The diameter of a sphere is measured as 100 ± 1 cm and the volume is calculated from this measurement. Estimate the percentage error in the volume calculation.

검증된 단계별 안내

First, recall the formula for the volume of a sphere: \( V = \frac{4}{3} \pi r^3 \), where \( r \) is the radius of the sphere.

Since the diameter is given as 100 cm, the radius \( r \) is half of the diameter: \( r = \frac{100}{2} = 50 \) cm.

The error in the radius \( \Delta r \) is half of the error in the diameter, so \( \Delta r = \frac{1}{2} = 0.5 \) cm.

Use the formula for the differential of the volume to estimate the error in the volume: \( \Delta V \approx 4 \pi r^2 \Delta r \). Substitute \( r = 50 \) cm and \( \Delta r = 0.5 \) cm into this formula.

Calculate the percentage error in the volume by dividing the estimated error in the volume \( \Delta V \) by the actual volume \( V \), and then multiply by 100 to convert it to a percentage: \( \text{Percentage Error} = \left( \frac{\Delta V}{V} \right) \times 100 \).

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주요 개념

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Volume of a Sphere

The volume of a sphere is calculated using the formula V = (4/3)πr³, where r is the radius of the sphere. Understanding this formula is crucial because the volume depends on the cube of the radius, which is half of the diameter. Any error in measuring the diameter will significantly affect the calculated volume.

Error Propagation

Error propagation refers to how uncertainties in measurements affect the calculated results. In this context, the error in the diameter measurement propagates to the volume calculation. Since volume is proportional to the cube of the radius, a small error in the diameter can lead to a larger error in the volume, necessitating careful estimation of percentage error.

Percentage Error

Percentage error quantifies the relative error in a measurement as a percentage of the true value. It is calculated by dividing the absolute error by the true value and multiplying by 100. In this problem, understanding how to compute the percentage error in the volume based on the error in the diameter measurement is essential for accurate estimation.