The pitch of the screw gauge is 1 mm and there are 100 divisions on the circular scale. When nothing

Q: Why is the zero error positive when the zero mark is 'below' the reference line?

Think about how the numbers on the circular scale increase. As you open the jaws, higher numbers cross the reference line. If the zero is below the line when closed, a positive number (like 8) is aligned with the reference line. The instrument is reading a value greater than zero, hence a positive zero error.

Q: How do we know the main scale reading is exactly 1 mm?

The problem states 'the first linear scale division is clearly visible.' Since the pitch is 1 mm, the divisions on the main scale are 1 mm apart. Seeing the first division means the thimble has moved past the 1 mm mark.

Q: Why do we subtract the zero error?

A positive zero error means the instrument is 'over-reading' by that amount. To get the true value, you must remove this extra amount from your final observed reading.

Q: I calculated 1.64 mm and it's in the options. Why is it wrong?

You successfully calculated the true *diameter* of the wire. However, the question specifically asks for the *radius*. Always double-check what the question is asking for before marking your answer!

Q: What if the zero mark was above the reference line?

If the zero mark was above the reference line, the instrument would be reading a value less than zero when closed. This is a negative zero error, and you would still subtract it (which mathematically means adding its magnitude) to the observed reading.

Question

The pitch of the screw gauge is $1	ext{ mm}$ and there are $100$ divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies $8$ divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while $72^{	ext{nd}}$ division on circular scale coincides with the reference line. The radius of the wire is

Options:

(A) $1.64	ext{ mm}$

(B) $0.82	ext{ mm}$

(C) $1.80	ext{ mm}$

(D) $0.90	ext{ mm}$

Sigma Percentile · Accepted Answer

The screw gauge is a beautiful instrument, a perfect marriage of rotational and linear motion that allows us to measure microscopic dimensions with ease. But like any real-world instrument, it has its quirks. Let's break down this problem step-by-step and uncover the hidden trap at the end!

### Understanding the Instrument's Precision

Before we can measure anything, we need to know the limits of our tool. The **Least Count (LC)** tells us the smallest value the screw gauge can reliably measure.

The problem states that the pitch is $1	ext{ mm}$ and there are $100$ divisions on the circular scale.

$$LC = \frac{	ext{Pitch}}{	ext{Number of divisions}} = \frac{1	ext{ mm}}{100} = 0.01	ext{ mm}$$

This means every single tick on the circular scale corresponds to an advancement of $0.01	ext{ mm}$.

### The Quirks of Reality: Zero Error

In an ideal world, when the jaws of the screw gauge are completely closed, the zero of the circular scale perfectly aligns with the main scale's reference line. But our instrument has a slight imperfection.

When closed, the zero of the circular scale lies $8$ divisions *below* the reference line. Imagine rotating the thimble: if the zero is below the line, it means the thimble hasn't been rotated fully back to the true zero. The instrument is already reading a positive value!

This is a **positive zero error**.

$$e = + 8 	imes LC = + 8 	imes 0.01	ext{ mm} = + 0.08	ext{ mm}$$

We must remember to subtract this "extra" reading from our final measurement.

### Taking the Measurement

Now, we place the wire between the jaws. The problem gives us two crucial pieces of information:

1. The first linear scale division is clearly visible. This gives us our **Main Scale Reading (MSR)**: $1	ext{ mm}$.
2. The $72^{	ext{nd}}$ division on the circular scale coincides with the reference line. This gives us our **Circular Scale Reading (CSR)**: $72 	imes 0.01	ext{ mm} = 0.72	ext{ mm}$.

Adding these together gives us the observed diameter:

$$	ext{Observed Reading} = MSR + CSR = 1	ext{ mm} + 0.72	ext{ mm} = 1.72	ext{ mm}$$

### Correcting the Measurement

We know our instrument reads $0.08	ext{ mm}$ too high. To find the true diameter, we subtract the zero error:

$$	ext{True Diameter} = 	ext{Observed Reading} - 	ext{Zero Error}$$

$$D = 1.72	ext{ mm} - (+0.08	ext{ mm}) = 1.64	ext{ mm}$$

### The Final Trap

If you look at the options, $1.64	ext{ mm}$ is sitting right there as option (a), practically begging you to select it. But wait! Read the last line of the question carefully: *"The radius of the wire is..."*

This is a classic exam trap. We found the diameter, but we need the radius.

$$r = \frac{D}{2} = \frac{1.64	ext{ mm}}{2} = 0.82	ext{ mm}$$

And there we have it! By staying vigilant and reading carefully, we arrive at the correct answer.

Select Answer:

Visualized Solution

$L e a s t C o u n t$

$Z er o E r r or$

$O b ser v e d R e a d in g$

$T r u e D iam e t er$

$R a d i u s o f W i r e$

The Sigma Insight: Vernier Calipers and Screw Gauge

Understanding the Instrument's Precision

The Quirks of Reality

Taking the Measurement

Correcting the Measurement

The Final Trap

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