The diameter of a wire as measured by a screw gauge of pitch 0.5 mm is 8.3 mm. The pitch scale reading is:

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Solution

The correct option is B 16
$P = 0.5 m m; d = 8.3 m m$
$P S R = P \times r e a d i n g$
$8 m m = 0.5 \times R$
Since $8.3 m m$ is $(8 m m + 0.3 m m)$ , $0.3 m m$ is less than pitch.
$R = \frac{8}{0.5} = 16$

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The diameter of a wire as measured by a screw gauge of pitch 0.5 mm is 8.3 mm. The pitch scale reading is:

The correct option is B 16P=0.5mm;d=8.3mmPSR=P×reading8mm=0.5×RSince 8.3mm is (8mm+0.3mm), 0.3mm is less than pitch.R=80.5=16

The correct option is B 16
$P = 0.5 m m; d = 8.3 m m$
$P S R = P \times r e a d i n g$
$8 m m = 0.5 \times R$
Since $8.3 m m$ is $(8 m m + 0.3 m m)$ , $0.3 m m$ is less than pitch.
$R = \frac{8}{0.5} = 16$