Home IGNOU Admission Guess Paper Previous Year Paper Solved Assignment Blog
No Preview
Session 2026 Verified Digital
Available Now

BPHCT 135 Solved Guess Paper 2026 for Thermal Physics and Statistical Mechanics in B.Sc. (General) BSCG

Our Price

โ‚น299.00 โ‚น399.00
25% OFF

Language

English

Session

2026

Delivery

PDF

Updated

Sep 2026

WHATSAPP

Value Highlights

BPHCT 135 Solved Guess Paper 2026 with 14 exam ready questions and complete model answers for Thermal Physics and Statistical Mechanics
Built after mapping six term end sessions from June 2023 to December 2025 question by question against all 14 units
Covers all 4 blocks with 9 answers written to the 5 mark shape and 5 answers written to the 10 mark shape with full derivations
5 full questions with answers and 7 real PDF pages are open on this page so quality can be judged before ordering
Answers follow the IGNOU self learning material with the definition first then the formula then the derivation and a closing line
25 page PDF with 18 pages of answers delivered on WhatsApp 9899436384 usually within a few working hours
Every question carries its block, unit, last appeared sessions and priority grade so revision can run top down
Targeted at both the June 2026 and December 2026 term end examinations

Course Overview

BPHCT 135 Solved Guess Paper 2026 is a fourteen question exam pack built for the Thermal Physics and Statistical Mechanics paper of the B.Sc. (General) BSCG programme at IGNOU. Every question in it was chosen after mapping the last six term end papers question by question, and every answer is written the way the examiner wants to read it, with the definition first, the governing formula next, the derivation in numbered steps and a one line conclusion at the end.

Physics students rarely fail this paper because the theory is beyond them. They lose marks because the derivation is half remembered, the symbols are never defined and the answer runs out of shape before the conclusion. That is the exact gap this pack was written to close for Thermal Physics and Statistical Mechanics.

BPHCT 135 key facts at a glance

Course code and titleBPHCT 135 Thermal Physics and Statistical Mechanics
ProgrammeBachelor of Science General BSCG, Physics discipline, third semester
Exam duration and marks2 hours and 50 maximum marks
Course structure4 blocks and 14 units
Questions in this pack14 with complete model answers
Answer pages18 pages of solutions inside a 25 page PDF
Sessions analysed6 term end papers from June 2023 to December 2025
Shown on this page5 full questions with answers and 7 page images
Target sessionsJune 2026 and December 2026 term end examinations

What the BPHCT 135 Solved Guess Paper 2026 contains

The pack holds 14 questions with complete model answers across 18 answer pages, covering all 4 blocks and all 14 units of BPHCT 135, targeted at the June 2026 and December 2026 term end examinations.

Nine of the fourteen questions carry 5 marks and are written to a 150 word shape. The remaining five carry 10 marks and are written to a 250 word shape, with the full derivation included rather than summarised. Kinetic theory, thermodynamics and statistical mechanics are all represented, so no block is left uncovered if the paper leans one way in a given session.

Each question card inside the PDF also carries three tags that most guess papers leave out: the block and unit it belongs to, the sessions in which the topic last appeared and a priority grade. A student revising three days before the exam can therefore work top down by priority instead of reading twenty five pages in order.

The answers stay inside the IGNOU self learning material. Where the book names a law, a scientist or a constant, the answer names it too, because that is what carries the mark. Nothing has been imported from outside textbooks that would read as unfamiliar to an evaluator marking BSCG physics scripts.

Five sample questions with model answers from the BPHCT 135 guess paper

Five of the fourteen questions are reproduced below in full, with the complete model answer exactly as it appears in the PDF, so quality can be judged before anything is paid for.

The five samples are drawn from four different blocks on purpose. Two are 5 mark answers, two are 10 mark answers and one is a derivation heavy question from statistical mechanics, which is where BSCG physics students usually lose the most marks in Thermal Physics and Statistical Mechanics.

Sample 1 ยท Guess Paper Q1 ยท Block 1 Unit 1 ยท 5 marks ยท Priority High

Question. Write van der Waals' equation of state for one mole of a gas, state the assumptions used to derive it and explain the terms.

Answer. The ideal gas equation fails to explain the behaviour of real gases at high pressure and low temperature because it assumes that gas molecules have no volume and no intermolecular forces. To overcome these limitations, van der Waals proposed a modified equation of state for real gases.

