As we know,
T = 2 \pi \sqrt{l\over g}
We observe that time period is inversely proportional to the square root of acceleration due to gravity.
Hence, when ‘g’ falls to one-fourth, time period increases.
When acceleration due to gravity is reduced to one-fourth, we see that
T = 2 \pi \sqrt{l\over g/4}
T = 2 \pi \sqrt{4l\over g}
T = 2×2 \pi \sqrt{l\over g}
Hence, we can say that when acceleration due to gravity is reduced to one-fourth, time period of a simple pendulum doubles.
As, the given pendulum is a seconds’ pendulum so T = 2s
∴ New T = 2 x 2 = 4s