In a Screw Gauge, fifth division of the circular scale coincides with the reference line when the ratchet is closed. There are 50 divisions on the circular scale, and the main scale moves by 0.5 mm on a complete rotation. For a particular observation the reading on the main scale is 5 mm and the 20th division of the circular scale coincides with reference line. Calculate the true reading.

First, we recall the basic relations for a screw-gauge.

The pitch is the distance advanced by the spindle in one full rotation of the circular scale. It is given as $$0.5\ \text{mm}$$.

The least count (LC) is defined by the formula $$\text{LC} = \dfrac{\text{pitch}}{\text{number of circular-scale divisions}}.$$

We have $$\text{number of divisions} = 50,$$ so

$$\text{LC} = \dfrac{0.5\ \text{mm}}{50} = 0.01\ \text{mm}.$$

Now we determine the zero error. When the ratchet is fully closed (no object between the anvils), the reading should be zero, but the fifth division of the circular scale coincides with the reference line. Hence the instrument shows an apparent reading of

$$\text{zero error} = 5 \times \text{LC} = 5 \times 0.01\ \text{mm} = 0.05\ \text{mm}.$$

This reading is greater than the true value (which should be $$0$$), so it is a positive zero error. The zero correction is therefore negative:

$$\text{zero correction} = -0.05\ \text{mm}.$$

For the given observation we note the two parts of the apparent reading:

• Main-scale reading (MSR): $$5\ \text{mm}.$$

• Circular-scale reading (CSR): $$20$$ divisions.

The circular-scale contribution equals

$$\text{CSR} \times \text{LC} = 20 \times 0.01\ \text{mm} = 0.20\ \text{mm}.$$

So the observed reading becomes

$$\text{observed reading} = \text{MSR} + (\text{CSR} \times \text{LC}) = 5\ \text{mm} + 0.20\ \text{mm} = 5.20\ \text{mm}.$$

To obtain the true reading we apply the zero correction:

$$\text{true reading} = \text{observed reading} + \text{zero correction} = 5.20\ \text{mm} + (-0.05\ \text{mm}) = 5.15\ \text{mm}.$$

Hence, the correct answer is Option C.