Answer
$\frac{66\pi}{17}\ sec$
Work Step by Step
Step 1. Assume $p\ (sec)$ is the period of a pendulum.
Step 2. Assume $L\ (cm)$ is the length of the pendulum and $g\ (cm/s^2)$ is the acceleration due to gravity.
Step 3. From the given conditions, we have $p=k\sqrt {\frac{L}{g}}$ and $6\pi=k\sqrt {\frac{289}{980}}$, thus $k=6\pi\sqrt {\frac{980}{289}}$
Step 4. For $L=121\ (cm)$, $g=980\ (cm/s^2)$, we have $p=6\pi\sqrt {\frac{980}{289}}\sqrt {\frac{121}{980}}=\frac{66\pi}{17}\ sec$
