Question

The period of oscillation of a simple pendulum is `T = 2pisqrt(L//g)`. Measured value of L is `20.0 cm` known to `1mm` accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. What is the accuracy in the determination of g?
A. `2%`
B. `3%`
C. `1%`
D. `5%`

Answer 1

Given `(Delta)/(L) = (0.1)/(20)`
`T = (90)/(100) sec DeltaT = (1)/(100)sec`
`(DeltaT)/(T) = (1)/(90)`
`g = ((1)/(4pi^(2))) (L)/(T^(2)) rArr (Deltag)/(g) xx 100 % = (DeltaL)/(L) xx 100 + (2DeltaT)/(T) xx 100`
`(Deltag)/(g)xx100%=((0.1)/(20))100+2((1)/(90))100=2.72%`
so nearest option is `3%` .