Formulae Sheets for Physics by Jitender Singh

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f - p h y s i c s . c om u

Speed of light Planck constant Gravitation constant Boltzmann constant Molar gas constant Avogadro’s number Charge of electron Permeability of vacuum Permitivity of vacuum Coulomb constant Faraday constant Mass of electron Mass of proton Mass of neutron Atomic mass unit Atomic mass unit Stefan-Boltzmann constant Rydberg constant Bohr magneton Bohr radius Standard atmosphere Wien displacement constant

c h hc G k R NA e µ0

Projectile Motion:

3 ⇥ 108 m/s 6.63 ⇥ 10 34 J s 1242 eV-nm 6.67 ⇥ 10 11 m3 kg 1 s 1.38 ⇥ 10 23 J/K 8.314 J/(mol K) 6.023 ⇥ 1023 mol 1 1.602 ⇥ 10 19 C 4⇡ ⇥ 10 7 N/A2

R

y = ut sin ✓ 12 gt2 g y = x tan ✓ x2 2u2 cos2 ✓ 2u sin ✓ u2 sin 2✓ u2 sin2 ✓ T = , R= , H= g g 2g x = ut cos ✓,

R1 µB a0 atm b

1.097 ⇥ 107 m 1 9.27 ⇥ 10 24 J/T 0.529 ⇥ 10 10 m 1.01325 ⇥ 105 Pa 2.9 ⇥ 10 3 m K

1.3: Newton’s Laws and Friction Linear momentum: p~ = m~v Newton’s first law: inertial frame. Newton’s second law: F~ =

Banking angle:

v2 rg

= tan ✓,

Pseudo force: F~pseudo =

Notation: ~a = ax ˆı + ay |ˆ + az kˆ q Magnitude: a = |~a| = a2x + a2y + a2z

~ a ⇥ ~b

ˆ k

~ a

ax bz )ˆ | + (ax by

m~a0 ,

v2 r mv 2 r

Fcentrifugal =

Average and Instantaneous Vel. and Accel.: ~vav =

~r/ t,

~vinst = d~r/dt

~aav =

~v / t

~ainst = d~v /dt

Motion in a straight line with constant a:

~vB

c 2016 by Jitender Singh. Ver. 2016

l l cos ✓ g

✓

✓ T

1.4: Work, Power and Energy ~ = F S cos ✓, Work: W = F~ · S

W =

Kinetic energy: K = 12 mv 2 =

p2 2m

Potential energy: F =

v2

q

mg

ay bx )kˆ

1.2: Kinematics

Relative Velocity: ~vA/B = ~vA

ac =

|ˆ

|~a ⇥ ~b| = ab sin ✓

s = ut + 12 at2 ,

µ+tan ✓ 1 µ tan ✓

ˆ ı

~b ✓

v = u + at,

=

mv 2 r ,

Conical pendulum: T = 2⇡

Dot product: ~a · ~b = ax bx + ay by + az bz = ab cos ✓

az by )ˆı + (az bx

v2 rg

fkinetic = µk N

Minimum speed to complete vertical circle: p p vmin, bottom = 5gl, vmin, top = gl

1.1: Vectors

~a ⇥~b = (ay bz

F~BA

Frictional force: fstatic, max = µs N,

MECHANICS

Cross product:

F~ = m~a

d~ p dt ,

Newton’s third law: F~AB =

Centripetal force: Fc =

1

x

H

2

F me mp mn u u

1 4⇡✏0

y

✓ u cos ✓

O

8.85 ⇥ 10 12 F/m 9 ⇥ 109 N m2 /C2 96485 C/mol 9.1 ⇥ 10 31 kg 1.6726 ⇥ 10 27 kg 1.6749 ⇥ 10 27 kg 1.66 ⇥ 10 27 kg 931.49 MeV/c2 5.67 ⇥ 10 8 W/(m2 K4 )

✏0

u sin ✓

0.1: Physical Constants

u2 = 2as

R

~ F~ · dS

@U/@x for conservative forces.

Ugravitational = mgh,

Uspring = 12 kx2

Work done by conservative forces is path independent and depends only on initial and final points: H F~conservative · d~r = 0. Work-energy theorem: W =

K

Mechanical energy: E = U + K. Conserved if forces are conservative in nature. Power Pav =

W t

,

Pinst = F~ · ~v

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1.5: Centre of Mass and Collision Centre of mass: xcm =

P Pxi mi , mi

mr 2

xcm =

CM of few useful configurations:

