The pitch of a screw gauge is 0.5mm and there are 100 divisions on its circular scale. The instruments 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. The diameter of the wire is (k/100) (in mm ), then find k .

Question

The pitch of a screw gauge is 0.5mm and there are 100 divisions on its circular scale. The instruments 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. The diameter of the wire is (k/100) (in mm ), then find k . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Least count =(0 . 5 mm/100) Positive error =2× (0 . 5 mm/100)=(1/100)mm Scale reading =8× 0.5mm +83× (0 . 5 mm/100)-(1/100)mm ⇒ 4.405mm≈4.40mm

Least count =1000.5mm​ Positive error =2×1000.5mm​=1001​mm Scale reading =8×0.5mm +83×1000.5mm​−1001​mm ⇒4.405mm≈4.40mm

Least count $= \frac{0.5 mm}{100}$
Positive error $= 2 \times \frac{0.5 mm}{100} = \frac{1}{100} mm$
Scale reading $= 8 \times 0.5 mm$ $+ 83 \times \frac{0.5 mm}{100} - \frac{1}{100} mm$
$\Rightarrow 4.405 mm \approx 4.40 mm$