Thermodynamics 31 – 37

14. Electrostatics 37 – 40

15. Current Electricity 41 – 47

16. Capacitance 47 – 51

17. Alternating Current 52 – 54

18. Magnetic Effect of Current & Magnetic

force on charge 54 – 56

19. Electromagnetic Induction 56 – 59

20. Geometrical Optics 59 – 66

21. Modern Physics 67 – 70

22. Wave Optics 70 – 73

23. Gravitation 73 – 75

24. Fluid Mechanics & Properties of Matter 75 – 77

25. Sound Wave 77 – 79

26. Electro Magnetic Waves 79 – 80

27. Error and Measurement 80 – 81

28. Principle of Communication 82 – 83

29. Semiconductor 84 – 85

Page # 1

PHYSICS
FORMULA BOOKLET

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
Metre m
1 Length Kilogram Kg
Second S
2 Mass Ampere A
Kelvin K
3 Time Candela Cd
Mole mol
4 Electric Current

5 Temperature

6 Luminous Intensity

7 Amount of Substance

* Supplementary Units :

S.No. Physical Quantity SI Unit Symbol
1 Plane Angle radian r
2 Solid Angle Steradian Sr

* Metric Prefixes :

S .N o . Prefix S ym b o l V a lu e
1 C e n ti c 10–2
2 M ili m 10–3
3 M icro µ 10–6
4 Nano n 10–9
5 P ic o p 10–12
6 K ilo K 103
7 Mega M 106

Page # 2

RECTILINEAR MOTION

Average Velocity (in an interval) :

vav = v = <v> = Total displacement = rf  ri
Total time taken t

Average Speed (in an interval)

Total distance travelled
Average Speed = Total time taken

Instantaneous Velocity (at an instant) :


  r 
v inst = lim  t 

t  0

Average acceleration (in an interval):

 = v = 
aav t vf  vi

t

Instantaneous Acceleration (at an instant):

  = lim  v 
a= dv 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.

xx

xi a<0
a>0 xi

0t 0t

x-t graph

 v is a linear polynomial in terms of t. Hence vt graph is a straight line of
slope a.

v t v
slope = a u slope = a

u a is negative
a is positive 0t

0

Page # 3

v-t graph

 at graph is a horizontal line because a is constant.

a a
positive
0
a acceleration negative

0 t a acceleration

a-t graph

Maxima & Minima

dy =0 d  dy  < 0 at maximum
dx & dx  dx 

dy d  dy 
and = 0 & dx  > 0 at minima.
dx  dx 

Equations of Motion (for constant acceleration)

(a) v = u + at

11 1
(b) s = ut + at2 s = vt  at2 x = x + ut + at2
22 fi 2

(c) v2 = u2 + 2as

(d) s = (u  v) t (e) a
sn = u + 2 (2n  1)
2

For freely falling bodies : (u = 0)

(taking upward direction as positive)

(a) v = – gt

11 1
(b) s = – gt2 s = vt  gt2 h = h – gt2

22 f i2
(c) v2 = – 2gs

g
(d) sn = – 2 (2n  1)

Page # 4

PROJECTILE MOTION & VECTORS

Time of flight : 2u sin 
T= g

Horizontal range : R = u2 sin 2
g

Maximum height : u2 sin2 
H = 2g

Trajectory equation (equation of path) :

gx2 x
y = x tan  – = x tan  (1 – )
2u2 cos2  R

Projection on an inclined plane

y


x Up the Incline Down the Incline
2u2 sin  cos(  ) 2u2 sin  cos(  )
Range
g cos2  gcos2 

Time of flight 2u sin  2u sin 
g cos  gcos 
Angle of projection with
incline plane for maximum  
42 42
range
u2 u2
Maximum Range g(1 sin) g(1 sin )

RELATIVE MOTION

 
v AB (velocity of A with respect to B)  v A  vB

 
a AB (acceleration of A with respect to B)  aA  aB

 
Relative motion along straight line - xBA  xB  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, vx = vR.
Velocity perpendicular to the river, vf = vmR

