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 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 diameter of a cylinder is measured using a vernier callipers with no zero error. It is found that the zero of the vernier scale lies between $5.10 \ cm$ and $5.15 \ cm$ of the main scale. The vernier scale has $50$ divisions equivalent to $2.45 \ cm$. The $24^{\text{th}}$ division of the vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is: (in $cm$)

In a screw gauge, when the circular scale is given five complete rotations, it moves linearly by $2.5 \text{ mm}$. If the circular scale has $100$ divisions, the least count of the screw gauge is . . . . . . $\text{mm}$.

$A$ 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

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