The periodic time of a simple pendulum is $\mathrm{T}_{1}$. Now if the point of suspension of this pendulum starts moving along the vertical direction according to the equation $\mathrm{y}=\mathrm{kt}^{2}$, the periodic time of the pendulum becomes $\mathrm{T}_{2}$ Therefore, $\left(\mathrm{T}_{1}^{2} / \mathrm{T}_{2}^{2}\right)=\ldots \ldots \ldots .\left(\mathrm{k}=1 \mathrm{~m} / \mathrm{s}^{2} \& \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$
(A) $6 / 5$
(B) $5 / 6$
(C) $4 / 5$
(D) 1