If the length of a seconds pendulum is first decreased by 10 cm and its time period is determined and then increased by 20 cm and the time period is once again determined, find the ratio of the time periods in two cases. (Take `g = 9.8 m s^(-2)`)

Explanation

The time period of given simple pendulum is = 2s
We know that `T 2pi sqrt((l)/(g)) rArr 2 = 2pi sqrt((l)/(9.8))`
`rArr l = (9.8)/((3.14)^(2)) = 0.99 m rArr l = 100 cm` (nearly)
Then, `(T_(1))/(T_(2)) = sqrt((L -10)/(L + 20)) = sqrt((100 - 10)/(100 + 20))`
(Since, `T propsqrtl`)
`rArr (T_(1))/(T_(2)) = sqrt((90)/(120)) = sqrt((9)/(12)) = sqrt((3)/(4)) = (sqrt3)/(2)`
`:. T_(1) : T_(2) = sqrt3 :2`