In a vernier callipers, $50$ vernier scale divisions are equal to $48$ main scale divisions. If one main scale division $= 0.05 \ mm$, then the least count of the vernier callipers is . . . . . . $mm$.

Answer the following:
$(a)$ You are given a thread and a metre scale. How will you estimate the diameter of the thread?
$(b)$ $A$ screw gauge has a pitch of $1.0\; mm$ and $200$ divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?
$(c)$ The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of $100$ measurements of the diameter expected to yield a more reliable estimate than a set of $5$ measurements only?

$10$ divisions on the main scale of a Vernier calliper coincide with $11$ divisions on the Vernier scale. If each division on the main scale is of $5$ units,the least count of the instrument is :

$A$ plano-convex lens of unknown material and unknown focal length is given. With the help of a Spherometer,we can measure the,

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

Thickness of a pencil measured by a screw gauge (least count $0.001 \ cm$) comes out to be $0.802 \ cm$. The percentage error in the measurement is (in $\%$)