The external and internal radii of a hollow cylinder are measured with the help of a vernier callipers. Their values are $(4.23 \pm 0.01) \, cm$ and $(3.87 \pm 0.01) \, cm,$ respectively. The thickness of the wall of the cylinder is

The least count of a vernier caliper is $\frac{1}{20N} \text{ cm}$. The value of one division on the main scale is $1 \text{ mm}$. Then the number of divisions of the main scale that coincide with $N$ divisions of the vernier scale is:

In a screw gauge, the zero of the circular scale lies $3$ divisions above the horizontal pitch line when their metallic studs are brought in contact. Using this instrument, the thickness of a sheet is measured. If the pitch scale reading is $1 \ mm$ and the circular scale reading is $51$, then the correct thickness of the sheet is . . . . . . $mm$. [Assume least count is $0.01 \ mm$]

The internal and external diameters of a hollow cylinder measured with vernier calipers are $(5.73 \pm 0.01) \text{ cm}$ and $(6.01 \pm 0.01) \text{ cm}$ respectively. Then the thickness of the cylinder wall is

In a Vernier Calipers,$10$ divisions of the Vernier scale are equal to $9$ divisions of the main scale. When both jaws of the Vernier calipers touch each other,the zero of the Vernier scale is shifted to the left of the zero of the main scale,and the $4^{\text{th}}$ Vernier scale division exactly coincides with a main scale division. One main scale division is equal to $1\,mm$. While measuring the diameter of a spherical body,the body is held between the two jaws. It is observed that the zero of the Vernier scale lies between $30$ and $31$ divisions of the main scale,and the $6^{\text{th}}$ Vernier scale division exactly coincides with a main scale division. The diameter of the spherical body is $.......\,cm$.

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 $cm$)