While measuring acceleration due to gravity by a simple pendulum , a student makes a positive error of `2%` in the length of the pendulum and a positive error of `1%` in the measurement of the value of `g` will be

Options

A. `3%`
B. `0%`
C. `4%`
D. `5%`

Answer

B

Explanation

`T = 2 pi sqrt((L)/( g))` or `T^(2) = 4 pi^(2) (L)/( g)`
`g = 4 pi^(2) (L)/(T^(2)) , (Delta g)/(g) = (Delta L)/(L) -2 (Delta T)/(T)`
`(Delta g)/(g) xx 100 = (Delta L)/(L) xx 100 -2 (Delta T)/(T) xx 100`
Actual % error in `g = (Delta L)/(L) xx 100 - 2(Delta T)/(T) xx 100`
`= + 2% -2 xx 1% = 0%`