In an experiment with 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 45t h division coincides with the main scale line and that the zero of the main scale is barely visible. Find the thickness of the sheet if the main scale reading is 0.5 mm and the 25t h division coincides with the main scale line?

Question

In an experiment with 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 45t h division coincides with the main scale line and that the zero of the main scale is barely visible. Find the thickness of the sheet if the main scale reading is 0.5 mm and the 25t h division coincides with the main scale line? (A) 0.70 mm (B) 0.50 mm (C) 0.75 mm (D) 0.80 mm

Tardigrade Admin · Accepted Answer

LC=(0.5/50)=0.01 mm Zero error =0.50-0.45=-0.05 Thickness =(0.5 + 25 × 0.01)+0.05 =0.5+0.25+0.05 =0.8 mm

0.70 mm

0.50 mm

0.75 mm

0.80 mm

LC=500.5​=0.01mm Zero error =0.50−0.45=−0.05 Thickness =(0.5+25×0.01)+0.05 =0.5+0.25+0.05 =0.8mm

$L C = \frac{0.5}{50} = 0.01 mm$
Zero error $= 0.50 - 0.45 = - 0.05$
Thickness $= (0.5 + 25 \times 0.01) + 0.05$
$= 0.5 + 0.25 + 0.05$
$= 0.8 mm$