Question

The sphericity of a cylindrical potato sample having diameter of 1.0 cm and length of 5.0 cm is closest to:

• A. 0.98
• B. 0.70
• C. 0.31
• D. 0.17

Hint:

When calculating sphericity, use the formulas for the volume and surface area of the object and then substitute the values into the sphericity formula.

Answer 1

The Correct Option is B

Sphericity (\( \phi \)) is a measure of how closely an object resembles a sphere and is given by the formula: \[ \phi = \frac{\pi^{1/3} (6V)^{2/3}}{A} \] Where:
\(V\) is the volume of the object,
\(A\) is the surface area of the object.
For a cylinder, the volume \(V\) is: \[ V = \pi r^2 h \] Where \(r\) is the radius of the base, and \(h\) is the height. The surface area \(A\) is: \[ A = 2\pi r^2 + 2\pi rh \] Given that the diameter \(d = 1.0 \, {cm}\), so the radius \(r = 0.5 \, {cm}\), and the length \(h = 5.0 \, {cm}\), we can now calculate: - Volume of the cylinder: \[ V = \pi (0.5)^2 \times 5 = 3.92699 \, {cm}^3 \] - Surface area of the cylinder: \[ A = 2\pi (0.5)^2 + 2\pi (0.5)(5) = 3.1416 + 15.7079 = 18.8496 \, {cm}^2 \] Now, substitute these values into the sphericity formula: \[ \phi = \frac{\pi^{1/3} (6 \times 3.92699)^{2/3}}{18.8496} \approx 0.70 \] Thus, the sphericity of the cylindrical potato sample is closest to 0.70.