estimate the percent error in measuring
a
a)distance of about 50 cm with a meter stick
b) a mass of about 1g with a chemical balance
c)time interval of about 4 mins with a stop watch
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1. the error in measurement could be 1mm. so percentage error is
(0.001 m / 0.50 m )*100 = 0.2%
2. chemical balance may have an accuracy of 0.1 mg or 0.2 mg , of that order.
So % error = (0.1 mg /1000 mg) * 100 = 0.01%
3. Let the minimum unit of measurement = 1 sec. in the stop watch.
% error = (1 sec / 240 sec)*100 = 0.4 %
if the stop watch has a resolution of 0.1 sec, then
% error = (0.1 / 240) * 100 = 0.04%
(0.001 m / 0.50 m )*100 = 0.2%
2. chemical balance may have an accuracy of 0.1 mg or 0.2 mg , of that order.
So % error = (0.1 mg /1000 mg) * 100 = 0.01%
3. Let the minimum unit of measurement = 1 sec. in the stop watch.
% error = (1 sec / 240 sec)*100 = 0.4 %
if the stop watch has a resolution of 0.1 sec, then
% error = (0.1 / 240) * 100 = 0.04%
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