First slide

Measurement

Question

In five complete rotations, the distance moved by screw on a linear scale is $0.5 cm$ . There are 100 divisions on circular scale. While measuring the thickness of a metal plate by the screw gauge, there are five divisions on the main scale and 35 divisions coincide with the reference line. Calculate the thickness of metal plate,

Moderate

A

5.35 cm
B

0.535 cm
C

0.6 cm
D

0.7 cm

Solution

Least count = $\frac{Pitch}{Number of divisions on circular scale}$
$= \frac{\frac{0.5}{5}}{100} = \frac{0.1}{100} cm = 0.001 cm$
Thickness = $5 \times 0.1 + 35 \times 0.001$
$= 0.535 cm$

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