One main scale division of a vernier callipers is a cm and nth division of the vernier scale coincid

Q: Why is the Least Count formula exactly 1 MSD - 1 VSD?

Think of the sliding mechanism. When the jaws are closed, the zero marks align. If you open the jaws just enough so that the 1st vernier mark aligns with the 1st main scale mark, the distance you opened the jaws is exactly the difference in length between one main scale division and one vernier scale division. This is the smallest measurable gap.

Q: What does 'nth division coincides with (n-1)th division' actually mean physically?

It means if you look at the scales, the total length of 'n' tiny blocks on the vernier scale is exactly equal to the total length of 'n-1' slightly larger blocks on the main scale. This allows us to mathematically relate the sizes of the two different blocks.

Q: I got a/n as my answer, but it's not in the options. What did I do wrong?

You fell into the classic unit trap! Your answer $a/n$ is perfectly correct, but it is in centimeters because the initial MSD was given in centimeters. The question specifically asked for the answer in millimeters. Multiplying by 10 gives the correct option.

Q: Can the vernier scale have more divisions than the main scale for the same length?

Yes, that is exactly how a standard vernier calliper is designed! By fitting 'n' divisions into the space of 'n-1' main scale divisions, we ensure that each vernier division is slightly smaller than a main scale division, which is what makes the vernier principle work.

Q: What if the question said (n+1) VSD coincides with n MSD?

The logic remains exactly the same! You would write $(n+1) \text{ VSD} = n \text{ MSD}$, find $1 \text{ VSD} = \frac{n}{n+1} \text{ MSD}$, and subtract it from $1 \text{ MSD}$. This is known as a retrograde vernier, but the mathematical approach is identical.

Question

One main scale division of a vernier callipers is $a$ cm and $n$th division of the vernier scale coincide with $(n - 1)$th division of the main scale. The least count of the callipers (in mm) is

Options:

(A) $\frac{10na}{(n-1)}$

(B) $\frac{10a}{(n-1)}$

(C) $\left(\frac{n-1}{10n}\right)a$

(D) $\frac{10a}{n}$

Sigma Percentile · Accepted Answer

### Mastering the Vernier Calliper: The Art of Finding the Least Count

Imagine you are holding a Vernier Calliper in your hands. It is a beautifully simple yet incredibly precise instrument. The secret to its precision lies in the slight mismatch between its two scales: the fixed **Main Scale** and the sliding **Vernier Scale**.

In this classic JEE problem, we are tasked with finding the **Least Count (LC)** of a custom-designed vernier calliper. The least count is the absolute smallest measurement the instrument can reliably make. Geometrically, it is simply the difference in length between one Main Scale Division (MSD) and one Vernier Scale Division (VSD).

### Analyzing the Setup

We are given two critical pieces of information:
1. The length of one Main Scale Division is $1 	ext{ MSD} = a 	ext{ cm}$.
2. The $n^{	ext{th}}$ division of the vernier scale perfectly coincides with the $(n - 1)^{	ext{th}}$ division of the main scale.

This coincidence is the mathematical heart of the instrument. It tells us exactly how compressed the vernier scale is compared to the main scale. We can write this relationship as an equation:

$$n 	ext{ VSD} = (n - 1) 	ext{ MSD}$$

### The Master Equation

To find the least count, we first need to know the exact length of a single Vernier Scale Division. By rearranging our coincidence equation, we get:

$$1 	ext{ VSD} = \frac{n - 1}{n} 	ext{ MSD}$$

Since we know that $1 	ext{ MSD} = a 	ext{ cm}$, we can substitute this value in:

$$1 	ext{ VSD} = \frac{n - 1}{n} a 	ext{ cm}$$

Now, we bring in the fundamental definition of the Least Count:

$$	ext{LC} = 1 	ext{ MSD} - 1 	ext{ VSD}$$

### Final Calculation

Let's substitute our known values into the Least Count formula:

$$	ext{LC} = a - \frac{(n - 1)a}{n}$$

To subtract these, we take a common denominator of $n$:

$$	ext{LC} = \frac{an - a(n - 1)}{n}$$

Expanding the numerator gives us $an - an + a$. The $an$ terms beautifully cancel each other out, leaving us with a clean, elegant expression:

$$	ext{LC} = \frac{a}{n} 	ext{ cm}$$

**The Unit Conversion Trap**

Many students will stop here, look at the options, and panic, or worse, pick a wrong option that looks similar. Why? Because they missed a crucial detail in the question: *"The least count of the callipers (in mm) is"*.

Our result is currently in centimeters. To convert centimeters to millimeters, we must multiply by 10.

$$	ext{LC} = \frac{a}{n} 	imes 10 	ext{ mm} = \frac{10a}{n} 	ext{ mm}$$

And there we have it! By carefully following the geometry of the scales and keeping a sharp eye on our units, we arrive at the correct answer.

Animated Solution for Physics - Physics and Measurement: One main scale division of a vernier callipers is $a$ cm and $n$ th division of the vernier scale coincide with $(n - 1)$ th division of the main scale. The least count of the callipers (in mm) is

Select Answer:

Visualized Solution

$1 MSD = a cm$

$n VSD = (n - 1) MSD$

$1 VSD = \frac{n - 1}{n} MSD$

$1 VSD = \frac{n - 1}{n} a cm$

$LC = 1 MSD - 1 VSD$

$LC = a - \frac{( n - 1 ) a}{n}$

$LC = \frac{a}{n} cm$

$LC = \frac{10 a}{n} mm$

The Sigma Insight: Vernier Calipers and Screw Gauge

Mastering the Vernier Calliper

Analyzing the Setup

The Master Equation

Final Calculation

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Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n VSD=(n−1) MSD

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1 VSD=nn−1​ MSD

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1 VSD=nn−1​a cm

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } LC=1 MSD−1 VSD

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } LC=a−n(n−1)a​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } LC=na​ cm

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } LC=n10a​ mm