There is a hole of area $A$ at the bottom of cylindrical vessel. Water is filled up to a height $ h$ and water flows out in $ t $ second. If water is filled to a height $4h,$ it will flow out in time equal to
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(c)Time required to emptied the tank $t = \frac{A}{{{A_0}}}\sqrt {\frac{{2H}}{g}} $
$\frac{{{t_2}}}{{{t_1}}} = \sqrt {\frac{{{H_2}}}{{{H_1}}}} = \sqrt {\frac{{4h}}{h}} = 2$ $\therefore \;\;{t_2} = 2t$
$\frac{{{t_2}}}{{{t_1}}} = \sqrt {\frac{{{H_2}}}{{{H_1}}}} = \sqrt {\frac{{4h}}{h}} = 2$ $\therefore \;\;{t_2} = 2t$
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