R R xdm dm

m1

ring

m2 r m1 +m2

2r ⇡

4. Semicircular disc: yc =

4r 3⇡

6. Solid Hemisphere: yc =

m1 r m1 +m2

h 3

C

h

C

r

r

3r 8

2r ⇡

r

C

4r 3⇡

~vcm = Impulse: J~ =

R

mi~vi , M

p~cm = M~vcm ,

F~ dt =

~acm

disk

shell

2 1 2 2 2 mr m(a +b ) 12

mr 2

sphere

rod

hollow

b

solid rectangle

Ik

Theorem of Parallel Axes: Ik = Icm + md2

z

Radius of Gyration: k =

C

r 2

C

3r 8

F~ext = M

p~

p

y x

I/m

~ = ~r ⇥ p~, Angular Momentum: L ~⌧ =

Ic d cm

Theorem of Perp. Axes: Iz = Ix + Iy

Torque: ~⌧ = ~r ⇥ F~ ,

7. Cone: the height of CM from the base is h/4 for the solid cone and h/3 for the hollow cone. P Motion of the CM: M = mi P

2 1 12 ml

h 3

r

r 2

5. Hemispherical shell: yc =

2 2 5 mr

m2

r C

3. Semicircular ring: yc =

2 2 3 mr

a

1. m1 , m2 separated by r:

2. Triangle (CM ⌘ Centroid) yc =

2 1 2 mr

f - p h y s i c s . c om

~ dL dt ,

~ = I~ L ! y

P ✓ ~ F ~ r x

⌧ = I↵ O

~ ~⌧ext = 0 =) L ~ = const. Conservation of L: P~ P Equilibrium condition: F = ~0, ~⌧ = ~0 Kinetic Energy: Krot = 12 I! 2 Dynamics: F~ext = m~acm , p~cm = m~vcm 2 2 1 1 ~ K = 2 mvcm + 2 Icm ! , L = Icm ! ~ + ~rcm ⇥ m~vcm

~⌧cm = Icm ↵ ~,

1.7: Gravitation Before collision After collision

Collision:

m1

m2

v1

m1

m2

v10

v2

v20

Momentum conservation: m1 v1 +m2 v2 = m1 v10 +m2 v20 2 2 Elastic Collision: 12 m1 v1 2+ 12 m2 v2 2 = 12 m1 v10 + 12 m2 v20 Coefficient of restitution: ⇢ (v10 v20 ) 1, e= = 0, v1 v2

If v2 = 0 and m1 ⌧ m2 then v10 = v1 . If v2 = 0 and m1 m2 then v20 = 2v1 . Elastic collision with m1 = m2 : v10 = v2 and v20 = v1 .

F F

m2

r

Potential energy: U =

GM m r

Gravitational acceleration: g = completely elastic completely in-elastic

m1

2 Gravitational force: F = G mr1 m 2

GM R2

Variation of g with depth: ginside ⇡ g 1 Variation of g with height: goutside ⇡ g 1

2h R h R

E↵ect of non-spherical earth shape on g: gat pole > gat equator (* Re Rp ⇡ 21 km) E↵ect of earth rotation on apparent weight: ! ~

1.6: Rigid Body Dynamics Angular velocity: !av = Angular Accel.: ↵av =

✓ t, ! t,

!= ↵=

d✓ dt ,

d! dt ,

~v = ! ~ ⇥ ~r

mg✓0 = mg

mg

m! 2 R cos2 ✓

m! 2 R cos ✓

✓ R

~a = ↵ ~ ⇥ ~r

Rotation about an axis with constant ↵: ! = !0 + ↵t,

✓ = !t + 12 ↵t2 ,

Moment of Inertia: I =

P

i

mi ri 2 ,

!2 I=

c 2016 by Jitender Singh. Ver. 2016

R

!0 2 = 2↵✓

Orbital velocity of satellite: vo =

r2 dm Escape velocity: ve =

q

q

GM R

2GM R

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Kepler’s laws: a

Compressibility: K = Poisson’s ratio:

1 B

F/A l/l ,

=

B=

P V

F A✓

, ⌘=

1 dV V dP

lateral strain longitudinal strain 1 2

V

=

D/D l/l

stress ⇥ strain ⇥ volume

kx (for small elongation x.)

Acceleration: a =

d2 x dt2

Time period: T =

2⇡ !

k mx

= = 2⇡

q

=

Surface tension: S = F/l

!2 x

Surface energy: U = SA

k m

Excess pressure in bubble:

Displacement: x = A sin(!t + ) p Velocity: v = A! cos(!t + ) = ±! A2

pair = 2S/R,

x2

Capillary rise: h = U

Potential energy: U = 12 kx2

0

A

x

A

K 0

A

x

A

Total energy: E = U + K = 12 m! 2 A2

Simple pendulum: T = 2⇡

q

Physical Pendulum: T = 2⇡

=

1 k1

2S cos ✓ r⇢g

Hydrostatic pressure: p = ⇢gh

l

v2

v1

Viscous force: F =

dv ⌘A dx

F

Stoke’s law: F = 6⇡⌘rv

I mgl

q

+

Equation of continuity: A1 v1 = A2 v2

Bernoulli’s equation: p + 12 ⇢v 2 + ⇢gh = constant p Torricelli’s theorem: ve✏ux = 2gh

l g

q

Torsional Pendulum T = 2⇡

1 keq

psoap = 4S/R

Buoyant force: FB = ⇢V g = Weight of displaced liquid

Kinetic energy K = 12 mv 2

Springs in series:

=

Elastic energy: U =

1.8: Simple Harmonic Motion Hooke’s law: F =

1.9: Properties of Matter Modulus of rigidity: Y =

First: Elliptical orbit with sun at one of the focus. ~ Second: Areal velocity is constant. (* dL/dt = 0). 4⇡ 2 3 Third: T 2 / a3 . In circular orbit T 2 = GM a .

f - p h y s i c s . c om

v

Poiseuilli’s equation:

Volume flow time

I k

1 k2

Terminal velocity: vt =

k1

⇡pr 4 8⌘l

r l

)g

k2

k2 k1

Springs in parallel: keq = k1 + k2

~ A

Superposition of two SHM’s:

2r 2 (⇢ 9⌘

=

✏

~2 A

~1 A

x1 = A1 sin !t,

x2 = A2 sin(!t + )

x = x1 + x2 = A sin(!t + ✏) q A = A1 2 + A2 2 + 2A1 A2 cos tan ✏ =

A2 sin A1 + A2 cos

c 2016 by Jitender Singh. Ver. 2016

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Waves

f - p h y s i c s . c om

4. 1st overtone/2nd harmonics: ⌫1 =

2.1: Waves Motion

2 2L

5. 2nd overtone/3rd harmonics: ⌫2 =

General equation of wave:

@2y @x2

=

2 1 @ y v 2 @t2 .

q

T µ

q

3 2L

6. All harmonics are present.

Notation: Amplitude A, Frequency ⌫, Wavelength , Period T , Angular Frequency !, Wave Number k, 1 2⇡ T = = , ⌫ !

v=⌫ ,

k=

2⇡

y = f (t

x/v),

+x;

String fixed at one end:

N

A

1. Boundary conditions: y = 0 at x = 0

x

y A

y = A sin(kx

x 2

!t) = A sin(2⇡ (x/

t/T ))

2.2: Waves on a String

2. Allowed Freq.: L = (2n + 1) 4 , ⌫ = 2n+1 4L 0, 1, 2, . . .. q 1 T 3. Fundamental/1st harmonics: ⌫0 = 4L µ q 3 T 4. 1st overtone/3rd harmonics: ⌫1 = 4L µ q 5 T 5. 2nd overtone/5th harmonics: ⌫2 = 4L µ

q

T µ,

n =

6. Only odd harmonics are present.

Speed of waves on a string with mass per unit length µ p and tension T : v = T /µ 2

2 2

Transmitted power: Pav = 2⇡ µvA ⌫ Interference: y1 = A1 sin(kx

A

N

/2

y = f (t + x/v),

Progressive sine wave:

T µ

L

Progressive wave travelling with speed v:

Sonometer: ⌫ /

1 L,

⌫/

p

T, ⌫ /

p1 . µ

⌫=

n 2L

q

T µ

2.3: Sound Waves !t),

y2 = A2 sin(kx

!t + )

y = y1 + y2 = A sin(kx !t + ✏) q A = A1 2 + A2 2 + 2A1 A2 cos

Standing Waves:

A

N

Displacement wave: s = s0 sin !(t Pressure wave: p = p0 cos !(t

x/v)

x/v), p0 = (B!/v)s0

Speed of sound waves: s s B Y vliquid = , vsolid = , ⇢ ⇢

A2 sin tan ✏ = A1 + A2 cos ⇢ 2n⇡, constructive; = (2n + 1)⇡, destructive. 2A cos kx

2

S h e e t f or

Intensity: I = A

N

A

2⇡ 2 B 2 2 v s0 ⌫

=

p0 2 v 2B

=

vgas =

s

P ⇢

p0 2 2⇢v

x

Standing longitudinal waves:

/4

p1 = p0 sin !(t y1 = A1 sin(kx

!t),

y2 = A2 sin(kx + !t)

x/v),

p2 = p0 sin !(t + x/v)

p = p1 + p2 = 2p0 cos kx sin !t

y = y1 + y2 = (2A cos kx) sin !t ⇢ n + 12 2 , nodes; n = 0, 1, 2, . . . x= n2, antinodes. n = 0, 1, 2, . . .

String fixed at both ends:

N

A

N

A

N

1. Boundary condition: y = 0 at x = 0

/2

1. Boundary conditions: y = 0 at x = 0 and at x = L q n T 2. Allowed Freq.: L = n 2 , ⌫ = 2L µ , n = 1, 2, 3, . . .. q 1 T 3. Fundamental/1st harmonics: ⌫0 = 2L µ c 2016 by Jitender Singh. Ver. 2016

L

Closed organ pipe:

L

v 2. Allowed freq.: L = (2n + 1) 4 , ⌫ = (2n + 1) 4L , n= 0, 1, 2, . . .