The net speed is given by vm = v 2  v 2
mR R

2. Shortest Path :
velocity along the river, v = 0

x

and velocity perpendicular to river v = v 2  v 2
y mR R

The net speed is given by vm = v 2  vR2
mR

at an angle of 90º with the river direction.
velocity vy is used only to cross the river,

Page # 6

dd

therefore time to cross the river, t = v y = v 2  v 2
mR R

and velocity vx is zero, therefore, in this case the drift should be zero.

 v – v sin  = 0 or v = v sin 
R mR R mR

or  = sin–1  vR 
v mR

RAIN PROBLEMS

  or vRm = vR2  v 2
vRm = vR – vm m

NEWTON'S LAWS OF MOTION

1. From third law of motion

FAB  FBA
FAB = Force on A due to B
FBA = Force on B due to A

2. From second law of motion

F= dPx = ma F = dPy = ma F= dPz = ma
x dt x y dt y z dt z

5. WEIGHING MACHINE :

A weighing machine does not measure the weight but measures the

force exerted by object on its upper surface.

6. SPRING FORCE F  kx

x is displacement of the free end from its natural length or deformation

of the spring where K = spring constant.

7. SPRING PROPERTY K ×  = constant

= Natural length of spring.

8. If spring is cut into two in the ratio m : n then spring constant is given
by

 m  = n. k = k11 = k22
1 = mn ; 2
mn

Page # 7

For series combination of springs 1  1  1  .......
For parallel combination of spring keq k1 k 2
k = k + k + k ............

eq 1 2 3

9. SPRING BALANCE:
It does not measure the weight. t measures the force exerted by the
object at the hook.
Remember :

Vp = V1  V2
2

aP = a1  a2
2

11. a  (m2  m1)g
m1  m2

T  2m1m2g
m1  m2

12. 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  m2a2  m3a3  ......

Fext  Net external force on the system.

m 1,m 2, m3 are the masses of the objects of the system and

a1,a2,a3 are the acceleration of the objects respectively.

14. NEWTON’S LAW FOR NON INERTIAL FRAME :


FReal F Pseudo  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 
aFrame

(a) Inertial reference frame: Frame of reference moving with con-

stant 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 : fk = 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 mo-
tion 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.

fmax = sN
0  fs  fsmax

Friction fstatic maximum
static friction
sN 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 + F3 + .....
 2

W = [ F ] . S ...(i)

    
W = F 1 . S + F 2 . S + F 3 . S + .....

or W = W + W + W + ..........
123

WORK DONE BY A VARIABLE FORCE


dW = F.ds

RELATION BETWEEN MOMENTUM AND KINETIC ENERGY

p2
K = 2m and P = 2 m K ; P = linear momentum

POTENTIAL ENERGY

 U2 dU   r2   i.e., r2   W
F  d r U2  U1   F dr 

U1 r1 r1

r 

U   F  d r  W


CONSERVATIVE FORCES
U

F= – r

WORK-ENERGY THEOREM

W +W +W = K
C NC PS

Modified Form of Work-Energy Theorem

W = U
C

WNC + WPS = K + U

WNC + WPS = E

Page # 10

POWER

The average power ( P or pav) delivered by an agent is given by P or

W
pav = t

 =    =  . 
P  F  dS F dS F v

dt dt

CIRCULAR MOTION

1. Average angular velocity 2  1 
 av = t2  t1 = t



d 
2. Instantaneous angular velocity   = dt

3. Average angular acceleration  av = 2  1 = 
t2  t1 t

d d
4. Instantaneous angular acceleration   = dt =  d

5. Relation between speed and angular velocity  v = r and    
v  r

7. Tangential acceleration (rate of change of speed)

dV d dr
 a= =r =  dt
t dt dt

8. Radial or normal or centripetal acceleration  a= v2 = 2r
r r

9. Total acceleration a at
 
O v
 a  at  ar  a = (at2 + ar2)1/2 P
ar or a c
Where      and     
at  r ar  v

Page # 11

10. Angular acceleration


 d 
  = dt (Non-uniform circular motion) ARCoWtation

v2 mv2  2  3 / 2
12. Radius of curvature R = = 1 
  dy 
a F
If y is a function of x. i.e. y = f(x)  R =   dx  
d2y