(p + a/V2)(V − b) = RT

where p is the pressure of the gas, V is the volume occupied by one mole of the gas, R is the universal gas constant, T is the absolute temperature, a is the constant representing intermolecular attraction and b is the constant representing the finite size of gas molecules.

Assumptions used to derive van der Waals' equation

  1. Gas molecules are not point particles; they have finite size and occupy a definite volume.
  2. Gas molecules attract one another through weak intermolecular forces.
  3. The pressure exerted by the gas is reduced because molecules near the container walls are pulled inward by neighbouring molecules.
  4. The available volume for molecular motion is less than the actual volume of the container because of the finite size of the molecules.
  5. The constants a and b depend only on the nature of the gas and remain constant for a given gas.

Explanation of the terms

Pressure correction a/V2. Intermolecular attraction reduces the force with which molecules strike the walls of the container, so the observed pressure is lower than the ideal pressure. To obtain the corrected pressure, the term a/V2 is added to the observed pressure. Here a measures the strength of intermolecular attractive forces.

Volume correction V − b. Since gas molecules have finite size, they occupy some space, so the free volume available for molecular motion is less than the actual volume of the container. The effective volume is therefore V − b, where b represents the excluded volume or co-volume of one mole of gas molecules.

Conclusion. Van der Waals' equation successfully explains the behaviour of real gases by considering both the finite size of gas molecules and the intermolecular attractive forces. At low pressure and high temperature, the correction terms become negligible and the equation reduces to the ideal gas equation PV = RT.

Sample 2 ยท Guess Paper Q4 ยท Block 1 Unit 3 ยท 5 marks ยท Priority High

Question. Define mean free path and obtain the survival equation for the distribution of free paths. Name the transport phenomena.

Answer. The mean free path is defined as the average distance travelled by a gas molecule between two successive collisions with other gas molecules. It is denoted by λ.

λ = 1 / (√2 π d2 n)

where λ is the mean free path, d is the diameter of a gas molecule and n is the number of molecules per unit volume. Thus the mean free path is inversely proportional to the molecular diameter and the number density of the gas.

Survival equation for the distribution of free paths

Consider a molecule travelling through a gas and let P(x) be the probability that the molecule travels a distance x without suffering a collision. If the molecule moves through an additional small distance dx, then the probability that it survives without collision decreases in proportion to dx. Hence

dP = − P(x) dx / λ   or   dP/P = − dx/λ

Integrating gives ln P = − x/λ + C. Since P = 1 when x = 0, we get C = 0. Therefore

P(x) = e−x/λ

This is called the survival equation. It represents the probability that a molecule travels a distance x without undergoing any collision.

Distribution of free paths. The probability that a molecule suffers its first collision between distances x and x + dx is dP = (1/λ) e−x/λ dx. This is known as the distribution function of free paths.

Transport phenomena. The transport phenomena in gases are viscosity, which is the transport of momentum; thermal conductivity, which is the transport of heat or energy; and diffusion, which is the transport of matter from a region of higher concentration to a region of lower concentration.

Conclusion. The mean free path is the average distance travelled between two successive collisions, the survival equation P(x) = e−x/λ describes the probability of a molecule travelling a distance x without collision, and the three important transport phenomena in gases are viscosity, thermal conductivity and diffusion.

Sample 3 ยท Guess Paper Q7 ยท Block 2 Units 6 and 7 ยท 10 marks ยท Priority High

Question. State the zeroth and first laws of thermodynamics. Starting from the first law, derive Mayer's relation Cp − Cv = R for an ideal gas.

Answer. Zeroth law of thermodynamics. If two systems are separately in thermal equilibrium with a third system, then they are also in thermal equilibrium with each other. This law introduces the concept of temperature.

First law of thermodynamics. The heat supplied to a system is used partly to increase its internal energy and partly to do external work. Mathematically δQ = dU + δW. For a gas δW = p dV, therefore δQ = dU + p dV.

Derivation of Mayer's relation

For one mole of an ideal gas, pV = RT.

At constant volume. When volume is constant, dV = 0, so no work is done and p dV = 0. From the first law δQ = dU. Heat supplied at constant volume is δQ = Cv dT, therefore dU = Cv dT.

At constant pressure. From the first law δQ = dU + p dV. Heat supplied at constant pressure is δQ = Cp dT, therefore Cp dT = dU + p dV.

