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A student uses a simple pendulum of exac...

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 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%

Text Solution

Verified by Experts

The correct Answer is:
(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%`
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