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. (a) Error `DeltaT` in measuring `T`, the time period, is `0.05 seconds`
B. (b) Error `DeltaT` in measuring `T`, the time period, is `1 seconds`
C. ( c ) Percentage error in the determination of `g` is `5%`
D. ( d) Percentage error in the determination of `g` is `2.5%`

Answer

A, C

Explanation

(a,c) As the length of the string of simple pendulum is exactly `1m` (given), therefore the error in length `DeltaI = 0`. Furhter the possibility of error in measurement time is `1s` in `40s`.
:. `(Deltat)/(t) = (DeltaT)/(T) = (1)/(40)`
The time period `T = (40)/(20) = 2 seconds`
:. `(DeltaT)/(T) = (1)/(40) rArr (DeltaT)/(2) = (1)/(40) rArr DeltaT = 0.05 sec`
we know that `T = 2pisqrt((1)/(g)` rArr `T^(2) = 4pi^(2)()/(g))`
:. `g = 4pi^(2)(I)/(T^(2)`
:. `(Deltag)/(g) xx 100 = (Deltal)/(l) xx 100 + 2(DeltaT)/(T) xx 100`
:. `(Deltag)/(g) xx 100 = 0 + 2 ((1)/(40)) xx 100 = 5`