Reasoning Aptitude CSIR 15

Question

A single-celled spherical organism contains $70$% water by volume. If it loses $10$% of its water content, how much would its surface area change by approximately?

1. $3\text{%}$
2. $5\text{%}$

3. $6\text{%}$

4. $7\text{%}$

Answer 1

Let $r$ be the original radius the original volume of water is $$V_1=\frac{70}{100}\times \frac{4}{3}\pi r^3=\frac{14}{15}\pi r^3$$ & volume of organic material $$V_2=\frac{30}{100}\times \frac{4}{3}\pi r^3=\frac{2}{5}\pi r^3$$ & surface area $S_0=4\pi r^2$

when it loses $10$% of water then remaining volume of water $$V_1'=\frac{90}{100}\times V_1= \frac{90}{100}\times\frac{70}{100} \times \frac{4}{3}\pi r^3=\frac{21}{25}\pi r^3$$ If $r'$ is the radius of remaining cell then new volume of cell $$V_1'+V_2=\frac{21}{25}\pi r^3+\frac{2}{5}\pi r^3$$ $$r'=\left(0.93 \right)^{1/3}r$$ hence the new surface area of cell $$S'=4\pi r'^2=4\pi \left(\left(0.93\right)^{1/3}r\right)^2=4\pi \left(0.93 \right)^{2/3}r^2$$

hence % change in surface area of the cell $$\frac{S'-S_0}{S_0}\times 100$$ $$=\frac{4\pi \left(0.93 \right)^{2/3}r^2-4\pi r^2}{4\pi r^2}\times 100=4.722877503\approx 5\ \text{%}$$

Answer 2

If $V$ is its volume and $S$ is its surface area, $S$ is proportional to $V^{2\over3}$ so $$ {dS\over S}={2\over3}{dV\over V}={2\over3}{0.1\times0.7V\over V}\approx0.05 $$