A seconds' pendulum is taken to a place where acceleration due to gravity falls to one forth. How is the time period of the pendulum affected, if at all? Give reason. What will be its new time period?

As we know,

$T = 2 π \sqrt{\dfrac{l}{g}}$

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 —

$T = 2 π \sqrt{\dfrac{l}{\dfrac{g}{4}}} \\[0.5em] T = 2 π \sqrt{\dfrac{4l}{g}} \\[0.5em] T = 2 \times 2 π \sqrt{\dfrac{l}{g}} \\[0.5em]$

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