Question

The main scale of a Vernier caliper reads \[10mm\] in \[10\]divisions. \[10\] divisions of the vernier scale coincide with \[9\] divisions of the main scale. When the two jaws of the calipers touch each other the right of zero of the main scale. When a cylinder is tightly placed between the two jaws, the zero of the Vernier scale lies slightly to the left of \[3.2cm\] and the fourth Vernier division coincides with the main scale division. The diameter of the cylinder is :
A. \[3.09cm\]
B. \[3.9cm\]
C. \[3.90cm\]
D. \[.39mm\]

Answer 1

Hint: It is a simple vernier caliper procedure. It is given that the main scale reading is \[10mm\] and \[10\] divisions. Calculate the least count of the vernier caliper when the jaws are together and use the same conditions when the cylinder is placed between the jaws.

Complete step by step answer:
It is given that in the Main scale, \[10mm\] is read in the 10th division of the main scale. This means that one main scale division can be identified as \[\dfrac{{10mm}}{{10}}\], which is equal to \[1mm\]. Now, It is given that the 10th division of the Vernier scale reading coincides with the 9th division of the Main scale reading when the jaws are closed. This means that,
\[ \Rightarrow 10VSD = 9MSD\]
\[ \Rightarrow 1VSD = \dfrac{9}{{10}}MSD\]
\[ \Rightarrow 1VSD = \dfrac{9}{{10}}mm\]
Now, the least count of the Vernier Caliper is given as the difference between one minimum scale division and One Vernier scale division.
\[ \Rightarrow LC = 1MSD - 1VSD\]
We know that \[1MSD\] is 1mm and \[1VSD\] as \[\dfrac{9}{{10}}\]mm. Using this we get,
\[ \Rightarrow LC = \dfrac{1}{{10}}mm = 0.1mm = 0.01cm\]
Now, it is also given that, when the jaws touch each other, the 5th division of the Vernier scale coincides with the 9th main scale division and the zero of the Vernier scale is to the right of the zero of the main scale. Hence, the zeroes are not coinciding and we have an error condition here. When the Zero is to the right of the main scale, it is taken as a positive zero error and the correction is negative. Hence, the correction value here is,
\[e = 5 \times LC = 5 \times 0.01cm = 0.05cm\]
Now, a cylinder of unknown diameter is tightly held between the jaws of the vernier caliper, and the reading is taken down. It is found that the zero of the Vernier scale coincides with a value slightly left of \[3.2cm\]. Now, we know that \[1MSD\]is equal to 1mm and hence the exact value of coincidence is \[3.1cm\].
Now, the final observed reading is given as ,
\[ \Rightarrow TSR = MSR + (VSD \times LC) - e\]
\[ \Rightarrow TSR = 3.1 + (4 \times 0.01) - 0.05\]
\[ \Rightarrow TSR = 3.14 - 0.05\]
\[ \Rightarrow TSR = 3.09cm\]
Hence, option(a) is the right answer for the given question.

Note: If the zero of the Vernier Scale is to the right of the zero of the main scale it is called a positive error and the correction done to the overall value should be negative. Subtract the error value from the final value to obtain an error-free dimension.