Question Number 45129 by Necxx last updated on 09/Oct/18
![](https://www.tinkutara.com/question/5513.png)
Commented by Necxx last updated on 09/Oct/18
![12 and 13 pls](https://www.tinkutara.com/question/Q45130.png)
$$\mathrm{12}\:{and}\:\mathrm{13}\:{pls} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 09/Oct/18
![13)v_0 =(√(2gh)) =(√(2ga)) here h=a height of liquid level above hole for firstfigure v_1 =(√(2gh_1 )) h_1 =((a(√2) )/2)=half of diagonal (v_1 /v_o )=(((√(2g)) ×(√h_1 ) )/( (√(2g)) ×(√h) ))=((a/( (√2)))×(1/a))^(1/2) =((1/2))^(1/4) v_1 =v_0 ×((1/2))^(1/4)](https://www.tinkutara.com/question/Q45143.png)
$$\left.\mathrm{13}\right){v}_{\mathrm{0}} =\sqrt{\mathrm{2}{gh}}\:=\sqrt{\mathrm{2}{ga}}\:\:\:{here}\:{h}={a}\:\:{height}\:{of}\:{liquid} \\ $$$${level}\:{above}\:{hole}\:{for}\:{firstfigure} \\ $$$${v}_{\mathrm{1}} =\sqrt{\mathrm{2}{gh}_{\mathrm{1}} }\:\:\:{h}_{\mathrm{1}} =\frac{{a}\sqrt{\mathrm{2}}\:}{\mathrm{2}}={half}\:{of}\:{diagonal} \\ $$$$\frac{{v}_{\mathrm{1}} }{{v}_{{o}} }=\frac{\sqrt{\mathrm{2}{g}}\:×\sqrt{{h}_{\mathrm{1}} }\:}{\:\sqrt{\mathrm{2}{g}}\:×\sqrt{{h}}\:}=\left(\frac{{a}}{\:\sqrt{\mathrm{2}}}×\frac{\mathrm{1}}{{a}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$${v}_{\mathrm{1}} ={v}_{\mathrm{0}} ×\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$ \\ $$
Commented by Necxx last updated on 09/Oct/18
![I am most grateful mr Tanmay](https://www.tinkutara.com/question/Q45145.png)
$${I}\:{am}\:{most}\:{grateful}\:{mr}\:{Tanmay} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Oct/18
![thank you...](https://www.tinkutara.com/question/Q45148.png)
$${thank}\:{you}… \\ $$