The diameter of a cylinder is measured using a vernier calipers with no zero error. It is found that

Q: Why is the Main Scale Reading 5.10 cm and not 5.15 cm?

The Main Scale Reading is always the mark on the main scale that is immediately to the left of the vernier scale's zero mark. Since the zero lies between 5.10 and 5.15, we take 5.10.

Q: How do we know 1 MSD is 0.05 cm?

The problem states the zero of the vernier scale lies between 5.10 cm and 5.15 cm. These are consecutive marks on the main scale, so the distance between them is 5.15 - 5.10 = 0.05 cm.

Q: What if there was a zero error?

If there was a zero error, we would subtract it from the final calculated reading (with its proper sign). Since the problem specifies 'no zero error', we don't need to adjust our final answer.

Q: Why do we multiply the coinciding division by the Least Count?

The coinciding division tells us how many Least Count units the vernier zero has moved past the main scale reading. Multiplying them gives the exact fractional distance to add to the main scale reading.

Q: Can the Least Count be calculated differently?

Yes, another common formula is LC = (Value of 1 MSD) / (Total number of divisions on the vernier scale). Let's check: 0.05 / 50 = 0.001 cm. This works perfectly here too!

Question

The diameter of a cylinder is measured using a vernier calipers with no zero error. It is found that the zero of the vernier scale lies between $5.10	ext{ cm}$ and $5.15	ext{ cm}$ of the main scale. The vernier scale has $50$ division equivalent to $2.45	ext{ cm}$. The $24^{	ext{th}}$ division of the vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is

Options:

(A) $5.112	ext{ cm}$

(B) $5.124	ext{ cm}$

(C) $5.136	ext{ cm}$

(D) $5.148	ext{ cm}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are holding a vernier caliper, carefully measuring the diameter of a cylinder. The first thing we always check is the zero error. Thankfully, the problem states there is **no zero error**, which makes our life a bit easier!

Now, let's look at the main scale. The zero of the vernier scale is resting between $5.10	ext{ cm}$ and $5.15	ext{ cm}$.

This immediately tells us our **Main Scale Reading (MSR)**. The MSR is always the reading just before the vernier zero. Therefore, our $	ext{MSR} = 5.10	ext{ cm}$.

### Decoding the Scales

To find the exact measurement, we need to understand the markings on both scales. Let's start with the main scale. The consecutive marks are at $5.10	ext{ cm}$ and $5.15	ext{ cm}$.

This means the value of one **Main Scale Division (MSD)** is simply the difference between these two marks:
$$1	ext{ MSD} = 5.15	ext{ cm} - 5.10	ext{ cm} = 0.05	ext{ cm}$$

Next, we turn our attention to the vernier scale. The problem provides a crucial piece of information: $50$ divisions on the vernier scale are equivalent to a length of $2.45	ext{ cm}$.

From this, we can easily find the value of one **Vernier Scale Division (VSD)**:
$$1	ext{ VSD} = \frac{2.45	ext{ cm}}{50} = 0.049	ext{ cm}$$

### The Master Equation: Least Count

The precision of our instrument is defined by its **Least Count (LC)**. The fundamental definition of Least Count is the difference between one main scale division and one vernier scale division.

Let's calculate it:
$$	ext{LC} = 1	ext{ MSD} - 1	ext{ VSD}$$
$$	ext{LC} = 0.05	ext{ cm} - 0.049	ext{ cm} = 0.001	ext{ cm}$$

*Pro-tip:* You can also calculate the Least Count using the shortcut formula $	ext{LC} = \frac{1	ext{ MSD}}{	ext{Total Vernier Divisions}} = \frac{0.05}{50} = 0.001	ext{ cm}$. Both methods yield the exact same result!

### Final Calculation

We have all the pieces of the puzzle. The formula for the total reading is:
$$	ext{Total Reading} = 	ext{MSR} + (	ext{Vernier Coinciding Division} 	imes 	ext{LC})$$

We know the $	ext{MSR}$ is $5.10	ext{ cm}$, the coinciding division is the $24^{	ext{th}}$ mark, and the $	ext{LC}$ is $0.001	ext{ cm}$. Let's substitute these values:
$$	ext{Total Reading} = 5.10 + (24 	imes 0.001)$$

Performing the multiplication first:
$$	ext{Total Reading} = 5.10 + 0.024$$

Adding them together gives us our final, precise measurement:
$$	ext{Total Reading} = 5.124	ext{ cm}$$

Since there is no zero error to account for, the diameter of the cylinder is exactly **$5.124	ext{ cm}$**.

Select Answer:

Visualized Solution

$Visualizing the Vernier Caliper$

$Main Scale Division (MSD)$

$Vernier Scale Division (VSD)$

$Least Count (LC)$

$Total Reading Setup$

$Final Calculation$

$Alternative LC Formula$

The Sigma Insight: Vernier Calipers and Screw Gauge

Analyzing the Setup

Decoding the Scales

The Master Equation

Final Calculation

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Select Answer:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Main Scale Division (MSD)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Vernier Scale Division (VSD)

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

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

.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; } Alternative LC Formula