Animated Solution for Physics - Physics and Measurement: The least count of the main scale of a screw gauge is $1\text{ mm}$. The minimum number of divisions on its circular scale required to measure $5\mu \text{m}$ diameter of a wire is

The universe is filled with dimensions so small that our eyes cannot perceive them. Yet, as physicists and engineers, we have devised ingenious tools to measure these microscopic realms with astonishing precision. One such masterpiece of mechanical engineering is the screw gauge.

Imagine trying to measure the thickness of a single strand of human hair or a delicate copper wire. A standard ruler is useless here. We need an instrument that translates tiny linear movements into large, readable rotational movements. This is exactly what a screw gauge does, and understanding its mechanics is a fundamental rite of passage in physics.

Pitch and Least Count

Before we dive into the mathematics, let's visualize the instrument. A screw gauge consists of two primary scales: the main scale (or sleeve) and the circular scale (or thimble).

When you rotate the thimble, the screw advances forward. The linear distance the screw travels in one complete rotation is called the pitch. In our problem, the least count of the main scale is given as $1\text{ mm}$. This means the smallest marking on the main scale is $1\text{ mm}$, which corresponds to the pitch. So, one full rotation moves the screw by exactly $1\text{ mm}$.

But we want to measure something much smaller than $1\text{ mm}$. We want to measure a diameter of $5\mu \text{m}$. This minimum measurable value is the least count of the entire instrument. The magic of the screw gauge lies in dividing that $1\text{ mm}$ pitch into many smaller fractions using the circular scale.

The governing principle is beautifully simple:

We are given the target least count and the pitch. Our mission is to find $N$, the number of divisions required on the circular scale.

First, we must establish harmony among our units. Physics is unforgiving when it comes to mismatched units. Let's convert everything into standard SI units (meters).

The pitch is $1\text{ mm}$, which is $10^{-3}\text{ m}$.

The required least count is $5\mu \text{m}$, which translates to $5\times 10^{-6}\text{ m}$.

Now, we substitute these values into our master equation:

Take a moment to appreciate this raw setup. On the left, we have the microscopic precision we demand. On the right, we have the mechanical constraints of our instrument. The variable $N$ is the bridge between them.

To find $N$, we simply rearrange the equation by cross-multiplying. We isolate $N$ on one side:

Now, we perform the algebraic computation. When we move $10^{-6}$ from the denominator to the numerator, the exponent flips its sign, becoming $10^6$.

Using the laws of exponents, we add the powers: $-3+6=3$.

Since $10^3$ is $1000$, the expression simplifies beautifully:

The math has spoken, and the answer is 200. But what does this mean physically?

It means that to measure a delicate $5\mu \text{m}$ wire accurately, the manufacturer must engrave exactly 200 equally spaced divisions around the circumference of the thimble. Every time you rotate the thimble by just one of those tiny divisions, the screw advances by a mere $5\mu \text{m}$.

This problem is a perfect illustration of how simple mathematical ratios govern the design of high-precision scientific instruments. Always remember to connect the algebraic steps back to the physical reality of the device in your hands!