Measurements and Experimentation — Question 3

As we know,

$\text T = 2 \text π \sqrt{\dfrac{\text l}{\text g}}$

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 —

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

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