A Sphere Contracts in Volume by 0.02 Percent
Imagine a perfectly round sphere, like a marble or a ball bearing. Now imagine that it shrinks in size, but only very slightly. Let's say it contracts in volume by 0.02 percent. What does this mean, and how does it affect the sphere's other properties, like its radius and surface area?
Understanding the Contraction
A 0.02 percent volume contraction means the sphere's volume has decreased by 0.02% of its original volume. This is a very small change, but it still has consequences for the sphere's other dimensions.
The Impact on Radius and Surface Area
To understand the impact on the radius and surface area, we need to consider the relationships between these quantities and the volume of a sphere:
- Volume of a sphere: V = (4/3)πr³
- Surface area of a sphere: A = 4πr²
Where:
- V is the volume
- A is the surface area
- r is the radius
Since the volume has decreased, the radius must also decrease. However, the relationship between volume and radius is cubic, meaning a small change in volume leads to a proportionally smaller change in radius.
Similarly, the surface area is affected by the change in radius, but the relationship is quadratic. This means the change in surface area will be even smaller than the change in radius.
Calculating the Changes
Let's assume the initial volume of the sphere is V₀ and the final volume is V₁. We know that:
V₁ = V₀ - 0.02%V₀ = 0.9998V₀
To find the new radius (r₁), we can use the volume formula:
0.9998V₀ = (4/3)πr₁³
Solving for r₁ gives us:
r₁ = (0.9998V₀/(4/3)π)^(1/3) = 0.9998^(1/3)r₀
This shows that the radius has also decreased by a small amount, but less than 0.02 percent.
To find the new surface area (A₁), we can use the surface area formula:
A₁ = 4πr₁² = 4π(0.9998^(1/3)r₀)² = 0.9998^(2/3)A₀
This shows that the surface area has decreased even less than the radius, by about 0.013 percent.
Conclusion
A 0.02 percent contraction in volume results in a small decrease in the sphere's radius and surface area. While the changes are proportionally smaller than the change in volume, they are still noticeable, especially for very precise measurements.
