One main scale division of a vernier callipers is $a \ cm$ and $n^{\text{th}}$ division of the vernier scale coincides with $(n-1)^{\text{th}}$ division of the main scale. The least count of the callipers in $mm$ is

The vernier constant of Vernier callipers is $0.1 \,mm$ and it has zero error of $(-0.05) \,cm$. While measuring the diameter of a sphere,the main scale reading is $1.7 \,cm$ and the coinciding vernier division is $5$. The corrected diameter will be ........... $\times 10^{-2} \,cm$.

Given below are two statements:
Statement $I$: In a vernier callipers,one vernier scale division is always smaller than one main scale division.
Statement $II$: The vernier constant is given by one main scale division multiplied by the number of vernier scale division.
In the light of the above statements,choose the correct answer from the options given below.

Answer the following:
$(a)$ You are given a thread and a metre scale. How will you estimate the diameter of the thread?
$(b)$ $A$ screw gauge has a pitch of $1.0\; mm$ and $200$ divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?
$(c)$ The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of $100$ measurements of the diameter expected to yield a more reliable estimate than a set of $5$ measurements only?

Identify the physical quantity that cannot be measured using a spherometer:

In a vernier callipers, $20$ $VSD$ coincide with $16$ $MSD$ (each division of length $1 \text{ mm}$). The least count of the vernier callipers is: (in $\text{ cm}$)