Question

5
In a workshop a worker measures the
length of a steel plate with a Vernier
callipers having a least count 0.01 cm.
Four such measurements of the length
yielded the following values: 3.11 cm,
3.13 cm, 3.14 cm, 3.14 cm. Find the
and the percentage error in the measured
vi)
mean length, the mean absolute error
value of the length.
[Ans: 3.13 cm, 0.01 cm, 0.32%]
​

Answer 1

Answer:

3.13 cm 0.01 cm 1.3 %

Explanation:

To get the mean length we usually add the values given then we divide them by the number of values given

Like in this case we have four values therefore

We add them then we divide by four

3.11 + 3.13 + 3.14 + 3.14 =12.52 cm

We then divide it by four

12.52 cm÷4 = 3.13cm

The mean absolute error is 0.01cm because it is the leat count

To calculate percentage error we first find the relative error

0.01÷3.13 = 0.031 cm

We then multiply the absolute error by 100 %

0.031 cm × 100 % = 3.1 %