A screw gauge with a pitch of 0.5mm and a circular scale with 50...

A screw gauge with a pitch of

0.5\text{mm}

and a circular scale with

50

divisions is used to measure the thickness of a thin sheet of aluminum. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the

{45}^{th}

division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness (in

mm

) of the sheet if the main scale reading is

0.5\text{mm}

and the

{30}^{th}

division coincides with the main scale line?

Correct answer

0.85

Step-by-step solution

Least count, $LC = \dfrac{0 . 5}{50} = 0 . 01mm$

Zero error, $ZE = \left(50 - 45\right) \times 0 . 01 = - 0 . 05mm$

Main Scale Reading \left(MSR\right) = 0.5\text{mm} $<br />Thickness,$ t = 0 . 5mm + 30 \times 0 . 01mm - \left(- 0 . 05mm\right) $</p><p>$ t = 0 . 85mm

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