In a Vernier Calipers. 10 divisions of Vernier scale is equal to 9 divisions of main scale. When bot

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1.Units, Dimensions and Measurement

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In a Vernier Calipers. $10$ divisions of Vernier scale is equal to the $9$ 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 $4^{\text {th }}$ Vernier scale division exactly coincides with the main scale reading. One main scale division is equal to $1\,mm$. 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 $30$ and $31$ divisions of main scale reading and $6^{\text {th }}$ Vernier scale division exactly. coincides with the main scale reading. The diameter of the spherical body will be $.......cm$

A$3.02$

B$3.06$

C$3.10$

D$3.20$

(JEE MAIN-2022)

Solution

$1 \text { M.S.D }=1\,mm$
$9 \text { M.S.D }=10 \text { V.S.D }$
$\text { 1 V.S.D }=0.9 \text { M.S.D }=0.9\,mm$
$\text { L.C of vernier caliper }=1-0.9=0.1\,mm =0.01\,cm$
$\text { zero error }=-(10-4) \times 0.1\,mm =-0.6\,mm$
$\text { Reading }=\text { M.S.R }+\text { V.S.R }-\text { Zero error }$
$=3\,cm +6 \times 0.01-[-0.06]$
$=3+0.06+0.06$
$=3.12\,cm$
Nearest given answer in the options is $3.10$

Standard 11

Physics

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