# DBT JRF 2020 — Question 20

Exam: DBT JRF 2020 · Section: A · Topic: Microbiology · Difficulty: Hard

## Question

During growth, the diameter of a Staphylococcus bacterial cell increases by 5%. The specific surface area (defined as surface area per unit volume)

## Options

- **A.** increases approximately by 5%
- **B.** increases approximately by $4\pi$ %
- **C.** decreases approximately by $4\pi$ %
- **D.** decreases approximately by 5%

## Answer

Option D

## Explanation

For a sphere of diameter $d$ : Surface Area $= \pi d^2$ , Volume $= \dfrac{\pi d^3}{6}$ . Specific surface area $= \dfrac{\text{SA}}{V} = \dfrac{\pi d^2}{\pi d^3/6} = \dfrac{6}{d}$ . If diameter increases by 5%: new diameter $= 1.05d$ . New specific SA $= \dfrac{6}{1.05d} = \dfrac{6}{d} \times \dfrac{1}{1.05} \approx \dfrac{6}{d} \times 0.952$ . This is a decrease of approximately $\dfrac{1}{1.05} - 1 \approx -4.76\% \approx -5\%$ . Therefore specific surface area decreases by approximately 5%.

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