In an experiment, the angles are required to be measured using an instrument. 29 divisions of the ma

Q: Why is the least count exactly $1 \text{ MSD} - 1 \text{ VSD}$?

The vernier scale works on the principle of small differences. The smallest measurable length is the tiny gap between a main scale mark and a vernier scale mark when they are slightly misaligned. This gap is exactly the difference between their respective division lengths.

Q: Can $1 \text{ VSD}$ ever be greater than $1 \text{ MSD}$?

Yes, in a 'retrograde' vernier scale, $1 \text{ VSD} > 1 \text{ MSD}$. However, in standard (direct) vernier scales like the one in this problem, the vernier divisions are slightly smaller, meaning $1 \text{ VSD} < 1 \text{ MSD}$.

Q: How do I convert degrees to minutes?

Just like hours and minutes on a clock, $1^\circ$ is divided into $60$ equal parts called minutes, denoted by $1'$. So, $1^\circ = 60'$. Therefore, $\frac{1}{60}^\circ$ is exactly $1'$.

Q: What if the question said $n$ VSD coincides with $n-1$ MSD?

That is the general formula for standard vernier calipers! The least count would always simplify to $\frac{1}{n} \text{ MSD}$. Here, $n=30$, so the least count is simply $\frac{1}{30} \text{ MSD}$.

Q: Is the least count always in the same unit as the main scale?

Initially, yes. The calculation gives you the least count in terms of the main scale's unit (degrees in this case). But you can, and often should, convert it to smaller units (like minutes or seconds) for convenience and standard reporting.

Question

In an experiment, the angles are required to be measured using an instrument. $29$ divisions of the main scale exactly coincide with the $30$ divisions of the vernier scale. If the smallest division of the main scale is half a degree ($= 0.5^\circ$), then the least count of the instrument is

Options:

(A) one minute

(B) half minute

(C) one degree

(D) half degree

Sigma Percentile · Accepted Answer

### The Anatomy of Precision

Imagine you are holding a highly precise scientific instrument designed to measure angles. At first glance, it looks like a standard protractor, but upon closer inspection, you notice a secondary, smaller scale sliding alongside the main one. This is the **Vernier Scale**, a brilliant mechanical invention that allows us to read measurements far more precisely than the naked eye could ever estimate between two tiny marks.

The core principle of any vernier instrument relies on a deliberate, slight mismatch between the divisions of the main scale and the divisions of the sliding vernier scale. By observing where the lines from both scales perfectly align, we can deduce the exact fractional measurement.

### The Mathematics of the Vernier Principle

To find the absolute limit of precision for this instrument—known as the **Least Count (LC)**—we must determine the smallest measurable difference. Mathematically, this is defined as the difference in length between one Main Scale Division (MSD) and one Vernier Scale Division (VSD):

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

The problem provides us with a crucial piece of information: $30$ divisions of the vernier scale exactly coincide with $29$ divisions of the main scale. We can write this relationship as an equation:

$$30 	ext{ VSD} = 29 	ext{ MSD}$$

From this coincidence, we can isolate the value of a single vernier division. By dividing both sides by $30$, we find:

$$1 	ext{ VSD} = \frac{29}{30} 	ext{ MSD}$$

### Executing the Calculation

Now, we bring this relationship back into our fundamental Least Count formula. Substituting the value of $1 	ext{ VSD}$, we get:

$$LC = 1 	ext{ MSD} - \frac{29}{30} 	ext{ MSD}$$

Factoring out the MSD, the subtraction becomes straightforward:

$$LC = \left(1 - \frac{29}{30}ight) 	ext{ MSD} = \frac{1}{30} 	ext{ MSD}$$

This tells us that the instrument can measure down to one-thirtieth of whatever a single main scale division happens to be.

### The Final Conversion

The problem states that the smallest division on the main scale is half a degree. Therefore:

$$1 	ext{ MSD} = 0.5^\circ = \frac{1}{2}^\circ$$

Let's plug this physical value into our simplified Least Count expression:

$$LC = \frac{1}{30} 	imes \frac{1}{2}^\circ = \frac{1}{60}^\circ$$

We have our answer, but it is in a fractional degree format. In angular measurements, just like on a clock, one full degree is divided into $60$ equal parts called **minutes** (denoted by the symbol $'$).

Since $1^\circ = 60'$, a fraction of $\frac{1}{60}^\circ$ is exactly equal to **$1$ minute**. Thus, the least count of this instrument is $1'$, making option (a) the correct choice. The elegance of the vernier scale lies in how it turns a simple mechanical mismatch into a powerful tool for extreme precision!

Select Answer:

Visualized Solution

Visualizing the Vernier Setup

The Least Count Formula

Establishing the Coincidence

Value of One VSD

Substituting into Least Count

Value of One MSD

Final Calculation

The Way Forward

The Sigma Insight: Vernier Calipers and Screw Gauge

The Anatomy of Precision

The Mathematics of the Vernier Principle

Executing the Calculation

The Final Conversion

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