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)

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Aicharya · Accepted Answer

From equation of continuity, $A v_1=a_2$ $\mathrm{A}\left(-\frac{\mathrm{dh}}{\mathrm{dt}}ight)=\mathrm{a} \sqrt{2 \mathrm{gh}}$ $\int_{\mathrm{h}_1}^{\mathrm{h}_2} \frac{-\mathrm{dh}}{\sqrt{\mathrm{h}}}=\int_0^{\mathrm{t}} \frac{\mathrm{a}}{\mathrm{A}} \sqrt{2 \mathrm{~g}} \mathrm{dt}$ $-[2 \sqrt{\mathrm{h}}]_{\mathrm{h}_1}^{\mathrm{h}_2}=\frac{\mathrm{a}}{\mathrm{A}} \sqrt{2 \mathrm{~g}} \mathrm{t}$ $2\left(\sqrt{\mathrm{h}_2}-\sqrt{\mathrm{h}_1}ight)=\frac{\mathrm{a}}{\mathrm{A}} \sqrt{2 \mathrm{gt}}$ $	herefore \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 $t_2=\frac{A}{A_0} \sqrt{\frac{8 h}{g}}$ $	herefore \quad \frac{t_2}{t_1}=2$ $	herefore \quad \mathrm{t}_2=2 \mathrm{t}_1$ $	herefore \quad \mathrm{t}_2=2 \mathrm{t}$ $\ldots\left(\because \mathrm{t}_1=\mathrm{t}ight)$

There is hole of area 'A' at the bottom of a cylindrical vessel. Water is filled to a height ' $h$ ' and…

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