# Question 0710f

Oct 5, 2015

The answer is (c) 2.0%

#### Explanation:

So, you use a vernier caliper with a least count of $\text{0.01 cm}$ to measure the thickness of a wire.

You measure the wire to have a thickness of $\text{0.45 cm}$. The precision of the measurement will be given by the least count of the vernier calipers.

The absolute uncertainty of the instrument is actually equal to half of the least count, which in this case would be

$\text{abs. uncertainty" = "0.01 cm"/2 = +- "0.005 cm}$

However, you need to report this uncertainty using the same number of decimal places as the measurement. In your example, the measurement is rounded to the hundredths place, shown here in red

$0.4 \textcolor{red}{5}$

This means that you need to round the uncertainty to the hundredths place as well, which in your case would be

$\pm \text{0.005 cm" ~~ +- 0.0color(red)(1)color(white)(x)"cm}$

In this respect, you always need to report the absolute uncertainty to the same decimal place as the measurement.

The percentage uncertainty can be calculated using the absolute uncertainty and the actual measurement

$\text{% uncertainty" = "absolute uncertainty"/"measured value} \times 100$

In your case, the percentage uncertainty will be equal to

"% uncertainty" = (0.01color(red)(cancel(color(black)("cm"))))/(0.45color(red)(cancel(color(black)("cm")))) xx 100 = 2.222%

You can only use two sig figs for the answer, so you have

"% uncertainty" = color(green)(2.0%)#

For two measurements, keep in mind that the measurement that has the smaller absolute uncertainty is more precise, and that the one that has the smallesr percent uncertainty is more accurate.