3. Fundamental/1st harmonics: ⌫0 =

v 4L

4. 1st overtone/3rd harmonics: ⌫1 = 3⌫0 =

3v 4L

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5. 2nd overtone/5th harmonics: ⌫2 = 5⌫0 =

Path di↵erence:

6. Only odd harmonics are present.

f - p h y s i c s . c om ✓

d S2

A N

Open organ pipe:

L

A N A

Phase di↵erence:

x=

v 2L

3. 1st overtone/2nd harmonics: ⌫1 = 2⌫0 =

2v 2L

4. 2nd overtone/3rd harmonics: ⌫2 = 3⌫0 =

3v 2L

2⇡

D

x

Interference Conditions: for integer n, ⇢ 2n⇡, constructive; = (2n + 1)⇡, destructive,

1. Boundary condition: y = 0 at x = 0 v Allowed freq.: L = n 2 , ⌫ = n 4L , n = 1, 2, . . . 2. Fundamental/1st harmonics: ⌫0 =

=

P y

S1

dy D

x=

⇢

n , n+

1 2

,

constructive; destructive

Intensity: p I = I1 + I2 + 2 I1 I2 cos , ⇣p ⇣p p ⌘2 Imax = I1 + I2 , Imin = I1

5. All harmonics are present.

p ⌘2 I2

l2 + d

l1 + d

I1 = I2 : I = 4I0 cos2 2 , Imax = 4I0 , Imin = 0

Resonance column:

Fringe width: w = Optical path:

D d

x0 = µ x

Interference of waves transmitted through thin film: l1 + d = 2 ,

l2 + d =

3 4

,

v = 2(l2

l1 )⌫

Beats: two waves of almost equal frequencies !1 ⇡ !2 p1 = p0 sin !1 (t

x/v),

p = p1 + p2 = 2p0 cos ! = (!1 + !2 )/2,

p2 = p0 sin !2 (t !(t

x/v) sin !(t

! = !1

!2

x = 2µd =

⇢

n , n+

1 2

,

constructive; destructive.

x/v) x/v)

Di↵raction from a single slit:

y ✓

b

y

(beats freq.)

D

For Minima: n = b sin ✓ ⇡ b(y/D)

Doppler E↵ect:

Resolution: sin ✓ =

v + uo ⌫= ⌫0 v us

1.22 b

Law of Malus: I = I0 cos2 ✓

where, v is the speed of sound in the medium, u0 is the speed of the observer w.r.t. the medium, considered positive when it moves towards the source and negative when it moves away from the source, and us is the speed of the source w.r.t. the medium, considered positive when it moves towards the observer and negative when it moves away from the observer.

✓ I0

I

2.4: Light Waves Plane Wave: E = E0 sin !(t Spherical Wave: E =

aE0 r

x v ),

sin !(t

I = I0 r v ),

I=

Young’s double slit experiment

c 2016 by Jitender Singh. Ver. 2016

I0 r2

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Optics

1 f

Lens maker’s formula:

f - p h y s i c s . c om

= (µ

h

1)

3.1: Reflection of Light

1 R1

1 R2

i f

Lens formula:

normal

Laws of reflection:

i r

incident

(i)

reflected

1 v

1 u

=

1 f,

v u

m=

u

Incident ray, reflected ray, and normal lie in the same plane (ii) \i = \r

v

Power of the lens: P = f1 , P in diopter if f in metre. Two thin lenses separated by distance d:

Plane mirror: d

d

1 1 1 = + F f1 f2

(i) the image and the object are equidistant from mirror (ii) virtual image of real object

d f1 f2

d f1

f2

I

Spherical Mirror:

O f u

3.3: Optical Instruments

v

Simple microscope: m = D/f in normal adjustment.

1. Focal length f = R/2 2. Mirror equation:

1 v

+

3. Magnification: m =

Eyepiece

Objective 1 u v u

=

1 f

1

O

Compound microscope:

u

v

3.2: Refraction of Light speed of light in vacuum speed of light in medium

Refractive index: µ = Snell’s Law:

sin i sin r

=

µ2

real depth apparent depth

Critical angle: ✓c = sin

=

1. Magnification in normal adjustment: m =

c v

=

incident µ1 i

µ2 µ1

Apparent depth: µ =

reflected

2. Resolving power: R =

d0 d I O

1 1 µ

µ

i

m

sin A+2 m , sin A2

= (µ

1)A,

i0

r0

r

Astronomical telescope:

i = i0 for minimum deviation

m 0

µ2

P

µ2 v

µ1 µ2 µ1 = , u R

m=

Q

O u

=

L = fo + fe

1 1.22

A

2

y

,

A>0

= (µy

1)A

2. Angular dispersion: ✓ = (µv Dispersive power: ! =

µv µr µy 1

⇡

Dispersion without deviation:

i

µ1

Refraction at spherical surface:

1 ✓

Cauchy’s equation: µ = µ0 +

1. Mean deviation:

i

fo fe ,

Dispersion by prism with small A and i:

general result

for small A

fe

3.4: Dispersion

µ

µ=

2µ sin ✓

1. In normal adjustment: m =

✓c

A

A,

=

refracted

d d0

Deviation by a prism:

