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 $\text{ cm}$)

There are two Vernier calipers,both of which have $1 \ cm$ divided into $10$ equal divisions on the main scale. The Vernier scale of one of the calipers $(C_1)$ has $10$ equal divisions that correspond to $9$ main scale divisions. The Vernier scale of the other caliper $(C_2)$ 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 $cm$) by calipers $C_1$ and $C_2$,respectively,are

Two full turns of the circular scale of a screw gauge cover a distance of $1 \ mm$ on its main scale. The total number of divisions on the circular scale is $50$. The screw gauge has a zero error of $-0.03 \ mm$. While measuring the diameter of a thin wire,a student notes the main scale reading of $3 \ mm$ and the $35^{\text{th}}$ division of the circular scale coincides with the reference line of the main scale. The diameter of the wire is: (in $mm$)

Diameter of a steel ball is measured using a Vernier callipers which has divisions of $0.1\,cm$ on its main scale $(MS)$ and $10$ divisions of its vernier scale $(VS)$ match $9$ divisions on the main scale. Three such measurements for a ball are given as:

$S$.No.	$MS\;(cm)$	$VS$ divisions
$(1)$	$0.5$	$8$
$(2)$	$0.5$	$4$
$(3)$	$0.5$	$6$

If the zero error is $-0.03\,cm,$ then the mean corrected diameter is ........... $cm$.

For the determination of the refractive index of a glass slab,a travelling microscope is used whose main scale contains $300$ equal divisions equal to $15 \ cm$. The vernier scale attached to the microscope has $25$ divisions equal to $24$ divisions of the main scale. The least count $(LC)$ of the travelling microscope is (in $cm$):

$A$ student measured the diameter of a small steel ball using a screw gauge of least count $0.001 \, cm$. The main scale reading is $5 \, mm$ and the circular scale division coinciding with the reference level is $25$. If the screw gauge has a zero error of $-0.004 \, cm$,the correct diameter of the ball is: (in $, cm$)