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Dual Nature Of Light

Matter waves (de-Broglie Waves).

According to de-Broglie a moving material particle sometimes acts as a wave and sometimes as a particle.


A wave is associated with moving material particle which control the particle in every respect.

The wave associated with moving particle is called matter wave or de-Broglie wave and it propagates in the form of wave packets with group velocity.





(1) de-Broglie wavelength

According to de-Broglie theory, the wavelength of de-Broglie wave is given by


Where h = Plank’s constant, m = Mass of the particle, v = Speed of the particle, E = Energy of the particle.

The smallest wavelength whose measurement is possible is that of -rays.

The wavelength of matter waves associated with the microscopic particles like electron, proton, neutron, a-particle etc. is of the order of m.

(i) de-Broglie wavelength associated with the charged particles.

The energy of a charged particle accelerated through potential difference V is

Hence de-Broglie wavelength

        Å, Å, Å, Å

(ii) de-Broglie wavelength associated with uncharged particles.

For Neutron de-Broglie wavelength is given as Å

Energy of thermal neutrons at ordinary temperature

; where k = Boltzman’s constant = Joules/kelvin , T = Absolute temp.

So Å


(2) Some graphs









Note : q    A photon is not a material particle. It is a quanta of energy.

  • When a particle exhibits wave nature, it is associated with a wave packet, rather then a wave.

(3) Characteristics of matter waves

(i) Matter wave represents the probability of finding a particle in space.

(ii) Matter waves are not electromagnetic in nature.

(iii) de-Brogile or matter wave is independent of the charge on the material particle. It means, matter wave of de-Broglie wave is associated with every moving particle (whether charged or uncharged).

(iv) Practical observation of matter waves is possible only when the de-Broglie wavelength is of the order of the size of the particles is nature.

(v) Electron microscope works on the basis of de-Broglie waves.

(vi) The electric charge has no effect on the matter waves or their wavelength.

(vii) The phase velocity of the matter waves can be greater than the speed of the light.

(viii) Matter waves can propagate in vacuum, hence they are not mechanical waves.

(ix) The number of de-Broglie waves associated with nth orbital electron is n.

(x) Only those circular orbits around the nucleus are stable whose circumference is integral multiple of de-Broglie wavelength associated with the orbital electron.

(4) Davision and Germer experiment

It is used to study the scattering of electron from a solid or to verify the wave nature of electron. A beam of electrons emitted by electron gun is made to fall on nickel crystal cut along cubical axis at a particular angle. Ni crystal behaves like a three dimensional diffraction grating and it diffracts the electron beam obtained from electron gun.








The diffracted beam of electrons is received by the detector which can be positioned at any angle by rotating it about the point of incidence. The energy of the incident beam of electrons can also be varied by changing the applied voltage to the electron gun.

According to classical physics, the intensity of scattered beam of electrons at all scattering angle will be same but Davisson and Germer, found that the intensity of scattered beam of electrons was not the same but different at different angles of scattering.





Intensity is maximum at 54 V potential difference and 50o diffraction angle.

If the de-Broglie waves exist for electrons then these should be diffracted as X-rays. Using the Bragg’s formula , we can determine the wavelength of these waves.

Where d = distance between diffracting planes, = glancing angle for incident beam = Bragg’s angle.

The distance between diffraction planes in Ni-crystal for this experiment is d = 0.91Å and the Bragg’s angle = 65o. This gives for n = 1, Å

Now the de-Broglie wavelength can also be determined by using the formula .

Thus the de-Broglie hypothesis is verified.

Heisenberg Uncertainty Principle.

According to Heisenberg’s uncertainty principle, it is impossible to measure simultaneously both the position and the momentum of the particle.

Let Dx and Dp be the uncertainty in the simultaneous measurement of the position and momentum of the particle, then ; where and h = 6.63 ´ 10–34
J-s is the Planck’s constant.

If Dx = 0 then Dp = ¥

and if Dp = 0 then Dx = ¥
i.e., if we are able to measure the exact position of the particle (say an electron) then the uncertainty in the measurement of the linear momentum of the particle is infinite. Similarly, if we are able to measure the exact linear momentum of the particle i.e., Dp = 0, then we can not measure the exact position of the particle at that time.


According to Eienstein’s quantum theory light propagates in the bundles (packets or quanta) of energy, each bundle being called a photon and possessing energy.

(1) Energy of photon

Energy of each photon is given by where c = Speed of light, h = Plank’s constant = 6.6 ´ 10–34
Jsec, n
= Frequency in Hz, l = Wavelength of light

Energy of photon in electron volt

(2) Mass of photon

Actually rest mass of the photon is zero. But it’s effective mass is given as

. This mass is also known as kinetic mass of the photon

(3) Momentum of the photon



(4) Number of emitted photons

The number of photons emitted per second from a source of monochromatic radiation of wavelength l and power P is given as ; where E = energy of each photon

(5) Intensity of light (I)

Energy crossing per unit area normally per second is called intensity or energy flux


At a distance r from a point source of power P intensity is given by


Photo-electric Effect.

It is the phenomenon of emission of electrons from the surface of metals, when light radiations (Electromagnetic radiations) of suitable frequency fall on them. The emitted electrons are called photoelectrons and the current so produced is called photoelectric current.

