The concentration of fluoride, lead, nitrate and iron in a water sample from an underground lake was found to be 1000 ppb, 40 ppb, 100 ppm and 0.2 ppm, respectively. This water is unsuitable for drinking due to high concentration of:

First, we rewrite every analytical concentration in the common unit “ppm” because the standard permissible limits for drinking water are normally quoted in $$\text{mg L}^{-1}$$, i.e. ppm. The conversion factor is stated as $$1\;\text{ppm}=10^3\;\text{ppb}.$$

We have:

$$ \begin{aligned} \text{Fluoride: }&1000\;\text{ppb}=1000\div10^3=1\;\text{ppm},\\[4pt] \text{Lead: }&40\;\text{ppb}=40\div10^3=0.040\;\text{ppm},\\[4pt] \text{Nitrate: }&100\;\text{ppm}\;\text{(already in ppm)},\\[4pt] \text{Iron: }&0.2\;\text{ppm}\;\text{(already in ppm)}. \end{aligned} $$

Now we recall the Bureau of Indian Standards (BIS) / WHO desirable limits for potable water:

$$ \begin{aligned} \text{Fluoride (F}^-)&:&1.0\;\text{ppm (desirable)},\;1.5\;\text{ppm (maximum)},\\[4pt] \text{Lead (Pb)}&:&0.05\;\text{ppm (maximum allowed)},\\[4pt] \text{Nitrate (NO}_3^-)&:&50\;\text{ppm (maximum allowed)},\\[4pt] \text{Iron (Fe)}&:&0.3\;\text{ppm (maximum allowed)}. \end{aligned} $$

We now compare each measured value with its respective maximum limit:

$$ \begin{aligned} \text{Fluoride: }&1;;\text{ppm}\;\lt\;1.5\;\text{ppm}\;\Rightarrow\;\text{acceptable},\\[4pt] \text{Lead: }&0.040\;\text{ppm}\;\lt\;0.05\;\text{ppm}\;\Rightarrow\;\text{acceptable},\\[4pt] \text{Nitrate: }&100\;\text{ppm}\;\gt\;50\;\text{ppm}\;\Rightarrow\;\text{exceeds limit},\\[4pt] \text{Iron: }&0.2\;\text{ppm}\;\lt\;0.3\;\text{ppm}\;\Rightarrow\;\text{acceptable}. \end{aligned} $$

Only the nitrate concentration surpasses its permissible limit, making the water unfit for drinking.

Hence, the correct answer is Option A.