dx 2

13. Normal reaction of road on a concave bridge
mv2

 N = mg cos  +
r

O

N

 V concave
mgcos bridge

mg

14. Normal reaction on a convex bridge
 N = mg cos  – mv2
r

NV

 mgcos convex
bridge
mg


O

15. Skidding of vehicle on a level road  vsafe  gr
16. Skidding of an object on a rotating platform  max = g / r

Page # 12

17. Bending of cyclist v2
 tan  = rg

v2
18. Banking of road without friction  tan  = rg

19. Banking of road with friction  v 2    tan
rg 1  tan 

20. Maximum also minimum safe speed on a banked frictional road

 rg(  tan ) 1/ 2  rg (tan   ) 1/ 2

Vmax  Vmin  

 (1  tan )   (1  tan) 
   

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

DO B

 P
AN

×

(1) (2) (3)

Vmin  4gL Vmin  4gL Vmin  4gL

(for completing the circle) (for completing the circle) (for completing the circle)

Page # 13

24. Conical pendulum :

fixed pointor
suspension
//////O///////

T cos  T  L
 h

r

mg

T cos  = mg
T sin  = m2 r

 Time period = 2 L cos 
g

25. Relations amoung angular variables :

0  Initial ang. velocity  = 0 + t

d,  or 

rO (Perpendicular
 ar to plane of paper
directed outwards
at or V for ACW rotation)

  Find angular velocity 1
 = 0t + 2 t2
  Const. angular acceleration 2 = 02 + 2 
  Angular displacement

CENTRE OF MASS


Mass Moment : M = m r
CENTRE OF MASS OF A SYSTEM OF 'N' DISCRETE PARTICLES

 
 m1r1  m2r2  ........  mnrn 
rcm = m1  m2  ........  mn ; rcm

Page # 14

n   1 n
mi ri mi ri
rcm = M
i 1 i 1

=n

mi

i1

CENTRE OF MASS OF A CONTINUOUS MASS DISTRIBUTION

x=  x dm ,y =  y dm ,z =  z dm

  cm dm cm dm cm 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)

b L
x= y=

c2 c2

Page # 15

 A triangular plate (By qualitative argument)
h

at the centroid : yc = 3

 A semi-circular ring 2R
y = x =O
c c

 A semi-circular disc 4R
yc = 3 xc = O

R
 A hemispherical shell y = x =O
 A solid hemisphere c2 c
 A circular cone (solid)
y= 3R x =O
c c
8

h
yc = 4

 A circular cone (hollow) h
yc = 3

Page # 16

MOTION OF CENTRE OF MASSAND CONSERVATION OF MOMENTUM:
Velocity of centre of mass of system

 m1 dr1  m2 dr2  m3 dr3 .......... ....  mn drn
vcm = dt dt dt dt

M

  
= m1 v1  m2 v 2  m3 v3 .......... mn vn
M

P System = M v cm
Acceleration of centre of mass of system

 m1 dv1  m2 dv 2  m3 dv 3 ..............  mn dv n
acm dt dt dt dt
=
M

  
= m1a1  m2a2  m3a3..........  mnan
M

Net forceon system Net External Force  Net internal Force
==
M M

Net External Force
=

M

Fext = M acm
IMPULSE
Impulse of a force F action on a body is defined as :-

 tf Fdt  (impulse - momentum theorem)
J  ΔP
J= ti

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)

Impulse of reformation Fr dt

e = Impulse of deformation = Fd dt

Velocity of separation along line of impact
= s Velocity of approach along line of impact

Page # 17

(a) e = 1  Impulse of Reformation = Impulse of Deformation
(b) e = 0  Velocity of separation = Velocity of approach
(c) 0 < e < 1  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 vrel (w.r.t. the system), then the force exerted by this mass


on the system has magnitude  vrel .


