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Exam

JEE Advanced

Conducting Body
One of the 7 zonal IITs on a rotational basis
Eligibility
Top rankers of the JEE Main exam
Subjects
Physics, Chemistry, Mathematics
Structure
Two compulsory papers on the same day
Question Types
Varies; includes MCQs, numerical, and matrix-match
Duration
3 hours per paper (6 hours total)
Marking
Complex scheme with negative and partial marking
Frequency
Once a year
Qualifies For
Undergraduate admission to the IITs

The Joint Entrance Examination (JEE) Advanced is the second and final stage of the selection process for admission to the Indian Institutes of Technology (IITs). It is exclusively for students who have secured a top rank in the first stage, JEE Main. Success in JEE Advanced opens the door to undergraduate programs in engineering, sciences, and architecture at any of the 23 IITs across India.

The exam is a rigorous, single-day event consisting of two compulsory papers, each three hours long. Both papers cover Physics, Chemistry, and Mathematics. The questions are designed to test a candidate's depth of understanding and analytical ability, rather than just rote memorization. They often require the application of multiple concepts from different topics to arrive at a solution.

What makes JEE Advanced particularly challenging is the complexity of its questions and the demanding marking scheme. The paper features a variety of question formats, including multiple choice questions with one or more correct answers, numerical answer types, and matrix-match questions. Negative marking is a common feature, penalizing incorrect guesses and forcing candidates to be confident in their answers. The intense competition and the need for both speed and accuracy under severe time pressure make it one of the most difficult undergraduate entrance examinations in the world.

One of these questions, worked through

JEE Advanced 2025

An ideal monatomic gas of n moles is taken through a cycle WXYZW consisting of consecutive adiabatic and isobaric quasi-static processes, as shown in the schematic V−TV-T diagram. The volume of the gas at W, X and Y points are 64 cm3,125^{3}, 125 cm3^{3} and 250 cm3,^{3}, respectively. If the absolute temperature of the gas TWT_{W} at the point W is such that nRTW=1J(R_{W} = 1 J (R is the universal gas constant), then the amount of heat absorbed (in J) by the gas along the path XY is ______.
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Step 1: Identify the processes and gas properties.

For an ideal monatomic gas, the ratio of specific heats is γ=53\gamma = \frac{5}{3}, which gives:

γ−1=23\gamma - 1 = \frac{2}{3}

The processes W→XW \to X and Y→ZY \to Z are adiabatic, while X→YX \to Y and Z→WZ \to W are isobaric.

Step 2: Analyze the adiabatic process W→XW \to X.

Using the relation TVγ−1=constantT V^{\gamma-1} = \text{constant} for an adiabatic process:

TWVW2/3=TXVX2/3T_W V_W^{2/3} = T_X V_X^{2/3} TX=TW(VWVX)2/3T_X = T_W \left(\frac{V_W}{V_X}\right)^{2/3}

Substitute the volumes VW=64 cm3V_W = 64\,\text{cm}^{3} and VX=125 cm3V_X = 125\,\text{cm}^{3}:

TX=TW(64125)2/3T_X = T_W \left(\frac{64}{125}\right)^{2/3} TX=1625TWT_X = \frac{16}{25} T_W

Step 3: Analyze the isobaric process X→YX \to Y.

For an isobaric process, the temperature is directly proportional to the volume:

TYVY=TXVX\frac{T_Y}{V_Y} = \frac{T_X}{V_X} TY=TX(VYVX)T_Y = T_X \left(\frac{V_Y}{V_X}\right)

Substitute the volumes VY=250 cm3V_Y = 250\,\text{cm}^{3} and VX=125 cm3V_X = 125\,\text{cm}^{3}:

TY=TX(250125)T_Y = T_X \left(\frac{250}{125}\right) TY=2TXT_Y = 2 T_X TY=3225TWT_Y = \frac{32}{25} T_W

Step 4: Calculate the heat absorbed along the path XYXY.

The heat absorbed during an isobaric process is given by Q=nCpΔTQ = n C_p \Delta T.

Cp=52RC_p = \frac{5}{2}R QXY=n(52R)(TY−TX)Q_{XY} = n \left(\frac{5}{2}R\right) (T_Y - T_X) QXY=52nR(3225TW−1625TW)Q_{XY} = \frac{5}{2} nR \left(\frac{32}{25} T_W - \frac{16}{25} T_W\right) QXY=52nR(1625TW)Q_{XY} = \frac{5}{2} nR \left(\frac{16}{25} T_W\right) QXY=85nRTWQ_{XY} = \frac{8}{5} nRT_W

Step 5: Substitute the given value of nRTWnRT_W.

Substitute nRTW=1 JnRT_W = 1\,\text{J} into the heat equation:

QXY=85(1 J)Q_{XY} = \frac{8}{5} (1\,\text{J}) QXY=1.6 JQ_{XY} = 1.6\,\text{J}

Answer: 1.61.6

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