Question

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 :
A. 0.02
B. 0.03
C. 0.01
D. 0.05

Answer 1

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