Animated Solution for Physics - Physics and Measurement: A screw gauge has $50$ divisions on its circular scale. The circular scale is $4$ units ahead of the pitch scale marking, prior to use. Upon one complete rotation of the circular scale, a displacement of $0.5\text{ mm}$ is noticed on the pitch scale. The nature of zero error involved and the least count of the screw gauge, are respectively

Imagine you are an engineer tasked with measuring the thickness of a human hair. A standard ruler won't help you here; you need an instrument of immense precision. Enter the Screw Gauge (or Micrometer). This beautiful piece of experimental physics relies on the simple principle of a screw rotating through a nut. Let's break down how to extract its secrets: the Pitch, the Least Count, and the elusive Zero Error.

The first thing we need to understand is the Pitch ($P$). Think of the pitch as the linear distance the screw travels when you give it exactly one full twist. The problem explicitly tells us that upon one complete rotation of the circular scale, a displacement of $0.5\text{ mm}$ is noticed on the pitch scale.

Therefore, our Pitch is:

Now, how precisely can this instrument measure? That's defined by the Least Count ($LC$). The circular scale is divided into $50$ equal parts. This means that one full rotation ($0.5\text{ mm}$) is broken down into $50$ tiny steps.

The formula for Least Count is:

Substituting our values:

Since our options are in micrometers ($\mu \text{m}$), we must convert this. Knowing that $1\text{ mm}=1000\mu \text{m}$, we get:

Now comes the tricky part—the Zero Error. Before we even place an object between the jaws, we must check if the instrument is perfectly calibrated. The problem states: "The circular scale is 4 units ahead of the pitch scale marking, prior to use."

What does "ahead" mean physically? It means that when the jaws are completely closed, the zero mark of the circular scale has already crossed the reference line and is sitting below it. The 4th division is currently aligning with the reference line.

Because the instrument is already reading a value of $+4$ divisions (or $+4\times 10\mu \text{m}$) without any object, it has a Positive Zero Error. To get a true reading later, you would have to subtract this extra value.

By carefully analyzing the mechanics of the screw gauge, we have deduced two critical properties: 1. The Zero Error is Positive. 2. The Least Count is $10\mu \text{m}$.

This perfectly aligns with our understanding of experimental measurements and leads us directly to the correct option.