The pitch and the number of divisions on the circular scale for a given screw gauge are $0.5\,mm$ 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 $5.5\,mm$ and $48$ respectively. The thickness of this sheet is: (in $,mm$)

Two full turns of the circular scale of a screw gauge cover a distance of $1 \ mm$ on its main scale. The total number of divisions on the circular scale is $50$. Further,it is found that the screw gauge has a zero error of $-0.03 \ mm$. While measuring the diameter of a thin wire,a student notes the main scale reading of $3 \ mm$ and the number of circular scale divisions in line with the main scale as $35$. The diameter of the wire is ....... $mm$

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 $\%$)

If $50$ Vernier divisions are equal to $49$ main scale divisions of a travelling microscope and one smallest reading of main scale is $0.5 \,mm$, the Vernier constant of travelling microscope is:

$A$ 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 $\pi = 3.2$.

$A$ 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