This effect is based on the principle of conservation of energy.

(1) Terms related to photoelectric effect

(i) Work function (or threshold energy) (W0) : The minimum energy of incident radiation, required to eject the electrons from metallic surface is defined as work function of that surface.

n0 = Threshold frequency; l0 = Threshold wavelength

Work function in electron volt W0(eV)

Note : q     By coating the metal surface with a layer of barium oxide or strontium oxide it’s work function is lowered.

(ii) Threshold frequency (n0) : The minimum frequency of incident radiations required to eject the electron from metal surface is defined as threshold frequency.

If incident frequency n < n0
Þ No photoelectron emission

(iii) Threshold wavelength (l0) : The maximum wavelength of incident radiations required to eject the electrons from a metallic surface is defined as threshold wavelength.

If incident wavelength l > l0
Þ No photoelectron emission

(2) Einstein’s photoelectric equation

According to Einstein, photoelectric effect is the result of one to one inelastic collision between photon and electron in which photon is completely absorbed. So if an electron in a metal absorbs a photon of energy E (= hn), it uses the energy in three following ways.

(i) Some energy (say W) is used in shifting the electron from interior to the surface of the metal.

(ii) Some energy (say W0) is used in making the surface electron free from the metal.

(iii) Rest energy will appear as kinetic energy (K) of the emitted photoelectrons.

Hence E = W + W0 + K

For the electrons emitting from surface W = 0 so kinetic energy of emitted electron will be max.

Hence E = W0 + Kmax
; This is the Einstein’s photoelectric equation

(3) Experimental arrangement to observe photoelectric effect

When light radiations of suitable frequency (or suitable wavelength and suitable energy) falls on plate P, photoelectrons are emitted from P.





  • If plate Q is at zero potential w.r.t. P, very small current flows in the circuit because of some electrons of high kinetic energy are reaching to plate Q, but this current has no practical utility.
  • If plate Q is kept at positive potential w.r.t. P current starts flowing through the circuit because more electrons are able to reach upto plate Q.
  • As the positive potential of plate Q increases, current through the circuit increases but after some time constant current flows through the circuit even positive potential of plate Q is still increasing, because at this condition all the electrons emitted from plate P are already reached up to plate Q. This constant current is called saturation current.
  • To increase the photoelectric current further we will have to increase the intensity of incident light.
  • Photoelectric current (i) depends upon
  • Potential difference between electrodes (till saturation)
  • Intensity of incident light (I)
  • (c) Nature of surface of metal

  • To decrease the photoelectric current plate Q is maintained at negative potential w.r.t. P, as the anode Q is made more and more negative, fewer and fewer electrons will reach the cathode and the photoelectric current decreases.
  • At a particular negative potential of plate Q no electron will reach the plate Q and the current will become zero, this negative potential is called stopping potential denoted by V0.
  • If we increase further the energy of incident light, kinetic energy of photoelectrons increases and more negative potential should be applied to stop the electrons to reach upto plate Q. Hence .


Stopping potential depends only upon frequency or wavelength or energy of incident radiation. It doesn’t depend upon intensity of light.

We must remember that intensity of incident light radiation is inversely proportional to the square of distance between source of light and photosensitive plate P
i.e., so )

Important formulae








(4) Different graphs

(i) Graph between potential difference between the plates P and Q and photoelectric current






(ii) Graph between maximum kinetic energy / stopping potential of photoelectrons and frequency of incident light





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Thermodynamics Theory



Thermodynamics: The branch of physics which deals with the energy transformations from one form to another. It is a macroscopic phenomenon where we deal with group of large number of particles rather than the individual particles. Thermodynamics deals with only energy changes and not how the energy changes are brought about [mechanism]


Thermodynamic System: A group of extremely large number of particles having certain value of pressure, volume and temperature is called thermodynamic system. For example large collection of gas molecules is a thermodynamic system.

Surroundings: Anything outside the system which can exchange energy with it and has direct effect on its behavior is called surrounding.

Thermodynamic Variable: The parameters [pressure, volume and temperature] which determines the state of the system are called variables. Other thermodynamic variables like internal energy [U], or entropy etc can be expressed in terms of P, V and T.

Equation Of State: A relation between pressure, volume and temperature for a system is called the equation of state. For example ideal gas equation PV =nRT is an example of equation of state for n moles of ideal gas.

Thermodynamic process: A thermodynamic process is said of taking place if the thermodynamic variables change with time

[a] Isothermal Process: The process in which the temperature of the system remains constant is called isothermal process. DT =0

[b] Isobaric process: the thermodynamic process in which the pressure of the system remains constant with time is called isobaric process. DP =0

[c] Isochoric Process: The thermodynamic process in which the volume of the system remains constant is called isochoric process. D V =0

Adiabatic Process: The thermodynamic process in which heat content of the system remains constant is called an adiabatic process. D Q =0


Conditions for perfect Isothermal and adiabatic Change:

Isothermal Change: When the cylinder containing some gas is compressed the temperature of the system increases, but if the temperature of the system is to remain constant then heat produced in compression must be equal to the heat lost to the surroundings. Similarly if the gas expands some heat is taken from the gas and gas cools. But if temperature is to be kept constant then heat lost should be equal to the heat gained from the surroundings. To acquire such a state

[a] The walls of the container in which the thermodynamic phenomenon takes place should be perfectly conducting so that heat can be exchanged with the surroundings.

