$A$ screw gauge has some zero error but its value is unknown. We have two identical rods. When the first rod is inserted in the screw gauge,the state of the instrument is shown by diagram $(I)$. When both the rods are inserted together in series,the state is shown by diagram $(II)$. What is the zero error of the instrument in $mm$? Given: $1 \, M.S.D. = 100 \, C.S.D. = 1 \, mm$.

$A$ tiny metallic rectangular sheet has a length and breadth of $5 \ mm$ and $2.5 \ mm$,respectively. Using a specially designed screw gauge which has a pitch of $0.75 \ mm$ and $15$ divisions on the circular scale,you are asked to find the area of the sheet. In this measurement,the maximum fractional error will be $\frac{x}{100}$ where $x$ is . . . . . .

The length of a cylinder measured by a Vernier caliper is given by the following observations: $3.29 \, cm, 3.28 \, cm, 3.29 \, cm, 3.31 \, cm, 3.28 \, cm, 3.27 \, cm, 3.29 \, cm, 3.30 \, cm$. The most accurate length of the cylinder is ........ $cm$.

$A$ screw gauge has a pitch of $1.5\; mm$ and there is no zero error. The linear scale has markings at $MSD = 1\; mm$ and there are $100$ equal divisions on the circular scale. When the diameter of a sphere is measured with this instrument,the $2\; mm$ mark is visible on the linear scale,but the $3\; mm$ mark is not visible. The $76^{th}$ division of the circular scale is in line with the linear scale. What is the diameter of the sphere in $mm$?

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 $cm$)

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 ........... $\times 10^{-2} \,cm$.