Figure $1$ shows the configuration of the main scale and Vernier scale before measurement. Figure $2$ shows the configuration corresponding to the measurement of the diameter $D$ of a tube. The measured value of $D$ is (in $cm$)

$A$ screw gauge of pitch $0.5\,mm$ is used to measure the diameter of a uniform wire of length $6.8\,cm$. The main scale reading is $1.5\,mm$ and the circular scale reading is $7$. The calculated curved surface area of the wire to appropriate significant figures is $......\,cm^2$. [Screw gauge has $50$ divisions on the circular scale]

$A$ screw gauge has $50$ divisions on its circular scale. The circular scale is $4$ units ahead of the pitch scale marking,prior to use. Upon one complete rotation of the circular scale,a displacement of $0.5\, mm$ is noticed on the pitch scale. The nature of zero error involved,and the least count of the screw gauge,are respectively

$A$ steel wire of diameter $0.5 \text{ mm}$ and Young's modulus $2 \times 10^{11} \text{ N m}^{-2}$ carries a load of mass $M$. The length of the wire with the load is $1.0 \text{ m}$. $A$ vernier scale with $10$ divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale,of least count $1.0 \text{ mm}$,is attached. The $10$ divisions of the vernier scale correspond to $9$ divisions of the main scale. Initially,the zero of vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by $1.2 \text{ kg}$,the vernier scale division which coincides with a main scale division is. . . . Take $g = 10 \text{ m s}^{-2}$ and $\pi = 3.2$.

$A$ 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 . . . . . .

$A$ 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: $1 \, M.S.D. = 100 \, C.S.D. = 1 \, mm$.