Question

A student uses a simple pendulum of exactly `1m` length to determine `g`, the acceleration due ti gravity. He uses a stop watch with the least count of `1sec` for this and record `40 seconds` for `20` oscillations for this observation, which of the following statement `(s) is (are)` true?
A. Error `Delta T` in measuring `T` , the time period , is `0.05 s`
B. Error `Delta T` in measuring `T` , the time period , is `1 s`
C. Percentage error in the determination of `g is 5%`
D. Percentage error in the determination of `g is 2.5%`

Answer 1

Correct Answer - A::C
`T = 40// 20 = 2 s`
`(Delta T)/( T) = ( Delta t)/( t) = (1)/( 40) rArr Delta T = (T)/( 40) = (2)/( 40) = 0.05 s`
` g = ( 4 pi^(2) L n^(2))/( t^(2)) where t = nT`
`( Delta g)/(g) = ( 2 Delta t)/(t) rArr %error = ( 2Delta t)/( t) xx 100 = 5%`