There are two vernier calipers both of which have 1cm divided into...

There are two vernier calipers both of which have $1cm$ divided into $10$ equal divisions on the main scale. The Vernier scale of one of the calipers \left({C}_{1}\right) $has$ 10 $equal divisions that correspond to$ 9 $main scale divisions. The Vernier scale of the other caliper \left({C}_{2}\right)$ has $10$ equal divisions that correspond to $11$ main scale divisions. The readings of the two calipers are shown in the figure. The measured values (in $cm$ ) by calipers ${C}_{1} and {C}_{2}$ respectively, are
There are two vernier calipers both of which have 1cm divided into 10 equal divisions on the main scale. The Vernier

Options

A $2 . 87$ and $2 . 86$
B $2 . 87$ and $2 . 82$
C $2 . 87$ and $2 . 83$
D $2 . 85$ and $2 . 82$

Correct answer

C. $2 . 87$ and $2 . 83$

Step-by-step solution

For caliper ${C}_{1}$

$10VSD = 9MSD$

$LC = 1MSD - 1VSD$

$LC = 0 . 01cm$

Measured value=Main scale reading $+$ vernier scale reading $\times$ least count

$= \left(2 .8 + 7 \times 0 .01\right)cm$

$= 2 . 87cm$

For caliper ${C}_{2}$

$10VSD = 11MSD$

$LC = - 0 . 01cm$ (Negative least count means we have to take $VSR$ from the right of the scale $10 ‐ 7$ )

Measured value $= \left\{2 .8 + \left(10 - 7\right) \times 0 .01\right\}cm$

$= 2 . 83cm$

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