The dimensions of a cone are measured using a scale with a least count of 2 mm. The diameter of the

Q: Why do we use diameter instead of radius in the volume formula?

The problem states that the *diameter* was measured directly. If we use radius $r = d/2$, we must remember that $\Delta r = \Delta d / 2$. Using diameter directly in the formula $V = \frac{\pi}{12} d^2 h$ prevents confusion and keeps the fractional error $\frac{\Delta d}{d}$ intact.

Q: Why do we multiply the fractional error of the diameter by 2?

In the volume formula, the diameter is squared ($d^2$). According to the rules of error propagation, the power of a measured quantity becomes a multiplier for its relative error. This is because errors can compound when a value is multiplied by itself.

Q: Does the constant $\frac{\pi}{12}$ affect the percentage error?

No. Constants are exact values with zero uncertainty. When calculating relative or percentage errors, constant multipliers drop out of the error equation entirely.

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

We are calculating the *maximum* possible error. Even if the actual errors might cancel each other out in reality, error analysis requires us to consider the worst-case scenario, which means we always add the absolute values of the relative errors.

Q: What if the least count was given in centimeters instead of millimeters?

If the least count was already in centimeters, we wouldn't need to convert it. The crucial step is ensuring that the absolute error ($\Delta d$, $\Delta h$) and the measured value ($d$, $h$) are in the exact same units before dividing them to find the fractional error.

Question

The dimensions of a cone are measured using a scale with a least count of $2	ext{ mm}$. The diameter of the base and the height are both measured to be $20.0	ext{ cm}$. The maximum percentage error in the determination of the volume is -

Sigma Percentile · Accepted Answer

Measurement is never perfect; it is always accompanied by a shadow of uncertainty. In this problem, we are tasked with finding the maximum percentage error in the volume of a cone, given the measurements of its base diameter and height. Let's break down the physics and mathematics behind this error analysis.

### Analyzing the Setup

We are given a cone whose base diameter $d$ and height $h$ are both measured to be $20.0	ext{ cm}$. The instrument used for these measurements has a least count of $2	ext{ mm}$.

The **least count** of a measuring instrument is the smallest value it can measure accurately. In error analysis, this least count represents the maximum possible absolute error in any single measurement. Therefore, the absolute error in both the diameter and the height is:

$$\Delta d = \Delta h = 2	ext{ mm}$$

Before we proceed, it is crucial to ensure all our units are consistent. Let's convert the absolute error from millimeters to centimeters:

$$\Delta d = \Delta h = 0.2	ext{ cm}$$

### The Master Equation

To find the error in the volume, we first need the formula for the volume of a cone. The standard formula is $V = \frac{1}{3} \pi r^2 h$. However, we measured the diameter $d$, not the radius $r$. To avoid unnecessary complications in error propagation, it is always best to express the formula in terms of the directly measured quantities. Substituting $r = \frac{d}{2}$, we get:

$$V = \frac{1}{3} \pi \left(\frac{d}{2}ight)^2 h = \frac{\pi}{12} d^2 h$$

Now, we apply the rules of error propagation for products and powers. When quantities are multiplied or divided, their fractional (or relative) errors add up. Furthermore, if a quantity is raised to a power, that power becomes a multiplier for its fractional error. Constants, like $\frac{\pi}{12}$, have zero uncertainty and drop out of the error equation. Thus, the maximum fractional error in volume is:

$$\frac{\Delta V}{V} = 2 \frac{\Delta d}{d} + \frac{\Delta h}{h}$$

### Final Calculation

Now, we substitute our known values into the error equation. The fractional error for both the diameter and the height is:

$$\frac{\Delta d}{d} = \frac{\Delta h}{h} = \frac{0.2}{20.0} = \frac{1}{100}$$

Plugging these into our master error equation, we get:

$$\frac{\Delta V}{V} = 2 \left(\frac{1}{100}ight) + \frac{1}{100} = \frac{3}{100}$$

To find the **percentage error**, we simply multiply the fractional error by 100:

$$	ext{Percentage Error} = \frac{\Delta V}{V} 	imes 100 = \left(\frac{3}{100}ight) 	imes 100 = 3\%$$

The maximum percentage error in the determination of the volume is exactly **3%**.

Animated Solution for Physics - Physics and Measurement: The dimensions of a cone are measured using a scale with a least count of $2 mm$ . The diameter of the base and the height are both measured to be $20.0 cm$ . The maximum percentage error in the determination of the volume is -

Enter Numerical Value:

Visualized Solution

$Visualizing the Measurements$

$Volume Formula$

$Error Propagation Rule$

$Calculating Fractional Errors$

$Percentage Error Setup$

$Final Calculation$

$Conceptual Takeaway$

The Sigma Insight: Errors in Measurement

Analyzing the Setup

The Master Equation

Final Calculation

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Enter Numerical Value:

Visualized Solution

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

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

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

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

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

.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; } Conceptual Takeaway