Reasoning Aptitude CSIR 15 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?


*

*$3\text{%}$

*$5\text{%}$

*$6\text{%}$

*$7\text{%}$
 A: If $V$ is its volume and $S$ is its surface area, $S$ is proportional to $V^{2\over3}$ so
$$
{dS\over S}={2\over3}{dV\over V}={2\over3}{0.1\times0.7V\over V}\approx0.05
$$
A: 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 $$\frac{S'-S_0}{S_0}\times 100$$
$$=\frac{4\pi \left(0.93 \right)^{2/3}r^2-4\pi r^2}{4\pi r^2}\times 100=4.722877503\approx 5\ \text{%}$$
