Question

The main scale of a vernier caliper reads in $mm$ and its vernier scale is divided into $8$ divisions, which coincides with $7$ divisions of the main scale. It was also observed that, when the two jaws are brought in contact, the zero of the vernier scale coincides with zero of main scale. The edge length of a cube is measured using this vernier caliper. The main scale reads $12mm$ and the $2^{\text{nd }}$ division of the vernier scale coincides with the main scale. The edge length (in $μm$ ) of the cube is

Answer 1

The given relation in question,
$8VSD=7MSD$
it means
$\Rightarrow 1VSD=\frac{7}{8}MSD$
so, the least count,
$LC=1MSD-1VSD$
Put the value of vernier scale reading from the above relation,
$LC=1MSD-\frac{7}{8}MSD$
$\Rightarrow LC=MSD\frac{1}{8}mm$
$\ell =12+2\times \frac{1}{8}=12.25mm=12250μm$