The pitch of a screw gauge having 50 divisions on its circular scale is 1 mm. When the two jaws of the screw gauge are in contact with each other, the zero of the circular scale lies 6 division below the line of graduation. When a wire is placed between the jaws, 3 linear scale divisions are clearly visible while 31st division on the circular scale coincides with the reference line. The diameter of the wire is mm

Question

The pitch of a screw gauge having 50 divisions on its circular scale is 1 mm. When the two jaws of the screw gauge are in contact with each other, the zero of the circular scale lies 6 division below the line of graduation. When a wire is placed between the jaws, 3 linear scale divisions are clearly visible while 31st division on the circular scale coincides with the reference line. The diameter of the wire is mm (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Δ l=1 mm N = 50 division Zero correction = - 6 divisions Diameter = Measured value + Zero correction =3×1+(6+31)×(1/50) =3+0.74=3.74 mm

Δl=1mm N=50 division Zero correction =−6 divisions Diameter = Measured value + Zero correction =3×1+(6+31)×501​ =3+0.74=3.74mm

$Δ l = 1 mm$
$N = 50$ division
Zero correction $= - 6$ divisions
Diameter = Measured value + Zero correction
$= 3 \times 1 + (6 + 31) \times \frac{1}{50}$
$= 3 + 0.74 = 3.74 mm$