# 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 %

psycho_physicist said:

## 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 %

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.

gneill said:
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 ?

psycho_physicist said:
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?