Is the oscillation of a simple pendulum at the centre of the earth possible?

The length of a second's pendulum on the surface of the earth is $1 \ m$. What will be the length of a second's pendulum on the moon?

An object of mass $3 \,kg$ is tied by a string of negligible mass to a ceiling and held such that the string is taut. The object is released suddenly such that the string remains taut. Its acceleration when released is (acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,ms^{-2}$)

The linear displacement $x$ of the bob of a simple pendulum from its mean position varies as $x = a \sin \left(\frac{\pi}{\sqrt{2}} t\right)$,where $a$ is its amplitude expressed in meters and $t$ is in seconds. The length of the simple pendulum is (Take $g = \pi^{2} \ m/s^{2}$): (in $m$)

The bob of a pendulum of length $l$ is pulled aside from its equilibrium position through an angle $\theta$ and then released. The bob will then pass through its equilibrium position with speed $v$,where $v$ equals

Two pendulums begin to swing simultaneously. If the ratio of the frequency of oscillations of the two is $7 : 8$,then the ratio of lengths of the two pendulums will be