A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure thick

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1.Units, Dimensions and Measurement

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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 wen the two jaws of the screw gauge are brought in contact, the $45^{th} $division coincides with the main scale line and the zero of the main scale is barely visible. What is the thickness of the sheet (in $mm$) if the main scale reading is $0.5\ mm$ and the $25^{th}$ division coincides with the main scale line

A

$0.70$

B

$0.50$

C

$0.75$

D

$0.80$

(JEE MAIN-2016)

Solution

$L.C = \frac{{0.5}}{{50}} = 0.001\,mm$
zero error $= 5 \times 0.001 = 0.05$ mm negative

Reading $= \left( {0.5 + 25 \times 0.01} \right) + 0.05 = 0.80$ mm

Standard 11

Physics

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