Embibe Experts Solutions for Chapter: Experiments in Physics, Exercise 3: Exercise-3

A student measured the diameter of a small steel ball using a screw gauge of the least count $0.001 \mathrm{cm}.$ The main scale reading is $5 \mathrm{mm}$ and zero of circular scale division coincides with $25$ divisions above the reference level. If screw gauge has a zero error of $-0.004 \mathrm{cm},$ the correct diameter of the ball is

In an experiment, the angles are required to be measured using an instrument. $29$ divisions of the main scale exactly coincide with the $30$ divisions of the vernier scale. If the smallest division of the main scale is half-a-degree $\left(0.5°\right)$, then the least count of the instrument is, 

A screw gauge gives the following reading when used to measure the diameter of a wire :
Main scale reading : $0\mathrm{mm}$
Circular scale reading : $52$ divisions
Given that $1$ mm on the main scale corresponds to $100$ divisions of the circular scale.
The diameter of the wire from the above data is

The pitch and the number of divisions, on the circular scale, for a given screw gauge are $0.5\mathrm{mm}$ and $100$ respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies $3$ divisions below the mean line. The readings of the main scale and the circular scale, for a thin sheet are $5.5\mathrm{mm}$ and $48$ respectively, the thickness of this sheet is:

The least count of the main scale of a screw gauge is $1 \mathrm{mm}$. The minimum number of divisions on its circular scale required to measure $5 \mu \mathrm{m}$ diameter of a wire is