Water flows from the base of rectangular tank, of depth 16 meters. The rate of flow of the water is proportional to the square root of depth at any time \(\mathrm{t}\). If depth is \(4 \mathrm{~m}\) when \(\mathrm{t}=2\) hours, then after 3.5 hours the depth (in meters) is

Given that \(\frac{\mathrm{d} x}{\mathrm{dt}} \propto \sqrt{x}\)
\(\therefore \quad \frac{\mathrm{d} x}{\mathrm{dt}}=\mathrm{a} \sqrt{x}\), for real number a
\(\therefore \quad \int \frac{\mathrm{d} x}{\sqrt{x}}=\int \mathrm{adt}\)
\(\therefore \quad 2 \sqrt{x}=\mathrm{at}+\mathrm{c}\)
When \(\mathrm{t}=0, x=16\)
\(\therefore \quad\) (i) \(\Rightarrow \mathrm{c}=8\)
\(\therefore \quad\) (i) becomes \(2 \sqrt{x}=\) at +8
When \(\mathrm{t}=2, x=4\)
\(\therefore \quad\) (ii) \(\Rightarrow \mathrm{a}=-2\)
\(\therefore \quad\) (ii) becomes \(2 \sqrt{x}=-2 t+8\)
\(\therefore \quad\) when \(\mathrm{t}=3.5\)
(iii) \(\Rightarrow x=0.25 \mathrm{~m}\)