When the light of frequency 2 ν 0 (where ν 0 is threshold frequency), is incident on a metal plate, the…

From Einstein's equation of Photoelectric effect, we can write
For case 1: $\frac{\mathrm{1}}{\mathrm{2}}{\mathrm{mv}}_{\mathrm{1}}^{\mathrm{2}}=2{\mathrm{h\nu }}_{\mathrm{0}}-{\mathrm{h\nu }}_{\mathrm{0}} \Rightarrow \frac{\mathrm{1}}{\mathrm{2}}{\mathrm{mv}}_{\mathrm{1}}^{\mathrm{2}}={\mathrm{hv}}_{\mathrm{0}}$ ... (i), where $m$ is mass of electron, $h$ is Planck's constant.
For case 2: $\frac{\mathrm{1}}{\mathrm{2}}{\mathrm{mv}}_{\mathrm{2}}^{\mathrm{2}}=5{\mathrm{h\nu }}_{\mathrm{0}}-{\mathrm{h\nu }}_{\mathrm{0}}\Rightarrow \frac{\mathrm{1}}{\mathrm{2}} {\mathrm{mv}}_{\mathrm{2}}^{\mathrm{2}}=4{\mathrm{h\nu }}_{\mathrm{0}}$ ... (ii)
From $(1)$ and $(2)$
$\frac{{\mathrm{v}}_{\mathrm{1}}}{{\mathrm{v}}_{\mathrm{2}}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$.

Hence, the ratio of $v_1$ and $v_2$ is $1:2$.