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. 0.02
B. 0.03
C. 0.01
D. 0.05

Answer

(b)

Explanation

Here, `I = 20 cm, Delta I =1mm = 0.1 cm, t = 90 sec.`
`Delta t =1 sec`
`T =2pi sqrt((I)/(g))`, squaring both sides, we get
`T^2 =4pi^2 xx(I)/(g) or g =4pi ^2(I)/(T^2)`
`:. (Deltag)/(g) xx100 + (2DeltaT)/(T)xx100`
As `(DeltaT)/(T) = (Deltat)/(t), therefore,`
`(Delta g)/(g)xx100 = (0.1)/(20) xx100 +2 xx(1)/(90)xx100`
`(Delta g)/(g) xx100 = (100)/(200) + (200)/(90) = (1)/(2) +(20)/(9) ~~3%`