[b] The process of expansion and compression should be slow so that time is available for heat exchange with the surroundings.


Adiabatic Change: When gas is compressed or expands its temperature can either rise or fall respectively. In both the cases the system has tendency to exchange heat with the surroundings so that the thermal equilibrium with surroundings could be established. But if the heat content of the system has to remain constant as in adiabatic process two conditions are required

[a] The walls of the container in which the thermodynamic process takes place should be perfectly insulating so that no heat is exchanged with the surroundings.

[b] The process of compression or expansion should take place fast so that no time is available for heat exchange with the surroundings.

Though perfectly adiabatic process are not realized in practice some examples are

[1] the sudden bursting of cycle tyre.

[2] the propagation of sound wave through the medium as it is fast phenomenon.



When two systems A and B are put into actual contact with each other or are separated by a diathermy wall, their state coordinates may or may not change. Eventually a stage is reached where no further change in the state coordinates of A and B takes place. The joint state of both the systems that exists when all changes in the coordinates ceases is called thermal equilibrium.

    Now suppose two systems a and B are separated from each other by an adiabatic wall but each is in contact with the system C through diathermic walls. The whole system is surrounded by an adiabatic wall. Experiment shows that the two system will be in thermal equilibrium with the third system C and no further change takes place if the adiabatic wall separated them is removed, which implies that the two system are in thermal equilibrium with each other also. This can be stated as ‘ two system in thermal equilibrium with third system are in thermal equilibrium with each other also.’ This principle is known as zeroth law of thermodynamics. The temperature is the property that determines whether or not it will be in equilibrium with other systems. If two systems are in thermal equilibrium they always have the same temperature.


First law of thermodynamics is the law of conservation of energy that ‘energy can neither be created nor destroyed it can only be transferred from one form to another’ or ‘when the mechanical work is spent in producing heat, a definite quantity of heat is produced for every unit of work spent; and conversely when heat is employed to do work, the same definite quantity of heat disappears for every unit of work obtained’

    The heat energy supplied to the system is partially used for increasing the internal energy and the remaining heat energy is used for doing the pressure volume work. If dQ is the heat energy supplied , then dU will be the change in internal energy and dW is the external work done. Then according to the first law of thermodynamics

dQ = dU + dW

Thus first law of thermodynamics is the relationship between the heat energy supplied and the external work done.


Sign Conventions: [a] The heat energy added to the system is taken as positive and heat energy removed from the system is taken as negative

[b] If internal energy of the system increases then dU is positive and if the temperature of the system decreases or internal energy decreases then dU is negative.

[c] Work done on the system is negative and work done by the system is positive i.e in the case of expanding gas work done is positive and in case of compressed gas work done is negative.



The amount of heat required to raise the temperature of given body through a given amount varies from body to body. If the heat DQ given to a body can increase the temperature of a given body through DT then heat capacity of the body is

Heat Capacity =

The heat capacity per unit mass of the body is called specific heat capacity.

C =

If m=1,DT=10C then C=DQ.

Thus, specific heat can also be defined as the amount of heat required to raise the temperature of unit mass of the substance through unit amount.

Units of Specific Heat: the units of specific heat are J/ kg0C (in SI) or ergl/gm0C. But the commonly used units are cal/gm0C. The specific heat of water is 1cal/gm0C or 4186 J/kg0C.

    The above definition of the specific heat is valid for solids or liquids but the specific heat of gases can vary from zero to infinity. For e.g. if the gas is compressed the temperature of the gas rises although no heat energy is supplied to it, thus specific heat for such a gas is zero. Similarly if the gas is supplied heat and the gas is allowed to expand so that their is no rise in the temperature then the specific heat of the gas becomes infinite as DT=0. Thus to find the specific heat of the gas either the pressure or the volume of the gas is kept constant.


Specific Heat Of Gases:

[a]Specific heat at constant volume: It is defined as the amount of heat energy required to raise the temperature of unit mass of the gas through 10C when the volume of the gas is constant. In is denoted by cv

[b]Specific heat at constant pressure: it is defined as the amount of heat energy required to raise the temperature of unit mass of the substance through 10C if the pressure of the gas is constant. It is denoted by cp.

Molar Specific Heat:

Molar specific heat is defined as the amount of heat required to raise the temperature of one mole of the substance through 10C. Molar specific heat is different if the gas is heated at constant pressure or if the gas is heated at constant volume. Molar specific heat at constant volume Cv and the molar specific heat at constant pressure is Cp.

If m denotes the molecular weigh of a gas then

Cp = M cp and Cv = M cv

Cp is greater than Cv: When gas is heated at constant volume the heat energy supplied is used for raising the temperature of the gas alone as no work is done for expansion. But if the gas is heated at constant pressure and its temperature is to be raised then some amount of heat supplied is wasted in doing the pressure volume work. Thus, heating a gas at constant pressure requires larger amount of heat as compared to constant volume.


