Two thin wires, Wire-1 of diameter 0.650 mm and Wire-2 of unknown diameter d are given. To obtain th

Q: Why do we subtract the zero error?

Zero error is the reading of the instrument when it should be zero. If it's positive, the instrument reads more than the actual value, so we must subtract it to get the true value. If it's negative, we subtract a negative value, which effectively adds to the measured value.

Q: How did we get the least count formula?

The least count is the smallest distance the spindle moves for a rotation of one circular scale division. Since one full rotation (100 divisions) moves the spindle by the pitch (0.5 mm), one division moves it by 0.5/100 mm.

Q: What if the zero error was negative?

If the measured diameter of Wire-1 was less than its true diameter, the zero error would be negative. We would still use the formula `True = Measured - Zero Error`, which would result in adding the magnitude of the error to the measured value.

Q: Why convert to micrometers at the end?

The question specifically asks for the value of $d$ in $\mu\text{m}$. Since $1\text{ mm}$ is equal to $1000\text{ }\mu\text{m}$, we multiply the result in mm by $1000$ to match the requested units.

Q: Can the main scale reading be different from the linear scale reading?

In the context of a screw gauge, the linear scale is the main scale. The terms are used interchangeably.

Question

Two thin wires, Wire-1 of diameter $0.650	ext{ mm}$ and Wire-2 of unknown diameter $d$ are given. To obtain the value of $d$, the diameters of the two wires are measured with a screw gauge. The screw gauge has a pitch of $0.5	ext{ mm}$ and there are $100$ divisions on the circular scale (CS). The smallest division on the linear scale (LS) is $0.5	ext{ mm}$. The table shows the readings of LS and CS for the measurements. The value of $d$ (in $\mu	ext{m}$) is:

$\begin{array}{|l|c|c|} \hline & \multicolumn{2}{c|}{	ext{Readings}} \ \cline{2-3} & 	ext{LS (mm)} & 	ext{CS} \ \hline 	ext{Wire-1} & 0.5 & 42 \ \hline 	ext{Wire-2} & 1.5 & 95 \ \hline \end{array}$

Sigma Percentile · Accepted Answer

Measuring the diameter of a thin wire requires precision, and that's exactly where a screw gauge comes into play. But what happens when your instrument isn't perfectly calibrated? This problem takes us on a journey of finding the true measurement by first uncovering the hidden flaws of our tool.

### Understanding the Instrument's Precision

Before we can trust any reading, we must determine the smallest value our screw gauge can accurately measure, known as the **Least Count (LC)**. The problem states that the pitch of the screw gauge is $0.5	ext{ mm}$ and the circular scale has $100$ divisions.

The least count is calculated by dividing the pitch by the number of circular scale divisions:

$$LC = \frac{	ext{Pitch}}{	ext{Number of CS divisions}} = \frac{0.5	ext{ mm}}{100} = 0.005	ext{ mm}$$

This means every single tick on the circular scale corresponds to a movement of $0.005	ext{ mm}$.

### Uncovering the Hidden Error

We are given a reference wire, Wire-1, with a known true diameter of $0.650	ext{ mm}$. Let's see what our screw gauge reads for it. The measured diameter is the sum of the Main Scale Reading (MSR) and the Circular Scale Reading (CSR) multiplied by the least count.

$$d_{	ext{measure, 1}} = 	ext{MSR}_1 + (	ext{CSR}_1 	imes 	ext{LC})$$

Substituting the values from the table:

$$d_{	ext{measure, 1}} = 0.5 + (42 	imes 0.005) = 0.5 + 0.21 = 0.71	ext{ mm}$$

Wait a minute! The true diameter is $0.650	ext{ mm}$, but our instrument reads $0.71	ext{ mm}$. This discrepancy means our screw gauge has a **Zero Error**. Since the measured value is greater than the true value, the error is positive.

$$	ext{Zero Error} = 	ext{Measured Value} - 	ext{True Value}$$

$$	ext{Zero Error} = 0.71	ext{ mm} - 0.65	ext{ mm} = +0.06	ext{ mm}$$

Our instrument consistently overestimates measurements by $0.06	ext{ mm}$. We must account for this in all future readings.

### Finding the True Diameter of Wire-2

Now, let's measure the unknown Wire-2. Using the same formula, we calculate its measured diameter:

$$d_{	ext{measure, 2}} = 	ext{MSR}_2 + (	ext{CSR}_2 	imes 	ext{LC})$$

$$d_{	ext{measure, 2}} = 1.5 + (95 	imes 0.005) = 1.5 + 0.475 = 1.975	ext{ mm}$$

To find the **true diameter**, we must subtract the zero error we discovered earlier from this measured value.

$$d_{	ext{true, 2}} = d_{	ext{measure, 2}} - 	ext{Zero Error}$$

$$d_{	ext{true, 2}} = 1.975	ext{ mm} - 0.060	ext{ mm} = 1.915	ext{ mm}$$

The question asks for the final answer in micrometers ($\mu	ext{m}$). Since $1	ext{ mm} = 1000	ext{ }\mu	ext{m}$, we multiply our result by $1000$:

$$d_{	ext{true, 2}} = 1.915 	imes 1000 = \mathbf{1915	ext{ }\mu	ext{m}}$$

This problem elegantly demonstrates the importance of calibration. By using a known standard, we were able to correct our instrument's inherent bias and arrive at the precise, true measurement.

Enter Numerical Value:

Visualized Solution

\text{Understanding the Screw Gauge}

\text{Least Count Formula}

\text{Calculating Least Count}

\text{Measured Value Formula}

\text{Measured Diameter of Wire-1}

\text{Calculating Measured Diameter of Wire-1}

\text{Zero Error Concept}

\text{Calculating Zero Error}

\text{Measured Diameter of Wire-2}

\text{Calculating Measured Diameter of Wire-2}

\text{True Diameter of Wire-2}

\text{Conclusion}

The Sigma Insight: Vernier Calipers and Screw Gauge

Understanding the Instrument's Precision

Uncovering the Hidden Error

Finding the True Diameter of Wire-2

Similar Questions

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