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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]

Respuesta :

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

[tex]= \frac{0.21}{9.88}\times100[/tex]

= +/-2.13

Percent difference with 9.88

Difference =  9.88 - 9.81

= 0.07

[tex]= \frac{0.07}{9.81}\times100[/tex]

= +/-0.71%

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

Mean of two values is

= [tex]\frac{9.67 + 9.88}{2}[/tex]

= 9.775

Difference = 9.81 - 9.775

= 0.035

Percentage difference is

[tex]= \frac{0.035}{9.81}\times 100[/tex]

= +/- 0.36%

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