In river water quality modeling, two rate constants govern oxygen balance:
• $k_r$ = re-aeration rate constant (oxygen replenishment)
• $k_d$ = de-oxygenation rate constant (oxygen consumption by pollutants)
Higher $k_r$ means the river absorbs oxygen from the atmosphere more quickly.
Higher $k_d$ indicates faster biochemical oxidation of pollutants.
Given values:
River X: $k_r = 0.92$, $k_d = 0.23$
River Y: $k_r = 1.12$, $k_d = 0.35$
Self-purification capacity depends on the ratio:
\[
\frac{k_r}{k_d}
\]
Compute the ratios:
River X: $\frac{0.92}{0.23} = 4.0$
River Y: $\frac{1.12}{0.35} \approx 3.2$
A higher ratio means better purification ability. Therefore:
$\frac{k_r}{k_d}$ for X>Y, meaning river X self-purifies more effectively than river Y.
Thus, option (B) is correct and (A) is incorrect.
For option (C):
River Y has a higher $k_r$ value (1.12>0.92), indicating:
• higher turbulence,
• or higher flow velocity,
• or higher algal activity (which increases oxygen levels).
Therefore, (C) is correct.
For option (D):
Higher $k_d$ means pollutants are more biodegradable.
River Y has $k_d = 0.35$ (higher) compared to 0.23 for X.
So pollutants are more biodegradable in Y, not X. Hence (D) is false.
Thus, the correct statements are (B) and (C).
Final Answer: (B), (C)