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 mentioned measurements has an uncertainly of `pm`0.01 cm. hence, the values can be written as
w=1.38 cm `pm`0.01 cm
`l=4.02 cm pm 0.01 cm`
The area obtained using the measured value is A=w x l
= 1.38 x 4.02
`A=5.55 cm^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 limits of the area ) comes out to be `A_"min"=4.01xx1.37 cm^2`
`A_"min"=5.49 cm^2`...(ii)
The upper limit can also be calculated similarly.
`therefore A_"max"=4.03xx1.39 cm^2`
`A_"max"=5.60 cm ^2` ...(iii)
using equation (i),(ii) and (iii) , we find that `A_"min"` is lower than A (the nominal measurement ) by the value `0.06 cm^2`
And `A_"max"` is higher than A by the value `0.05 cm^2`
As a practical rule, we choose the higher of these two deviations (from the measured value ) as the uncertainly, in our result.
Therefore, in this particular example, we can write the area of plate as `5.55 cm^2 pm 0.06 cm^2`