When a cell of `2mum` diameter grows to doyble its diameter, its surface area : volume relationship will

Options

A. Remain the same
B. Become Half
C. Become double
D. Become undetermined

Answer

To solve the problem of how the surface area to volume relationship changes when a cell's diameter doubles, we can follow these steps:

### Step 1: Determine the initial dimensions of the cell
- The initial diameter of the cell is 2 micrometers.
- Therefore, the initial radius \( r_1 \) is half of the diameter:
\[ r_1 = \frac{2 \text{ micrometers}}{2} = 1 \text{ micrometer} \]

### Step 2: Calculate the initial surface area and volume
- The formula for the surface area \( A \) of a sphere is:
\[ A = 4\pi r^2 \]
- Substituting \( r_1 \):
\[ A_1 = 4\pi (1)^2 = 4\pi \text{ square micrometers} \]

- The formula for the volume \( V \) of a sphere is:
\[ V = \frac{4}{3}\pi r^3 \]
- Substituting \( r_1 \):
\[ V_1 = \frac{4}{3}\pi (1)^3 = \frac{4}{3}\pi \text{ cubic micrometers} \]

### Step 3: Calculate the initial surface area to volume ratio
- The surface area to volume ratio \( R_1 \) is given by:
\[ R_1 = \frac{A_1}{V_1} = \frac{4\pi}{\frac{4}{3}\pi} = 3 \]

### Step 4: Determine the new dimensions of the cell after growth
- The new diameter of the cell is double the original:
\[ \text{New diameter} = 2 \times 2 \text{ micrometers} = 4 \text{ micrometers} \]
- Therefore, the new radius \( r_2 \) is:
\[ r_2 = \frac{4 \text{ micrometers}}{2} = 2 \text{ micrometers} \]

### Step 5: Calculate the new surface area and volume
- The new surface area \( A_2 \) is:
\[ A_2 = 4\pi (2)^2 = 4\pi \times 4 = 16\pi \text{ square micrometers} \]

- The new volume \( V_2 \) is:
\[ V_2 = \frac{4}{3}\pi (2)^3 = \frac{4}{3}\pi \times 8 = \frac{32}{3}\pi \text{ cubic micrometers} \]

### Step 6: Calculate the new surface area to volume ratio
- The new surface area to volume ratio \( R_2 \) is:
\[ R_2 = \frac{A_2}{V_2} = \frac{16\pi}{\frac{32}{3}\pi} = \frac{16 \times 3}{32} = \frac{48}{32} = \frac{3}{2} \]

### Step 7: Compare the ratios
- The initial ratio \( R_1 \) was 3, and the new ratio \( R_2 \) is \( \frac{3}{2} \).
- This shows that the new ratio is half of the initial ratio.

### Conclusion
- Therefore, the surface area to volume ratio becomes half when the cell doubles its diameter. The correct answer is **B) it becomes half**.

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