1 d

v D u fe

fo r

2. Resolving power: R =

= i + i0

fe D

(µy

1)A + (µ0y

✓ y

µr )A (if A and i small) A

µ0

µ

A0

1)A0 = 0

Deviation without dispersion: (µv µr )A = (µ0v µ0r )A0

v

µ1 v µ2 u

c 2016 by Jitender Singh. Ver. 2016

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Heat and Thermodynamics

4.4: Theromodynamic Processes First law of thermodynamics:

4.1: Heat and Temperature Temp. scales: F = 32 + 95 C,

f - p h y s i c s . c om

n : number of moles

a V2

van der Waals equation: p +

(V

F A

Thermal stress of a material:

=Y

W = p V,

Z

W =

b) = nRT Wisothermal = nRT ln

Thermal expansion: L = L0 (1 + ↵ T ), A = A0 (1 + T ), V = V0 (1 + T ),

U+

W

Work done by the gas:

K = C + 273.16

Ideal gas equation: pV = nRT ,

Q=

= 2 = 3↵

V2

pdV ◆

V1

✓

V2 V1

Wisobaric = p(V2 V1 ) p1 V 1 p2 V 2 Wadiabatic = 1 Wisochoric = 0

l l

4.2: Kinetic Theory of Gases General: M = mNA , k = R/NA

T1 Q1

Efficiency of the heat engine:

n

W Q2

Maxwell distribution of speed:

T2 vp v ¯ vrms

RMS speed: vrms = Average speed: v¯ =

q

q

3kT m

=

8kT ⇡m

=

Most probable speed: vp = 2 Pressure: p = 13 ⇢vrms

q

q

q

v

3RT M

work done by the engine Q1 Q2 = heat supplied to it Q1 Q2 T2 ⌘carnot = 1 =1 Q1 T1 ⌘=

8RT ⇡M

Coe↵. of performance of refrigerator:

2kT m

T2

COP =

Internal energy of n moles of an ideal gas is U = f2 nRT .

Entropy:

Q2 W

Q2 Q1 Q2

=

Q T ,

S=

Const. T :

Sf

S=

Adiabatic process: 4.3: Specific Heat

Si =

Q T,

Rf

Q T

i

Varying T :

Q = 0, pV

Conduction:

Latent heat: L = Q/m

Q t

=

Tf Ti

= constant

T x

KA

x KA

Thermal resistance: R =

Specific heat at constant volume: Cv =

Q n T

Specific heat at constant pressure: Cp = Relation between Cp and Cv : Cp

V

Q n T

Rseries = R1 + R2 =

1 A

p 1 Rparallel

Cv = R

=

1 R1

+

1 R2

=

⇣

1 x

x1 K1

+

x2 K2

⌘

K1

K2

x1

x2

(K1 A1 + K2 A2 )

A2

K1

A1

x

U = nCv T

Kirchho↵ ’s Law:

emissive power absorptive power

=

Ebody abody

Specific heat of gas mixture: n1 Cv1 + n2 Cv2 , n1 + n2

A

K2

= Cp /Cv

Relation between U and Cv :

Cv =

S = ms ln

4.5: Heat Transfer

Q m T

Ratio of specific heats:

W Q2

Equipartition of energy: K = 12 kT for each degree of freedom. Thus, K = f2 kT for molecule having f degrees of freedoms.

Specific heat: s =

T1 Q1

= Eblackbody E

=

n1 Cp1 + n2 Cp2 n1 Cv1 + n2 Cv2

Molar internal energy of an ideal gas: U = f2 RT , f = 3 for monatomic and f = 5 for diatomic gas.

c 2016 by Jitender Singh. Ver. 2016

Wien’s displacement law:

mT = b m

Stefan-Boltzmann law: Newton’s law of cooling:

Q t

= eAT 4

dT dt

=

bA(T

T0 )

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Electricity and Magnetism

5.3: Capacitors Capacitance: C = q/V

5.1: Electrostatics Coulomb’s law: F~ =

1 q1 q2 ˆ 4⇡✏0 r 2 r

~ r) = Electric field: E(~

q1

1 q ˆ 4⇡✏0 r 2 r

Electrostatic potential: V = ~ · ~r, E

r

q

q2

q

Parallel plate capacitor: C = ✏0 A/d

A

+q d

~ E

~ r

A

r2

1 q1 q2 4⇡✏0 r

Electrostatic energy: U =

dV =

f - p h y s i c s . c om

Spherical capacitor: C =

4⇡✏0 r1 r2 r 2 r1

q +q

r1

1 q 4⇡✏0 r

V (~r) =

Z

~ r 1

~ · d~r E

Cylindrical capacitor: C =

2⇡✏0 l ln(r2 /r1 )

r2

l

r1

p ~

Electric dipole moment: p~ = q d~

q

+q

d

Capacitors in parallel: Ceq = C1 + C2

A C1

C2

B 1 p cos ✓ 4⇡✏0 r 2

Potential of a dipole: V =

V (r)

✓ r

Capacitors in series:

p ~ Er

Field of a dipole:

✓ r

E✓

p ~

Er =

1 2p cos ✓ , 4⇡✏0 r3

E✓ =

~ U= Pot. energy of a dipole placed in E:

+

1 C2

q

Drift speed: vd =

x

Q2 2C

= 12 QV

✏0 KA d

1 eE 2 m⌧

=

i neA

Resistance of a wire: R = ⇢l/A, where ⇢ = 1/ P

~ E

Temp. dependence of resistance: R = R0 (1 + ↵ T ) Ohm’s law: V = iR

r

R

r

R

Kirchho↵ ’s Laws: (i) The Junction Law: The algebraic sum of all the currents directed towards a node is zero i.e., ⌃node Ii = 0. (ii)The Loop Law: The algebraic sum of all the potential di↵erences along a closed loop in a circuit is zero i.e., ⌃loop Vi = 0. Resistors in parallel:

E and V of a uniformly charged spherical shell: ⇢ 0, for r < R E E= 1 Q , for r R 2 4⇡✏0 r O R ( 1 Q 4⇡✏0 R , for r < R V V = 1 Q , for r R 4⇡✏0 r O

B

Current density: j = i/A = E

a

E and V (of a uniformly charged sphere: 1 Qr 4⇡✏0 R3 , for r < R E E= 1 Q for r R 4⇡✏0 r 2 , O ( 2 1 Qr V 4⇡✏0 R3 , for r < R V = 1 Q for r R O 4⇡✏0 r ,

C2

5.4: Current electricity

Field of a uniformly charged ring on its axis:

r

1 Req

=

1 R1

+

1 R2

Resistors in series: Req = R1 + R2

A R1

A

R1

R2

R1

R

Wheatstone bridge:

R2

B

r

2⇡✏0 r

Field of an infinite sheet: E =

C1 A

Force between plates of a parallel plate capacitor: Q2 F = 2A✏ 0

Capacitor with dielectric: C =

~ p~ · E

5.2: Gauss’s Law and its Applications H ~ · dS ~ Electric flux: = E H ~ · dS ~ = qin /✏0 Gauss’s law: E

Field of a line charge: E =

1 C1

Energy density in electric field E: U/V = 12 ✏0 E 2

~ ~⌧ = p~ ⇥ E ~ Torque on a dipole placed in E:

EP =

=

Energy stored in capacitor: U = 12 CV 2 =

1 p sin ✓ 4⇡✏0 r 3

qx 1 4⇡✏0 (a2 +x2 )3/2

1 Ceq

B R2

" G R3

R4 V

Balanced if R1 /R2 = R3 /R4 . 2✏0

Field in the vicinity of conducting surface: E =

c 2016 by Jitender Singh. Ver. 2016

Electric Power: P = V 2 /R = I 2 R = IV ✏0

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i

Galvanometer as an Ammeter:

i

i

ig S

ig G = (i

ig )S R

G

A ig

y x

z

5.6: Magnetic Field due to Current C

R

e

w d

B

Charging of capacitors: q(t) = CV 1

~ B

l

Bi ned i

"

VAB = ig (R + G)

h

~ Energy of a magnetic dipole placed in B: ~ U= µ ~ ·B Hall e↵ect: Vw =

Galvanometer as a Voltmeter:

f - p h y s i c s . c om

t RC

~ = Biot-Savart law: dB

✓ d~l

V

i

~ ⌦B

i

µ0 i d~l⇥~ r 4⇡ r 3

~ r

✓2 C

Discharging of capacitors: q(t) = q0 e

t RC

Field due to a straight conductor:

d

i

q(t)

~ ⌦B

✓1

R

B=

Time constant in RC circuit: ⌧ = RC

µ0 i 4⇡d (cos ✓1

cos ✓2 )

Field due to an infinite straight wire: B = H Q

Peltier e↵ect: emf e =

=

Peltier heat charge transferred .

Force between parallel wires:

dF dl

=

i1

µ 0 i1 i2 2⇡d

i2 d

e

Seeback e↵ect:

T0

Tn

T

Ti

a

Field on the axis of a ring:

1. Thermo-emf: e = aT + 12 bT 2

P

i

3. Neutral temp.: Tn = 4. Inversion temp.: Ti = Thomson e↵ect: emf e =

H Q

BP =

a/b.

=

=

T.

Field at the centre of an arc: B =

a

µ0 i✓ 4⇡a

~ B

i

✓ a

Faraday’s law of electrolysis: The mass deposited is 1 F

µ0 ia2 2(a2 +d2 )3/2

2a/b. Thomson heat charge transferred

~ B

d

2. Thermoelectric power: de/dt = a + bT .

m = Zit =

µ0 i 2⇡d

Field at the centre of a ring: B =

Eit Ampere’s law:

where i is current, t is time, Z is electrochemical equivalent, E is chemical equivalent, and F = 96485 C/g is Faraday constant.

