JEE Main 2026 · PhysicsMultiple choiceSingle correctHardMulti-step

JEE Main 4 April 2026, Shift 1, Physics Q36

Question 36 of 75 in this shift, Physics question 11 of 25, Section A. NTA dropped this question from the final key; it is left out of every count on this site.

An ideal gas undergoes a process maintaining relation between pressure (

P

) and volume (

V

) as

P=P_o\left[1+\left(\frac{V_o}{V}\right)^2\right]^{-1}

, where

P_o

and

V_o

are constants. If two samples

A

and

B

(two moles each) with initial volumes

V_o

and

3V_o

respectively undergo above mentioned process and attain same pressure, then the difference at the temperatures of these samples,

T_B-T_A

is ______. (

R=

gas constant)

(1) $\frac{9P_oV_o}{8R}$
(2) $\frac{11P_oV_o}{10R}$
(3) $\frac{7P_oV_o}{6R}$
(4) $\frac{13P_oV_o}{11R}$

Official answer

Dropped by NTA (marks given to all)

NTA final key.

Chapter: Thermodynamics
Topic: Thermodynamic Processes
Idea tested: No close match in our taxonomy

Same topic in other shifts

All Thermodynamic Processes questions

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.