The main scale of a vernier calliper 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

Question

The main scale of a vernier calliper 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 coincided with the zero of the main scale. The edge length of a cube is measured using this vernier calliper. The main scale reads $12 mm$ and the $2^{nd}$ division of the vernier scale coincides with the main scale. The edge length (in $mm$ ) of the cube is

Answer:

12.25

Solution:

$8 VSD = 7 MSD$

$\Rightarrow 1 VSD = \frac{7}{8} MSD$

$LC = 1 MSD - 1 VSD$

$LC = 1 MSD - \frac{7}{8} MSD = \frac{1}{8} MSD = \frac{1}{8} mm$

$l = 12 + 2 \times \frac{1}{8} = 12.25 mm$

TestHub · Accepted Answer

8 VSD = 7 MSD ⇒ 1 VSD = 7 8 MSD LC = 1 MSD - 1 VSD LC = 1 MSD - 7 8 MSD = 1 8 MSD = 1 8 mm l = 12 + 2 × 1 8 = 12 . 25 mm

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