The period of oscillation of a simple pendulum is `T = 2pisqrt((L)/(g))`. Meaured value of `L` is `20.0 cm` know to `1mm` accuracy and time for `100` oscillation of the pendulum is found to be `90 s` using a wrist watch of `1 s` resolution. The accracy in the determinetion of `g` is :

Options

A. `3%`
B. `1%`
C. `5%`
D. `2%`

Answer

A

Explanation

We can write the acceleration due to gravity as :
`g=(4pi^(2) L)/(T^2)`
Time period,
`T=(90)/(100)s`
Accuracy in time period,
`triangleT= (1)/(100)s`
Accuracy in length, `triangleL= 1mm =0.1cm`
`(triangleg)/(g)xx100=(2triangleT)/(T)xx100+(triangleL)/(L)xx100`
`=(2xx0.01)/(0.9)xx100+(0.1)/(20)xx100`
`=2.23+0.5`
`approx 2.73`
Accuracy in determination of g is approximately 3%