Answer
(a) increase
(b) $2.45s$
Work Step by Step
(a) We know that $T=2\pi\sqrt{\frac{L}{g}}$. This equation shows that the time period is inversely proportional to 'g'. Thus, if the pendulum is taken to the Moon, then its time period will increase because the value of 'g' on the Moon is less than that of its value on Earth.
(b) The time period for the Moon is given as
$T_{Moon}=2\pi \sqrt{\frac{L}{\frac{1}{6}g_{Earth}}}$
$\implies T_{Moon}=\sqrt 6.2\pi \sqrt{\frac{L}{g_{Earth}}}$
$\implies T_{Moon}=\sqrt 6.T_{Earth}$
We plug in the known values to obtain:
$T_{Moon}=\sqrt6(1.00s)$
$T_{Moon}=2.45s$
