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JEE Main 2024 · PhysicsMultiple choiceSingle correctEasyCalculation

JEE Main 6 April 2024, Shift 2, Physics Q39

Question 39 of 90 in this shift, Physics question 9 of 30, Section A.

A total of 48 J heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by

2^\circ\mathrm{C}

. The work done by the gas is : Given,

R=8.3\ \mathrm{J\,K^{-1}\,mol^{-1}}

.

(1)24.9 J
(2)48 J
(3)72.9 J
(4)23.1 JOfficial answer

Official answer

Option 4

NTA final key.

Chapter: Thermodynamics
Topic: First Law of Thermodynamics
Idea tested: First Law Energy Balance

Same idea in other shifts

Asked 2× in all

6 Apr 2023, Shift 1 · Q44A source supplies heat to a system at the rate of 1000 W. If the system performs work at a rate of 200 W. The rate at which internal…EasySingle correct

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.