Given,
Temperature \((T)\) is same for most probable velocity ( \(v_{\text {mp }}\)) and for root mean square velocity \(\left[v_{\text {rms }}\right]\)
Also,
\(\begin{aligned}
& v_{(\mathrm{mp})}=400 \mathrm{~ms}^{-1} \\
& v_{(\mathrm{mp})}=\sqrt{\frac{2 R T}{M}} \text { and } \\
& v_{(\mathrm{mp})}=\sqrt{\frac{2 R T}{M}} \text { or, } v_{(\mathrm{mp})}^2=\frac{2 R T}{M}
\end{aligned}\)
or \(M=\frac{2 R T}{v_{(\mathrm{mp})}^2}=\frac{2 R T}{(400)^2}\)...(i)
Similarly \(\quad M=\frac{3 R T}{v_{(\mathrm{rms})}^2} \quad\left[\because v_{\text {rms }}=\sqrt{\frac{3 R T}{M}}\right]\)...(ii)
On comparing (i) and (ii), we have
\(\frac{2 R T}{(400)^2}=\frac{3 R T}{v_{(\mathrm{rms})}^2}\)
or \(v_{(\mathrm{rms})}^2=\frac{3}{2} \times(400)^2\)
\(\begin{aligned}
v_{\text {(rms) }} & =1.2247 \times 400 \\
& =489.89 \approx 490.00 \mathrm{~ms}^{-1}
\end{aligned}\)
Alternate method
Ratio of
\(\therefore \quad v_{\text {rms }}: v_{\text {avg }}: v_{\text {most probable }}\)
is \(1: 0.92: 0.82\), therefore,
\(\begin{aligned}
v_{\mathrm{rms}} & =\frac{v_{\text {most probable }}}{0.82} \\
& =\frac{400}{0.82}=487.8 \cong 490.0
\end{aligned}\)
Hence, option (b) is the correct answer.