Using dU = Cv dT we get Cp dT = Cv dT + p dV. For one mole of an ideal gas pV = RT, and at constant pressure p dV = R dT. Substituting,

Cp dT = Cv dT + R dT

Dividing by dT gives Cp = Cv + R, hence

Cp − Cv = R

Conclusion. Mayer's relation for one mole of an ideal gas is Cp − Cv = R. It shows that the molar heat capacity at constant pressure is greater than that at constant volume by the universal gas constant R.

Sample 4 ยท Guess Paper Q9 ยท Block 3 Unit 9 ยท 5 marks ยท Priority High

Question. Define entropy and obtain the expression for the change in entropy of an ideal gas in an isothermal change.

Answer. Entropy is a thermodynamic property that measures the degree of disorder or randomness of a system. It is also a measure of the unavailability of heat energy for conversion into useful work. For a reversible process the change in entropy is defined as dS = δQrev / T, where δQrev is the heat absorbed reversibly and T is the absolute temperature. The SI unit of entropy is J K−1.

Change in entropy of an ideal gas in an isothermal process

Consider one mole of an ideal gas expanding reversibly and isothermally from volume V1 to V2 at constant temperature T. Since the process is isothermal, ΔU = 0, and from the first law of thermodynamics Q = W. The work done during reversible isothermal expansion is W = RT ln (V2/V1), therefore Q = RT ln (V2/V1).

Using the definition of entropy ΔS = Qrev / T and substituting the value of Q,

ΔS = R ln (V2/V1)

Using the ideal gas equation P1V1 = P2V2 we get V2/V1 = P1/P2, hence ΔS = R ln (P1/P2).

Important observations. During isothermal expansion where V2 is greater than V1, entropy increases. During isothermal compression where V2 is less than V1, entropy decreases. Entropy remains constant only in a reversible adiabatic or isentropic process.

Conclusion. For one mole of an ideal gas undergoing a reversible isothermal process the change in entropy is ΔS = R ln (V2/V1) = R ln (P1/P2). This shows that the entropy change depends only on the initial and final states of the system and not on the path followed.

Sample 5 ยท Guess Paper Q13 ยท Block 4 Unit 14 ยท 10 marks ยท Priority High

Question. Compare the Maxwell-Boltzmann, Bose-Einstein and Fermi-Dirac distribution functions, and obtain the Fermi-Dirac distribution function.

Answer. The average number of particles in an energy state E is given by the three distribution functions below.

Maxwell-Boltzmann distribution. fMB(E) = A e−E/kT. It applies to classical particles which are distinguishable.

Bose-Einstein distribution. fBE(E) = 1 / (e(E−μ)/kT − 1). It applies to bosons such as photons and helium-4 atoms. Any number of bosons can occupy the same energy state.

Fermi-Dirac distribution. fFD(E) = 1 / (e(E−μ)/kT + 1). It applies to fermions such as electrons, protons and neutrons. No two fermions can occupy the same quantum state.

At high temperature and low density, both Bose-Einstein and Fermi-Dirac distributions reduce approximately to the Maxwell-Boltzmann distribution.

Derivation of the Fermi-Dirac distribution

Consider an energy level Ei having degeneracy gi, and let ni fermions occupy this level. For fermions, according to Pauli's exclusion principle, each quantum state can contain either 0 or 1 particle. The number of ways of arranging ni particles in gi states is

Wi = gi! / [ ni! (gi − ni)! ]

Taking the product over all energy levels and then the logarithm, and using Stirling's approximation ln n! ≈ n ln n − n, we get

ln W = ∑ [ gi ln gi − ni ln ni − (gi − ni) ln (gi − ni) ]

For equilibrium, ln W is maximum subject to the conditions ∑ ni = N and ∑ ni Ei = E. Using Lagrange multipliers we maximise ln W − α ∑ ni − β ∑ ni Ei, which gives

ln [ (gi − ni) / ni ] = α + βEi

Therefore (gi − ni)/ni = eα+βEi, so gi/ni − 1 = eα+βEi and ni/gi = 1 / (eα+βEi + 1). Hence the Fermi-Dirac distribution function is

fFD(Ei) = ni/gi = 1 / (e(Ei−μ)/kT + 1)

where μ is the chemical potential.