Thrust Force ( Ft )

    dm 
Ft v rel  dt 

Rocket propulsion :

If gravity is ignored and initial velocity of the rocket u = 0;

v = v ln  m0  .
r m

RIGID BODY DYNAMICS

1. RIGID BODY :

A 1 VA VAsin1
B A

VB 2 VAcos1

VBsin2 B
VBcos2

Page # 18

If the above body is rigid

VA cos  = VB cos 
1 2

V = relative velocity of point B with respect to point A.
BA

A

B
VBA

Types of Motion of rigid body

Pure Translational Pure Rotational Combined Translational and
Motion Motion Rotational Motion

2. 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.

 = mr 2 + m r 2 +.........................
1 22

=  +  +  +.........................
S units of Moment of Inertia is Kgm2.

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

 = miri2
i1

2.3 For a continuous object :

 = dmr2

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
[Onlyapplicableto plane lamina(thatmeansfor2-Dobjectsonly)].

z = x + y (when object is in x-y plane).

3.2 Parallel Axis Theorem
(Applicable to any type of object):
 =  + Md2

 cm

List of some useful formula :

Object Moment of Inertia

2 MR2 (Uniform)
5

Solid Sphere

2 MR 2 (Uniform)
3

Hollow Sphere

MR2 (Uniform or Non Uniform)

Page # 20

Ring. MR2 (Uniform)
Disc 2
Hollow cylinder
Solid cylinder MR2 (Uniform orNonUniform)

MR2 (Uniform)
2

ML2 (Uniform)
3

ML2 (Uniform)
12

Page # 21

2m2
(Uniform)

3

AB = CD = EF = Ma2 (Uniform)
12

Square Plate Ma2 (Uniform)
Square Plate 6
Rectangular Plate
 = M(a2  b2 ) (Uniform)
Cuboid 12

M(a2  b2 ) (Uniform)
12
Page # 22

4. RADIUS OF GYRATION :
 = MK2

5. TORQUE :

 

  rF

5.5 Relation between '' & '' (for hinged object or pure rotation)

Hinge = Hinge 
ext 

Where  = net external torque acting on the body about Hinge

ext Hinge

point

Hinge = moment of Inertia of body about Hinge point

F1t

F1c r1 F2t
x F2c
r2

F1t = M1a1t = M1r1

F2t = M2a2t = M2r2

resultant = F r + F r + ........
1t 1 2t 2
= M1  r12 + M2  r22 + ............

) = 
resultant external

Rotational Kinetic Energy = 1 ..2
2

 
P  MvCM  Fexternal  MaCM

Net external force acting on the body has two parts tangential and

centripetal.

 FC = maC = m v2 = m2 rCM  Ft = mat = m rCM
rCM

Page # 23

6. ROTATIONAL EQUILIBRIUM :
For translational equilibrium.

Fx  0 ............. (i)

and 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 = rpsin
L r P


L =r ×P



L = P× r

7.3 Angular momentum of a rigid body rotating about fixed axis :

 = H
LH

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 P VQ
r r
r
Q wr

Q  VQ

VP  VQ2  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 : 
Fext
  
cm  cm  ,  Macm


Psystem  Mvcm ,

Total K.E. = 1 Mvcm2 + 1 cm 2
2 2


Angular momentum axis AB = L about C.M. + L of C.M. about AB

 
L AB  cm   rcm  Mvcm

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
 = T = 2f

Time period (T) : T= 2 m
 = 2 k
k
m

Speed : v   A2  x2
Acceleration : a = 2x

11 1
Kinetic Energy (KE) : 2 mv2 = 2 m2 (A2 – x2) = 2 k (A2 – x2)

Potential Energy (PE) : 1
Kx2

2

Total Mechanical Energy (TME)

1 11
= K.E. + P.E. = k (A2 – x2) + Kx2 = KA2 (which is constant)

2 22

SPRING-MASS SYSTEM

km
(1)   T = 2

k

smooth surface
m

(2)

T = 2  m1m2
K where  = (m1  m2 ) known as reduced mass

Page # 26

COMBINATION OF SPRINGS 1/keq = 1/k1 + 1/k2
Series Combination : k =k +k
Parallel combination :
eq 1 2


SIMPLE PENDULUM T = 2 g = 2 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) :  where, C = Torsional constant
T = 2 C

Superposition of SHM’s along the same direction

x = A sin t & x = A sin (t + )
11 22

A2

A

A1
If equation of resultant SHM is taken as x = A sin (t + )

