First slide

Measurement

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 divisions 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. Find diameter of the wire.

Moderate

A

3.40 mm
B

3.50 mm
C

3.60 mm
D

3.70 mm

Solution

L C $= \frac{Pitch}{Number of divisions on circular scale}$
$= \frac{1 mm}{50} = 0.02 mm$
Positive zero error $= n_{1} (LC)$
$e = 6 \times 0.02 = 0.12 mm$
Linear scale reading =3 (pitch) = $3 mm$
Circular scale reading = $n_{2} (LC) = 31 \times 0.02$
$= 0.62 mm$
Measured diameter of wire $= (3 + 0.62) mn$
$= 3.62 mm$
Actual diameter of wire
$= 3.62 mm - 0.12 mm = 3.50 mm$

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