The density of a material in the shape of a cube is determined by measuring three sides of the cube

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

To find the *maximum* possible error, we always add the absolute values of the relative errors. If we subtracted them, we might accidentally calculate a case where the errors cancel each other out, which doesn't give us the worst-case scenario.

Q: Why is the error in length multiplied by 3?

Because the volume of a cube is length cubed ($L^3$). In error analysis, when a measured quantity is raised to a power $n$, its relative error is multiplied by $n$ in the final calculation.

Q: What if the error was given as absolute error instead of percentage?

You would first convert the absolute error ($\Delta x$) into a percentage error by dividing it by the measured value ($x$) and multiplying by $100$, i.e., $(\frac{\Delta x}{x}) \times 100$.

Q: Does the shape of the object matter for the error formula?

Yes, because the volume formula changes. For a sphere, the volume is $\frac{4}{3}\pi R^3$, so the error is still $3$ times the error in radius. But for a cylinder ($V = \pi R^2 H$), the error would be $2$ times the error in radius plus the error in height.

Q: Are constants like $\pi$ or pure numbers included in error calculation?

No, exact mathematical constants like $\pi$, $e$, or pure fractions like $\frac{4}{3}$ have absolutely zero uncertainty, so they contribute nothing to the final error.

Question

The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively $1.5\%$ and $1\%$, the maximum error in determining the density is

Options:

(A) $2.5\%$

(B) $3.5\%$

(C) $4.5\%$

(D) $6\%$

Sigma Percentile · Accepted Answer

In the world of experimental physics, no measurement is absolutely perfect. Every time we use a ruler, a weighing scale, or a stopwatch, a tiny bit of uncertainty creeps into our data. But what happens when we use these slightly imperfect measurements to calculate something else, like density? This is where the beautiful mathematics of **error propagation** comes into play.

### Analyzing the Setup

Imagine you are holding a solid block of material shaped perfectly like a cube. To determine its density, you need two fundamental properties: its mass ($M$) and its volume ($V$).

The density $ho$ is defined simply as mass per unit volume:

$$ho = \frac{M}{V}$$

Since our object is a cube with side length $L$, its volume is the cube of its side length ($V = L^3$). Substituting this into our density equation gives us our master formula for this experiment:

$$ho = \frac{M}{L^3}$$

### The Master Equation of Errors

Now, here is the catch. We are given the relative errors (which are essentially percentage errors when expressed with a $\%$ sign) for both the mass and the length. The error in mass is $1.5\%$, and the error in length is $1\%$.

When quantities are multiplied or divided, their **relative errors add up** to give the maximum possible relative error in the final result. Furthermore, if a quantity is raised to a power, that power becomes a multiplier for its relative error.

Applying these rules to our density equation $ho = M \cdot L^{-3}$, we get the maximum percentage error formula:

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

Notice how the power of $3$ from the volume ($L^3$) drops down to multiply the error in length. This means that any small mistake in measuring the length will be magnified three times in the final density calculation!

### Final Calculation

We are now ready to plug in our numbers. We know the percentage error in mass is $1.5\%$, and the percentage error in length is $1\%$.

Substituting these into our master equation:

$$\frac{\Delta ho}{ho} 	imes 100 = 1.5\% + 3(1\%)$$

$$\frac{\Delta ho}{ho} 	imes 100 = 1.5\% + 3\%$$

$$\frac{\Delta ho}{ho} 	imes 100 = 4.5\%$$

And there we have it! The maximum percentage error in determining the density of the cube is **$4.5\%$**. This problem beautifully illustrates why precision is so critical when measuring dimensions that will be raised to higher powers in formulas.

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Visualized Solution

$Density of a Cube$

$Volume of Cube$

$Maximum Percentage Error$

$Substituting Values$

$Final Calculation$

$What if it was a Sphere?$

The Sigma Insight: Errors in Measurement

Analyzing the Setup

The Master Equation of Errors

Final Calculation

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Visualized Solution

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting 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; } What if it was a Sphere?