If $50$ Vernier divisions are equal to $49$ main scale divisions of a travelling microscope and one smallest reading of main scale is $0.5 \mathrm{~mm}$, the Vernier constant of travelling microscope is:

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

A student measured the length of a rod and wrote it as $3.50\;cm$. Which instrument did he use to measure it?

A student measures the thickness of a human hair by looking at it through a microscope of magnification $100$. He makes $20$ observations and finds that the average width of the hair in the field of view of the microscope is $3.5 \;mm$. What is the estimate on the thickness(in $mm$) of hair ?

Diagrams show readings of a screw gauge. figure $(i)$ shows the zero error reading when the screw gauge is closed, figure $(ii)$ the reading when the screw gauge is being used to measure the diameter of a ball-bearing. What is the diameter of the ball-bearing in $mm$? There are $50$ divisions on circular scale

The diameter of a cylinder is measured using a vernier callipers with no zero error. It is found that the zero of the vernier scale lies between $5.10 \ cm$ and $5.15 \ cm$ of the main scale. The vernier scale has $50$ division equivalent to $2.45 \ cm$. The $24^{\text {th }}$ division of the vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is :