Relation Between Cp and Cv [Cp – Cv = R]

Let us take one mole of a gas and the gas under isochoric conditions so that temperature of the gas increases from T to T + dT. As the gas is heated at constant volume therefore the pressure volume work is zero. Therefore whole of the heat energy supplied should be used for increasing the internal energy of the gas.

dU = CvdT

If the same gas is heated at constant p[pressure through a temperature dT. If dQ is the amount of heat supplied to the gas in this case,

dQ = CpdT

The heat energy supplied at the constant volume will be used for increasing the internal energy of the gas and the remaining energy is used for doing the pressure volume work. The pressure volume work done by the gas is

dW = PdV

According to first law of thermodynamics,

dQ = dU + dW

CpdT = CvdT +PdV

CpdT = CvdT + RdT

Cp – Cv = R


Boiling Process: Let us heat a liquid of mass m and at its boling point it starts changing into vapors. Let V1 be the volume of the liquid and V2 the volume of vapours formed. The pressure volume work done is

dW= PdV= P ( V2 – V1)

If the latent heat of vaporization is Lv, then heat energy absorbed in the boiling process is

dQ = mLv

If Ui denoted internal energy in liquid phase and Uf denotes final internal energy then

dU = Uf – Ui

From first law of thermodynamics , we get

mLv = (Uf – Ui) + P (V2 – V1)


Melting Process: If heat energy is supplied to mass m of solid so that it begins to melt in liquid state, the process is referred as melting. In melting process when solid changes into liquid the volume of the system remains constant. Thus dV = 0. Thus, no work is done in the melting phenomenon by the heat supplied. Using first law of thermodynamics we get

dW = dU + PdV

mLf = dU

here Lf denotes the latent heat of fusion of the solid.


Equation for adiabatic Change:

According to first law of thermodynamics the total heat energy supplied to the system is used for changing the internal energy and doing the pressure volume work

dQ = dU +PdV    …….[1]

If one mole of is heated at constant volume so that its temperature increases by dT then the total heat energy supplied is equal to dU such that

dU = CvdT     ………. [2]

From [1] and [2], we have

dQ = CvdT + PdV    ….[3]

In adiabatic change the heat content of the system remains constant, dQ=0.

CvdT + PdV =0    …..[4]

According to the ideal gas equation for 1 mole of a gas PV=RT. Differentiating this equation, we get

PdV + VdP= RdT

dT =     …….[5]

Substitute [5] in [4] we obtain

CvPdV + CvVdP + RPdV=0

dividing both sides by CvVP, we get

where is the ratio of two specific heats of a gas i.e. g=

Integrating both sides,

logeP + glogeV = C

logePVg= C

taking antilog

PVg = K


Adiabatic Relation between volume and temperature: the pressure volume relation for adiabatic process is

PVg = constant

Also for one mole of ideal gas P =

T Vg-1 = constant

Or T1


Adiabatic Relation Between pressure and temperature: as for ideal gas V =



Work done in isothermal expansion: Isothermal phenomena is one in which the temperature of the system remains constant. Thus the change in internal energy for an isothermal process is zero. From first law of thermodynamics

dQ = dU + dW

As the change in internal energy is zero, therefore dQ = dW

dW = PdV

dW =

Total work done to change the volume from V1 to V2 is

W = RT

W = RT [ln V2 – ln V1]

W = RT ln

Thus for n moles of an ideal gas,

W = 2.303 nRT log

Work Done id adiabatic process: According first law of thermodynamics

dQ = dW + dU

In an adiabatic process, there is no transfer of heat between system and surroundings, thus dQ = 0

dW = -dU

dW = -nCvdT

the total work done to change the temperature of the system from T1 to T2 is

W =

W = -nCv ( T2 – T1)

For one mole of ideal gas,

W = -Cv (T2-T1)

Also, we know that Cv =

W =


Limitations of first Law of thermodynamics: First law of thermodynamics is the law of conservation of energy, but it has two major limitations

[1] It doe not indicate the direction of heat transfer.

[b] It does not indicate the extent to which the heat change takes place.

For e.g.

[1] This law explains the heating of bullet when it strikes a target but fails to explain why this heat can’t be converted back into kinetic energy.

[2] Similarly, if the brakes are applied and cycle stops, the kinetic energy getting converted into heat energy. Why this heat energy can’t be converted back into kinetic energy of rotation of wheel.

[3] The first law gives no idea how much or what percentage of heat energy supplied can be converted into work and if there is some limitation in this conversion.


Reversible Process: Reversible process is one in which the process can be retraced in opposite direction passing through the same intermediate process for e.g slow expansion or contraction of the spring. The conditions for reversible process are;

[a]the process should take place slowly so that the system should remain in thermal, chemical and mechanical equilibrium at all stages of the process

[b]there should be no frictional losses. This is so because energy spend against such dissipative forces can’t be recovered back. There is loss of energy due to friction.

Some more examples of reversible process are

[1] the working substance taken along complete cannot cycle

[2] All thermal processes taking place at infinitely slow rate.

[3] All mechanical processes taking place under conservative forces.


Irreversible Process: It is process in which the system cannot be made to proceed in the reverse direction through the same intermediate steps as in the case of direct process. A part of energy of the system does work against dissipative forces which can’t be recovered back. In the irreversible process there is always some loss of energy due to heat energy generated in friction and the fast thermodynamics process. For e.g.

[1] Dissolving of sugar into water is an irreversible process

[2] diffusion of gases is an irreversible process.

[3] Adiabatic expansion of compression of gas is an irreversible process.


