Previous in Properties of Matter Next in Properties of Matter
Question Number 45129 by Necxx last updated on 09/Oct/18
Commented by Necxx last updated on 09/Oct/18
$$\mathrm{12}\:{and}\:\mathrm{13}\:{pls} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 09/Oct/18
$$\left.\mathrm{13}\right){v}_{\mathrm{0}} =\sqrt{\mathrm{2}{gh}}\:=\sqrt{\mathrm{2}{ga}}\:\:\:{here}\:{h}={a}\:\:{height}\:{of}\:{liquid} \\ $$$${level}\:{above}\:{hole}\:{for}\:{firstfigure} \\ $$$${v}_{\mathrm{1}} =\sqrt{\mathrm{2}{gh}_{\mathrm{1}} }\:\:\:{h}_{\mathrm{1}} =\frac{{a}\sqrt{\mathrm{2}}\:}{\mathrm{2}}={half}\:{of}\:{diagonal} \\ $$$$\frac{{v}_{\mathrm{1}} }{{v}_{{o}} }=\frac{\sqrt{\mathrm{2}{g}}\:×\sqrt{{h}_{\mathrm{1}} }\:}{\sqrt{\mathrm{2}{g}}\:×\sqrt{{h}}\:}=\left(\frac{{a}}{\sqrt{\mathrm{2}}}×\frac{\mathrm{1}}{{a}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$${v}_{\mathrm{1}} ={v}_{\mathrm{0}} ×\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$ \\ $$
Commented by Necxx last updated on 09/Oct/18
$${I}\:{am}\:{most}\:{grateful}\:{mr}\:{Tanmay} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Oct/18
$${thank}\:{you}... \\ $$