If the length and time period of an oscillating pendulum have errors of 1% and 2% respectively. The error in the estimation of 'g' is

Options

A. 0.01
B. 0.02
C. 0.03
D. 0.05

Answer

D

Explanation

`(Delta L)/(L) = 1% " and " (Delta T)/(T) = 2%`
now we have ` T = 2pi sqrt(T/g) , g = 4pi (1)/(T^2)`
` (Delta g)/(g) xx 100 = ((Delta L)/(L) xx 100) + 2 ( (Delta T)/(T) xx 100) = 1% + 2(2%)`
` (Delta g)/(g) xx 100 = 5%`