Question

Consider the following data: 10 MSDs = 1cm, 10 VSDs = 9 MSD, zero of the Vernier scale is to the right of the main scale with 6 VSDs coinciding with MSDs and the actual reading for length measurement is 4.3 cm with 2 VSDs coinciding with main scale graduations. Estimate the length.

Answer 1

Hint: Here we need to calculate the value of the Vernier constant of the instrument and then we have to calculate the zero error of the scale. After calculating the zero error find the value of the true length using the given value of actual reading and the value of the zero error

Complete step-by-step answer:
The value of the Vernier constant can be calculated as below:
The Vernier constant is the difference between the smallest division on the main scale and Vernier scale
So we are given the value of 10 VSD = 9 MSD so we get 1 VSD = $ \dfrac{9}{10} $ MSD
So the difference becomes MSD – VSD = $(1- \dfrac{9}{10} )$MSD = 0.1 MSD
Now we know the value of the MSD is $ \dfrac{1}{10} $ cm so we get the 1 MSD = 0.1 cm = 1mm
So the value of the Vernier constant is 0.1 MSD = 0.1 mm
Hence we go the least count value as 0.1 mm
Now the zero error is positive and we get since 6VSD coinciding with the MSDs
We get zero error is : 6 $ \times $ 0.1 = 0.6 mm
Since the error is positive we get the correction is negative
So the correction is -0.6 mm = -0.06cm
So we get the true reading as
The actual reading is 4.3 + 2 $ \times $ 0.01 = 4.32cm
Now combing the corectionw ege the true reading as 4.32 + correction = 4.32 + (-0.06) = 4.26 cm
Hence we got the true reading value as 4.26 cm.

Note: There is a possibility of error when one is doing the correction part. Often we may think that the zero error is the value to be added to the actual reading we get the wrong results. We need to take care that the correction is always of the opposite sign to the zero error.