[wadeekha]

I'm confused with some problems listed below, could you help me with them? I'm so sorry about this!

There was this way about finding the actual value with uncertainty given the percentage uncertainties of measurements, other than the fractional uncertainty method. However, I'm not sure if it is correct.

1.) Jane needs to calculate the volume of her pool, so that she knows how much water she'll need to fill it. She measures the length, width, and height:
           length  L  =  5.56    +/-  0.14 meters
                      =  5.56 m  +/-  2.5%

           width   W  =  3.12    +/-  0.08 meters
                      =  3.12 m  +/-  2.6%

           depth   D  =  2.94    +/-  0.11 meters
                      =  2.94 m  +/-  3.7%

To calculate the volume, she multiplies together the length, width and depth:
              volume =  L * W * D =  (5.56 m) * (3.12 m) * (2.94 m)
                   
                     =  51.00 m^3

    percentage uncertainty in volume =   (percentage uncertainty in L) +
                                         (percentage uncertainty in W) +
                                         (percentage uncertainty in D)

                                     =  2.5% + 2.6% + 3.7%

                                     =  8.8%
Therefore, the uncertainty in the volume (expressed in cubic meters, rather than a percentage) is
      uncertainty in volume  =( (volume) * (percentage uncertainty in volume) )/100

                             = ( (51.00 m^3) * (8.8%) )/100

                             = 4.49 m^3
Therefore,
       volume  =  51.00  m3  +/-  4.49  m^3
               =  51.00 m3  +/-  8.8%

Then I did the way taught, given only absolute uncertainties where,
  length  L  =  5.56    +/-  0.14 meters
  width   W  =  3.12    +/-  0.08 meters
  depth   D  =  2.94    +/-  0.11 meters
Uncertainty in the volume is found by,
(0.14)/(5.56) + (0.08)/(3.12) + (0.11)/(2.94) = 0.0882358477 ( Do I round off here? )
0.0882358477 x 51.00 m3 ( actual calculated volume ) = 4.500028233 = 5 (1 s.f)
Therefore, volume  =  51  m3  +/-  5  m^3

The resultant answers are a bit different, one is 51  m3  +/-  5  m^3 when done the fractional uncertainty way, the other is 51.00 m3  +/-  8.8% when done the adding up of percentage uncertainties way. Shouldn't they be the same?

1a.) 51.00 m3  +/-  8.8% <--- In the case of percentage uncertainty, is it right to say that the value measured should follow the significant figures of  the percentage uncertainty instead of d.p?

3.)Does percentage deviation = percentage error?

4) In general mole/ uncertainty calculations, is it that the final answer should be have the same number of significant figures as the value with the least significant figures given in the question?

Thank you!

[juanrga]

I'm confused with some problems listed below, could you help me with them? I'm so sorry about this!

There was this way about finding the actual value with uncertainty given the percentage uncertainties of measurements, other than the fractional uncertainty method. However, I'm not sure if it is correct.

1.) Jane needs to calculate the volume of her pool, so that she knows how much water she'll need to fill it. She measures the length, width, and height:
           length  L  =  5.56    +/-  0.14 meters
                      =  5.56 m  +/-  2.5%

           width   W  =  3.12    +/-  0.08 meters
                      =  3.12 m  +/-  2.6%

           depth   D  =  2.94    +/-  0.11 meters
                      =  2.94 m  +/-  3.7%

To calculate the volume, she multiplies together the length, width and depth:
              volume =  L * W * D =  (5.56 m) * (3.12 m) * (2.94 m)
                    
                     =  51.00 m^3

    percentage uncertainty in volume =   (percentage uncertainty in L) +
                                         (percentage uncertainty in W) +
                                         (percentage uncertainty in D)

                                     =  2.5% + 2.6% + 3.7%

                                     =  8.8%
Therefore, the uncertainty in the volume (expressed in cubic meters, rather than a percentage) is
      uncertainty in volume  =( (volume) * (percentage uncertainty in volume) )/100

                             = ( (51.00 m^3) * (8.8%) )/100

                             = 4.49 m^3
Therefore,
       volume  =  51.00  m3  +/-  4.49  m^3
               =  51.00 m3  +/-  8.8%

Then I did the way taught, given only absolute uncertainties where,
  length  L  =  5.56    +/-  0.14 meters
  width   W  =  3.12    +/-  0.08 meters
  depth   D  =  2.94    +/-  0.11 meters
Uncertainty in the volume is found by,
(0.14)/(5.56) + (0.08)/(3.12) + (0.11)/(2.94) = 0.0882358477 ( Do I round off here? )
0.0882358477 x 51.00 m3 ( actual calculated volume ) = 4.500028233 = 5 (1 s.f)
Therefore, volume  =  51  m3  +/-  5  m^3

The resultant answers are a bit different, one is 51  m3  +/-  5  m^3 when done the fractional uncertainty way, the other is 51.00 m3  +/-  8.8% when done the adding up of percentage uncertainties way. Shouldn't they be the same?

1a.) 51.00 m3  +/-  8.8% <--- In the case of percentage uncertainty, is it right to say that the value measured should follow the significant figures of  the percentage uncertainty instead of d.p?

3.)Does percentage deviation = percentage error?

4) In general mole/ uncertainty calculations, is it that the final answer should be have the same number of significant figures as the value with the least significant figures given in the question?

Thank you!

In the first place, ##(5.56 \pm 0.14) \mathrm{m}##  is not equal to ##(5.56 \mathrm{m} \pm 2.5\%)##, ##(3.12 \pm 0.08) \mathrm{m}##  is not equal to ##(3.12 \mathrm{m} \pm 2.6\%)## and so on. In fact,
$$L = (5.56 \pm 0.14) \mathrm{m} = 5.56 · (1 \pm 2.5\%) \mathrm{m} $$
In the second place, volume V is length times width times depth, but ##(5.56 \mathrm{m}) · (3.12 \mathrm{m}) · (2.94 \mathrm{m})## is not volume because length is not 5.56 but ##(5.56 \pm 0.14) \mathrm{m}##, for instance.

The above ##51.00 \mathrm{m}^3## computed by Jane is really an average volume
$$\langle V \rangle =  \langle L \rangle · \langle W \rangle · \langle D \rangle =  (5.56 \mathrm{m}) · (3.12 \mathrm{m}) · (2.94 \mathrm{m}) = 51.00 \mathrm{m}^3$$
The volume is the average volume plus the uncertainty
$$V = \langle V \rangle + \Delta V$$
We can compute the uncertainty in volume as
$$\Delta V = \frac{\partial V}{\partial L} \Delta L + \frac{\partial V}{\partial W} \Delta W + \frac{\partial V}{\partial D} \Delta D$$
This would solve your doubts.

Regarding the 0.0882358477, the absolute uncertainties given have two figures. Therefore this would be rounded to 0.088.