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In a Vernier Calipers, $\mathrm{10}$ divisions of Vernier scale is equal to the $\mathrm{9}$ divisions of main scale. When both jaws of Vernier calipers touch each other, the zero of the Vernier scale is shifted to the left of zero of the main scale and $\mathrm{4^{\text {th }}}$ Vernier scale division exactly coincides with the main scale reading. One main scale division is equal to $\mathrm{1 \mathrm{~mm}. }$ While measuring diameter of a spherical body, the body is held between two jaws. It is now observed that zero of the Vernier scale lies between $\mathrm{30}$ and $\mathrm{31 }$ divisions of main scale reading and $\mathrm{6^{\text {th }}}$ Vernier scale division exactly coincides with the main scale reading. The diameter of the spherical body will be :
Option: 1
$\mathrm{3.02 \mathrm{~cm}}$

Option: 2
$\mathrm{3.06 \mathrm{~cm}}$

Option: 3
$\mathrm{3.10 \mathrm{~cm}}$

Option: 4
$\mathrm{3.20 \mathrm{~cm}}$

Answers (1)

best_answer

$\mathrm{10 \mathrm{VSD}=9 \mathrm{MSD}}$

$\mathrm{\text { Least count }(L \cdot C .)=1\: M S D-1 \text { VSD }}$

$\mathrm{=\frac{1}{10} M S D}$

$\mathrm{=0.1 \mathrm{~mm}}$

$\mathrm{\text { zero error }=(V S D) \times L C}$

$\mathrm{=(4) \times 0.1}$

$\mathrm{=0.4 \mathrm{~mm}}$

$\mathrm{\text { Reading }=\text { MSR }+\text { VSR }+\text { Zero correction }}$

$\mathrm{=30+6(0.1)+(-0.4)}$

$\mathrm{ =30.2 \mathrm{~mm}}$

$\mathrm{=3.02 \mathrm{~cm}}$

Hence (1) is correct option.

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