Question

Calculate percentage error in determination of time period of a pendulum.
`T = 2pi sqrt(l/g)`
where, l and g are measured with `+-1%` and `+-2%`.

Answer 1

`(DeltaT)/(T)xx100=pm((1)/(2)xx(Deltal)/(l)xx100+(1)/(2)xx(Deltag)/(g)xx100)`
`=pm((1)/(2)xx1+(1)/(2)xx2)=pm1.5%`.