The pitch of a screw gauge is $1 mm$ and there are 100 divisions on circular scale. When faces A and B are just touching each other without putting anything between the studs 32 nd division of the circular scale coincides with the reference line. When a glass plate is placed between the studs, the linear scale reads 4 divisions and the circular scale reads 16 divisions. Find the thickness of the glass plate. Zero of linear scale is not hidden from circular scale when A and B touches each other.

Moderate

Solution

Least count (L C) $= \frac{Pitch}{Number of divisions on circular scale} = \frac{1}{100} m m$
= 0.01 mm
As zero is not hidden from circular scale when A and B touches each other. Hence, the screw gauge has positive error.
$e = + n (LC) = 32 \times 0.01 = 0.32 mm$
Linear scale reading $= 4 \times (1 mm) = 4 mm$
Circular scale reading $= 16 \times (0.01 mm) = 0.16 mm$
Measured reading $= (4 + 0.16) mm = 4.16 mm$
Absolute reading = Measured reading -e
$= (4.16 - 0.32) mm = 3.84 mm$
Therefore, thickness of the glass plate is $3.84 mm$ .

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