Question

A vernier callipers is designed by dividing 19 M.S.D into 20 V.S.D & value of 1
M.S.D=1mm. When two jaws are kept in contact, zero of vernier is right of zero of
mains scale & 124 vernier division is coinciding with one of main scale division. If
this vernier callipers is used to find the length of the object, it is observed that zero of
vernier lies between 10th & 11th M.S.D & 10 hvernier livision is coinciding with one of
the main scale division. Then find length of cylinder?​

Answer 1

Given that 19 Main Scale Division coincides with 20 Vernier Scale Division.

$\displaystyle\sf{\longrightarrow 19M=20V}$

$\displaystyle\sf{\longrightarrow V=\dfrac {19M}{20}}$

Since $\displaystyle\sf {1M=1\ mm,}$

$\displaystyle\sf{\longrightarrow V=\dfrac {19}{20}\ mm}$

Then least count is,

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

$\displaystyle\sf{\longrightarrow LC=1\ mm-\dfrac {19}{20}\ mm}$

$\displaystyle\sf{\longrightarrow LC=\dfrac {1}{20}\ mm}$

$\displaystyle\sf{\longrightarrow LC=0.05\ mm}$

When two jaws are kept in contact, zero of vernier is right of zero of mains scale and 124th vernier division is coinciding with one of the main scale divisions.

So the vernier callipers has a zero error which is 124 times the least count.

And since the zero of vernier is right to that of main, the zero error is positive.

$\displaystyle\sf{\longrightarrow Z=124(LC)}$

$\displaystyle\sf{\longrightarrow Z=124\times 0.05\ mm}$

$\displaystyle\sf{\longrightarrow Z=6.2\ mm}$

On finding length of cylinder, the zero of vernier lies between 10th and 11th main scale division, i.e., right of $\displaystyle\sf {10\times1\ mm=10\ mm.}$

And 10th vernier division is coinciding with one of main scale divisions.

So reading for length of cylinder will be,

$\displaystyle\sf{\longrightarrow L=10\ mm+10(LC)-Z}$

$\displaystyle\sf{\longrightarrow L=10\ mm+10\times0.05\ mm-6.2\ mm}$

$\displaystyle\sf{\longrightarrow\underline {\underline {L=4.3\ mm}}}$