Step 1: RMS speed formula.
The root mean square (rms) speed of gas molecules is given by
\[
V_{rms} = \sqrt{\frac{3kT}{m}}
\]
where \(T\) is temperature and \(m\) is molecular mass.
Step 2: Ratio of rms speeds.
\[
\frac{V_H}{V_{He}} = \sqrt{\frac{T_H / m_H}{T_{He} / m_{He}}} = \sqrt{\frac{T_H \, m_{He}}{T_{He} \, m_H}}
\]
Given \(T_H = 500\text{ K}, T_{He} = 2000\text{ K}\), and \(m_{He} = 4m_H\):
\[
\frac{V_H}{V_{He}} = \sqrt{\frac{500 \times 4m_H}{2000 \times m_H}} = \sqrt{1} = 1
\]
Step 3: Conclusion.
Hence, \(V_H = V_{He}\).
For a particular ideal gas, which of the following graphs represents the variation of mean squared velocity of the gas molecules with temperature?

At a particular temperature T, Planck's energy density of black body radiation in terms of frequency is \(\rho_T(\nu) = 8 \times 10^{-18} \text{ J/m}^3 \text{ Hz}^{-1}\) at \(\nu = 3 \times 10^{14}\) Hz. Then Planck's energy density \(\rho_T(\lambda)\) at the corresponding wavelength (\(\lambda\)) has the value \rule{1cm}{0.15mm} \(\times 10^2 \text{ J/m}^4\). (in integer)
[Speed of light \(c = 3 \times 10^8\) m/s]
(Note: The unit for \(\rho_T(\nu)\) in the original problem was given as J/m³, which is dimensionally incorrect for a spectral density. The correct unit J/(m³·Hz) or J·s/m³ is used here for the solution.)