Second law of thermodynamics:

Lord Kelvin statement: it is impossible to get a continuous supply of work form a body by cooling it to temperature lower than the temperature of the surroundings. This law applies to the heat engine, which absorbs heat from the source does some work and reject remaining heat to the sink at low temperature. It’s not possible to make a heat engine, which converts all of the heat energy absorbed into useful work.


Clausius Statement: It is impossible to make heat flow from the body at lower temperature to the body at higher temperature without doing any external work on the working substance. This applies to the refrigerator, if heat energy is to be removed from interiors of refrigerator at lower temperature and released into the surrounding atmosphere t higher temperature then some external work must be done.



Heat Engine:

Heat engine is device used for converting heat energy into mechanical heat engine.heat engines are of two main types:

External combustion engine: these engines are those in which the burning of fuel is done outside the main body of the cylinder as in the case of the steam engine.

Internal combustion engine: internal combustion engine is one in which the burning of fuel takes place inside the body of the engine as in the case of petrol engine or diesel engine.

    For any heat engine there are three essential requirements:

[a]Source: a hot body at fixed higher temperature T1 from which the heat engine can draw heat, is called source.

[b]Sink: A cold body at lower temperature T2 to which any amount of heat can be rejected is called sink.

[c]Working substance: The material which on being supplied heat performs work is called the working substance.


Efficiency of Heat engine: It is defined as the ratio of external work done to the amount of heat energy absorbed from the heat source.

If Q1 is the heat energy absorbed from the source and Q2 is the heat energy released into sink after doing external work then the work done in the process is given by

W = Q1 – Q2

The efficiency is thus given by

Carnot Heat Engine: heat engine is a practical arrangement to convert heat into mechanical work. Sadi Carnot devised an ideal heat engine free from all imperfectness of the actual heat engines and hence it is not possible to generate such an engine in actual practice. Carnot’s heat engine consists of four main parts:

[a]Source : the source is maintained at a fixed higher temperature and has infinite thermal capacity. By infinite thermal capacity we mean that any amount of heat can be taken out of it without changing the temperature of the source.

[b]Sink: It is the reservoir at lower temperature T2 and it also has infinite thermal capacity i.e. any amount of heat can be added to it without changing its temperature.

[c]Working substance: the working substance in the Carnot engine is the ideal gas which absorbs heat from the source does some mechanical work and rejects the remaining amount of heat into sink. It is placed in a cylinder with insulating base but perfectly conducting bottom.

[d]Insulating pad: The pad is used in Carnot cycle for adiabatic expansion and contraction of the gas.


Carnot Cycle

Carnot cycle consists of four main steps:

[a]ISOTHERMAL EXPANSION: For isothermal expansion of the gas the cylinder is placed in contact with the source so that acquires the temperature of the source T1. The gas allowed to expand slowly by the outward motion of the piston. The expansion results in cooling and the decrease in temperature is compensated by gaining the required amount of heat from the source. Thus overall temperature of the gas remains constant during the expansion. The pressure and volume of the gas are (P1, V1) and the final pressure and volume are (P2,V2). The work done in the process is given by:

Q1= W1= RT1 loge
= Area ABMKA

[b]ADIABATIC EXPANSION: For adiabatic expansion of the gas, the cylinder is removed from the source and placed on the insulating pad. The gas is allowed to expand further from (P2, V2) to (P3,V3). The process is adiabatic because the cylinder is thermally insulated from all the sides. Thus, because of the expansion of the gas temperature of the gas falls from T1 to T2.the work done in the process is given by:

W2 = = Area BCNMB


[c]ISOTHERMAL COMPRESSION: the cylinder is removed from the insulating pad and placed on the sink. The gas is then compressed by the inward motion of the piston. Compression of the gas results in generation of heat and the temperature of the gas is kept constant by releasing the heat generated to the sink. The pressure and volume of the gas changes from (P3, V3) to (P4, V4). The work done on the gas is given by:

Q2= W3= RT2 loge
= Area CNLDC

[d]ADIABATIC COMPRESSION: The cylinder containing the gas is placed on the insulating pad and compressed so that the pressure and the volume of the gas returns to the initial value (P1,V1). The temperature of the gas returns increases to T1. The work done in the process is given by:

W4= =Area DLKAD

In the first two steps the work done by the gas is positive as the gas is expanding whereas the work done by the gas in the compression is negative. Thus total work done during Carnot cycle is:

W= W1 + W2 -W3 -W4

W= W1 – W3

As W1 is equal to the heat energy absorbed Q1 and W3 is the heat energy released in the sink Q2 thus net work done is

W= Q1 – Q2


    The efficiency of Carnot engine is the ratio of amount of work done by the gas to the total heat energy absorbed by the gas

h =

also as, thus efficiency of the Carnot engine is given by:

h =

thus we can conclude that the efficiency of the Carnot engine:

[a]depends upon the temperature of the source and the sink

[b]it is always less than 100% as the temperature of source can’t be infinite and temperature of the sink can’t be 00K

[c]is directly proportional to the temperature difference between the source and sink.


