A student performs an experiment of measuring the thickness of a slab with a vernier calliper whose $50$ divisions of the vernier scale are equal to $49$ divisions of the main scale. He noted that zero of the vernier scale is between $7.00\; cm$ and $7.05 \;cm$ mark of the main scale and $23^{rd}$ division of the vernier scale exactly coincides with the main scale. The measured value of the thickness of the given slab using the calliper will be

In a vernier calliper, when both jaws touch each other, zero of the vernier scale shifts towards left and its $4^{\text {th }}$ division coincides exactly with a certain division on main scale. If $50$ vernier scale divisions equal to $49$ main scale divisions and zero error in the instrument is $0.04 \mathrm{~mm}$ then how many main scale divisions are there in $1 \mathrm{~cm}$ ?

The vernier scale of a travelling microscope has $50$ divisions which coincide with $49$ main scale divisions. If each main scale division is $0.5\, mm$, calculate the minimum inaccuracy in the measurement of distance.

$N$ divisions on the main scale of a vernier calliper coincide with $(N + 1 )$ divisions of the vernier scale. If each division of main scale is $a$ units , then the least count of the instrument is

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

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