Figure 1 shows the configuration of main scale and Vernier scale before measurement. Fig. 2 shows th

Q: Why isn't the least count just $0.01\text{ cm}$ like it always is?

Because this is a custom-calibrated Vernier caliper! In a standard caliper, $10\text{ VSD} = 9\text{ MSD}$. But if you look at Figure 1, $10\text{ VSD}$ perfectly aligns with $7\text{ MSD}$. You must always calculate the least count based on the given visual evidence.

Q: How do I determine the Main Scale Reading (MSR)?

Look at the zero mark on the Vernier scale. See which main scale mark it has just passed. In Figure 2, the Vernier zero is between $0.1\text{ cm}$ and $0.2\text{ cm}$, so the MSR is $0.1\text{ cm}$.

Q: What does the solution mean by 'Reading = $2\text{ MSD} - \text{VSD}$'?

This is a purely geometric way to read the scale. Look at Figure 2: the $1^{\text{st}}$ Vernier mark aligns perfectly with the $2^{\text{nd}}$ main scale mark. The distance we want to measure is from the main scale zero to the Vernier zero. Geometrically, this distance is exactly $2\text{ MSD}$ minus the length of $1\text{ VSD}$.

Q: Can I always use the formula $\text{MSR} + \text{VSR} \times \text{LC}$?

Yes! The formula $\text{Total Reading} = \text{MSR} + (\text{VSR} \times \text{LC})$ is universal and will always work, provided you have calculated the correct Least Count for that specific instrument.

Q: Why do we subtract $1\text{ VSD}$ from $1\text{ MSD}$ to get the least count?

The least count is the smallest difference in length that the instrument can resolve. When the Vernier zero moves past the main scale zero by exactly one least count, the $1^{\text{st}}$ Vernier division will align with the $1^{\text{st}}$ main scale division. The distance it moved is exactly $1\text{ MSD} - 1\text{ VSD}$.

Question

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:

Options:

(A) 0.12 cm

(B) 0.11 cm

(C) 0.13 cm

(D) 0.14 cm

Sigma Percentile · Accepted Answer

This problem is a beautiful reminder of why we must never blindly trust memorized formulas in physics. It tests your fundamental understanding of how a Vernier caliper is constructed and calibrated.

### Decoding the Main Scale
Let's start by looking at the main scale. Between the $0$ and $1	ext{ cm}$ marks, there are exactly $10$ divisions. This is standard. It tells us that one Main Scale Division (MSD) is simply:

$$1	ext{ MSD} = \frac{1	ext{ cm}}{10} = 0.1	ext{ cm}$$

### The Vernier Trap (Figure 1)
Now, look at Figure 1. This figure shows the zero-error state, where the jaws of the caliper are fully closed. In a standard Vernier caliper, $10$ Vernier Scale Divisions (VSD) coincide with $9$ Main Scale Divisions. But look closely at this specific instrument!

The $10^{	ext{th}}$ division of the Vernier scale aligns perfectly with the $7^{	ext{th}}$ division of the main scale. This is the catch! We must calculate the length of one VSD based on this specific geometry:

$$10	ext{ VSD} = 7	ext{ MSD}$$

$$1	ext{ VSD} = \frac{7}{10}	ext{ MSD} = 0.7 	imes 0.1	ext{ cm} = 0.07	ext{ cm}$$

With this, we can find the true Least Count (LC) of this instrument. The least count is the smallest measurable difference, which is the difference between one main scale division and one Vernier scale division:

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

$$	ext{LC} = 0.1	ext{ cm} - 0.07	ext{ cm} = 0.03	ext{ cm}$$

### Taking the Measurement (Figure 2)
Now we move to Figure 2 to take the actual measurement of the tube's diameter.

First, we find the Main Scale Reading (MSR). We look at where the zero mark of the Vernier scale lands. It has crossed the $1^{	ext{st}}$ mark ($0.1	ext{ cm}$) but hasn't reached the $2^{	ext{nd}}$ mark ($0.2	ext{ cm}$). Therefore:

$$	ext{MSR} = 0.1	ext{ cm}$$

Next, we find the Vernier Scale Reading (VSR) by identifying which Vernier division perfectly coincides with any main scale division. Scanning across, we see that the very $1^{	ext{st}}$ Vernier division aligns perfectly with the $2^{	ext{nd}}$ main scale mark. Thus:

$$	ext{VSR} = 1$$

### The Final Calculation
We now plug these values into our universal reading formula:

$$	ext{Total Reading} = 	ext{MSR} + (	ext{VSR} 	imes 	ext{LC})$$

$$	ext{Total Reading} = 0.1	ext{ cm} + (1 	imes 0.03	ext{ cm}) = 0.13	ext{ cm}$$

**Alternative Geometric Method:**
If you look at Figure 2 purely geometrically, the distance from the main scale zero to the Vernier zero is exactly the distance to the $2^{	ext{nd}}$ main scale mark minus the length of $1$ Vernier division.

$$	ext{Reading} = 2	ext{ MSD} - 1	ext{ VSD}$$

$$	ext{Reading} = 0.2	ext{ cm} - 0.07	ext{ cm} = 0.13	ext{ cm}$$

Both methods yield the exact same result, proving the elegance of the Vernier principle. The correct option is **(C)**.

Animated Solution for Physics - Physics and Measurement: 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:

Select Answer:

Visualized Solution

$M ain S c a l e C a l ib r a t i o n$

$V er ni er S c a l e C a l ib r a t i o n$

$V a l u e o f 1 V S D$

$L e a s t C o u n t (L C)$

$M ain S c a l e R e a d in g$

$C o in c i d in g D i v i s i o n$

$T o t a l R e a d in g S e t u p$

$F ina l A n s w er$

$T h e W a y F or w a r d$

The Sigma Insight: Vernier Calipers and Screw Gauge

Decoding the Main Scale Let's start by looking at the main scale

The Vernier Trap (Figure 1) Now, look at Figure 1

Taking the Measurement (Figure 2)

The Final Calculation

Similar Questions

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

Select Answer:

Visualized Solution

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Value of 1 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; } Main Scale Reading

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

.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 Answer

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