The edge of a cube is measured using a vernier caliper. (9 divisions of the main scale is equal to 1

Q: Why is the least count $0.1 \text{ mm}$?

Because $10$ vernier divisions equal $9$ main scale divisions ($9 \text{ mm}$). So, $1 \text{ VSD} = 0.9 \text{ mm}$. The least count is $1 \text{ MSD} - 1 \text{ VSD} = 1 \text{ mm} - 0.9 \text{ mm} = 0.1 \text{ mm}$.

Q: Why do we round the volume to $3$ significant figures?

The edge length ($1.01 \text{ cm}$) has $3$ significant figures. When multiplying or dividing, the result must have the same number of significant figures as the measurement with the fewest significant figures.

Q: Why is the final density rounded to $3$ significant figures?

The mass has $4$ significant figures ($2.736$) and the volume has $3$ ($1.03$). The rule for division dictates that the result must be rounded to the lesser number of significant figures, which is $3$.

Q: What if the $2^{\text{nd}}$ vernier division coincided?

Then the vernier scale reading would be $2$, and the edge length would be $10 \text{ mm} + 2 \times 0.1 \text{ mm} = 10.2 \text{ mm}$.

Q: Can we calculate density without rounding the volume first?

Yes, it's actually better practice to keep intermediate values unrounded to avoid compounding errors. If you use $1.030301 \text{ cm}^3$, the density is $2.6555... \text{ g/cm}^3$, which still rounds to $2.66 \text{ g/cm}^3$ for $3$ significant figures.

Question

The edge of a cube is measured using a vernier caliper. (9 divisions of the main scale is equal to 10 divisions of vernier scale and 1 main scale division is 1 mm). The main scale division reading is 10 and 1 division of vernier scale was found to be coinciding with the main scale. The mass of the cube is $2.736	ext{ g}$. Calculate the density in $	ext{g/cm}^3$ upto correct significant figures.

Sigma Percentile · Accepted Answer

### The Beauty of Precision

Imagine you are a scientist trying to determine the exact density of a mysterious cube. You have a vernier caliper and a highly precise weighing scale. The mass is easy, but measuring the edge length requires a keen eye and an understanding of how a vernier caliper works. This problem is a beautiful blend of instrumentation and the golden rules of significant figures. Let's dive into it!

### Decoding the Vernier Caliper

The first step in using any vernier caliper is to determine its least count. The least count is the smallest measurement the instrument can accurately make. We are given a crucial piece of information: $9$ divisions of the main scale are equal to $10$ divisions of the vernier scale.

Since $1 	ext{ MSD} = 1 	ext{ mm}$, we can easily find the length of one vernier scale division:

$$1 	ext{ VSD} = \frac{9}{10} 	ext{ MSD} = 0.9 	ext{ mm}$$

The least count is simply the difference between one main scale division and one vernier scale division:

$$	ext{Least Count (LC)} = 1 	ext{ MSD} - 1 	ext{ VSD} = 1 	ext{ mm} - 0.9 	ext{ mm} = 0.1 	ext{ mm}$$

### Measuring the Edge Length

Now that we know our instrument's precision, let's read the measurement. The main scale reading (MSR) is $10 	ext{ mm}$. We are also told that the $1^{	ext{st}}$ division of the vernier scale coincides perfectly with a main scale mark.

The total reading for the edge length $a$ is given by:

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

Substituting our values:

$$a = 10 	ext{ mm} + (1 	imes 0.1 	ext{ mm}) = 10.1 	ext{ mm}$$

To make our future calculations easier, let's convert this to centimeters:

$$a = 1.01 	ext{ cm}$$

Notice that this measurement has exactly **three significant figures**. This tiny detail will dictate the precision of our final answer!

### The Volume and the Golden Rule of Significant Figures

To find the density, we first need the volume of the cube. The volume $V$ is the cube of the edge length:

$$V = a^3 = (1.01 	ext{ cm})^3 = 1.030301 	ext{ cm}^3$$

But wait! We cannot claim this level of precision. The golden rule of significant figures in multiplication states that our result can only have as many significant figures as the measurement with the fewest significant figures. Since our edge length has $3$ significant figures, our volume must also be rounded to $3$ significant figures:

$$V = 1.03 	ext{ cm}^3$$

### The Final Density

Finally, we calculate the density $ho$ by dividing the mass by the volume. The mass is given as $2.736 	ext{ g}$, which has $4$ significant figures.

$$ho = \frac{	ext{Mass}}{	ext{Volume}} = \frac{2.736 	ext{ g}}{1.03 	ext{ cm}^3} = 2.6563... 	ext{ g/cm}^3$$

Once again, we must apply the rule of significant figures for division. We are dividing a number with $4$ significant figures by a number with $3$ significant figures. Our final answer must be restricted to $3$ significant figures.

Rounding $2.6563...$ to $3$ significant figures, we get:

$$ho = 2.66 	ext{ g/cm}^3$$

And there we have it! By carefully reading our instrument and strictly adhering to the rules of significant figures, we've found the true density of the cube.

Enter Numerical Value:

Visualized Solution

$Least Count$

$Edge Length Formula$

$Edge Length Calculation$

$Volume Calculation$

$Density Calculation$

$Final Answer$

The Sigma Insight: Significant Figures

The Beauty of Precision

Decoding the Vernier Caliper

Measuring the Edge Length

The Volume and the Golden Rule of Significant Figures

The Final Density

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Visualized Solution

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

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

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

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

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

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