A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the t

Q: Why is the zero error negative and not positive?

When the jaws are fully closed, if the zero of the circular scale has crossed the reference line (meaning a higher division like 45 is coinciding), it means the instrument is reading less than zero. This is a negative zero error.

Q: How do we calculate the magnitude of a negative zero error?

You subtract the coinciding division from the total number of divisions (e.g., 50 - 45 = 5), and then multiply by the least count. The error is -0.05 mm.

Q: Why do we subtract the zero error from the measured value?

The measured value includes the initial error. By subtracting the error (True = Measured - Error), a negative error mathematically adds to the measured value, compensating for the instrument starting 'behind' zero.

Q: The book's solution says the answer is 0.75 mm. Is that correct?

No, the book's solution incorrectly ignores the zero error calculation despite it being explicitly described in the problem. The true rigorous answer, which accounts for the -0.05 mm zero error, is 0.80 mm.

Q: What does 'zero of the main scale is barely visible' mean?

It confirms that the circular scale hasn't completed a full rotation backwards. It is just slightly behind the zero mark, perfectly consistent with a small negative zero error of -0.05 mm.

Question

A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet, if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line?

Options:

(A) 0.75 mm

(B) 0.80 mm

(C) 0.70 mm

(D) 0.50 mm

Sigma Percentile · Accepted Answer

The problem of the screw gauge is a classic test of not just your ability to plug numbers into a formula, but your fundamental understanding of how measuring instruments work. It is a beautiful exercise in precision, error analysis, and spatial reasoning. Let's embark on this journey to decode the true thickness of the aluminium sheet.

### The Anatomy of a Screw Gauge

Before we take any measurements, we must understand the tool in our hands. A screw gauge operates on a simple principle: rotational motion is converted into linear motion. The **pitch** is the linear distance the main scale advances in one complete rotation of the circular scale.

In our case, the pitch is given as $0.5 	ext{ mm}$. The circular scale is divided into $50$ equal divisions. The smallest measurement this instrument can accurately make, known as the **Least Count (LC)**, is the pitch divided by the total number of divisions.

$$ 	ext{LC} = \frac{0.5 	ext{ mm}}{50} = 0.01 	ext{ mm} $$

This means every single tick mark on the circular scale represents a linear movement of exactly $0.01 	ext{ mm}$.

### Decoding the Zero Error

In an ideal world, when the jaws of the screw gauge are perfectly closed with nothing between them, the zero of the circular scale should perfectly align with the reference line of the main scale. But instruments in the real world suffer from wear and tear.

The problem states that when the jaws are brought in contact, the $45^{	ext{th}}$ division coincides with the reference line, and the zero of the main scale is barely visible. Imagine what this means physically. The circular scale has rotated *past* the true zero mark. It hasn't stopped at $0$; it has kept going until the $45^{	ext{th}}$ division hit the reference line.

Because the zero of the circular scale is now *above* the reference line, the instrument is reading a negative value when it should be reading zero. This is a **negative zero error**.

To calculate the exact magnitude of this error, we see how many divisions it has gone past zero. It went from $50$ (which is $0$) down to $45$, which is a difference of $5$ divisions.

$$ e = - (50 - 45) 	imes 	ext{LC} $$

$$ e = - 5 	imes 0.01 	ext{ mm} = -0.05 	ext{ mm} $$

Our screw gauge has a built-in error of $-0.05 	ext{ mm}$. It starts every measurement with a "handicap" of $-0.05 	ext{ mm}$.

### The Measurement Phase

Now, we place the thin aluminium sheet between the jaws. The main scale reading (MSR) is clearly visible as $0.5 	ext{ mm}$. The circular scale reading (CSR) shows the $25^{	ext{th}}$ division perfectly aligned with the reference line.

Let's calculate the raw, uncorrected measurement.

$$ 	ext{Measured Value} = 	ext{MSR} + (	ext{CSR} 	imes 	ext{LC}) $$

$$ 	ext{Measured Value} = 0.5 	ext{ mm} + (25 	imes 0.01 	ext{ mm}) $$

$$ 	ext{Measured Value} = 0.5 	ext{ mm} + 0.25 	ext{ mm} = 0.75 	ext{ mm} $$

If we were to stop here, we would conclude the thickness is $0.75 	ext{ mm}$. But remember our instrument's handicap!

### The Final Correction (And a Common Trap)

This is where many students—and surprisingly, even some textbook solutions—fall into a trap. They see the raw measurement of $0.75 	ext{ mm}$ and immediately select option (a). But we must account for the zero error to find the **true reading**.

The fundamental equation for any instrument is:

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

Let's substitute our values carefully, paying close attention to the signs.

$$ 	ext{True Reading} = 0.75 	ext{ mm} - (-0.05 	ext{ mm}) $$

$$ 	ext{True Reading} = 0.75 	ext{ mm} + 0.05 	ext{ mm} = 0.80 	ext{ mm} $$

Because the instrument started at $-0.05 	ext{ mm}$, it had to travel $0.05 	ext{ mm}$ just to get to true zero, and *then* it traveled another $0.75 	ext{ mm}$ to measure the sheet. The total thickness of the sheet is the sum of these distances.

Therefore, the rigorous, physically accurate thickness of the aluminium sheet is **$0.80 	ext{ mm}$**. Always trust the physics, respect the signs, and never forget to correct for your zero error!

Select Answer:

Visualized Solution

$Screw Gauge Parameters$

$Least Count Formula$

$Calculating Least Count$

$Identifying Zero Error$

$Negative Zero Error$

$Calculating Zero Error$

$Taking the Measurement$

$Measured Reading Formula$

$Calculating Measured Value$

$True Reading Formula$

$Final Calculation$

$Conclusion$

The Sigma Insight: Vernier Calipers and Screw Gauge

The Anatomy of a Screw Gauge

Decoding the Zero Error

The Measurement Phase

The Final Correction (And a Common Trap)

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Screw Gauge Parameters

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Least Count Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Least Count

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Identifying Zero Error

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Negative Zero Error

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Zero Error

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Taking the Measurement

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Measured Reading Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Measured Value

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } True Reading Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion