Question

A Vernier caliper with $0.02mm$ least count is to be constructed. If $1MSD = 0.5mm$ how many divisions would the slide scale have?
$\left( A \right)10$
$\left( B \right)25$
$\left( C \right)24$
$\left( D \right)20$

Answer 1

Hint: The Vernier scale is $9mm$ long and it consists of $10$ divisions on it. To find the number of divisions of the slide scale we have to use the formula form for it, where the slide scale has a relation with MSD and the least count of the Vernier caliper.

Complete answer:
Vernier caliper:
It is a measuring device that consists of a main scale with a fixed jaw and also a sling jaw with an attached Vernier in it.
From the problem we have a Vernier caliper with $0.02mm$ least count to be constructed. If $1MSD = 0.5mm$.
We have to find the number of divisions on a sliding scale.
We know that,
Number of divisions of the sliding scale or we can say the Vernier scale is equal to the ratio of the MSD is the least count of the scale.
Mathematically we can write,
${D_{Vernier}} = \dfrac{{1MSD}}{{LS}}$
Where,
$1MSD$ is the main scale division which is given as $0.5mm$.
$LS$ is the least count of the given scale which is equal to $0.02mm$.
Hence on putting the given values we will get,
${D_{Vernier}} = \dfrac{{0.5mm}}{{0.02mm}} = \dfrac{{\dfrac{5}{{10}}mm}}{{\dfrac{2}{{100}}mm}}$
Cancelling the common terms and unit we will get,
${D_{Vernier}} = \dfrac{{50}}{2} = 25$
So there is a $25$ division of the Vernier scale.
Therefore the correct option is (B).

Note: Remember that the least count of Vernier scale is counted by subtracting the value of one Vernier scale division from the value of one main scale division. Vernier scale is used to measure the inner and outer dimension of objects that are made up of steel.