H

~ · d~l = µ0 Iin B

Field inside a solenoid: B = µ0 ni, n =

Field inside a toroid: B =

5.5: Magnetism

µ0 i 2a

N l

l

µ0 N i 2⇡r

r

~ + qE ~ Lorentz force on a moving charge: F~ = q~v ⇥ B ~2 B

Charged particle in a uniform magnetic field: v q

r=

mv qB ,

T =

Field of a bar magnet:

~1 B

d

~⌦ r B

B1 = ~ B

Force on a current carrying wire:

µ0 2M 4⇡ d3 ,

B2 =

µ0 M 4⇡ d3

Angle of dip: Bh = B cos

~l ~ F

Bv

Tangent galvanometer: Bh tan ✓ =

Magnetic moment of a current loop (dipole): ~ µ ~ A ~ µ ~ = iA i

µ0 ni 2r ,

~ = µH ~ Permeability: B

B

i = K tan ✓

Moving coil galvanometer: niAB = k✓, Time period of magnetometer: T = 2⇡

~ ~⌧ = µ ~ Torque on a magnetic dipole placed in B: ~ ⇥B

Bh

Horizontal

i

~ F~ = i ~l ⇥ B

c 2016 by Jitender Singh. Ver. 2016

d N

S

2⇡m qB

i= q

k nAB ✓

I M Bh

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C

5.7: Electromagnetic Induction H ~ · dS ~ Magnetic flux: = B

RC circuit: Z=

d dt

Faraday’s law: e =

Lenz’s Law: Induced current create a B-field that opposes the change in magnetic flux.

Z=

Motional emf: e = Blv

l

~ v

p

p

˜

+

(1/!C)2 ,

tan

Self inductance of a solenoid: L = µ0 n2 (⇡r2 l) h Growth of current in LR circuit: i = Re 1 e L

t L/R

i

R

= L

i

Z=

R

R !L

˜

Z

e0 sin !t

R 2 + ! 2 L2 ,

tan L

e=

R

1 !CR

i

LCR Circuit: = Li,

Z

1 !C

e0 sin !t

R2

=

~ ⌦B

di L dt

R

i

LR circuit:

+

Self inductance:

f - p h y s i c s . c om

q

C

!L R

R

˜

e0 sin !t

R2 +

⌫resonance =

1 !C

Z

!L

R

i

1 !C q

1 2⇡

2

!L ,

tan

=

1 !C

1 !C

!L

!L R

1 LC

Power factor: P = erms irms cos

e 0.63 R

e i

S

t

L R

Transformer:

Decay of current in LR circuit: i = i0 e L

t L/R

=

e1 e2 ,

e 1 i1 = e2 i2

e1

˜

N1 i1

N2 i2

˜

e2

p Speed of the EM waves in vacuum: c = 1/ µ0 ✏0

i

R

N1 N2

i0 0.37i0

S

i

t

L R

Time constant of LR circuit: ⌧ = L/R Energy stored in an inductor: U = 12 Li2 U V

Energy density of B field: u = Mutual inductance:

= M i,

=

e=

B2 2µ0 di M dt

EMF induced in a rotating coil: e = N AB! sin !t i

Alternating current:

t T

i = i0 sin(!t + ),

T = 2⇡/! RT Average current in AC: ¯i = T1 0 i dt = 0 RMS current: irms =

h R 1 T T

0

i2 dt

i1/2

=

i0 p 2

i2 t T

Energy: E = irms 2 RT Capacitive reactance: Xc =

1 !C

Inductive reactance: XL = !L Imepedance: Z = e0 /i0 Visit www.concepts-of-physics.com to buy “IIT JEE Physics (1978-2015: 38 Year) Topic-wise Complete Solutions”. Foreword by Prof HC Verma.

c 2016 by Jitender Singh. Ver. 2016

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f - p h y s i c s . c om

Modern Physics

N N0

Population at time t: N = N0 e

6.1: Photo-electric e↵ect

t

N0 2

t1/2

O

Photon’s energy: E = h⌫ = hc/ Photon’s momentum: p = h/ = E/c

Half life: t1/2 = 0.693/

Max. KE of ejected photo-electron: Kmax = h⌫

Average life: tav = 1/

Threshold freq. in photo-electric e↵ect: ⌫0 = /h

Population after n half lives: N = N0 /2n . Mass defect:

V0

Stopping potential: Vo =

hc e

hc e

1 e

1

hc e

de Broglie wavelength:

m = [Zmp + (A

Binding energy: B = [Zmp + (A Q-value: Q = Ui

= h/p

Z)mn ]

t

M M ] c2

Z)mn

Uf

Energy released in nuclear reaction: where m = mreactants mproducts .

