The density of a solid ball is to be determined in an experiment. The diameter of the ball is measur

Q: Why is the absolute error in the diameter taken as the least count?

In any measuring instrument, the maximum possible instrumental error in a single reading is equal to its least count. Since no other error information is given, we assume the absolute error $\Delta D$ is the least count.

Q: Why do we multiply the percentage error of the diameter by 3?

Density is mass divided by volume, and the volume of a sphere is proportional to the cube of its diameter ($D^3$). According to the rules of error propagation, when a quantity is raised to a power, its relative error is multiplied by that power.

Q: Can we use the radius instead of the diameter for the error calculation?

Yes, but you must be careful. The radius is $D/2$. The relative error in radius $\frac{\Delta r}{r}$ is exactly the same as the relative error in diameter $\frac{\Delta D}{D}$ because the constant factor of $1/2$ cancels out. It's safer to stick with the measured quantity, which is the diameter.

Q: What if the screw gauge had a zero error?

If a zero error was present, you would subtract it (with its proper sign) from the measured reading to get the true diameter. However, the absolute error $\Delta D$ would still remain the least count unless specified otherwise.

Q: Why do we add the errors instead of subtracting them?

When calculating the maximum possible error, we always add the relative errors, even if the quantities are being divided (like mass and volume). Errors always compound; they never cancel each other out.

Question

The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is $0.5 	ext{ mm}$ and there are $50$ divisions on the circular scale. The reading on the main scale is $2.5 	ext{ mm}$ and that on the circular scale is $20$ divisions.
If the measured mass of the ball has a relative error of $2\%$, the relative percentage error in the density is

Options:

(A) $0.9\%$

(B) $2.4\%$

(C) $3.1\%$

(D) $4.2\%$

Sigma Percentile · Accepted Answer

### The Setup: Measuring the Unseen

Imagine you are in a physics lab, holding a small solid ball. Your goal? To determine its density.

But here is the catch: every measurement you make has a tiny bit of uncertainty. The mass scale might be slightly off, and your screw gauge has a limit to how precisely it can measure.

In this problem, we are given the relative percentage error in the mass, which is **$2\%$**.

To find the error in the density, we first need to understand the error in our diameter measurement.

### Decoding the Screw Gauge

Before we can calculate the diameter, we must determine the **Least Count (LC)** of our screw gauge. The least count is the smallest value the instrument can measure accurately.

It is calculated by dividing the pitch by the total number of divisions on the circular scale.

$$ 	ext{LC} = \frac{	ext{Pitch}}{	ext{Circular Scale Divisions}} $$

Plugging in our values, we get:

$$ 	ext{LC} = \frac{0.5 	ext{ mm}}{50} = 0.01 	ext{ mm} $$

This $0.01 	ext{ mm}$ is crucial. It represents the absolute error ($\Delta D$) in our diameter measurement.

Now, let's calculate the actual diameter ($D$) using the main scale reading (MSR) and the circular scale reading (CSR).

$$ D = 	ext{MSR} + (	ext{CSR} 	imes 	ext{LC}) $$

$$ D = 2.5 	ext{ mm} + (20 	imes 0.01 	ext{ mm}) = 2.70 	ext{ mm} $$

### The Volume Trap: Why Powers Matter

Density ($ho$) is defined as mass ($m$) divided by volume ($V$). For a solid sphere, the volume is $\frac{4}{3}\pi r^3$.

Since the radius $r$ is half of the diameter $D$, we can rewrite the density formula entirely in terms of the diameter:

$$ ho = \frac{m}{\frac{4}{3}\pi \left(\frac{D}{2}ight)^3} = \frac{6m}{\pi D^3} $$

Here is where the magic of error propagation happens. When calculating the maximum relative percentage error, constants like $6$ and $\pi$ are ignored because they have no uncertainty.

However, the powers of the variables become multipliers for their respective relative errors. Since $D$ is cubed, its relative error is multiplied by $3$.

$$ \frac{\Delta ho}{ho} 	imes 100 = \left( \frac{\Delta m}{m} ight) 	imes 100 + 3 \left( \frac{\Delta D}{D} ight) 	imes 100 $$

### Bringing It All Together

We already know the percentage error in mass is $2\%$.

For the diameter, the absolute error $\Delta D$ is $0.01 	ext{ mm}$, and the measured diameter $D$ is $2.70 	ext{ mm}$. Let's substitute these into our error equation:

$$ \frac{\Delta ho}{ho} 	imes 100 = 2\% + 3 \left( \frac{0.01}{2.70} ight) 	imes 100 $$

Let's simplify the diameter's error term:

$$ 3 	imes \left( \frac{1}{270} ight) 	imes 100 = \frac{300}{270}\% = \frac{10}{9}\% \approx 1.11\% $$

Finally, we add this to the mass error:

$$ \frac{\Delta ho}{ho} 	imes 100 = 2\% + 1.11\% = 3.11\% $$

The total relative percentage error in the density is **$3.11\%$**.

This perfectly matches option (c). Notice how the seemingly tiny error in the diameter measurement was magnified by a factor of three because of the volume formula. Always respect the powers in your equations!

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Visualized Solution

$Measurement Setup$

$Least Count Formula$

$Calculating Least Count$

$Diameter Formula$

$Calculating Diameter$

$Density Formula$

$Percentage Error Formula$

$Substituting Error Values$

$Final Calculation$

$Conclusion$

The Sigma Insight: Errors in Measurement

The Setup

Decoding the Screw Gauge

The Volume Trap

Bringing It All Together

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Visualized Solution

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

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

.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; } Diameter Formula

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

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

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

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

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

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