Conclusion. Maxwell-Boltzmann statistics applies to classical particles, Bose-Einstein statistics applies to bosons and Fermi-Dirac statistics applies to fermions. The Fermi-Dirac distribution reflects the Pauli exclusion principle.

The remaining nine questions of the fourteen, including the Carnot cycle with its indicator diagram, Maxwell's thermodynamic relations with both TdS equations, the Clausius-Clapeyron derivation, Planck's law with Wien's and Stefan's laws, and the single particle partition function, are inside the complete pack.

How the BPHCT 135 guess paper was built from previous year papers

Papers from the last five years were collected, and the six most recent sessions from June 2023 to December 2025 were mapped topic by topic against the 14 units before a single answer was written.

The method is simple to describe and slow to execute. Each question in each paper was tagged to its unit, then to a topic label, then counted. Topics that surfaced in four or more of the six sessions became high priority, and topics appearing twice or three times became medium priority.

A topic that has been quiet for two sessions but repeats on a three session cycle was also pulled forward, because gaps matter as much as repeats when predicting a paper. Anyone who wants to run this exercise independently can start from our IGNOU Previous Year Question Paper archive.

The December 2025 paper was the last one added, and it changed two calls. Mean free path and the survival equation moved up after appearing again, and the Clausius-Clapeyron derivation was reinstated after a long quiet stretch. Both are inside the fourteen.

Answers were then written unit by unit against the IGNOU self learning material rather than from memory, checked once by a physics reviewer for symbol accuracy and once more for answer shape, since a correct derivation that overshoots 250 words on a 10 mark question still costs a student time in a 2 hour paper.

BPHCT 135 block and unit index for Thermal Physics and Statistical Mechanics

BPHCT 135 is organised into 4 blocks and 14 units, and the 2026 pack draws at least one question from every block.

Block and unit index of IGNOU BPHCT 135 Thermal Physics and Statistical Mechanics with the topics carried into the 2026 pack
BlockUnitUnit titleTopics carried into the 2026 questions
Block 1 Kinetic Theory of GasesUnit 1Ideal and Real GasesVan der Waals' equation, pressure and volume corrections, kinetic interpretation of temperature, equipartition of energy
Block 1Unit 2Molecular Velocity Distribution FunctionMaxwell speed distribution function, average speed, root mean square speed, most probable speed and their ratio
Block 1Unit 3Mean Free Path and Transport PhenomenaMean free path expression, survival equation, free path distribution, viscosity, thermal conductivity, diffusion
Block 1Unit 4Brownian MotionCharacteristics of Brownian motion, Einstein's 1905 theory, Avogadro number determination, Perrin's verification
Block 2 Zeroth and First LawsUnit 5Thermodynamic Description of a SystemOpen, closed and isolated systems, boundaries, intensive against extensive variables
Block 2Unit 6The Zeroth Law of ThermodynamicsStatement of the zeroth law, thermal equilibrium and the concept of temperature
Block 2Unit 7The First Law of Thermodynamics and its ApplicationsDifferential form of the first law, internal energy, Mayer's relation Cp minus Cv equals R
Block 3 Second and Third LawsUnit 8Carnot CycleIndicator diagram, two isothermals and two adiabatics, efficiency expression, independence from working substance
Block 3Unit 9Entropy and the Laws of ThermodynamicsDefinition of entropy, isothermal entropy change, entropy in expansion and compression, third law statement
Block 3Unit 10The Thermodynamic PotentialsInternal energy, enthalpy, Helmholtz and Gibbs functions, four Maxwell relations, TdS equations, Clausius-Clapeyron equation
Block 3Unit 11Theory of RadiationPlanck's law, Wien's distribution and displacement laws, Stefan's law by integration
Block 4 Statistical MechanicsUnit 12Basic Concepts of Statistical MechanicsMicrostates and macrostates, thermodynamic probability, phase space, Boltzmann relation S equals k ln W
Block 4Unit 13Classical StatisticsMaxwell-Boltzmann statistics, single particle partition function of a monatomic ideal gas, thermal de Broglie wavelength
Block 4Unit 14Quantum StatisticsBose-Einstein and Fermi-Dirac distributions, Pauli exclusion principle, Fermi-Dirac derivation by Lagrange multipliers

Topic frequency across the last six BPHCT 135 question papers

Of the 14 questions in the pack, 10 rest on topics that appeared in four or more of the six sessions between June 2023 and December 2025.

