The time period of a simple pendulum is $T_1$ when its point of suspension is at rest with respect to the ground. When the point of suspension moves vertically upwards according to the law $y=k{t}^2$ where $k=1 {\mathrm{ms}}^{-2}$, its new time period is $T_2$, then the value of $\frac{T_1^2}{T_2^2}$ is $[$Take $g=10 {\mathrm{ms}}^{-2}]$

• A

  $\frac{4}{5}$
• Correct answer
  B

  $\frac{6}{5}$
• C

  $\frac{5}{6}$
• D

  $1$

Hence, the correct answer is option: 'B'.