In an experiment of simple pendulum, time period measured was `50s` for `25` vibrations when the length of the simple pendulum was taken to be `100cm`. If the lest count of stop watch is `0.1s` and that of metre scale is `0.1 cm`, calculate the maximum possible percentage error in the measurement of value of `g`.

Answer

0.5

Explanation

The time period of a simple pendulum is given by
`T=2pisqrt((l)/(g))` or `T^(2)=(4pi^(2)l)/(g)` or `g=(4pi^(2)l)/(T^(2))`
As `4` and `pi` are constants, maximum permissible error in `g` is given by `(Deltag)/(g)=(Deltal)/(l)+(2DeltaT)/(T)`
Here, `Deltal=0.1cm`, `l=1m=100cm`,
`DeltaT=0.1s`, `T=50s`
`:. (Deltag)/(g)=(0.1)/(100)+2((0.1)/(50))=(0.1)/(100)+((0.1)/(25))`
or `(Deltag)/(g)xx100=((0.1)/(100)+(0.1)/(25))xx100`
`=0.1+0.4=0.5%`