The least count of the main scale of a vernier callipers is 1 mm. Its vernier scale is divided into

Q: Why do we subtract the zero error? Shouldn't we add it if it's an error?

Think of a weighing scale that shows $2 \text{ kg}$ when there is nothing on it. If you step on it and it shows $62 \text{ kg}$, your actual weight is $60 \text{ kg}$. You have to subtract the initial 'extra' reading. A positive zero error means the instrument is over-reading, so we must subtract it to find the true value.

Q: How did we know the Main Scale Reading (MSR) was exactly $3.1 \text{ cm}$?

The MSR is always the main scale mark that is immediately to the left of the vernier scale's zero mark. Since the problem states the vernier zero lies between $3.1 \text{ cm}$ and $3.2 \text{ cm}$, the mark immediately to its left is $3.1 \text{ cm}$.

Q: What if the zero of the vernier scale was to the left of the main scale zero?

That would be a Negative Zero Error. It means the instrument is under-reading. The formula $\text{True Value} = \text{Observed} - \text{Zero Error}$ still applies, but since the error itself is negative, subtracting a negative number effectively adds it to the observed reading.

Q: Why is $1 \text{ VSD}$ equal to $0.9 \text{ mm}$?

The problem states that $10$ divisions on the vernier scale occupy the exact same physical length as $9$ divisions on the main scale. Since $1 \text{ MSD} = 1 \text{ mm}$, $9 \text{ MSD} = 9 \text{ mm}$. Therefore, $10 \text{ VSD} = 9 \text{ mm}$, which means $1 \text{ VSD} = \frac{9}{10} \text{ mm} = 0.9 \text{ mm}$.

Q: Can the least count be calculated differently?

Yes! A faster formula for standard vernier calipers is $\text{LC} = \frac{1 \text{ MSD}}{\text{Total number of divisions on Vernier Scale}}$. In this case, $\text{LC} = \frac{1 \text{ mm}}{10} = 0.1 \text{ mm} = 0.01 \text{ cm}$. This works perfectly when $n \text{ VSD} = (n-1) \text{ MSD}$.

Question

The least count of the main scale of a vernier callipers is 1 mm. Its vernier scale is divided into 10th division and coincide with 9th division of the main scale. When jaws are touching each other, the 7th division of vernier scale coincides with a division of main scale and the zero of vernier scale is lying right side of the zero of main scale. When this vernier is used to measure length of a cylinder the zero of the vernier scale lies between 3.1 cm and 3.2 cm and 4th VSD coincides with a main scale division. The length of the cylinder is (VSD is vernier scale division)

Options:

(A) 3.2 cm

(B) 2.99 cm

(C) 3.07 cm

(D) 3.21 cm

Sigma Percentile · Accepted Answer

### The Anatomy of the Vernier Caliper

Imagine you are holding a precision instrument designed to measure lengths down to a fraction of a millimeter. The Vernier Caliper achieves this magic not with complex electronics, but through pure, elegant geometry. It consists of two scales: a fixed **Main Scale** and a sliding **Vernier Scale**.

In our specific problem, we are given the fundamental DNA of this caliper: the main scale is marked in millimeters, meaning one Main Scale Division ($1 	ext{ MSD}$) is exactly $1 	ext{ mm}$. The sliding vernier scale has $10$ divisions, but these $10$ divisions are squeezed into the exact same length as $9$ divisions on the main scale.

This slight mismatch is the secret to its precision. Mathematically, we can write:
$$10 	ext{ VSD} = 9 	ext{ MSD}$$
$$1 	ext{ VSD} = \frac{9}{10} 	ext{ MSD} = 0.9 	ext{ mm}$$

### Uncovering the Least Count

The **Least Count (LC)** is the smallest measurement the instrument can accurately resolve. It is defined as the difference in length between one main scale division and one vernier scale division.

Let's calculate it:
$$	ext{LC} = 1 	ext{ MSD} - 1 	ext{ VSD}$$
$$	ext{LC} = 1 	ext{ mm} - 0.9 	ext{ mm} = 0.1 	ext{ mm}$$

Since our final options are in centimeters, let's convert this right away. 
$$	ext{LC} = 0.01 	ext{ cm}$$

### The Hidden Flaw: Zero Error

Before we measure anything, we must calibrate our instrument. When the jaws of the caliper are completely closed, the zero mark of the vernier scale *should* align perfectly with the zero mark of the main scale.

However, the problem states that when the jaws touch, the zero of the vernier scale is lying to the **right** of the main scale's zero. This means the caliper is already reading a small positive value even when holding nothing! This is called a **Positive Zero Error**.

To find exactly how much this error is, we look for the vernier division that perfectly aligns with any main scale mark. The problem tells us it's the $7^{	ext{th}}$ division.

$$	ext{Zero Error} = + (	ext{Coinciding Division} 	imes 	ext{LC})$$
$$	ext{Zero Error} = + (7 	imes 0.01 	ext{ cm}) = + 0.07 	ext{ cm}$$

### The Final Measurement

Now, we place the cylinder between the jaws. The zero of the vernier scale slides past the $3.1 	ext{ cm}$ mark but doesn't quite reach $3.2 	ext{ cm}$. This locks in our **Main Scale Reading (MSR)** at $3.1 	ext{ cm}$.

To find the fractional part, we again look for the coinciding division. This time, the $4^{	ext{th}}$ Vernier Scale Division (VSD) aligns perfectly.

Let's calculate the raw, **Observed Value**:
$$	ext{Observed Value} = 	ext{MSR} + (	ext{VSR} 	imes 	ext{LC})$$
$$	ext{Observed Value} = 3.1 	ext{ cm} + (4 	imes 0.01 	ext{ cm}) = 3.14 	ext{ cm}$$

I know it is tempting to stop here, but remember our hidden flaw! The caliper is inherently adding an extra $0.07 	ext{ cm}$ to every measurement. To find the **True Value**, we must subtract this positive zero error from our observed reading.

$$	ext{True Value} = 	ext{Observed Value} - 	ext{Zero Error}$$
$$	ext{True Value} = 3.14 	ext{ cm} - (+ 0.07 	ext{ cm})$$
$$	ext{True Value} = 3.07 	ext{ cm}$$

And there we have it. The exact length of the cylinder is **$3.07 	ext{ cm}$**, which corresponds to option (c).

Select Answer:

Visualized Solution

$The Vernier Setup$

$Calculating Least Count$

$Identifying Zero Error$

$Calculating Zero Error$

$Taking the Measurement$

$Calculating Observed Value$

$Finding the True Length$

$The Way Forward$

The Sigma Insight: Vernier Calipers and Screw Gauge

The Anatomy of the Vernier Caliper

Uncovering the Least Count

The Hidden Flaw

The Final Measurement

Similar Questions

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

A vernier calipers has $1 mm$ marks on the main scale. It has $20$ equal divisions on the vernier scale which match with $16$ main scale divisions. For this vernier calipers, the least count is

Figure 1 shows the configuration of main scale and Vernier scale before measurement. Fig. 2 shows the configuration corresponding to the measurement of diameter $D$ of a tube. The measured value of $D$ is:

$N$ divisions on the main scale of a vernier calipers coincide with $(N + 1)$ divisions on the vernier scale. If each division on the main scale is of $a$ units, determine the least count of instrument.

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

Select Answer:

Visualized Solution

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

.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; } Identifying Zero Error

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding the True Length

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