$A$ vernier callipers used by a student has $20$ divisions in $1\;cm$ on the main scale. $10$ vernier divisions coincide with $9$ main scale divisions. When the jaws are closed,the zero of the main scale is to the left of the zero of the vernier scale,and the $6^{th}$ division of the vernier scale coincides with a main scale division. The student places a wooden cylinder between the jaws to measure its length. The zero of the vernier scale is to the right of the $3.20\;cm$ mark,and the $8^{th}$ vernier division coincides with a main scale division. When measuring the thickness (diameter) of the cylinder,the zero of the vernier scale lies to the right of the $1.50\;cm$ mark,and the $6^{th}$ vernier division coincides with a main scale division. The correct values of the measured length and diameter are respectively:

  • A
    $3.21\;cm, 1.50\;cm$
  • B
    $3.210\;cm, 1.500\;cm$
  • C
    $3.27\;cm, 1.93\;cm$
  • D
    $3.270\;cm, 1.560\;cm$

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Similar Questions

In a screw gauge,$5$ complete rotations of the circular scale give a $1.5 \, mm$ reading on the linear scale. The circular scale has $50$ divisions. The least count of the screw gauge is: (in $, mm$)

The circular divisions of a screw gauge are $50$. It moves $0.5 \ mm$ on the main scale in one rotation. When the diameter of a wire is measured,the main scale reading is $3.5 \ mm$ and the circular scale reading is $32$. If the zero error (positive) in the screw gauge is $0.06 \ mm$,then the diameter of the wire is:

The pitch of the screw gauge is $1\, mm$ and there are $100$ divisions on the circular scale. When nothing is put in between the jaws,the zero of the circular scale lies $8$ divisions below the reference line. When a wire is placed between the jaws,the first linear scale division is clearly visible while $72^{nd}$ division on the circular scale coincides with the reference line. The radius of the wire is.........$mm$

In a vernier callipers, $20$ $VSD$ coincide with $16$ $MSD$ (each division of length $1 \text{ mm}$). The least count of the vernier callipers is: (in $\text{ cm}$)

Assertion $A$: If in five complete rotations of the circular scale,the distance travelled on the main scale of the screw gauge is $5 \, mm$ and there are $50$ total divisions on the circular scale,then the least count is $0.001 \, cm$.
Reason $R$: $\text{Least Count} = \frac{\text{Pitch}}{\text{Total divisions on circular scale}}$
In the light of the above statements,choose the most appropriate answer from the options given below:

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