(D) Given: $9 \, {MSD} = 10 \, {VSD}$.Since $1 \, {MSD} = 1 \, {mm}$,then $10 \, {VSD} = 9 \, {mm}$,so $1 \, {VSD} = 0.9 \, {mm}$.Least Count $({LC})$

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(D) Given: $9 \, {MSD} = 10 \, {VSD}$.Since $1 \, {MSD} = 1 \, {mm}$,then $10 \, {VSD} = 9 \, {mm}$,so $1 \, {VSD} = 0.9 \, {mm}$.Least Count $({LC})$ = $1 \, {MSD} - 1 \, {VSD} = 1 \, {mm} - 0.9 \, {mm} = 0.1 \, {mm} = 0.01 \, {cm}$.Observed Diameter = ${MSR} + ({VSR} \times {LC}) = 10 \, {mm} + (8 \times 0.1 \, {mm}) = 10.8 \, {mm} = 1.08 \, {cm}$.Positive zero error = $0.04 \, {cm}$.Corrected Diameter = $\text{Observed Reading} - \text{Zero Error} = 1.08 \, {cm} - 0.04 \, {cm} = 1.04 \, {cm}$.Radius = $\frac{\text{Diameter}}{2} = \frac{1.04 \, {cm}}{2} = 0.52 \, {cm}$.Expressing in $10^{-2} \, {cm}$: $0.52 \, {cm} = 52 \times 10^{-2} \, {cm}$.Thus,the value is $52$.

Class 11 Physics Units and Measurements Vernier Calipers, Micrometer screw gauge

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The diameter of a spherical bob is measured using a vernier callipers. $9$ divisions of the main scale,in the vernier callipers,are equal to $10$ divisions of vernier scale. One main scale division is $1\, {mm}$. The main scale reading is $10\, {mm}$ and $8^{\text{th}}$ division of vernier scale was found to coincide exactly with one of the main scale division. If the given vernier callipers has positive zero error of $0.04\, {cm}$,then the radius of the bob is $...... \,\times 10^{-2} \,{cm}$.

JEE MAIN 2021 All JEE MAIN

A
$0.52$
B
$520$
C
$5.2$
D
$52$

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Vernier Calipers, Micrometer screw gauge

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$A$ screw gauge with a pitch of $0.5 \ mm$ and a circular scale with $50$ divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement,it is found that when the two jaws of the screw gauge are brought in contact,the $45^{th}$ division coincides with the main scale line and the zero of the main scale is barely visible. What is the thickness of the sheet (in $mm$) if the main scale reading is $0.5 \ mm$ and the $25^{th}$ division coincides with the main scale line?

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