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The one division of main scale of vernier callipers reads $1\,mm$ 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 $4.1\,cm$ and $4.2\,cm$ and $6^{\text {th }}$ Vernier division coincides with a main scale division. The diameter of the bob will be $.............10^{-2}\,cm$
$413$
$411$
$141$
$412$
Solution
$10\,VSD =9\,MSD$
$1\,VST =.9\,MSD$
$L.C.$ $=1\,mm =.01\,cm$
$+ve$ zero error $=4\,mm$
$=0.04\,cm$
Negative zero error $=4.1\,cm +6 \times .01$
$=4.12\,cm$
$=412 \times 10^{-2}\,cm$