Question

The vernier constant of a travelling microscope is 0.001 cm. If 49 main scale divisions coincide with 50 vernier scale divisions, then the value of 1 main scale division is:
A. $0.01mm$
B. $0.02cm$
C. $0.05cm$
D. $0.0005cm$

Answer 1

We know that the travelling microscope is being classified into two halves and there are $50$ divisions on a vernier scale. The least count (LC) of a travelling microscope is $0.001cm$. Adding on, MSD is the smallest division on the main scale of travelling microscopes.

Complete answer:
We know that the main scale division of a travelling microscope is the smallest division on the main scale.
Now to find the value of 1 MSD, we will be using the formula of vernier constant, where,
$VC=\dfrac{MSD}{divisions}$
Where, VC is the Vernier constant or basically the least count which is defined as the ratio of the smallest division of the main scale (MSD) divided by the number of divisions on the vernier scale.
We have,

& LC=0.001cm \\\ & divisions=50 \\\ \end{aligned}$$ Substituting that in the formula we get, $$\begin{aligned} & \Rightarrow 0.001=\dfrac{MSD}{50} \\\ & \Rightarrow MSD=0.001\times 50 \\\ & \therefore MSD=0.05cm \\\ \end{aligned}$$ **Hence, option C is the correct answer.** **Additional Information:** In order to find the length of the vernier scale instrument we can use the formula, $$length=MSR+(VSR\times LC)$$ Where, MSR is the main scale reading VSR is the vernier scale reading LC is the least count of the equipment While reading the measurements, we have to see if the instrument has an error or not. That is vernier scales can have two types of error, positive or negative zero error. Positive zero error is the additional readings right to the zero which we will have to subtract from the obtained measurement. While negative zero error is the additional readings present left to the zero when the jaws are in contact. This we have to add to the obtained measurement. **Note:** The least count of the equipment becomes very important when going from large scale to small scale readings. Excluding any errors in the equipment will result in the wrong answer. The combined result of the main scale and the vernier scale will give us accurate values.