E

mc2

=

6.2: The Atom 6.4: Vacuum tubes and Semiconductors

Energy in nth Bohr’s orbit: mZ 2 e4 , 8✏0 2 h2 n2

En =

En =

13.6Z 2 eV n2

Half Wave Rectifier:

a0 = 0.529 ˚ A

Full Wave Rectifier:

D

Radius of the nth Bohr’s orbit: rn =

✏ 0 h2 n 2 , ⇡mZe2

rn =

n2 a 0 , Z

Quantization of the angular momentum: l = Photon energy in state transition: E2 E2

˜

Output

nh 2⇡

E1 = h⌫

Grid

Triode Valve:

Cathode Filament

Plate

E2 h⌫

E1

R Output

˜

h⌫ E1 Absorption

Emission

Wavelength of emitted radiation: for from nth to mth state:  1 1 1 2 = RZ n2 m2

a

transition

min

=

p

Vp Vg

Vg =0 ip Vg

Vp =0

ip =0

Relation between rp , µ, and gm : µ = rp ⇥ gm K

hc eV

K↵ Ie

Ic

Current in a transistor: Ie = Ib + Ic min

Moseley’s law:

Transconductance of a triode: gm = Amplification by a triode: µ =

I

X-ray spectrum:

Vp ip

Plate resistance of a triode: rp =

↵

Ib

⌫ = a(Z

b)

↵ and

X-ray di↵raction: 2d sin ✓ = n

Ic Ib ,

Heisenberg uncertainity principle: p x h/(2⇡), E t h/(2⇡)

parameters of a transistor: ↵ = = 1 ↵↵

Transconductance: gm =

Ic Ie ,

=

Ic Vbe

Logic Gates: 6.3: The Nucleus Nuclear radius: R = R0 A1/3 , Decay rate:

dN dt

=

R0 ⇡ 1.1 ⇥ 10

15

m

A 0 0 1 1

B 0 1 0 1

AND AB 0 0 0 1

OR A+B 0 1 1 1

NAND AB 1 1 1 0

NOR A+B 1 0 0 0

XOR ¯ + AB ¯ AB 0 1 1 0

N

c 2016 by Jitender Singh. Ver. 2016

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IIT JEE Physics Book 7.1: Book Description Two IIT batch-mates have worked together to provide a high quality Physics problem book to Indian students. It is an indispensable collection of previous 38 years IIT questions and their illustrated solutions for any serious aspirant. The success of this work lies in making the readers capable to solve complex problems using few basic principles. The readers are also asked to attempt variations of the solved problems to help them understand the concepts better. Key features of the book are: • 1300+ solved problems in 2 volumes • Concept building by problem solving • IIT preparation with school education • Topic and year-wise content arrangement • Promotes self learning • Quality typesetting and figures The readers can use the book as a readily available mentor for providing hints or complete solutions as per their needs. 7.2: About Contents The volume 1 of the book covers three parts: Mechanics, Waves, and Optics. The list of chapters in this volume are: 1. Units and Measurements 2. Rest and Motion: Kinematics 3. Newton’s Laws of Motion 4. Friction 5. Circular Motion 6. Work and Energy 7. Centre of Mass, Linear Momentum, Collision 8. Rotational Mechanics 9. Gravitation 10. Simple Harmonic Motion 11. Fluid Mechanics 12. Some Mechanical Properties of Matter 13. Wave Motion and Waves on a String 14. Sound Waves 15. Light Waves 16. Geometrical Optics 17. Optical Instruments 18. Dispersion and Spectra 19. Photometry The volume 2 of the book covers three parts: Thermodynamics, Electromagnetism, and Modern Physics. The list of chapters in this volume are: 20. Heat and Temperature 21. Kinetic Theory of Gases 22. Calorimetry 23. Laws of Thermodynamics 24. Specific Heat Capacities of Gases 25. Heat Transfer 26. Electric Field and Potential 27. Gauss’s Law 28. Capacitors 29. Electric Current in Conductors 30. Thermal and Chemical E↵ects of Electric Current 31. Magnetic Field 32. Magnetic Field due to a Current 33. Permanent Magnets

c 2016 by Jitender Singh. Ver. 2016

34. 35. 36. 37. 38. 39. 40. 41. 42.

f - p h y s i c s . c om

Electromagnetic Induction Alternating Current Electromagnetic Waves Electric Current through Gases Photoelectric E↵ect and Wave-Particle Duality Bohr’s Model and Physics of the Atom X-rays Semiconductors and Semiconductor Devices The Nucleus

7.3: About Authors Jitender Singh is working as a Scientist in DRDO. He has a strong academic background with Integrated M. Sc. (5 years) in Physics from IIT Kanpur and M. Tech. in Computational Science from IISc Bangalore. He is All India Rank 1 in GATE and loves to solve physics problems. He is a member of Prof. H.C. Verma’s team which is dedicated to improve the quality of physics education in the country. Shraddhesh Chaturvedi holds a degree in Integrated M. Sc. (5 years) in Physics from IIT Kanpur. He is passionate about problem solving in physics and enhancing the quality of texts available to Indian students. His career spans many industries where he has contributed with his knowledge of physics and mathematics. An avid reader and a keen thinker, his philosophical writings are a joy to read. 7.4: Where to Buy Please visit www.concepts-of-physics.com to find links to the books page on Amazon. This website also give additional information about the book.

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