The time for a tank to drain through a small hole depends on the height of water.

The rate of flow is proportional to the square root of the height (h), given by Torricelli's law: $\mathrm{v} = \sqrt{\left(2\mathrm{gh}\right)}$.

Therefore, the time (t) to drain is proportional to the square root of the initial height (H).

Mathematically, $\mathrm{t}\propto \sqrt{\mathrm{H}}$. So, for two heights

 ${\mathrm{t}}_2/{\mathrm{y}}_1= \sqrt{({\mathrm{H}}_2/{\mathrm{H}}_1)}$

Given: t₁ = t₀ when H₁ = H. Find t₂ when H₄ = 4H

${\mathrm{t}}_2/{\mathrm{t}}_0=\sqrt{(4\mathrm{H}/\mathrm{H})}=\sqrt{4}=2$

Thus, t₂ = 2t₀

Final answer: 2t₀