Question

Find the time taken for the liquid levels in the two tanks to become equal.

• A. $t=\frac{8\pi r^2}{5a}\sqrt{\frac{H}{g}}$
• B. $t=\frac{4\pi r^2}{5a}\sqrt{\frac{H}{g}}$
• C. $t=\frac{\pi r^2}{5a}\sqrt{\frac{H}{g}}$
• D. $t=\frac{8{\pi }^{2}r}{5a}\sqrt{\frac{2H}{g}}$

Answer 1

The correct option is A $t=\frac{8\pi r^2}{5a}\sqrt{\frac{H}{g}}$

From touicellis law we know that,

velocity of the flow$4=\sqrt{2g(H-h)}[\therefore$ where $(H-h)$ is difference in water level at both side ]

Area so smaller tank$=\pi r^2$

Area so smaller tank$=4\pi r^2$

So,

If $H$ height decreases in small tank$\frac{H}{4}$height increase in larger tank.

$h$ runs from $0$ to $h=\frac{8H}{5}$ [when level equals]

Let $h$ be decrease in smaller tank and $\frac{h}{4}$ be rise in larger one.

So,

$Q=\frac{d(3H-h)A}{dt}=a\sqrt{2g(3H-h-H-\frac{h}{4})}$

$\Rightarrow -\frac{dh}{dt}=\frac{a}{A}\sqrt{\frac{2g(8H-5h)}{4}}$

$\Rightarrow {\int }_{\frac{8H}{5}}^0\frac{-2dh}{\sqrt{2g(8H-5h)}}={\int }_t^0\frac{a}{A}dt$

$\Rightarrow \frac{2}{5}\frac{{(8H-5h)}^{\frac{1}{2}}}{\sqrt{2g}}=\frac{a}{\pi r^2}t$

$\Rightarrow \frac{a}{\pi r^2}$

t$=\frac{8}{5}\sqrt{\frac{H}{g}}$

$t=\frac{8\pi r^2}{5a}\sqrt{\frac{H}{g}}$

$=\frac{dh}{\sqrt{8H-3h}}$

$=8H-\frac{24H}{5}$

$=\frac{40-24H}{5}$

$=\sqrt{\frac{16H}{5}}$

$A\frac{dh}{dt}$

$3H-h=H+\frac{h}{4}$

$\Rightarrow 2H=\frac{5h}{4}$

$\Rightarrow h=\frac{8H}{5}$