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 secondsB. Error `DeltaT` in measuring T, the time period, is 1 secondC. Percentage error in the determination of g is 5%D. Percentage error in the detremination of g is 2.5%

A student uses a simple pendulum of exactly `1m` length to determine `

1.	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 secondsB. Error `DeltaT` in measuring T, the time period, is 1 secondC. Percentage error in the determination of g is 5%D. Percentage error in the detremination of g is 2.5%
Answer» Correct Answer - (a,c) Here `Delta t = 1 sec, t = 40 s, I = 1m,` `T = 2 sec., Delta T =?` `(Delta T)/(T) = (Delta t)/(t) = (1s)/(40s) = (1)/(40)` `DeltaT = ((1)/(40)xx T) = (1)/(40)xx2 = 0.05 sec` Also, `t = 2pi sqrt((I)/(g))` (time period of simple pendulum) `:. g= (4pi^2L)/(t^2)` `:. (Deltag)/(g) =(Delta L)/(L) + (2 Delta t)/(t)` As L is exactly known, `DeltaL = 0` `:. (Delta g)/(g) = (2 Delta t)/(t) = 2xx (1)/(40) = (2)/(40)` `% error (Delta g)/(g)xx100 = (2)/(40)xx100 = 5%`