A screw gauge with a pitch of $0.5 mm$ and a circular scale with $50$ divisions is used to measure the thickness of a thin sheet of aluminium. 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 of the sheet if the main scale reading is $0.5 mm$ and the $25^{th}$ division coincides with the main scale line?

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Solution

Least count $= \frac{Pitch}{circular scale divisions}$
$= \frac{0.5 mm}{50}$
$= 0.01 mm$

Negative zero error $= - 5 \times LC = - 5 \times 0.01 = - 0.05 mm$

Final reading $=$ main scale reading $+$ circular scale reading $-$ zero error
$= 0.5 + (25 \times 0.01) - (- 0.05)$
$= 0.80 mm$

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Least count =Pitchcircular scale divisions =0.5 mm50 =0.01 mm Negative zero error =−5×LC=−5×0.01=−0.05 mm Final reading = main scale reading + circular scale reading − zero error =0.5+(25×0.01)−(−0.05) =0.80 mm

Least count $= \frac{Pitch}{circular scale divisions}$
$= \frac{0.5 mm}{50}$
$= 0.01 mm$

Negative zero error $= - 5 \times LC = - 5 \times 0.01 = - 0.05 mm$

Final reading $=$ main scale reading $+$ circular scale reading $-$ zero error
$= 0.5 + (25 \times 0.01) - (- 0.05)$
$= 0.80 mm$