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

If the screw on a screw-gauge is given six rotations, it moves by $3\; \mathrm{mm}$ on the main scale. If there are $50$ divisions on the circular scale the least count of the screw gauge is

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

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

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