JEE Advanced 2021 Paper 2 Physics: Paragraph 2

In steady state inflow equals outflow, so $Q=a\sqrt{2gh}$ by Torricelli's law. Squaring, $h=\dfrac{Q^2}{2g\,a^2}$. With $Q=4.0\times10^{-4}\ \mathrm{m^3\,s^{-1}}$, $a=2.0\times10^{-4}\ \mathrm{m^2}$ and $g=10$, $h=\dfrac{16\times10^{-8}}{2\times10\times4\times10^{-8}}=\dfrac{16}{80}=0.20\ \mathrm{m}$. The tank area does not appear because the level is stationary. The exit speed is then $\sqrt{2gh}=2\ \mathrm{m\,s^{-1}}$, and $a\times2=4\times10^{-4}\ \mathrm{m^3\,s^{-1}}$ equals the inflow, confirming the result. This steady level is stable, because a higher level gives a larger outflow and a lower level a smaller outflow. Therefore the answer is $0.20$ metres above the hole, a small value compared with the initial filling height used in the next part, so the level would otherwise keep falling towards it.

With the tap closed, volume conservation gives $-A\dfrac{dh}{dt}=a\sqrt{2gh}$. Separating variables, $\displaystyle t=\dfrac{A}{a\sqrt{2g}}\int_{h_2}^{h_1}\dfrac{dh}{\sqrt h}=\dfrac{2A}{a\sqrt{2g}}\left(\sqrt{h_1}-\sqrt{h_2}\right)=\dfrac{A}{a}\sqrt{\dfrac2g}\left(\sqrt{h_1}-\sqrt{h_2}\right)$. Here $A/a=0.50/(2.0\times10^{-4})=2500$ and $\sqrt{2/g}=0.4472$. Also $\sqrt{1.8}-\sqrt{0.2}=1.3416-0.4472=0.8944$. Then $t=2500\times0.4472\times0.8944=2500\times0.4=1000\ \mathrm{s}$. The product $0.4472\times0.8944$ equals exactly $0.4$, so the time is exactly $1000$ s, about $16.7$ minutes. Treating the surface speed as zero is justified because $a/A=4\times10^{-4}$. The same result follows by noting that the surface speed is $v=\sqrt{2gh}\,(a/A)$, so $h$ falls with a time dependence $\sqrt h=\sqrt{h_1}-\frac{a}{A}\sqrt{\frac g2}\,t$, which is linear in time for the square root of height; setting $\sqrt h=\sqrt{0.2}$ gives the same $t$.