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

  • [JEE MAIN 2021]
  • A

    $\frac{10 na }{( n -1)}$

  • B

    $\frac{10 a }{( n -1)}$

  • C

    $\left(\frac{ n -1}{10 n }\right) a$

  • D

    $\frac{10 a }{ n }$

Similar Questions

Asseretion $A:$ If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is $5\, {mm}$ and there are $50$ total divisions on circular scale, then least count is $0.001\, {cm}$.

Reason $R:$ Least Count $=\frac{\text { Pitch }}{\text { Total divisions on circular scale }}$

In the light of the above statements, choose the most appropriate answer from the options given below:

  • [JEE MAIN 2021]

A specially designed Vernier calliper has the main scale least count of $1 \,mm$. On the Vernier scale, there are $10$ equal divisions and they match with $11$ main scale divisions. Then, the least count of the Vernier calliper is ........... $mm$

  • [KVPY 2019]

The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is $0.5 \ mm$ and there are $50$ divisions on the circular scale. The reading on the main scale is $2.5 \ mm$ and that on the circular scale is $20$ divisions. If the measured mass of the ball has a relative error of $2 \%$, the relative percentage error in the density is

  • [IIT 2011]

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 mean corrected diameter is  ........... $cm$

  • [JEE MAIN 2015]

A screwgauage has pitch $1.5\;  mm$ and there is no zero error. Linear scale has marking at $MSD = 1\; mm$  and there are $100$ equal division of circular scale. When diameter of a sphere is measured with instrument, main scale is having $2\; mm$ mark visible on linear scale, but $3\; mm$ mark is not visible, $76^{th}$ division of circuler scale is in line with linear scale. .......... $mm$ is the diameter of sphere.