Embibe Experts Solutions for Chapter: Experiments in Physics, Exercise 1: Exercise-1

The pitch of a screw gauge is $0.5 \mathrm{mm}$ and there are $100$ divisions on it circular scale. The instrument reads $+2$ circular scale divisions when nothing is put in-between its jaws. In measuring the diameter of a wire, there are $8$ divisions on the main scale and $83\mathrm{rd}$ circular scale division coincides with the reference line. Then the diameter of the wire is

The pitch of a screw gauge having $50$ divisions on its circular scale is $1 \mathrm{mm}$. When the two jaws of the screw gauge are in contact with each other, the zero of the circular scale lies $6$ divisions below the line of graduation. When a wire is placed between the jaws, $3$ linear scale divisions are clearly visible while the ${31}^{\mathrm{st}}$ division on the circular scale coincides with the reference line. The diameter of the wire is:

The smallest division on the main scale of a vernier caliper is $1 \mathrm{mm}$, 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 $2.0$ and $2.1 \mathrm{cm}$ and the fifth division of the vernier scale coincides with a scale division. The diameter of the sphere is,

The main scale of a vernier calipers reads $10 \mathrm{mm}$ in $10$ divisions. $10$ divisions of Vernier scale coincide with $9$ divisions of the main scale. When a cylinder is tightly placed between the two jaws, the zero of vernier scale lies slightly behind $3.2 \mathrm{cm}$ and the fourth vernier division coincides with a main scale division. The diameter of the cylinder is: