Errors/ Uncertainty in measurement problem

[psycho_physicist]

Homework Statement

[/B]
The surface tension (T) is measured by capillary rise formula T = \frac {rh ρg}{2 cos\Theta} . The quantities of ρ, g and θ are taken from the table of constants while the height and diameter are measured as
h = (3.00 + 0.01)cm and
D = (0.250 ± 0.001)cm
Find the percentage error in T
My answer came out to be 1.1 % but doesn't coincide with the answer given (0.7 %).
Thanks for any help!

Homework Equations

According to the book I refer, the equation for error in product or quotient is -[/B]
\frac{\Delta Z} {Z}= \frac {\Delta A} {A} +\frac {\Delta B} {B}
where Δz/z, Δa/a and Δb/b are relative errors (letters in deltas present uncertainty in measurement)

The Attempt at a Solution

I have attempted the solution but have failed to acquire the given answer. Here's what I tried
Acc. to \frac{\Delta Z} {Z}= \frac {\Delta A} {A} +\frac {\Delta B} {B}
we ignore constants (ρ, Θ and g)
h = (3.00 + 0.01)cm
D = (0.250 ± 0.001)cm
so r = D/2 or r = (0.125 ± 0.001)cm
so
\frac{\Delta T} {T}= \frac {\Delta r} {r} +\frac {\Delta h} {h}
which gives,
\frac{\Delta T} {T} =\frac {0.001} {0.125} +\frac {0.01} {3.00}
or,
\frac{\Delta T} {T} =0.0080 + 0.0034
or,
\frac{\Delta T} {T} =0.0113
where
\frac{\Delta T} {T} is relative error. When we multiply that by 100, we get percent error
therefore percent error = 1.13 %
while the answer given is 0.7 %
please help

 

[Samy_A]

Homework Statement

[/B]
The surface tension (T) is measured by capillary rise formula T = \frac {rh ρg}{2 cos\Theta} . The quantities of ρ, g and θ are taken from the table of constants while the height and diameter are measured as
h = (3.00 + 0.01)cm and
D = (0.250 ± 0.001)cm
Find the percentage error in T
My answer came out to be 1.1 % but doesn't coincide with the answer given (0.7 %).
Thanks for any help!

Homework Equations

According to the book I refer, the equation for error in product or quotient is -[/B]
\frac{\Delta Z} {Z}= \frac {\Delta A} {A} +\frac {\Delta B} {B}
where Δz/z, Δa/a and Δb/b are relative errors (letters in deltas present uncertainty in measurement)

The Attempt at a Solution

I have attempted the solution but have failed to acquire the given answer. Here's what I tried
Acc. to \frac{\Delta Z} {Z}= \frac {\Delta A} {A} +\frac {\Delta B} {B}
we ignore constants (ρ, Θ and g)
h = (3.00 + 0.01)cm
D = (0.250 ± 0.001)cm
so r = D/2 or r = (0.125 ± 0.001)cm
so
\frac{\Delta T} {T}= \frac {\Delta r} {r} +\frac {\Delta h} {h}
which gives,
\frac{\Delta T} {T} =\frac {0.001} {0.125} +\frac {0.01} {3.00}
or,
\frac{\Delta T} {T} =0.0080 + 0.0034
or,
\frac{\Delta T} {T} =0.0113
where
\frac{\Delta T} {T} is relative error. When we multiply that by 100, we get percent error
therefore percent error = 1.13 %
while the answer given is 0.7 %
please help

Check your uncertainty on r.

 

[gneill]

When you divided the diameter by two, the error in the resulting figure should not remain the same as the original. The percentage error of the result should remain the same. So, scale both by the same factor: 0.250/2 = 0.125; 0.001/2 = 0.0005.

 

[psycho_physicist]

When you divided the diameter by two, the error in the resulting figure should not remain the same as the original. The percentage error of the result should remain the same. So, scale both by the same factor: 0.250/2 = 0.125; 0.001/2 = 0.0005.

Thanks, that correction seems to nail it!
One Q - the instrument (say vernier calipers) was used to measure diameter and radius is just half of diameter; then why do we halve the error? Shouldn't the uncertainty remain same ?

 

[gneill]

Thanks, that correction seems to nail it!
One Q - the instrument (say vernier calipers) was used to measure diameter and radius is just half of diameter; then why do we halve the error? Shouldn't the uncertainty remain same ?

What's the rule when multiplying or dividing by a constant? Does the percent error change?

 

[Dr Transport]

Read this file, https://www.wmo.int/pages/prog/gcos/documents/gruanmanuals/UK_NPL/mgpg11.pdf

You really should be taking the derivative of the original formula with respect to each of the variables and do the root square addition for them to get the uncertainty. Each of the variables in your equation has an uncertainty, even the ones pulled from a reference table and it needs to be included.

 

[Ray Vickson]

Thanks, that correction seems to nail it!
One Q - the instrument (say vernier calipers) was used to measure diameter and radius is just half of diameter; then why do we halve the error? Shouldn't the uncertainty remain same ?

The diameter lies between .250+.001 cm = 0.251 cm and .250 - .001 cm = 0.249 cm, so the radius lies between .251/2 = 0.1255 cm and 0.249/2 = 0.1254 cm. Those figures are .125 ± 0.0005 cm. The percentage errors in diameter and radius are exactly the same.