A screw gauge gives the following reading when used to measure the diameter of a wire. Main scale re

Q: Why did we divide by 100 to find the least count?

The least count is the smallest distance the screw gauge can measure. Since 1 full rotation (100 divisions) moves the screw by 1 mm (the pitch), moving it by just 1 division moves it by 1 mm / 100 = 0.01 mm.

Q: What if the main scale reading wasn't 0?

You would simply add that value to the circular scale contribution. For example, if MSR was 2 mm, the total reading would be 2 mm + 0.52 mm = 2.52 mm.

Q: Why did we convert the final answer to cm?

Always look at the options! The calculated diameter was 0.52 mm, but all the given options were in cm. Converting units is a crucial final step to avoid silly mistakes.

Q: How do we know there is no zero error in this question?

The question does not mention any initial reading when the jaws are closed. In such cases, we assume the instrument is ideal and has zero error.

Q: What does '1 mm on main scale corresponds to 100 divisions of the circular scale' mean?

It means that when you give the circular scale one complete rotation (turning it by 100 divisions), it advances by exactly 1 mm on the main scale. This 1 mm is the pitch of the screw gauge.

Question

A screw gauge gives the following reading when used to measure the diameter of a wire. Main scale reading $0 	ext{ mm}$ Circular scale reading $52$ divisions Given that $1 	ext{ mm}$ on main scale corresponds to $100$ divisions of the circular scale. The diameter of wire from the above data is

Options:

(A) $0.052 	ext{ cm}$

(B) $0.026 	ext{ cm}$

(C) $0.005 	ext{ cm}$

(D) $0.52 	ext{ cm}$

Sigma Percentile · Accepted Answer

### The Art of Precise Measurement

Imagine you are an engineer tasked with measuring the exact thickness of a delicate copper wire. A standard ruler won't cut it; you need something that can measure down to fractions of a millimeter. Enter the **screw gauge**, a beautiful mechanical instrument designed to amplify tiny linear movements into readable rotational movements.

In this problem, we are given a set of readings from a screw gauge and asked to determine the diameter of a wire. Let's break down the mechanics of the instrument and the mathematics of the measurement step-by-step.

### Decoding the Screw Gauge

A screw gauge consists of two primary scales: the **Main Scale** (a linear scale on the sleeve) and the **Circular Scale** (a rotating scale on the thimble).

The problem states that the Main Scale Reading (MSR) is $0 	ext{ mm}$. This means the edge of the circular scale hasn't even crossed the first millimeter mark on the main scale. The wire is very thin!

Next, we look at the Circular Scale Reading (CSR), which is given as $52$ divisions. This means the $52^{	ext{nd}}$ mark on the rotating thimble perfectly aligns with the central reference line of the main scale.

### Finding the Least Count

Before we can calculate the total reading, we need to know the **Least Count (LC)** of our specific screw gauge. The least count is the absolute smallest distance the instrument can measure, which corresponds to rotating the circular scale by exactly one division.

We are given a crucial piece of information: "$1 	ext{ mm}$ on the main scale corresponds to $100$ divisions of the circular scale." This tells us two things:
1. The **Pitch** (the linear distance moved in one full rotation) is $1 	ext{ mm}$.
2. The total number of divisions on the circular scale is $100$.

The formula for the least count is:

$$ 	ext{LC} = \frac{	ext{Pitch}}{	ext{Total Circular Divisions}} $$

Substituting our values:

$$ 	ext{LC} = \frac{1 	ext{ mm}}{100} = 0.01 	ext{ mm} $$

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

### Calculating the Total Reading

Now we have all the pieces of the puzzle. The master formula for the total reading (which is the diameter $d$ of the wire) is:

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

Let's substitute the values we've gathered:

$$ d = 0 	ext{ mm} + (52 	imes 0.01 	ext{ mm}) $$

Executing the multiplication:

$$ d = 0 + 0.52 	ext{ mm} = 0.52 	ext{ mm} $$

So, the diameter of the wire is $0.52 	ext{ mm}$.

### The Final Trap: Unit Conversion

We have our answer, but there is a catch here. If you look at the options provided in the question, they are all in **centimeters (cm)**, not millimeters (mm). This is a classic trap set by examiners to test your presence of mind.

To convert millimeters to centimeters, we must divide by $10$:

$$ d = \frac{0.52}{10} 	ext{ cm} = 0.052 	ext{ cm} $$

And there we have it! The correct diameter of the wire is **$0.052 	ext{ cm}$**, which perfectly matches option (a). Always remember to double-check your final units before marking the answer!

Select Answer:

Visualized Solution

$Given Data$

$Least Count Formula$

$Calculating Least Count$

$Total Reading Formula$

$Substituting Values$

$Calculating Diameter in mm$

$Converting to cm$

$What if there was a Zero Error?$

The Sigma Insight: Vernier Calipers and Screw Gauge

The Art of Precise Measurement

Decoding the Screw Gauge

Finding the Least Count

Calculating the Total Reading

The Final Trap

Similar Questions

The least count of the main scale of a screw gauge is $1 mm$ . The minimum number of divisions on its circular scale required to measure $5 μ m$ diameter of a wire is

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.

If the screw on a screw gauge is given six rotations, it moves by 3 mm on the main scale. If there are 50 divisions on the circular scale, the least count of the screw gauge is

Select Answer:

Visualized Solution

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

.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; } Total Reading Formula

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if there was a Zero Error?