Question

In an experiment to measure the acceleration g due to gravity, two values 9.96 ms-2, and 9.72 ms-2 are determined. Find(i) the percentage difference of the measurements, (ii) the percentage error of each measurement and(iii) the percentage error of their mean. (accepted value of g = 9.81 ms-2)

Answer 1

i) difference =9.96-9.72=0.24

% difference in measurement$=0.24/9.72\times 100\%=2.5\%$

ii) % difference with 9.72

difference=9.81-9.72=0.09

% difference in measurement=$0.09/9.81\times 100=0.92\%$

% difference with 9.96

difference =9.96-9.81=0.15

% difference in measurement$=0.15/9.81\times 100=1.53\%$

iii) mean of two values$=(9.96+9.72)/2=9.84$

difference = 9.84-9.81=0.03

% difference $=0.03/9.81\times 100=0.31\%$