Question

A tank of cross-section A contains a liquid of density up to a height $H$. there is a hole of cross-section area $a$, at the bottom of the tank. The time in which the liquid level is reduced to half is

• A. $\left(A/a\right)(\sqrt{2}-1)\sqrt{H/g}$
• B. $\left(A/a\right)\sqrt{H/g}$
• C. $(\sqrt{2}-1)\sqrt{H/g}$
• D. $Aa(\sqrt{2}-1)\sqrt{H/g}$

Answer 1

The correct option is A $\left(A/a\right)(\sqrt{2}-1)\sqrt{H/g}$

velocity of ejection of water at any height $X$ is $\sqrt{2gx}$

and $A{V}_2=a{V}_1$

acceleration $a_1=\frac{d{v}_1}{dt}=\frac{d}{dt}\sqrt{2gx}$

$=\frac{\sqrt{2g}1}{2x^{1/2}}\times \frac{dx}{dt}$

$but\frac{dx}{dt}=V_2$

$\Rightarrow a_1=\frac{\sqrt{2g}}{2x^{1/2}}{V}_2=\frac{\sqrt{2g}}{2x^{1/2}}(\frac{a}{A}{)}^{V_1}$

$a_1=g\left(\frac{a}{A}\right)$

$v-u=-a_{1}t$

$\Rightarrow \sqrt{2h\frac{H}{2}}-\sqrt{2gH}=-g\left(\frac{a}{A}\right)b$

$\Rightarrow t=\frac{A}{a}(\sqrt{2}-1)\sqrt{\frac{H}{g}}$