An experiment is performed to find the refractive index of glass using a travelling microscope. In this experiment,distances are measured by

  • A
    a meter scale provided on the microscope
  • B
    a vernier scale provided on the microscope
  • C
    a screw gauge provided on the microscope
  • D
    a standard laboratory scale

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The main scale division of a vernier caliper is $1 \ mm$. The vernier scale divisions are in an arithmetic progression $(A.P.)$; the $1^{st}$ division is $0.95 \ mm$, the $2^{nd}$ division is $0.90 \ mm$, and so on. When an object is placed between the jaws of the vernier caliper, the zero of the vernier scale lies between $3.1 \ cm$ and $3.2 \ cm$, and the $4^{th}$ division of the vernier scale coincides with a main scale division. The reading of the vernier caliper is .......... $cm$.

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Two full turns of the circular scale of a screw gauge cover a distance of $1 \ mm$ on its main scale. The total number of divisions on the circular scale is $50$. The screw gauge has a zero error of $-0.03 \ mm$. While measuring the diameter of a thin wire,a student notes the main scale reading of $3 \ mm$ and the $35^{\text{th}}$ division of the circular scale coincides with the reference line of the main scale. The diameter of the wire is: (in $mm$)

Student $A$ and Student $B$ used two screw gauges of equal pitch and $100$ equal circular divisions to measure the radius of a given wire. The actual value of the radius of the wire is $0.322 \, \text{cm}$. The absolute value of the difference between the final circular scale readings observed by the students $A$ and $B$ is .... .
Given pitch $= 0.1 \, \text{cm}$.

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 $, mm$)

Consider the diameter of a spherical object being measured with the help of a Vernier callipers. Suppose its $10$ Vernier Scale Divisions $(V.S.D.)$ are equal to its $9$ Main Scale Divisions $(M.S.D.)$. The least division on the $M.S.$ is $0.1 \ cm$ and the zero of $V.S.$ is at $x=0.1 \ cm$ when the jaws of the Vernier callipers are closed. If the main scale reading for the diameter is $M=5 \ cm$ and the number of the coinciding vernier division is $8$,the measured diameter after zero error correction is: (in $cm$)

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