Kinetic Theory

JEE Main level, single correct, medium

28 of 168

Question 28 of 168

If the root mean square velocity of hydrogen molecule at a given temperature and pressure is 2kms−1,2km{s}^{-1}, the root mean square velocity of oxygen at the same condition in kms−1km{s}^{-1} is:

Show answer and solution

Answer: (B) 0.50.5

Root mean square velocity is given by, Vrms=3RTM{V}_{rms}=\sqrt{\dfrac{3RT}{M}}
Therefore, V1V2=M2M1⇒2V2=322\dfrac{{V}_{1}}{{V}_{2}}=\sqrt{\dfrac{{M}_{2}}{{M}_{1}}}\Rightarrow \dfrac{2}{{V}_{2}}=\sqrt{\dfrac{32}{2}}
⇒V2=0.5kms−1\Rightarrow {V}_{2}=0.5km{s}^{-1}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.

More Physics practice

JEE Main Physics practice question on Kinetic Theory (Speed of Gas), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.