The table below is the short version, mapped to the fourteen questions. The full analysis inside the PDF tracks 26 topics, including the ones deliberately left out such as the Joule-Thomson effect and the third law, which are worth a 2 mark statement rather than a full answer. Students who want to verify a row can pull the matching BPHCT 135 Question paper and check it themselves.

Topic appearance count across six IGNOU BPHCT 135 term end sessions from June 2023 to December 2025 with assigned priority
Question in the packTopicUnitSessions of sixPriority
Q1Van der Waals' equation of stateUnit 14High
Q2Kinetic interpretation of temperature and equipartitionUnit 13High
Q3Maxwell speed distribution and the three speedsUnit 25High
Q4Mean free path and survival equationUnit 34High
Q5Brownian motion and Einstein's theoryUnit 45High
Q6Thermodynamic systems and boundariesUnit 54High
Q7Zeroth and first laws with Mayer's relationUnits 6 and 75High
Q8Carnot cycle and efficiencyUnit 85High
Q9Entropy and isothermal entropy changeUnit 93High
Q10Maxwell's relations and TdS equationsUnit 104High
Q11Clausius-Clapeyron equationUnit 102Medium
Q12Planck's law with Wien's and Stefan's lawsUnit 115High
Q13Maxwell-Boltzmann, Bose-Einstein and Fermi-Dirac distributionsUnit 143High
Q14Single particle partition function and Boltzmann relationUnits 12 and 132Medium

BPHCT 135 exam pattern and marks distribution for 2026

The BPHCT 135 term end paper runs for 2 hours and carries 50 marks, all questions are compulsory with internal choices, and a calculator is allowed.

Question wise marks distribution of the IGNOU BPHCT 135 term end examination paper
QuestionPatternMarksAnswer shape to aim for
Question 1Attempt any five of eight short parts5 into 2 equals 10Two or three crisp lines carrying the key formula
Question 2Answer any two parts2 into 5 equals 10About 150 words with definition, formula and conclusion
Question 3Answer any two parts2 into 5 equals 10About 150 words with a short derivation where asked
Question 4Answer any two parts2 into 5 equals 10About 150 words with the diagram if the question names one
Question 5One long question with an internal choice1 into 10 equals 10About 250 words with the full derivation written out

Two habits carry marks that most students give away. Define every symbol immediately after writing a formula, and close every answer with a single sentence stating the physical significance. Both are built into all fourteen model answers.

BPHCT 135 practical file position and the BPHCL 136 laboratory course

BPHCT 135 is a theory paper with no practical file attached to it, because the laboratory work for Thermal Physics and Statistical Mechanics is assessed separately under the course code BPHCL 136.

This confuses a fair number of third semester BSCG physics students, who search for a BPHCT 135 practical file and find nothing. The two courses are registered together in the same semester but they are examined differently. Marks for BPHCT 135 come from the assignment and the term end theory paper, while BPHCL 136 is evaluated through laboratory work at the study centre.

So the practical index a student needs for the laboratory sits with BPHCL 136, not here. This guess paper does not attempt to cover experiments, and it should not be bought as a substitute for laboratory records. It is built for the descriptive theory paper alone, which is exactly where the fifty marks of BPHCT 135 are decided.

How to order the complete BPHCT 135 Solved Guess Paper

Send the course code BPHCT 135 on WhatsApp to 9899436384 and the complete 14 question pack is shared as a PDF, usually within a few working hours on the same day.

Five questions are already open on this page. The other nine, including the Carnot indicator diagram, both TdS equations and the Clausius-Clapeyron derivation, come with the complete pack. Message the number, say which session you are appearing in, and you will be told the current price before anything is confirmed. If you are also collecting material for your other papers this semester, the wider IGNOU Solved Guess Paper shelf carries the same format across other course codes.

One thing worth saying plainly. A guess paper is a prediction, not a leaked question paper. Ten of these fourteen topics have repeated in four or more of the last six sessions, which is a strong base, but the study material still deserves a read. Treat this pack as your revision spine and the IGNOU book as the reference beside it.

Students from other programmes ask us for the same thing every session, and the format does travel. A humanities student preparing for a theory heavy paper works from the same three part answer shape, which is why the M.A. in History guess Paper set used by M.A. in History (MAH / MAHI) students is built on the identical question card layout.

