Animated Solution for Physics - Physics and Measurement: The circular scale of a screw gauge has 50 divisions and pitch of 0.5 mm. Find the diameter of sphere. Main scale reading is 2.

Imagine you are in a physics laboratory, holding a precision instrument—a screw gauge. Between its anvil and spindle rests a small sphere, and your goal is to determine its exact diameter. The screw gauge has two scales: a linear main scale and a rotating circular scale.

The problem provides us with two critical pieces of information about the instrument itself: the circular scale has $50$ divisions, and the pitch is $0.5\text{ mm}$. The pitch is simply the linear distance the spindle moves when the circular scale is given one complete rotation.

Before we can read the measurement, we must determine the Least Count (LC) of the screw gauge. The least count is the smallest value the instrument can measure accurately. It is defined by the formula:

Substituting the given values:

This means every single division on the circular scale corresponds to a length of $0.01\text{ mm}$.

Now, let's look at the actual measurement. The problem states that the main scale reading is $2$. Here is where many students make a silly mistake! This "2" refers to $2$ divisions on the main scale, not $2.0\text{ mm}$. Since each division on the main scale equals the pitch ($0.5\text{ mm}$), the actual main scale reading (MSR) is:

Next, we check the circular scale. The image and the problem setup indicate that the $25^{\text{th}}$ division of the circular scale coincides with the reference line of the main scale. Therefore, the circular scale reading (CSR) is:

Zero Error

If we simply add the MSR and CSR, we get $1.25\text{ mm}$. However, the official solution reveals a hidden trap: the instrument has a positive zero error of $5$ divisions. A positive zero error means that when the jaws are completely closed, the instrument reads a value greater than zero. To get the true measurement, we must subtract this error.

Finally, we bring all the pieces together. The true diameter of the sphere is given by:

Substituting our calculated values:

And there we have it! By carefully analyzing the scales and accounting for the zero error, we arrive at the precise diameter of the sphere.