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?

Options

A. Error `Delta T` in measuring T, the time period, is 0.05 seconds
B. Error `DeltaT` in measuring T, the time period, is 1 second
C. Percentage error in the determination of g is 5%
D. Percentage error in the detremination of g is 2.5%

Answer

(c )

Explanation

Let t be the time for n oscillations and T be the time period of oscillations of the simple pendulum.
Then `(DeltaT)/(T) = (Delta t)/(t) = (1)/(40) = 0.05`
`g = (4pi^2I)/(T^2) = (4pi^2 I n^2)/(t^2)`
`:. (Delta g)/(g) = (2Delta t)/(t) = 2xx (1)/(40) = (1)/(20)`
% error in determination of g
`=(Delta g)/(g)xx100 = (1)/(20)xx100 = 5%`