The period of oscillation of a simple pendulum is `T=2pisqrt(L/(g))` . `L` is about `10` cm and is known to `1` mm accurancy. The period of oscillation is about `0.5` second. The time of `100` oscillation is measured with a wrist watch of `1` s resolution. What is the accurancy in the determination of `g` ?

Options

A. `3%`
B. `2%`
C. `5%`
D. `4%`

Answer

C

Explanation

`(Deltag)/(g)=(DeltaL)/(L)+2xx(DeltaT)/(T)`
In terms of percentage,
`(Deltag)/(g)xx100=(DeltaL)/(L)xx100+2xx(DeltaT)/(T)xx100`
Percentage error in `L=100xx(DeltaL)/(L)=100xx(0.1)/(10)=1%`
Percentage error in `T=100xx(DeltaT)/(T)=100xx(1)/(50)=2%`
Percentage error in `g=100(Deltag)/(g)=1%+2xx2%=5%`
`(c )` is correct.