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 ........... $cm$.

  • A
    $0.52$
  • B
    $0.59$
  • C
    $0.56$
  • D
    $0.53$

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In an experiment to find out the diameter of a wire using a screw gauge,the following observations were noted:
$(A)$ Screw moves $0.5 \ mm$ on the main scale in one complete rotation.
$(B)$ Total divisions on the circular scale $= 50$.
$(C)$ Main scale reading is $2.5 \ mm$.
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Then the diameter of the wire is: (in $mm$)

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Consider the diameter of a spherical object being measured with the help of a Vernier callipers. Suppose its $10$ Vernier Scale Divisions $(V.S.D.)$ are equal to its $9$ Main Scale Divisions $(M.S.D.)$. The least division on the $M.S.$ is $0.1 \ cm$ and the zero of $V.S.$ is at $x=0.1 \ cm$ when the jaws of the Vernier callipers are closed. If the main scale reading for the diameter is $M=5 \ cm$ and the number of the coinciding vernier division is $8$,the measured diameter after zero error correction is: (in $cm$)

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

Given below are two statements:
Statement $I$: In a vernier callipers,one vernier scale division is always smaller than one main scale division.
Statement $II$: The vernier constant is given by one main scale division multiplied by the number of vernier scale division.
In the light of the above statements,choose the correct answer from the options given below.

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