In an experiment of simple pendulum , the time period measured was `50 s for 25` vibrations when the length of the simple pendulum was taken `100 cm`. If the least count of stop watch is `0.1 s` and that of meter scale is `0.1 cm`. Calculate the maximum possible error in the measurement of value of `g`. If the actual value of `g` at the place of experiment is `9.7720 m s^(-2) ` , Calculate the percentage error.

Explanation

The time period of a simple pendulum is given by
`T = 2 pi sqrt((l)/(g))` or `T^(2) = ( 4 pi^(2) l) /( g)` or `g = ( 4 pi^(2) l)/( T^(2))`
As `4 and pi` are constants , the maximum permissible error in `g`
is given by `( Delta g)/(g) = ( Delta l) /( l) + ( 2 Delta T)/( T)`
Here `Delta L = 0.1 cm , L = 1 m = 100 cm , Delta T = 0.1 s , T = 50 s`.
:. `( Delta g)/( g) = (0.1)/( 100) + 2((0.1)/(50)) = ( 0.1)/( 100) + ((0.1)/(25))`,
`( Delta g)/( g) xx 100 = [ ( 0.1)/(100) + (0.1) /(25) ] xx 100 = 0.1 + 0.4 = 0.5%`
`g = ( 4 pi^(2) l^(2))/( T^(2)) . Here T = ( 50)/(25) = 2 ` . Therefore,
`g' = ( 4 xx (3.14)^(2) xx (1)^(2))/(( 2)^(2)) = 9.8596 m s ^(-2)`,
Actual value `g = 9.7720 m s^(-1)` . Therefore,
Percentage error `= ( g' - g)/( g) xx 100`
`= ( 9.8596 - 9.7720)/(9.7720) xx 100 = 0.8964%`