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The one division of main scale of Vernier callipers reads $\mathrm{1 \mathrm{~mm}}$ and 10 divisions of Vernier scale is equal to the 9 divisions on main scale. When the two jaws of the instrument touch each other, the zero of the Vernier lies to the right of zero of the main scale and its fourth division coincides with a main scale division. When a spherical bob is tightly placed between the two jaws, the zero of the Vernier scale lies in between $\mathrm{4.1 \mathrm{~cm}}$ and $\mathrm{4.2 \mathrm{~cm}}$ and $\mathrm{6^{\text {th }}}$ Vernier division coincides with a main scale division. The diameter of the bob will be ___________ $\mathrm{ \times 10^{-2} \mathrm{~cm}}.$
Option: 1
412

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

$\mathrm{1 M S D=1 \mathrm{~mm}}$

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

$\mathrm{1 V S D=\left(\frac{9}{10}\right) M S D}$

$\mathrm{L \cdot C=1M S D-1V S D=\frac{1}{10} M S D}$

$\mathrm{\text { Zero error }=\text { (MSD) (LC) }}$

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

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

$\mathrm{=0.04 \mathrm{~cm}}$

$\mathrm{\text { Zero correction }=-(0.04) \mathrm{cm}}$

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

$\mathrm{=4.1 \mathrm{~cm}+(6 \times 0.1 \mathrm{~mm})+\left ( -0.04 \right )cm}$

$\mathrm{\text { Reading }=4.12 \mathrm{~cm}}$

$\mathrm{=412 \times 10^{-2} \mathrm{~cm}}$

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