The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line. The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is:

We are told that the pitch of the screw gauge is $$0.5\ \text{mm}$$ and that the circular scale has $$100$$ equal divisions.

First, we obtain the least count. We recall the definition: the least count (L.C.) of a screw gauge is given by

$$\text{Least Count}=\frac{\text{Pitch}}{\text{Number of circular-scale divisions}}.$$

Substituting the given values, we have

$$\text{L.C.}=\frac{0.5\ \text{mm}}{100}=0.005\ \text{mm}.$$

Next, we analyse the zero error. When the screw is fully tightened and no object is placed, the zero of the circular scale stands $$3$$ divisions below the main (reference) line. The reading shown by the instrument in that situation is therefore

$$\text{Zero-reading}=3 \times \text{L.C.}=3 \times 0.005\ \text{mm}=0.015\ \text{mm}.$$

This positive reading appears even though the true length should be zero, so the instrument possesses a positive zero error of $$+0.015\ \text{mm}$$. Accordingly, the required zero correction is the negative of this, namely

$$\text{Zero correction}=-0.015\ \text{mm}.$$

Now we consider the measurement made with the thin sheet. The main-scale reading (M.S.R.) is given as $$5.5\ \text{mm}$$. The circular-scale division that coincides is $$48$$, so the circular-scale reading (C.S.R.) is

$$\text{C.S.R.}=48 \times \text{L.C.}=48 \times 0.005\ \text{mm}=0.240\ \text{mm}.$$

Adding the two contributions gives the observed reading:

$$\text{Observed thickness}= \text{M.S.R.} + \text{C.S.R.} =5.5\ \text{mm}+0.240\ \text{mm}=5.740\ \text{mm}.$$

Finally, we apply the zero correction. Because the correction is $$-0.015\ \text{mm}$$, we add it algebraically to the observed reading:

$$\text{True thickness}=5.740\ \text{mm}+(-0.015\ \text{mm})=5.725\ \text{mm}.$$

Hence, the correct answer is Option C.