In the examples below the measurements of a and b arc used t

In the examples below, the measurements of a and b arc used to calculate the product q=ab. Calculate both the quantity q with error, and calculate the percent error based on the exact values given. Careful: Please state the % error with the correct number of significant figures! Use the rules stated earlier to figure this out. a = 11.5 plus minus 0.2 cm and b = 25.4 plus minus 0.2 cm. Exact value = 315 cm^2 a = l0. Plus minus 1 cm and, b = 27.2 plus minus 0.1sec. Exact value 250 cm*sec. a = 3.0 ft plus minus 8% and b = 4.0 lb plus minus2%. Exact value = 12 ft*lb a = 1.0 plus minus 0.2 kg and b = 3.4 plus minus 0.1 m/sec. Exact value = 3.2 kg.m/sec

Solution

percent error=|(estimated value-actual value)/(actual value)|*100% (actual value that is exact value)

(a) if a=11.5+0.2 and b=25.4+0.2 then q=ab=299.52

and if a=11.5-0.2 and b=25.4-0.2 then q=ab=284.76

to find the percent error we consider the small error that isconsider the estimated value near to exact value

here estimated value=299.52cm^2 actual value=315cm^2

therefore percent error=|(299.52-315)/(315)|*100%=4.91%

(b)a=10.+1cm and b=27.2+0.1sec then q=ab=300.3cm*sec

and if a=10.-1cm and b=27.2-0.1sec then q=ab=243.9cm*sec

here estimated value=243.9cm*sec actual value=250cm*sec

therefore percent error=|(243.9-250)/(250)|*100%=2.44%

(c)a=3.0ft+8% and b=4.0lb+2% then q=ab=12.3816ft*lb

and if a= 3.0ft-8% and b=4.0lb-2% then q=ab=11.6216

here(you can take any value) estimated value=12.3816ft*lb actual value=12ft*lb

therefore percent error=|(12.38-12)/(12)|*100%=3.16%

(d)a=1.0+0.2kg and b=3.4+0.1m/sec then q=ab=4.2kgm/sec

and if a=1.0-0.2kg and b=3.4-0.1m/sec then q=ab=2.64kg.m/sec

here estimated value=2.64kg.m/sec actual value=3.2kg.m/sec

therefore percent error=|(2.64-3.2)/(3.2)|*100%=17.5%