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

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

If in a Vernier callipers $10 \,VSD$ coincides with $8 \,MSD$, then the least count of Vernier calliper is ………… $m$  [given $1 \,MSD =1 \,mm ]$

There are two Vernier calipers both of which have $1 \mathrm{~cm}$ divided into $10$ equal divisions on the main scale. The Vernier scale of one of the calipers $\left(C_1\right)$ has $10$ squal divisions that correspond to $9$ main scale divisions. The Vernier scale of the other caliper $\left(C_2\right)$ has $10$ equal divisions that correspond to $11$ main scale divisions. The readings of the two calipers are shown in the figure. The measured values (in $\mathrm{cm}$ ) by calipers $C_1$ and $C_2$, respectively, are

In a screw gauge, $5$ complete rotations of the screw cause it to move a linear distance of $0.25\, cm$. There are $100$ circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of $4$ main scale divisions and $30$ circular scale divisions . Assuming negligible zero error, the thickness of the wire is