A= A12  A 2  2A1A2 cos & tan  = A 2 sin 
2 A1  A 2 cos 

1. 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.

 x = A0e–bt/ 2m sin [1t +  ], where '   k  –  b 2
 m   2m 

For small b

 angular frequency '  k / m,  0

–bt

 Amplitude A  A0e 2m

l

 Energy E (t) = 1 KA2 e –bt /m
2

 Quality factor or Q value , Q = 2 E = '
| E | 2Y

where , '  k . b2 b
m 4m2 , Y  2m

2. Forced Oscillations And Resonance
External Force F(t) = F0 cos d t
x(t) = A cos (dt + )

F0
 A  tan   v0
 m2 2  d2 2  d2 b2  and d x0
 
 

A F0
m 2  2d
 (a) Small Damping

(b) Driving Frequency Close to Natural Frequency A  F0
db

Page # 28

STRING WAVES

GENERAL EQUATION OF WAVE MOTION :

2y 2y
t2 = v2 x2

x
y(x,t) = f (t ± )

v
where, y (x, t) should be finite everywhere.

 f  t  x  represents wave travelling in – ve x-axis.
 v

 f  t  x  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.

T T  Tension
v =  where   mass per unit length

POWER TRANSMITTED ALONG THE STRING BY A SINE WAVE
Average Power P = 22 f2 A2 v

Intensity I = P = 22 f2 A2 v
s

REFLECTION AND REFRACTION OF WAVES
yi = Ai sin (t – k1x)

y t  A t sin (t  k2 x) 
yr   Ar sin (t  k1x) if incident from rarer to denser medium (v2 < v1)

Page # 29

y t  A t sin (t – k 2x) if incident from denser to rarer medium. (v > v )
 21
yr  Ar sin (t  k1x) 

(d) Amplitude of reflected & transmitted waves.

A= k1  k2 Ai &A = 2k1 Ai
r k1  k2 t k1  k2

STANDING/STATIONARY WAVES :-

(b) y1 = A sin (t – kx + 1)

y= A sin (t + kx + 2)
2

y1 + y2 = 2 A cos  kx  2  1  sin t  1  2 
  2   2

The quantity 2A cos kx  2  1  represents resultant amplitude at
 2

x. At some position resultant amplitude is zero these are called nodes.

At some positions resultant amplitude is 2A, these are called antin-

odes.

(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 trans-
mitted across these.

VIBRATIONS OF STRINGS ( STANDING WAVE)
(a) Fixed at both ends :
1. Fixed ends will be nodes. So waves for which

L=  2 L = 3
2 L= 2

2

are possible giving 2L
n or  = n where n = 1, 2, 3, ....

L= 2 nT
T fn = 2L  , n = no. of loops

as v =  Page # 30

(b) String free at one end :


1. for fundamental mode L = 4 = or  = 4L

fundamental mode

First overtone L = 3 Hence  = 4L
4 3

 first overtone

3T
so f =  (First overtone)
1 4L

5T
Second overtone f2 = 4L 

n  1  T  (2n  1) T
 2  4L 
so fn = 2L

HEAT & THERMODYNAMICS

1 33
Total translational K.E. of gas = M < V2 > = PV = nRT

2 22

3P V= 3P 3RT 3KT
< V2 > = rms = Mmol = m


Important Points :

– V  T V 8KT KT KT
rms m = 1.59 m V = 1.73

rms m

2KT KT
Most probable speed V = = 1.41  V > >V
p m m rms V mp

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
Internal energy U = nRT

2

Workdone in isothermal process : W = [2.303 nRT log10 Vf ]
Vi

Internal energy in isothermal process : U = 0

Work done in isochoric process : dW = 0

Change in int. energy in isochoric process :

f
U = n 2 R T = heat given

Isobaric process :

Work done W = nR(T – T)
f i

change in int. energy U = nCv T

heat given Q = U + W

Specific heat : f Cp =  f  1 R
Cv = 2 R  2 

Molar heat capacity of ideal gas in terms of R :

(i) for monoatomic gas : Cp
Cv = 1.67

(ii) for diatomic gas : Cp
Cv = 1.4

(iii) for triatomic gas : Cp
Cv = 1.33

In general : = Cp = 1  2
Cv f 

Mayer’s eq.  Cp – Cv = R 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 :

n1M1  n2M2
Mol. wt. = n1  n2

C= n1Cv 1  n2Cv 2
v n1  n2

= Cp (mix) = n1Cp  n2Cp  .....
Cv (mix) 12

n1Cv1  n2Cv2  ....

