Assuming erythrocytes are spherical, what is the approximate volume of a typical erythrocyte with a diameter of 7 µm?

Assuming erythrocytes are spherical, what is the approximate volume of a typical erythrocyte with a diameter of 7 µm?

• A. 10 µm3
• B. 180 µm3
• C. 250 µm3
• D. 300 µm3

Answer: (B)

To find the volume of a typical erythrocyte, which is modeled as a sphere, we can use the formula for the volume of a sphere: \[ V = \frac{4}{3} \pi r^3 \] In this scenario, the diameter of the erythrocyte is given as 7 µm, which means the radius (r) is half of that, or 3.5 µm. Substituting the radius into the volume formula gives us: \[ V = \frac{4}{3} \pi (3.5 \, \mu m)^3 \] Calculating \( (3.5 \, \mu m)^3 \): \[ (3.5)^3 = 42.875 \, \mu m^3 \] Now substituting this value back into the volume formula: \[ V = \frac{4}{3} \pi (42.875) \] Approximating \( \pi \) as 3.14, we get: \[ V \approx \frac{4}{3} \times 3.14 \times 42.875 \] \[ V \approx \frac{4 \times 3.14 \

To find the volume of a typical erythrocyte, which is modeled as a sphere, we can use the formula for the volume of a sphere:

[ V = \frac{4}{3} \pi r^3 ]

In this scenario, the diameter of the erythrocyte is given as 7 µm, which means the radius (r) is half of that, or 3.5 µm.

Substituting the radius into the volume formula gives us:

[ V = \frac{4}{3} \pi (3.5 , \mu m)^3 ]

Calculating ( (3.5 , \mu m)^3 ):

[ (3.5)^3 = 42.875 , \mu m^3 ]

Now substituting this value back into the volume formula:

[ V = \frac{4}{3} \pi (42.875) ]

Approximating ( \pi ) as 3.14, we get:

[ V \approx \frac{4}{3} \times 3.14 \times 42.875 ]

[ V \approx \frac{4 \times 3.14 \