Let the distance moved by the screw of a screw gauge is 2 mm in four rotations and there are 50 divisions on its cap. When nothing is put between its jaws, `20^"th"` division of circular scale coincides with reference line and zero of linear scale is hidden from circular scale when two jaws touch each other or zero of circular scale is lying above the reference line. If plate is placed between the jaws, the main scale reads 2 divisions and circular scale reads 20 divisions. What is the thickness of plate ?

Options

A. `1.1mm`
B. `1.2 mm`
C. `1.4 mm`
D. `1.5 mm`

Answer

D

Explanation

Distance moved in one rotation
= 0.5 mm
Least count,
LC = "0.5 mm"/"50 divisions"=0.01 mm`
Screw gauge has negative zero error.
This error is (50 - 20) 0.01 mm or (30) (0.01) mm.
Thickness of plate
= (2 x 0.5 mm) + (30 + 20)
(0.01 mm)
= 1.5 mm