The density of a solid metal sphere is determined by measuring its mass and its diameter. The maximu

Q: Why do we add the errors even though diameter is in the denominator?

When calculating the **maximum possible error**, we must assume the worst-case scenario where all individual errors compound together. Therefore, we always add the absolute values of the relative errors, regardless of whether the variable is in the numerator or denominator.

Q: Why did we change radius to diameter in the volume formula?

The problem explicitly gives us the percentage error in the **diameter**, not the radius. While $\frac{\Delta r}{r} = \frac{\Delta d}{d}$, it is safer and more direct to express the formula entirely in terms of the measured quantities to avoid any confusion during error propagation.

Q: What happens to the constants $\frac{4}{3}$ and $\pi$ during error analysis?

Exact mathematical constants like $\frac{4}{3}$, $\pi$, or $6$ have absolute certainty. Their measured error is zero ($\Delta k = 0$). Therefore, they disappear when we take the relative error of the equation.

Q: Why is the power of diameter multiplied by 3 in the error equation?

In the density formula, diameter is cubed ($d^3$) because volume is a three-dimensional quantity. According to the power rule of error propagation, if a quantity is raised to a power $n$, its relative error is multiplied by $n$.

Q: I got 10.5 as the answer. Why is it 1050?

Read the question carefully! The error is given as $\left(\frac{x}{100}\right)\%$. You calculated the total error as $10.5\%$. Equating them gives $\frac{x}{100} = 10.5$, which means $x = 10.5 \times 100 = 1050$. It's a classic trap to test your attention to detail.

Question

The density of a solid metal sphere is determined by measuring its mass and its diameter. The maximum error in the density of the sphere is $\left(\frac{x}{100}\right)\%$. If the relative errors in measuring the mass and the diameter are $6.0\%$ and $1.5\%$ respectively, the value of $x$ is ....... .

Sigma Percentile · Accepted Answer

The journey to solving this problem begins in the laboratory. Imagine you are handed a solid metal sphere and asked to determine its density. You can't measure density directly; instead, you must measure its mass and its physical dimensions.

In this scenario, you measure the mass $M$ and the diameter $d$. However, every measurement comes with a tiny bit of uncertainty or error. The core of this problem is figuring out how those small errors in mass and diameter combine to create a larger error in your final calculated density.

### The Master Equation

To connect our measurements to density, we start with the fundamental definition:
$$ ho = \frac{	ext{Mass}}{	ext{Volume}} $$

We know the volume of a sphere is given by $V = \frac{4}{3}\pi r^3$. But our instrument measured the diameter $d$, not the radius $r$. Since $r = \frac{d}{2}$, we can rewrite the volume entirely in terms of diameter:
$$ V = \frac{4}{3}\pi \left(\frac{d}{2}ight)^3 = \frac{\pi d^3}{6} $$

Substituting this back into our density equation, we get our master formula:
$$ ho = \frac{M}{\frac{\pi d^3}{6}} = \frac{6M}{\pi d^3} $$

### Unleashing Error Propagation

Now, we apply the principles of error analysis. When quantities are multiplied or divided, their **relative errors add up**. Furthermore, if a quantity is raised to a power, its relative error is multiplied by that power.

It is crucial to remember that exact constants like $6$ and $\pi$ have absolute certainty, meaning their error is zero. They drop out of our error equation.

Applying the power rule to our master formula, the maximum percentage error in density becomes:
$$ \frac{\Delta ho}{ho} 	imes 100 = \left( \frac{\Delta M}{M} 	imes 100 ight) + 3 \left( \frac{\Delta d}{d} 	imes 100 ight) $$

Notice that even though diameter is in the denominator, we **add** its error. In error analysis, we always assume the worst-case scenario where errors compound, so we take the absolute values.

### Final Calculation

The problem provides us with the percentage errors for our measurements:
- Percentage error in mass: $\frac{\Delta M}{M} 	imes 100 = 6.0\%$
- Percentage error in diameter: $\frac{\Delta d}{d} 	imes 100 = 1.5\%$

Let's substitute these values into our error equation:
$$ \frac{\Delta ho}{ho} 	imes 100 = 6.0\% + 3(1.5\%) $$

Performing the arithmetic:
$$ \frac{\Delta ho}{ho} 	imes 100 = 6.0\% + 4.5\% = 10.5\% $$

So, the maximum percentage error in the density is $10.5\%$.

But wait, the question sets a final trap! It states that the maximum error is $\left(\frac{x}{100}ight)\%$. To find $x$, we simply equate our result to this expression:
$$ \frac{x}{100} = 10.5 $$

Multiplying both sides by $100$, we arrive at our final destination:
$$ x = 1050 $$

And there we have it! By carefully tracking our variables and respecting the rules of error propagation, we've successfully navigated the problem.

Enter Numerical Value:

Visualized Solution

Visualizing the Sphere

Density Formula

Error Propagation

Substituting Given Values

Calculating Total Error

Finding $x$

The Sigma Insight: Errors in Measurement

The Master Equation

Unleashing Error Propagation

Final Calculation

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