In a vernier callipers,$(N+1)$ divisions of vernier scale coincide with $N$ divisions of main scale. If $1 \text{ MSD}$ represents $0.1 \text{ mm}$,the vernier constant (in $\text{cm}$) is:

The smallest division on the main scale of a vernier callipers is $1 \ mm$,and $10$ vernier divisions coincide with $9$ main scale divisions. While measuring the diameter of a sphere,the zero mark of the vernier scale lies between $2.0 \ cm$ and $2.1 \ cm$ of the main scale,and the fifth division of the vernier scale coincides with a main scale division. The diameter of the sphere is:

In an experiment,the angles are required to be measured using an instrument where $29$ divisions of the main scale exactly coincide with $30$ divisions of the vernier scale. If the smallest division of the main scale is half a degree $(= 0.5^\circ)$,then the least count of the instrument is:

If $50$ Vernier divisions are equal to $49$ main scale divisions of a travelling microscope and one smallest reading of main scale is $0.5 \,mm$, the Vernier constant of travelling microscope is:

While measuring the diameter of a wire using a screw gauge, the following readings were noted. The main scale reading is $1 \,mm$ and the circular scale reading is equal to $42$ divisions. The pitch of the screw gauge is $1 \,mm$ and it has $100$ divisions on the circular scale. The diameter of the wire is $\frac{x}{50} \,mm$. The value of $x$ is:

In an experiment to find out the diameter of a wire using a screw gauge,the following observations were noted:
$(A)$ Screw moves $0.5 \ mm$ on the main scale in one complete rotation.
$(B)$ Total divisions on the circular scale $= 50$.
$(C)$ Main scale reading is $2.5 \ mm$.
$(D)$ $45^{\text{th}}$ division of the circular scale is in the reference line.
$(E)$ Instrument has $0.03 \ mm$ negative zero error.
Then the diameter of the wire is: (in $mm$)