Answer
$ T=2 \sqrt{ 2} \pi\ sec= 8.89\ sec
.$
Work Step by Step
The period $T$ of a pendulum is given by
$$
T=2\pi\sqrt{\frac{l}{32}}=2\pi\sqrt{\frac{64}{32}}=2 \sqrt{ 2} \pi\ sec
.$$
Thus, we have:
$$ T=2 \sqrt{ 2} \pi\ sec= 8.89\ sec
.$$
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