The pitch of a screw gauge is 0.5 mm and there are 100 divisions on it circular scale. The instrument reads +2 divisions when nothing is put in between its jaws. In measuring the diameter of a wire, there are 8 divisions on the main scale and 83rd division coincides with the reference line. Then the diameter of the wire is

Question

The pitch of a screw gauge is 0.5 mm and there are 100 divisions on it circular scale. The instrument reads +2 divisions when nothing is put in between its jaws. In measuring the diameter of a wire, there are 8 divisions on the main scale and 83rd division coincides with the reference line. Then the diameter of the wire is (A) 4.05 mm (B) 4.405 mm (C) 3.05 mm (D) 1.25 mm

Tardigrade Admin · Accepted Answer

Δ l= 0.5 mm N= 100 divisions Zero correction = 2 divisions Reading = Measured value + Zero correction = (8×0.5)mm + (83-2)×(0.5/100) =4mm+81×(0.5/100) =4.405 mm

$4.05 mm$

$4.405 mm$

$3.05 mm$

$1.25 mm$

$Δ l = 0.5 mm$
$N = 100$ divisions
Zero correction $= 2$ divisions
Reading = Measured value + Zero correction
$= (8 \times 0.5) mm + (83 - 2) \times \frac{0.5}{100}$
$= 4 mm + 81 \times \frac{0.5}{100}$
$= 4.405 mm$

4.05mm

4.405mm

3.05mm

1.25mm

Δl=0.5mm N=100 divisions Zero correction =2 divisions Reading = Measured value + Zero correction =(8×0.5)mm+(83−2)×1000.5​ =4mm+81×1000.5​ =4.405mm

$4.05 mm$

$4.405 mm$

$3.05 mm$

$1.25 mm$

$Δ l = 0.5 mm$
$N = 100$ divisions
Zero correction $= 2$ divisions
Reading = Measured value + Zero correction
$= (8 \times 0.5) mm + (83 - 2) \times \frac{0.5}{100}$
$= 4 mm + 81 \times \frac{0.5}{100}$
$= 4.405 mm$