Refrigerator works in a way opposite to the heat engine. It absorbs heat q2 from the sink does some work W on iy and release the heat Q1 to the source at higher temperature. An amount of work W is done on it by some external means. In actual refrigerator the vapors of Freon (dichlorodifluoromethane) gas acts as the working substance. The interior of the refrigerator acts as the sink. A certain amount of substance W is performed by the compressor of the refrigerator on the gas. The source in this case is the atmosphere or surrounding air at room temperature T1 to which the heat Q1 is rejected by the radiator fixed at the back of the refrigerator. thus work done in the process is the difference of heat rejected to the source to the heat absorbed from sink.

W= Q1 – Q2


Coefficient of performance:

It is defined as the ratio of quantity of heat removed per cycle from the contents of the refrigerator to the energy spent per cycle to remove this heat.

b = Q2/W

b = Q2/(Q1 – Q2)

also as Q2/Q1 = T2/T1


b = T2/ (T1 – T2)

Greater the temperature difference between the source and the sink smaller is the coefficient of performance. The value of the coefficient of performance is the measure of the efficiency of the refrigerator and smaller the value of temperature difference smaller will be the efficiency.

















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Books recommended

Physics Books recommended for IIT-JEE preparation

For basic concepts students can use either of the two books

  1. University Physics By Sears And Zemansky  12th Edition By Pearson Publications
  2. Fundamentals Of Physics By Resnick/Halliday And Walker Publisher: John Wiley And Sons

For Numericals

  1. Concepts Of Physics By Dr. H. C. verma [Volume 1 And 2]
  2. New pattern IIT-JEE [For MCQ’s] By D.C. Pandey [Level 1 problems for JEE mains and level 2 for JEE Advance]
  3. Objective Physics by Parmod Agarwal GRB Publications [For JEE mains]
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Rotational Motion Formula Sheet

Rotational Motion:

The rotational motion is different from other mechanics concepts because it cannot be studied by assuming the body to be a point body. In rotational motion all particles of the body undergo different linear displacements unlike translational motion, therefore we will assume the body to be rigid body.


Angular displacement: when a body rotates about a fixed axis, each particle travels in circular path. Although different particles will have different linear displacements there angular displacement is same. If the angular displacement is θ, then

where s is the distance traveled while moving in a circular path of radius r.

    Angular displacement is a scalar quantity if θ is large and is vector if θ is small. This is so because for large angular displacements it does not satisfy the laws of vector addition. For small angular displacements, its direction is given by right hand curl rule.


Angular velocity: It is rate of change of angular displacement of the body with respect to time. i.e.

As s = θr, therefore we can also write

V = rω

In the vector form

Angular acceleration is an axial vector and its direction is always along the axis of rotation and is given by right hand curl rule. For rigid body linear velocity of particles is different but angular velocity of all bodies is same.


Angular Acceleration: It is rate of change of angular velocity w.r.t. time.

As v = rω, therefore we can also write a = r α

The direction of angular acceleration is same as angular velocity if angular velocity is increasing and opposite to angular velocity if angular velocity is decreasing.


Equation of kinematics: the equation of kinematics can be written analogous to the linear motion as

ω = ω0 + αt

θ = θ0 + ω0t + αt2/2

ω2 = ω02 + 2 α (θ-θ0)

θ = θ0 + t

These equation of motion can be applied if the angular acceleration of the body is constant.

Moment of Inertia: Inertia is the property of the body to resist the changes in the state of motion and moment of inertia is the property, which resists changes in state of rotational motion of the body.

For point mass

I = mr2, where m is the mass and r is the distance from axis of rotation.

For n particles

I =

Where ri is the distance from the axis of rotation of mass mi.

For rigid body:

I =

Where r is the perpendicular distance from the axis of rotation of mass dm.

MI depends upon

[a] Axis of rotation selected

[b] distribution of mass about the axis of rotation.

    It does not depend upon the dimension parallel to the axis of rotation. For e.g. if MI of cylinder is calculated about is axis of rotation it does not depend on the length, but only on the mass and radius.


Parallel Axis Theorem: According to parallel axis theorem the MI of the body about given axis is equal to the MI of the body about an axis parallel to given axis and passing through the center of mass of the body IG plus Ma2 where M is the mass of the body and ‘a’ is he distance between the two axis.

I= IG + Ma2

Theorem Of Perpendicular Axis: According to it the sum of MI of the plane lamina about two mutually perpendicular axis lying in its plane is equal to the MI about a axis perpendicular to the plane of lamina and passing through the point of intersection of two axes.

IZ = IX + IY



Important Note regarding MI: [1] Theorem of parallel axis is applicable for any type of rigid body whereas perpendicular axis theorem is applicable to 2 dimensional bodies only.

[2] Moment of inertia of the rigid body [symmetrically cut from whole mass] is same as that of the body e.g. a small part of circular disc will have same moment of inertia as that of the disc]


Radius of Gyration: Radius of gyration of the body about an axis is the effective distance from the axis where the whole mass can be assumed to be concentrated so that the moment of inertia remains the same

If K is radius of gyration and M is the mass then I= MK2

K can also be said to be the rms distance from the axis of rotation


Radius of gyration is not a constant quantity it changes with change in axis of rotation of the body.


Angular Momentum: The momentum associated with rotational motion is called angular momentum, it can be associated with a particle or a rigid body.

Suppose a particle of mass m moving with momentum p. Its angular momentum is

For a rigid body of moment of inertia I rotating with angular velocity ω the angular momentum is given by

If a body is in translational as well as rotational motion then its angular momentum is sum of both the components.


