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 45th 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 25th division coincides with the main scale line ?

Question

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 45th 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 25th division coincides with the main scale line ? (A) 0.75 mm (B) 0.80 mm (C) 0.70 mm (D) 0.50 mm

Tardigrade Admin · Accepted Answer

L.C. =( text pitch / text No. of division on circular scale )=(0.5/50)=0.001 mm -ve zero error = 5 × L.C. = - 0.005 mm ∴ Measured value = main scale reading + screw gauge reading - zero error = 0.5 mm + 25 × 0.001 - (- 0.05) mm = 0.80 mm

0.75 mm

0.80 mm

0.70 mm

0.50 mm

L.C.= No. of division on circular scale pitch ​=500.5​=0.001mm -ve zero error = 5×L.C.=−0.005mm ∴ Measured value = main scale reading + screw gauge reading − zero error =0.5mm+{25×0.001−(−0.05)}mm=0.80mm

$L . C . = \frac{pitch}{No. of division on circular scale} = \frac{0.5}{50} = 0.001 mm$
-ve zero error = $5 \times L . C . = - 0.005 mm$
$∴$ Measured value
= main scale reading $+$ screw gauge reading $-$ zero error
$= 0.5 mm + {25 \times 0.001 - (- 0.05)} mm = 0.80 mm$