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?

Options

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

Answer

b

Explanation

`g = 4 pi^(2) (I)/(T^(2))`
`rArr (Delta g)/(g) xx 100 = (Delta I)/(I) xx 100 + 2 (Delta T)/(T) xx 100`
` = (Delta I)/(I) xx 100 + 2 (Delta t)/(t) xx 100`
` = (0.1)/(20.0)xx 100 + 2 xx(1)/(90) xx 100 `
`= (100)/(200) + (200)/(90) = (1)/(2) + (20)/(9) = 3%`