Let the rate of change of height of the tank be denoted by $\frac{dH}{dt}$
Then according to the question, $\frac{dH}{dt} =\frac{K}{H}$ -----$\left ( 1 \right )$
Givent at $t=0$, $H=10$ and at $t=1$, $H=12$
So, integrating by limits mentioned above, we get $\int_{10}^{12}HdH=K\int_{0}^{1}dt$
$\Rightarrow$ $K=\frac{144-100}{2}=22$
To find the complete time to fill up the tank, we will use again equation $\left ( 1 \right )$
Now, $\int_{10}^{20}HdH=22\times \int_{0}^{t}dt$
$\therefore$ $t=\frac{\left ( 400-100 \right )}{44}$ $\Rightarrow$ $t=\frac{75}{11}$
Option A is correct