There is hole of area 'A' at the bottom of a cylindrical vessel. Water is filled to a height ' \(h\) ' and water flows out in ' \(t\) ' second. If water is filled to a height ' \(4 \mathrm{~h}\) ', it will flow out in time (in second)

From equation of continuity,
\(A v_1=a_2\)
\(\mathrm{A}\left(-\frac{\mathrm{dh}}{\mathrm{dt}}\right)=\mathrm{a} \sqrt{2 \mathrm{gh}}\)
\(\int_{\mathrm{h}_1}^{\mathrm{h}_2} \frac{-\mathrm{dh}}{\sqrt{\mathrm{h}}}=\int_0^{\mathrm{t}} \frac{\mathrm{a}}{\mathrm{A}} \sqrt{2 \mathrm{~g}} \mathrm{dt}\)
\(-[2 \sqrt{\mathrm{h}}]_{\mathrm{h}_1}^{\mathrm{h}_2}=\frac{\mathrm{a}}{\mathrm{A}} \sqrt{2 \mathrm{~g}} \mathrm{t}\)
\(2\left(\sqrt{\mathrm{h}_2}-\sqrt{\mathrm{h}_1}\right)=\frac{\mathrm{a}}{\mathrm{A}} \sqrt{2 \mathrm{gt}}\)
\(\therefore \quad t=\frac{2 A}{a} \sqrt{\frac{H}{2 g}}=\frac{A}{a} \frac{\sqrt{2 H}}{\sqrt{g}}\)
For height \(h\), the time is \(t_1=\frac{A}{A_0} \sqrt{\frac{2 h}{g}}\)
For height \(4 \mathrm{~h}\), the time is \(t_2=\frac{A}{A_0} \sqrt{\frac{8 h}{g}}\)
\(\therefore \quad \frac{t_2}{t_1}=2\)
\(\therefore \quad \mathrm{t}_2=2 \mathrm{t}_1\)
\(\therefore \quad \mathrm{t}_2=2 \mathrm{t}\)
\(\ldots\left(\because \mathrm{t}_1=\mathrm{t}\right)\)