Internal energy of ${\mathrm{n}}_1$ moles of hydrogen of temperature T is equal to the internal energy of ${\mathrm{n}}_2$ mole of helium at temperature 2T.Then the ratio $\frac{{\mathrm{n}}_1}{{\mathrm{n}}_2}$ is

Internal energy of a n moles of an ideal gas at temperature T is given by

  $\mathrm{U} = \frac{\mathrm{f}}{2}\mathrm{nRT} [\mathrm{f} = \mathrm{degress} \mathrm{of} \mathrm{freedom}]$

${\mathrm{U}}_1 = {\mathrm{U}}_2$

${\mathrm{f}}_1{\mathrm{n}}_1{\mathrm{T}}_1 = {\mathrm{f}}_2{\mathrm{n}}_2{\mathrm{T}}_2$

$\therefore \frac{{\mathrm{n}}_1}{{\mathrm{n}}_2} = \frac{{\mathrm{f}}_2{\mathrm{T}}_2}{{\mathrm{f}}_1{\mathrm{T}}_1} = \frac{3\times 2}{5\times 1} = \frac{6}{5}$

Here, ${\mathrm{f}}_2 = \mathrm{degrees} \mathrm{of} \mathrm{freedom} \mathrm{of} \mathrm{He} = 3$and ${\mathrm{f}}_1 = \mathrm{degrees} \mathrm{of} \mathrm{freedome} \mathrm{of} {\mathrm{H}}_2 = 5$