Heat Engines

Efficiency ,   work done by the engine
heat sup pliedtoit

= W  QH – QL  1– QL
QH QH QH

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 perform-
ing an equivalent amount of work.

Page # 33

 Rudlope Classius Statement
It is impossible to make heat flow from a body at a lower
temperature to a body at a higher temperature without doing external work
on the working substance

Entropy

Q f Q
T i T
 change in Sf
entropy of the system is S = – Si 

 In an adiabatic reversible process, entropy of the system remains con-
stant.

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  Q2  T2   1– T2
V1 V4 Q1 T1 T1

Page # 34

Refrigerator (Heat Pump)

Refrigerator

Hot (T2) Hot (T1)

Q2 Q1

W

 Coefficient of performance,   Q2 =  1 =  1
W –1 T1 – 1
T1

T2 T2

Calorimetry and thermal expansion
Types of thermometers :

(a) Liquid Thermometer : T=    0  × 100
(b) Gas Thermometer :  100  
  0 

Constant volume : T=  P  P0  × 100 ; P = P + g h
 P100  P0  0
 

V
Constant Pressure : T =  V  V T0
(c) Electrical Resistance Thermometer :

T=  Rt  R0  × 100
 R100  R 
 0 

Thermal Expansion :
(a) Linear :

L or = L (1
 = L0T L 0 + T)

Page # 35

(b) Area/superficial :

A or A = A (1 + T)
 = A0T 0
(c) volume/ cubical :

V or V = V0 (1 +  T)
r = V0T

 
23

Thermal stress of a material :

F  Y 
A
Energy stored per unit volume :

1 or E  1 AY (L)2
E = K(L)2 2L

2

Variation of time period of pendulum clocks :

1
T = 2 T
T’ < T - clock-fast : time-gain
T’ > T - clock slow : time-loss

CALORIMETRY : dQ dT
Q dt = – KA dx

Specific heat S = 
m.T R = KA
Q

Molar specific heat C = n.T
Water equivalent = mWSW
HEAT TRANSFER

Thermal Conduction :

Thermal Resistance :

Page # 36

Series and parallel combination of rod :

(i) Series :  eq = 1   2  ....... (when A1 = A2 = A3 = .........)
(ii) Parallel : K eq K1 K2

Keq Aeq = K1 A1 + K2 A2 + ...... (when  1 =  =  = .........)
2 3

for absorption, reflection and transmission
r+t+a=1

Emissive power : U
E = A t

Spectral emissive power : dE
E=

 d

Emissivity : E of a body at T temp.
e = E of a black body at T temp.

E (body )
Kirchoff’s law : a(body ) = E (black 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 – u = e A (T4 – T 4)
0 0

d
Newton’s law of cooling : dt = k ( – 0) ;  = 0 + (i – 0) e–k t

ELECTROSTATICS

Coulomb force between two point charges

  1 |qr1q|32  =  1 |qr1q|22 rˆ
F 4 0  r r 40r

 The electric field intensity at any point is the force experienced

by unit positive charge, given by  
E F

q0

 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

(i) Point charge E= Kq  rˆ = Kq  V = Kq
 r3 r, r
| r |2

(ii) Infinitely long line charge  rˆ = 2Krˆ
2 0r r

V = not defined, v B – v = –2K ln (r / r )
A BA

(iii) Infinite nonconducting thin sheet  nˆ ,
20

V = not defined, vB  vA  rB  rA 


20

(iv) Uniformly charged ring

KQx

 E = R2 3/2 , E =0
axis centre
 x2

Vaxis = KQ KQ
R2  x2 , Vcentre = R

x is the distance from centre along axis.

