The pitch and the number of divisions,
Question:

The pitch and the number of divisions, on the circular scale for a given screw gauge are $0.5 \mathrm{~mm}$ and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 division below the mean line.

The readings of the main scale and the circular scale, for a thin sheet, are $5.5 \mathrm{~mm}$ and 48 respectively, the thickness of the sheet is:

  1. (1) $5.755 \mathrm{~mm}$

  2. (2) $5.950 \mathrm{~mm}$

  3. (3) $5.725 \mathrm{~mm}$

  4. (4) $5.740 \mathrm{~mm}$


Correct Option: 3

Solution:

(3) Least count of screw gauge,

$\mathrm{LC}=\frac{\text { Pitch }}{\text { No. of division }}$

$=0.5 \times 10^{-3}=0.5 \times 10^{-2} \mathrm{~mm}$

$+$ ve error $=3 \times 0.5 \times 10^{-2} \mathrm{~mm}$

$=1.5 \times 10^{-2} \mathrm{~mm}=0.015 \mathrm{~mm}$

Reading $=$ MSR $+$ CSR $-(+$ ve error $)$

$=5.5 \mathrm{~mm}+\left(48 \times 0.5 \times 10^{-2}\right)-0.015$

$=5.5+0.24-0.015=5.725 \mathrm{~mm}$

Administrator

Leave a comment

Please enter comment.
Please enter your name.