In a vernier callipers, $10$ divisions of vernier scale coincides with $9$ divisions of main scale, the least count of which is $0.1\,cm$. If in the measurement of inner diameter of cylinder zero of vernier scale lies between $1.3\,cm$ and $1.4\, \ cm$ of main scale and $2^{nd}$ division of vernier scale coincides with main scale division then diameter will be $.......... cm$
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$\mathrm{L} \mathrm{C} =1 \mathrm{MSD}-1 \mathrm{VSD}$$=1 \mathrm{MSD}-\frac{9}{10} \mathrm{MSD}$
$=\left(1-\frac{9}{10}\right) \times 1 \mathrm{MSD} \left[\begin{array}{l}{\text { Also }} \\ {10 \mathrm{VSD}=9 \mathrm{MSD}} \\ {1 \mathrm{VSD}=\frac{9}{10} \mathrm{MSD}}\end{array}\right]$
$=0.1 \times 0.1 \ \mathrm{cm}$
$=0.01 \ \mathrm{cm}$
Diameter $=\text{MSR+LC} \times \text{VSR}$
$=1.3+0.01 \times 2$
$=1.3+0.02$
$=1.32 \ \mathrm{cm}$
$=\left(1-\frac{9}{10}\right) \times 1 \mathrm{MSD} \left[\begin{array}{l}{\text { Also }} \\ {10 \mathrm{VSD}=9 \mathrm{MSD}} \\ {1 \mathrm{VSD}=\frac{9}{10} \mathrm{MSD}}\end{array}\right]$
$=0.1 \times 0.1 \ \mathrm{cm}$
$=0.01 \ \mathrm{cm}$
Diameter $=\text{MSR+LC} \times \text{VSR}$
$=1.3+0.01 \times 2$
$=1.3+0.02$
$=1.32 \ \mathrm{cm}$
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