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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 `triangleT` in measuring T, the time period, is 0.05s.

B

Error `triangleT` in measuring T, the time period, is 1s.

C

percentage error in the determination of g is 5%

D

percentage error in the determination of g is 2.5%

Text Solution

Verified by Experts

The correct Answer is:
A, C
We can see that time period is `T=(40)/(20)=2s`
A total error of 1s will get divided among 20 oscillations, hence error in the time period is given by
`triangle T=(1)/(20)=0.05s`
We can see that option a is correct.
Further, we know that time period of oscillation is given by
`T=2pi sqrt((l)/(g)) implies g=(4pi^(2)l)/(t^2)`
Here length is measured to be exact, hence it will not be included for error calculation. For above relation for acceleration due to gravity we can write the following :
`(triangleg)/(g)xx100=2(triangleT)/(T)xx100= 2(0.05)/(2)xx100=5%`
We can see that option c is also correct.
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