# Question-68510

Question Number 68510 by oyemi kemewari last updated on 12/Sep/19
Answered by mr W last updated on 14/Sep/19
$${v}=\sqrt{\mathrm{2}{gh}} \\$$$${base}\:{area}\:{of}\:{tanks}={A}=\mathrm{1}\:{ft}^{\mathrm{2}} \\$$$${A}\frac{{dh}}{\mathrm{2}}=−\frac{\pi{d}^{\mathrm{2}} }{\mathrm{4}}{vdt} \\$$$${dh}=−\frac{\pi{d}^{\mathrm{2}} \sqrt{\mathrm{2}{gh}}}{\mathrm{2}{A}}{dt} \\$$$$\frac{{dh}}{\:\sqrt{{h}}}=−\frac{\pi{d}^{\mathrm{2}} \sqrt{\mathrm{2}{g}}}{\mathrm{2}{A}}{dt} \\$$$$\int_{{h}_{\mathrm{0}} } ^{{h}} \frac{{dh}}{\:\sqrt{{h}}}=−\frac{\pi{d}^{\mathrm{2}} \sqrt{\mathrm{2}{g}}}{\mathrm{2}{A}}\int_{\mathrm{0}} ^{{t}} {dt} \\$$$$\mathrm{2}\left[\sqrt{{h}}−\sqrt{{h}_{\mathrm{0}} }\right]=−\frac{\pi{d}^{\mathrm{2}} \sqrt{\mathrm{2}{g}}}{\mathrm{2}{A}}{t} \\$$$$\sqrt{{h}_{\mathrm{0}} }−\sqrt{{h}}=\frac{\pi{d}^{\mathrm{2}} \sqrt{\mathrm{2}{g}}}{\mathrm{4}{A}}{t} \\$$$$\Rightarrow{t}=\frac{\mathrm{4}{A}\left(\sqrt{{h}_{\mathrm{0}} }−\sqrt{{h}}\right)}{\pi{d}^{\mathrm{2}} \sqrt{\mathrm{2}{g}}} \\$$$${h}_{\mathrm{0}} =\mathrm{1}\:{ft} \\$$$${d}=\mathrm{0}.\mathrm{5}\:{in}=\frac{\mathrm{1}}{\mathrm{24}}\:{ft} \\$$$$\\$$$${for}\:{h}=\mathrm{0}.\mathrm{25}\:{ft}: \\$$$${t}=\frac{\mathrm{4}×\mathrm{1}\left(\sqrt{\mathrm{1}}−\sqrt{\mathrm{0}.\mathrm{25}}\right)}{\pi\left(\frac{\mathrm{1}}{\mathrm{24}}\right)^{\mathrm{2}} \sqrt{\mathrm{2}{g}}}=\mathrm{82}\:{seconds} \\$$$$\\$$$${for}\:{h}=\mathrm{0}: \\$$$${t}=\frac{\mathrm{4}×\mathrm{1}\left(\sqrt{\mathrm{1}}−\sqrt{\mathrm{0}}\right)}{\pi\left(\frac{\mathrm{1}}{\mathrm{24}}\right)^{\mathrm{2}} \sqrt{\mathrm{2}{g}}}=\mathrm{164}\:{seconds} \\$$
Commented by mr W last updated on 14/Sep/19
Commented by oyemi kemewari last updated on 17/Sep/19
$$\\$$
Commented by oyemi kemewari last updated on 28/Sep/19
thank you sir