Question

A Vernier caliper has 25 divisions on the Vernier scale. An object is held between the external jaws of this instrument. If the zeroth division of the Vernier scale is to the left of the ${32}^{\mathrm{nd}}$division of the main scale and the ${20}^{\mathrm{th}}$division of the Vernier scale coincides with the main scale division, determine the measurement of the object.$(\mathrm{Take}1\mathrm{M}.\mathrm{S}.\mathrm{D}.=0.5\mathrm{mm})$

Answer 1

Step1:- Given data

$\mathrm{Vernier}\mathrm{scale}\mathrm{divisions}\left(\mathrm{N}\right)=25$

$\mathrm{Vernier}\mathrm{coincides}\mathrm{division}(\mathrm{V}.\mathrm{C}.\mathrm{D}.)=20$

Also, $1\mathrm{M}.\mathrm{S}.\mathrm{D}=0.5\mathrm{mm}$

Step2:- Finding the least count.

$$\begin{gathered} \mathrm{Least}\mathrm{count}(\mathrm{L}.\mathrm{C}.)=\frac{1\mathrm{M}.\mathrm{S}.\mathrm{D}.}{\mathrm{N}} \\ =\frac{0.5}{25} \\ =0.02\mathrm{mm} \end{gathered}$$

Step3:- Finding main scale reading$(\mathrm{M}.\mathrm{S}.\mathrm{R}.)$

$$\begin{gathered} 1\mathrm{M}.\mathrm{S}.\mathrm{D}=0.5\mathrm{mm} \\ \therefore 32\mathrm{M}.\mathrm{S}.\mathrm{D}=0.5\times 32 \\ =16\mathrm{mm} \end{gathered}$$

So, the main scale reading is $(\mathrm{M}.\mathrm{S}.\mathrm{R})=$The first $(\mathrm{M}.\mathrm{S}.\mathrm{D})$to the left of the zero of the Vernier scale.

$$\begin{gathered} 1\mathrm{M}.\mathrm{S}.\mathrm{R}=16-0.5 \\ =15.5\mathrm{mm} \end{gathered}$$

Step4:- Finding the length of the object.

$$\begin{gathered} \mathrm{length}\mathrm{of}\mathrm{object}=\mathrm{M}.\mathrm{S}.\mathrm{R}.+\mathrm{V}.\mathrm{C}.\mathrm{D}.\times \mathrm{L}.\mathrm{C}. \\ =15.5+20\times 0.02 \\ =15.9\mathrm{mm} \end{gathered}$$

Hence the length of the object is $15.9\mathrm{mm}$