Question

One cm on the main scale of vernier callipers is divided into ten equal parts. If 0 divisions of vernier so coincide with 8 small divisions of the main scale. What will be the least count of callipers ?

Answer 1

Question:-

1 cm on the main scale of vernier callipers is divided into 10 equal parts. If 10 divisions of vernier scale coincide with 8 divisions of main scale, what will be the least count of callipers?

Answer:-

$\displaystyle\Large\boxed {\sf {0.2\ mm}}$

Solution:-

• $\displaystyle\sf {M=}$ main scale division (least count of main scale)

• $\displaystyle\sf {V=}$ vernier scale division (least count of vernier scale)

Since 1 cm of main scale is divided into 10 equal parts, the main scale division,

$\displaystyle\longrightarrow\sf{M=\dfrac {1}{10}\ cm}$

$\displaystyle\longrightarrow\sf{M=1\ mm}$

Given that 10 vernier scale divisions coincide with 8 main scale divisions, i.e.,

$\displaystyle\longrightarrow\sf{10V=8M}$

$\displaystyle\longrightarrow\sf{V=\dfrac {4M}{5}}$

$\displaystyle\longrightarrow\sf{V=\dfrac {4\times1\ mm}{5}}$

$\displaystyle\longrightarrow\sf {V=0.8\ mm}$

Now, least count,

$\displaystyle\longrightarrow\sf{LC=M-V}$

$\displaystyle\longrightarrow\sf{LC=1\ mm-0.8\ mm}$

$\displaystyle\longrightarrow\sf {\underline {\underline {LC=0.2\ mm}}}$

In centimetres,

$\displaystyle\longrightarrow\sf {\underline {\underline {LC=0.02\ cm}}}$