A vernier callipers is used to measure the width and the length of a rectangular plate. The measured values are 1.38 cm and 4.02 cm respectively.Find the uncertainly in the value of its area.

Explanation

Given that, width w = 1.38 cm
length, l = 4.02 cm
Since the least count of a vernier callipers is 0.01 cm, each of the above mentional measurements has an uncertainty of `pm0.01cm`. Hence, the values can be written as
`w=1.38cmpm0.01cm`
`l=4.02cmpm0.01cm`
The area obtained using the measured values is
`A=wxxl`
= `1.38xx4.02`
`A=5.55cm^(2)" "...(i)`
If we take the lower limits of both the measured values i.e.
`l_(min)=4.02-0.01`
= 4.01 cm
and `w_(min)=1.38-0.01`
= 1.37 cm
The minimum possible area (lower limit of the area) comes out to be
`A_(min)=4.01xx1.37cm^(2)`
`A_(min)=5.49cm^(2)" "...(ii)`
The upper limit can also be calculated similarly.
`thereforeA_(max)=4.03xx1.39cm^(2)`
`A_(max)=5.60cm^(2)" "...(iii)`
Using equation (i), (ii) and (iii), we find that `A_(min)` is lower than A (the nominal measurement) by the value `0.06cm^(2)`.
And `A_(max)` is higher than A by the value `0.05cm^(2)`.
As a practical rule, we choose the higher of these two deviations (from the measured value) as the uncertainty, in our result.
Therefore, in this particular example, we can write the area of the plate as `5.55cm^(2)pm0.06cm^(2)`