Law of Conservation of angular momentum:

Uniform Pure Rolling: Pure rolling means no relative motion at point of contact between the two bodies. For example, consider a disc of radius R moving with linear velocity v and angular velocity ω on a horizontal ground. The disc is said to be moving without slipping if the velocities of point P and Q are equal

Vp = vQ

v- Rω=0

v = Rω

If Vp >vQ or v > Rω, the motion is said to be forward slipping and if v< Rω is case of backward slipping.

    Thus v = Rω is condition for pure rolling on a stationary ground. Sometimes it is simply said rolling. Suppose the base over which the body moves is moving with some velocity say v0, then the condition of pure rolling are different. For example in the above figure vp=vq

v – Rω = v0

Thus in this condition will be different. By uniform pure rolling we mean that v and ω are constant.


Note: [1] distance moved by the center of mass of the rigid body in one full rotation is 2πR. This can be shown as

S = v T = [ωR] [2π/ω] = 2πR

In forward slipping s > 2πR

In backward slipping s < 2πR

[2] Instantaneous axis of rotation passes through the bottommost point, as it is a point of zero velocity. Thus the combined motion of rotation and translation can be assumed to be pure rotational motion about the bottommost point with same angular speed ω.

[3] The speed of point on the circumference of the body at the instant is shown in figure is 2 v sin

[4] The path of point on the circumference is a cycloid and the distance moved by this point in 1 full rotation is 8R.

Instantaneous center of rotation: Since rolling body does not slip, therefore the point of contact O with the surface is instantaneously at rest and ring momentarily rotates about this point. All the particles appear to describe a circular paths with angular velocity ω about O as center. The point O is then called instantaneous center of rotation.

Velocity of any point: The magnitude of linear velocity of any point is given by the product of angular velocity and the distance of that point from the point of contact the direction of velocity vector is perpendicular to the line joining the point and the point of contact.


Accelerated pure rolling:

So far we were discussing the uniform pure rolling in which v and ω were constants. Now suppose an external force acts on the rigid body, the motion will no longer remain uniform. The condition on a stationary ground is,

v = R ω

Differentiating this equation with respect to time we have

a = R α

Thus in addition to v = R ω at every instant of time, linear acceleration= R α for pure rolling to take place. Here, friction plays an important role in maintaining role in maintaining pure rolling. The friction may sometimes act in forward direction, sometimes in backward direction or under certain conditions it may be zero. Here, we should not forget the basic nature of friction, which is self adjusting force and has tendency to stop the relative motion between the two bodies in contact. Let us take example illustrating the above theory.

    Suppose a force F is applied at the topmost point of the rigid body of radius R, mass M and moment of inertia I about an axis passing through the center of the mass. Now the applied force can produce by itself,

[a] linear acceleration ‘a’

[b] an angular acceleration α

If a = R α, then there is no need of friction and force of friction =0. If a < Rα, then to support the linear motion the force of friction f will act in forward direction. Similarly if a > R α then to support the angular motion force of friction will act in backward direction. So, in this case force of friction can be in forward, backward or zero also. Let us assume it in forward direction


α =

For pure rolling to take place

a = R α

solving the above equations we get

f =

From equation following conclusions can be drawn

[a] if I= MR2 then friction force is zero.

[b] if I < MR2, like in the case of solid or hollow sphere, f is positive i.e. force of friction will be forward

[c] If I > MR2, f will be negative.

Here, it should be noted that force of friction F obtained above should be less than μMg for pure rolling to take place. Further, we saw that I < MR2 force of friction acts in forward direction. This is because if I is small α will be more i.e. to support the linear motion friction acts in forward direction.



Rolling on inclined plane:


As we said earlier the force of friction in this case is backward. Equation of motion are



α =

For pure rolling to take, place

a = R α

Solving the equations, we get

f =


Further, the force of friction calculated in equation for pure rolling to take place should be less than or equal to the maximum friction force

Velocity of the body at the bottom of inclined plane is

v =

As here acceleration is constant and body starts from rests so time to taken to reach the bottom of inclined plane is


Important Points for Inclined plane motion: [1] As factor β= depends only on the shape and not on the mass and radius there same bodies with different mass and radius are allowed to roll down an inclined plane they will reach the bottom at the same time.

[2] Velocity, acceleration and time of descend depends upon β. Lesser the MI lesser will be β so greater will be velocity and acceleration and lesser will be time of descend.

[3] The velocity is independent of the inclination of the plane and depends only on the height from which the body descends. Whereas acceleration and time of descend depends upon inclination larger the inclination larger will be acceleration and smaller will be time of descend.



Note: [a] the torque equation can be applied about only two points

[1] the center of mass of the body

[2] point about which the body is in pure rotation


[b]In cases where the pulley is having friction, the tension on the two sides of the pulley is different and it is essential to consider the rotational motion of the pulley.


[c] Work done in pure rolling on a stationary ground is zero as the point of application of force is at rest. Therefore, mechanical energy can be conserved if all other dissipative forces are ignored.


[d] the equation ζ = I α holds only in the inertial reference frame. However there exists a special case when the axis of rotation passes through the center of mass otherwise pseudo forces produces pseudo torque on the body.

