Question

A 35 gm/l solution of tracer was discharged into stream at a constant rate of $12c{m}^3/s$. The background concentration of tracer in stream water was found to be zero. If the discharge of stream is $50m^3/s$ then the equilibrium concentration of the tracer at a downstream section sufficiently far away is____ $\times {10}^{6}gm/l$.

• A. 4.8
• B. 6.6
• C. 7.2
• D. 8.4

Answer 1

The correct option is D 8.4

For a constant rate injection method $Q=\frac{Q_1(C_1-C_2)}{(C_2-C_0)}$

$C_0$ = background concentration before injection of tracer.
$C_1$ = Concentration of tracer
$C_2$ = Equilibrium concentration at downstream

$\therefore 50=\frac{12\times {10}^{-6}}{C_2}\times (35-C_2)$

$\Rightarrow C_2=8.399\times {10}^{-6}gm/l$