Question

In an experiment to measure the acceleration due to gravity g, two independent equally reliable measurements gave 9.67 m/s2 and 9.88 m/s2. Determine (i) the percent difference of the measurements (ii) the percent error of their mean. [Take the theoretical value of g to be 9.81 m/s2]

Answer 1

Answer and Explanation:

a. The computation of the percent difference between the measurements is shown below:-

The first value of g is 9.67 and the second value is 9.88

So, difference = 9.88 - 9.67

= 0.21

Percentage difference in measurement is

`= (0.21)/(9.88)*100`

= +/-2.13

Percent difference with 9.88

Difference = 9.88 - 9.81

= 0.07

`= (0.07)/(9.81)*100`

= +/-0.71%

b. The Computation of percent error of their mean is shown below:-

Mean of two values is

=
`(9.67 + 9.88)/(2)`

= 9.775

Difference = 9.81 - 9.775

= 0.035

Percentage difference is

`= (0.035)/(9.81)* 100`

= +/- 0.36%