(v) Infinitely large charged conducting sheet  nˆ
0

V = not defined, vB  vA   rB  rA 
0

(vi) Uniformly charged hollow conducting/ nonconducting /solid

conducting sphere

(a) for   kQ rˆ , r R, V= KQ
E   r
| r |2

(b)  for r < R, V= KQ
E0 R

Page # 38

(vii) Uniformly charged solid nonconducting sphere (insulating material)

 kQ rˆ KQ
E  r
(a) | r |2 for r  R,V=

(b)      for r  R, 
E KQ r r V = 60 (3R2–r2)
R3 30

(viii) thin uniformly charged disc (surface charge density is )

  x  R2  x2  x
1   20 
Eaxis = 20  Vaxis =
R2  x2 

 Work done by external agent in taking a charge q from A to B is

(W ) = q (V – V ) or (W ) = q (V – V ) .
ext AB BA el AB AB

 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) ....

1
 Energy Density = E2

2

 Self Energy of a uniformly charged shell = Uself  KQ2
2R

 Self Energy of a uniformly charged solid non-conducting sphere

= Uself  3KQ2
5R

 Electric Field Intensity Due to Dipole

 2KP
(i) on the axis E = r3

 
KP
(ii) on the equatorial position : E = – r3

(iii) Total electric field at general point O (r,) is E= KP 1 3 cos 2 
res
r3

Page # 39

 Potential =En- epr.gEy of an Electric Dipole in External Electric Field:
U

 Electric Dipole in Uniform Electric Field :
 
torque   p x E; F =0


 Electric Dipole in Nonuniform Electric Field:

torque   p x   Net force |F| = p E
 E; r
U =  p E ,

 Electric Potential Due to Dipole at General Point (r, ) :
p . r
Pcos
V = 40r2  40r3

 The electric flux over the whole area is given by


 E =
E.dS = SEndS

S

 Flux using Gauss's law, Flux through a closed surface

E =  qin .
E  dS = 0

 Electric field intensity near the conducting surface

=  nˆ
0

 Electric pressure : Electric pressure at the surface of a conductor is
given by formula

2
P = 20 where  is the local surface charge density.
 Potential difference between points A and B

B 

VB – VA = – E.dr

A

 =  ˆi  V  ˆj  V  kˆ  V  = –  ˆi   ˆj   kˆ   V
E x x z x x z 

= – V = –grad V

Page # 40

CURRENT ELECTRICITY

1. ELECTRIC CURRENT
q

Iav = t and instantaneous current

i =. Lim q  dq
t dt
t0

2. ELECTRIC CURRENT IN A CONDUCTOR
I = nAeV.

vd  
,

1  eE 2
2 m  = 1 eE  ,
vd   2m

I = neAVd
3. CURRENT DENSITY

 dI 
J n
ds

4. ELECTRICAL RESISTANCE

I = neAVd = neA  eE  =  ne2  AE
 2m   2m 

V I =  ne2   A  V =  A  V = V/R  V = IR
E=  so  2m     

 is called resistivity (it is also called specific resistance) and

2m 1
= ne2 =  ,  is called conductivity. Therefore current in conductors
is proportional to potential difference applied across its ends. This is
Ohm's Law.
Units:

R  ohm(),   ohm  meter(  m)

also called siemens,   1m1.

Page # 41

Dependence of Resistance on Temperature :

R = R (1 + ).
o

Electric current in resistance

I= V2  V1
R

5. ELECTRICAL POWER

P = V

Energy = pdt

V2
P = I2R = V = .

R

H = Vt = 2Rt = V 2 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 = R + R + R +................ + Rn (this means Req is greater then any
123

resistor) ) and

V = V1 + V2 + V3 +................ + Vn .

