Uncertainty
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Give 3 different ways we can reduce uncertainty.
• Increase resolution of instrument/ ie. use instrument with smaller intervals. • Take repeated readings • Size of measurement taken.
How do we work out the uncertainty of equipment where one judgement is made?
• ±1/2 of smallest interval
How do we work out the uncertainty of equipment where two judgements are made (ie. 2 sources of uncertainty)?
• ±1/2 of smallest interval + - ±1/2 of smallest interval
Reading: 1660mg, without any info on resolution of equipment, what would the uncertainty be?
Reading: 1660mg, without any info on resolution of equipment, what would the uncertainty be? - ±10 - Look at what place value the last significant digit is, in this case, its the tens.
Reading: 1663mg, without any info on resolution of equipment, what would the uncertainty be?
-±1 - Look at what place value the last significant digit is, in this case its in the ones.
If we have 10 leaves that need to be measured, what would lead to lowest uncertainty 1.) Measuring all the leaves together (gives length of 2.1mm) 2.) Measuring one leaf at a time and WHY? Assume, that the uncertainty for measuring 1 leaf is ±0.01cm
• Measure length all leaves stacked on top of each other. • 10 leaves together: 2.1 ±0.1 • 1 leaf: 0.21 ±0.01(reduces the uncertainty.) • We can do this because percentage uncertainty in measuring 1 leaf = same as 10 leaves.
How can uncertainty be calculated when we have repeated measurements?
• Calculate mean of repeated measurments. • Range/ 2 = uncertainty of the repeated measurements.
How do we calculate percentage uncertainty in repeated measurements?
uncertainty/ mean value x 100.
June 2019 Paper 1 I have measured a structure on microscope as 3mm. The intervals on the ruler are +1mm. - Magnification = x500 Calculate the maximum length of the structure. Use a ruler with mm intervals. Use the uncertainty in your measurement to determine the uncertainty calculated in the maximum length. Assume there is no uncertainty in the magnification. V. important skill!! You got this wrong 1st time!
Uncertainty = +/- 0.5mm Image = actual x mag. MAXIMUM image = 3mm +0.5 = 3.5mm MINIMUM image length = 3-0.5=2.5mm • 3.5mm/500= 7x10^-6 = 7micrometers • 2.5mm/500= 5x10^6 = 5 micrometers. Maximum length (MP1): 7 micrometers. Uncertainty of calculated maximum length (MP2): 7-5 =2micrometers.
Paper 3 2017 A scientist measured the size of each gill lamella of gills of 40 damselfly larvae. Mean width: 1.61. Uncertainty: (+/-0.19) Mean length: 6.12. Uncertainty: (+/- 0.41) Calculate mean surface area of one side of gill lamella. Assume that a gill lamella is rectangular and give your answer to the appropriate number of significant figures. Include the percentage error (uncertainty) of surface area in your answer.
1.61mm x 6.12mm = 9.85mm^2 Uncertainty: (0.19/ 1.61x100) + (0.41/6.12 x100) = 18.5%