For a material to float on the surface of water, the material must have a density less than that of water \((1.0 \mathrm{~g} / \mathrm{mL})\) and must not react with the water or dissolve in it. A spherical ball has a radius of \(0.50 \mathrm{~cm}\) and weighs \(2.0 \mathrm{~g} .\) Will this ball float or sink when placed in water? (Note: Volume of a sphere \(=\frac{4}{3} \pi r^{3} .\) )

Short Answer

Expert verified
The spherical ball has a volume of approximately \(1.57 \mathrm{cm^3}\) and a density of approximately \(1.27 \mathrm{g/cm^3}\). Since the density of the sphere is greater than the density of water \((1.0 \mathrm{g/cm^3})\), the sphere will sink when placed in water.

Step by step solution

01

Find the Volume of the Sphere

We know the radius of the spherical ball (r) is 0.50 cm, and the formula to calculate the volume of a sphere is \(\frac{4}{3} \pi r^3\). Substitute the given value of radius in the formula to find the volume: \[V = \frac{4}{3} \pi (0.50)^3\]
02

Calculate the Volume

Now, let's compute the volume: \[V = \frac{4}{3} \pi (0.125)\] \[V = \frac{1}{2} \pi\] \[V \approx 1.57 \mathrm{cm^3}\] So, the volume of the sphere is approximately 1.57 cm³.
03

Calculate the Density of the Sphere

Now that we have the volume of the sphere, we can find its density. The density is the mass of an object divided by its volume. The mass of the sphere (m) is 2.0 g, and the volume (V) is 1.57 cm³. Use the formula: \[\rho = \frac{m}{V}\] Substitute the given values: \[\rho = \frac{2.0}{1.57}\]
04

Compute the Density

Now, let's compute the density: \[\rho \approx 1.27 \mathrm{g/cm^3}\] So, the density of the sphere is approximately 1.27 g/cm³.
05

Determine if the Sphere Will Float or Sink

Now that we have the density, compare it to the density of water (1.0 g/mL or g/cm³). If the density of the sphere is less than the density of water, it will float; otherwise, it will sink. Since the density of the sphere (1.27 g/cm³) is greater than the density of water, the sphere will sink when placed in water.

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