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 .......... $cm$.

The smallest division on the main scale of a vernier callipers is $1 \ mm$,and $10$ vernier divisions coincide with $9$ main scale divisions. While measuring the diameter of a sphere,the zero mark of the vernier scale lies between $2.0 \ cm$ and $2.1 \ cm$ of the main scale,and the fifth division of the vernier scale coincides with a main scale division. The diameter of the sphere is:

The vernier constant of Vernier callipers is $0.1 \,mm$ and it has zero error of $(-0.05) \,cm$. While measuring the diameter of a sphere,the main scale reading is $1.7 \,cm$ and the coinciding vernier division is $5$. The corrected diameter will be ........... $\times 10^{-2} \,cm$.

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 $cm$)

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

Identify the physical quantity that cannot be measured using a spherometer: