For the tank shown in figure, if time of decreasing the level of water in the tank from height of 16m to 4m is 't', then choose the correct option(s).  $(g=10m/s^2)$s 

 

$\begin{array}{l}\pi r^2\sqrt{2gy}=-\pi x^2\frac{dy}{dt}\\ r^2\sqrt{2gy}dt=-\frac{y^{\frac{2}{a}}}{k^{{ }^{\frac{2}{a}}}}dy\\ k^{{ }^{\frac{2}{a}}}{r}^2\sqrt{2g}\underset{0}{\overset{t}{\int }}dt=+\underset{4}{\overset{16}{\int }}{y}^{\frac{2}{a}-\frac{1}{2}}dy\\ \left(A\right)k=1,a=4\\ \sqrt{5}\times {10}^{-4}\sqrt{20}t=12\\ t=12000\mathrm{s}\\ \left(B\right)k=1,a=2\\ r^2\sqrt{2g}t=t\underset{4}{\overset{16}{\int }}{y}^{\frac{1}{2}}dy\\ r^2\sqrt{2g}t=\frac{2({16}^{\frac{3}{2}}-4^{\frac{3}{2}})}{3}\\ \sqrt{5}\times {10}^{-4}\sqrt{20}t=\frac{2}{3}\left(56\right)\\ t=\frac{2\left(56\right)\times {10}^4}{3\sqrt{5}\sqrt{20}}\\ =\frac{2}{3}\left(56\right)\times 1000\\ \left(C\right)\left(4\right)\sqrt{5}\times {10}^{-4}\sqrt{20}t=\frac{2}{3}\left(56\right)\\ t=\frac{2\left(56\right)\times 1000}{3\times 4}\\ \left(D\right)\sqrt{4}\sqrt{5}\times {10}^{-4}\sqrt{20}t=12\\ t=6000\mathrm{s}\end{array}$