In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm . There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, what is the thickness (in mm) of the wire?

Question

In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm . There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, what is the thickness (in mm) of the wire? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Least count =(0 .25/5 × 100) cm=5× 10- 4 cm Reading =4× 0.05 cm+30× 5× 10- 4 cm =(0 .2 + 0 .0150) cm=2.15 mm

Least count =5×1000.25​cm=5×10−4cm Reading =4×0.05cm+30×5×10−4cm =(0.2+0.0150)cm=2.15mm

Least count $= \frac{0.25}{5 \times 100} c m = 5 \times 1 0^{- 4} c m$
Reading $= 4 \times 0.05 c m + 30 \times 5 \times 1 0^{- 4} c m$
$= (0.2 + 0.0150) c m = 2.15 mm$