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Full Syllabus XII Class

Q.1         An electron is moving with a velocity of 107m/s enters a uniform magnetic field of 1T. The direction of velocity of the electron is perpendicular to the magnetic field. What would be the trajectory of electron in this field?

Q.2         An equiconvex lens of focal length 15cm is cut into two halves along the plane perpendicular to the principle axis. What is the focal length of each half?

Q,3         for circuits used for transporting electric power, a low power factor implies large power less in transmission. Explain?

Q.4         What is the shortest wavelength present in Brackett series of the hydrogen spectra?

Q.5         Show graphically how the stopping potential for a given photosensitive surface varies with frequency of incident radiation.

Q.6         Show by diagram that two infinitely long parallel wires carrying equal currents will attract each other. Mark the direction of current, magnetic field and force.

Q.7         Calculate the refractive index of the material of equilateral prism for which the angle of minimum deviation is 600.

Q,8         The electric field at a point due to point charge is 30N/C and the electric potential at that point is 15J/C. Find the magnitude of the charge and the distance of the point from the point charge.


Q.8         A uniform electric field of 5000N/C along positive x axis exists in space, find the flux of this field through a square of 10cm on side whose plane is parallel to the y-z plane. What would be the flux through the same square if the plane makes an angle of 300 with the x axis.

Q.9         A charge particle q is placed at a distance x from the center of short electric dipole on its axial line. Find the work done to move the charge from that point to the center of electric dipole.

Q.10       Out of the two magnetic material , A has relative permeability slightly less than unity and B has relative permeability very much greater than one. Identify the nature of magnetic materials A and B. How will the magnetic lines of force of uniform magnetic field distort if these materials are placed in uniform field.

Q.11       State the condition for maxima and minima in single slit diffraction pattern and thus using this relation find the width of central maxima for slit width ‘d’ on a screen placed at a distance D using wavelength λ.

Q.12       How can we use NAND gate to form OR gate. Write truth table for OR gate.

Q.13       State Kirchoff’s rules. Explain briefly the conservation laws on which these rules are based.

Q.14       An athlete paddles a stationary tricyle whose pedals are attached to a coil having 100 turns each of area 0.1m2. the coil is lying in XY plane and is rotated in this plane at 50rpm about y axis where uniform magnetic field B=0.01T along positive z axis is present. Find the [a] maximum emf [b] the average emf generated in the coil over one complete revolution.

Q.15       State the principle on which Cyclotron is based. Explain briefly how this machine is used to accelerate charge particles to high energies. Why it can’t accelerate neutron.

Q.16       Use Gauss law to find electric field intensity of large sheet. Use the result and diagrammatically find the electric field intensity for two large oppositely charge sheets having surface charge density σ and –σ.

Q.17       A proton and deuteron are accelerated through same accelerating potential. Which one of the two has [a] greater value of de-Broglie wavelength associated with it [b] less momentum . Give reasons to justify your answers.

Q.18       [a]Monochromatic light of frequency 6 x 1014Hz is produced by a laser. The power emitted is 2mW. Estimate the number of photons emitted per second by the source.

[b] Draw a plot showing variation of photoelectric current versus the intensity of incident radiation on a given photosensitive surface.

Q.19       Use mirror formula to show that a convex mirror always produces a virtual image independent of the location of the object.

[b] Use lens equation to deduce algebraically that the object placed within focus of convex lens produces a virtual and enlarged image.

Q.20       Draw plot showing variation of amplitude versus frequency for an amplitude modulated wave. Define modulation index. State its importance in effective amplitude modulation.

Q.21       [a] Give the ratio of number of holes and number of conduction electrons in an intrinsic semiconductor.

[b] How does the energy gap in intrinsic semiconductor vary when doped with trivalent impurity

[c] Drae I vs V graph for forward biased pn junction diode.

Q.22       With the necessary circuit diagram, describe briefly how npn transistor in CE configuration amplifies a small sinusoidal voltage. Write expression for ac current gain and voltage gain.

Q.23       When Aman a physics student of class XII came to know that her parents are planning to rent out top floor of the house to a mobile company he protested. He tried hard to convince his parents that this would be health Hazard and his parents agreed. [a] In what way setting up of transmission power by a mobile company prove injurious to health? [b] By objecting to the plans of his parents, what values does he display? [c] Estimate the range of em waves which can transmitted by antenna of height 20m.

Q.24       [a]State and prove principle of potentiometer. How it be used to find the internal resistance of the cell.

[b]A potentiometer wire of length 1m has a resistance of 5 ohm. It is connected to a 8V battery in series with a resistance of 15 ohm. Determine the emf of the primary cell which gives balance point at 60cm.

Q.25       Draw a labeled diagram showing formation of final image at least distance of distinct vision by a compound microscope.

[b] A convex lens of focal length 20cm is placed coaxially with convex mirror of radius of curvature 20cm. The two are kept 15cm apart. A point object is placed 40cm in front of the convex lens. Find the position of image formed by this combination. Draw the ray diagram showing the image formation.

Q.26       Draw a schematic diagram of an AC generator describing its basic elements. State briefly its working and principle. Show a plot of variation of [a] magnetic flux [b] alternating emf versus time generated by a loop of wire rotating in magnetic field .

[c] Why is choke coil needed in the use of fluorescent tubes with ac mains.