Get the complete BPHCT 135 pack on WhatsApp 9899436384

Who prepared and reviewed the BPHCT 135 guess paper

The BPHCT 135 pack was compiled by Prateek Talwar, founder of Unnati Educations, and reviewed for subject accuracy by Sheetal Kirola, M.Ed., before release.

Prateek Talwar has spent years working through IGNOU term end papers session after session, and he built the frequency method that ranks the fourteen questions here. The paper mapping, the priority grades and the answer shaping against the 150 and 250 word targets are his work.

Sheetal Kirola checked the physics that carries marks: symbol definitions, the direction of every correction term, the sign conventions in the first law and the limits used to pull Wien's law out of Planck's law. Where the review disagreed with a draft answer, the IGNOU self learning material settled it.

BPHCT 135 guess paper questions students ask us

The eight answers below cover validity for both 2026 sessions, block coverage, delivery format and the practical file position of BPHCT 135.

Is the BPHCT 135 Solved Guess Paper valid for both June 2026 and December 2026?

Yes. The pack was prepared for both 2026 term end sessions of BPHCT 135. The syllabus, the block structure and the paper pattern stay the same across June and December, so the same fourteen questions apply. If the December 2025 paper had shifted the pattern, the analysis would have been rebuilt, and it did not shift.

How many questions does the BPHCT 135 pack contain and how many are shown here?

The BPHCT 135 pack contains fourteen questions with complete model answers across eighteen answer pages. Five full questions with their answers are open on this page as samples, which leaves nine that come with the complete file. Nine of the fourteen are 5 mark answers and five are 10 mark answers with full derivations written out.

Does the BPHCT 135 guess paper cover all four blocks of Thermal Physics and Statistical Mechanics?

Yes. All four blocks and all fourteen units are represented. Block 1 on kinetic theory carries five questions, Block 2 on the zeroth and first laws carries two, Block 3 on the second and third laws carries five and Block 4 on statistical mechanics carries two. No block is left without at least one prepared answer.

Are the BPHCT 135 answers written from the IGNOU book or from outside sources?

Every BPHCT 135 answer is written from the IGNOU self learning material for the course. Where the book names a law, a scientist or a constant, the answer names it in the same way, because that is what an evaluator looks for. No outside textbook notation has been imported that would read as unfamiliar in a BSCG physics script.

Will the BPHCT 135 guess paper help with the assignment as well as the term end exam?

It helps, though it is not built for that. The BPHCT 135 assignment repeats several theory topics such as Mayer's relation, the kinetic theory assumptions, the Fermi-Dirac derivation and the Clausius-Clapeyron equation, so those model answers transfer across almost word for word. Numerical assignment questions sit outside this pack, since all fourteen questions target the descriptive part of the term end theory paper rather than calculation practice.

Is there a practical file included with the BPHCT 135 guess paper?

No. BPHCT 135 is a theory course and carries no practical file of its own. Laboratory work for Thermal Physics and Statistical Mechanics is registered and examined separately under the course code BPHCL 136, which is evaluated through experiments at the study centre. This pack covers the two hour fifty mark theory paper only and makes no attempt to substitute for laboratory records or observation books.

In what format is the BPHCT 135 guess paper delivered and how long does it take?

The BPHCT 135 pack is delivered digitally as a PDF on WhatsApp at 9899436384, normally within a few working hours of confirmation on the same day. It is a twenty five page file with eighteen pages of answers, readable on a phone, and it can be printed for revision without the equations breaking across lines.

Is this BPHCT 135 guess paper an official IGNOU publication?

No. This BPHCT 135 guess paper is prepared independently by Unnati Educations as a study aid. Unnati Educations is not affiliated with, endorsed by or connected to IGNOU in any way. The answers follow the IGNOU self learning material for the course, but the prediction, the priority grades and the wording are entirely our own.

Ask a BPHCT 135 doubt on WhatsApp 9899436384

Unnati Educations is an independent academic support platform and is not affiliated with, endorsed by or connected to IGNOU. All course codes, programme names and paper titles belong to IGNOU and are used for identification only. This guess paper is a study aid prepared from the IGNOU self learning material and is not an official IGNOU publication.

Ratings & Reviews

0.0
โ˜… โ˜… โ˜… โ˜… โ˜…
No reviews yet
Be the first to review this material.

Price

โ‚น299.00