Explanation

(b) If $A_{0}$ in the area orifice at the bottom below the free surface and A that of vessel, then time t taken to be emptied the tank is given as
$t = \frac{A}{A_{0}}\sqrt{\frac{2H}{g}}$
$\therefore\ \frac{t_{1}}{t_{2}} = \sqrt{\frac{H_{1}}{H_{2}}}$
$\Longrightarrow \ \ \frac{t}{t_{2}} = \sqrt{\frac{H_{1}}{H_{1}/2}}$
$\Longrightarrow \ \ \ \frac{t}{t_{2}} = \sqrt{2}$
$\therefore\ \ \ \ t_{2} = \frac{t}{\sqrt{2}} = \frac{10}{\sqrt{2}}$
$= 5\sqrt{2} = 7\min$