V1 = R1  R2 R1 V ; V2 = R2 V ;
 ......... Rn R1  R2  ......... Rn

2. Resistances in Parallel :

Page # 42

11. WHEATSTONE NETWORK : (4 TERMINAL NETWORK)

When current through the galvanometer is zero (null point or balance

PR
point) Q = S , then PS = QR
13. GROUPING OF CELLS
13.1 Cells in Series :



Equivalent EMFEeq = E1  E2  .......  En [write EMF's with polarity]
Equivalent internal resistance req = r1  r2  r3  r4  ....  rn

13.2 Cells in Parallel:

Eeq  1  2 r2  ....  n rn [Use emf with polarity]
r1 
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

IG . RG = (I – IG)S

 G .R G S = G RG when  >> G.
S =   G 

where  = Maximum current that can be measured using the given

ammeter.

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

V If V

R= G –R R << R  R 
S G GS S G

17. POTENTIOMETER


=

rR

V – V =  .R
A B Rr

Potential gradient (x)  Potential difference per unit length of wire

x = VA  VB =  R
.
L Rr L

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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
in case ,
 = x ....(1)
11

In figure (3) is joint to (2) then balance length = 2

 = x ....(2)
22

1  1
2 2

If any one of 1 or 2 is known the other can be found. If x is known then
both  and  can be found

12

(b) To find current if resistance is known

VA – VC = x
1
x1
IR =
1

x 1
 = R1

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.

Ist arrangement 2nd arrangement

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by first arrangement ’ = x1 ...(1)
by second arrangement IR = x

2

 = x2 , '
R also  = r'R

 ' = x 2  x1 = x 2
r'R R r'R R

r’ =  1   2  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  

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Applying the condition for balanced Wheatstone bridge, we get R Q = P X

Q 100  
 x=R or X =  R
P

Since R and  are known, therefore, the value of X can be calculated.

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 = 1 Q2 = QV
CV2 =
2 2C 2

(iv) Energy density = 1 r E2 = 1
2 2 K E2

r = Relative permittivity of the medium.
K= r : Dielectric Constant

For vacuum, energy density = 1 E2
2

(v) Types of Capacitors :

(a) Parallel plate capacitor

C = 0r A = K 0A
d d

A : Area of plates

d : distance between the plates( << size of plate )

(b) 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

ab b 12
C= 4 (b  a)
 C = 40K2ab a

(b  a) b aK1 K2 K3

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(c) Cylindrical Capacitor :  >> {a,b}

Capacitance per unit length = 2 F/m  b
n(b / a)


(vi) Capacitance of capacitor depends on
(a) Area of plates
(b) Distance between the plates
(c) Dielectric medium between the plates.

(vii) Electric field intensity between the plates of capacitor

V
E = 0  d
Surface change density

q2
(viii) Force experienced by any plate of capacitor : F = 2A0

2. 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= C1V1  C2V2 = Total charge
C1  C2 Total capacitance

(b) Q1' = C1V = C1 (Q1 + Q2)
C1  C2

C2
Q2' = C2 V = C1  C2 (Q1 +Q2)

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(c) Heat loss during redistribution :

H = U – U = 1 C1C2 (V – V )2
i f C1  C2 12
2

The loss of energy is in the form of Joule heating in the wire.

3. Combination of capacitor :
(i) Series Combination

1 111 V1 : V2 : V3  1 : 1 : 1
 C1 C2 C3

Ceq C1 C2 C3

+Q –Q +Q –Q +Q –Q
C1 C2
C3
V1 V2 V3

(ii) Parallel Combination :

Q+ –Q
C1

Q+ –Q
C2

Q+ –Q
C3

V Q1: Q2 :Q3 = C1 : C2 : C3

Ceq = C1 + C2 + C3

4. Charging and Discharging of a capacitor :
(i) Charging of Capacitor ( Capacitor initially uncharged ):
q = q0 ( 1 – e– t /)

R

VC

q0 = Charge on the capacitor at steady state
q0 = CV

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Time constant = CReq.

I = q0 V
e – t /  e– t / 
R

(ii) Discharging of Capacitor :
q = q e –t/

0

q0 = Initial charge on the capacitor

I = q0 e – t / 


q

R
q0

C
0.37v0

t

5. Capacitor with dielectric :
(i) Capacitance in the presence of dielectric :

C= K0A = KC0
d

+ +
– – – – – – – – – – – – b

V 0 b0

+ + + + + + + + + + + + + + +b
– –

C = Capacitance in the absence of dielectric.
0

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