Question

A container is filled with liquid upto height $H$ as shown in figure. There are $2$ holes in the container such that water coming out from them have same range. If the range of water coming out of the holes is half of maximum range, then the distance $x$ between these holes is

• A. $\frac{\sqrt{5}H}{4}$
• B. $\frac{\sqrt{3}H}{4}$
• C. $\frac{2H}{3}$
• D. $\frac{\sqrt{3}H}{2}$

Answer 1

The correct option is D $\frac{\sqrt{3}H}{2}$

If the orifice is at a depth of $y$ below the surface,
$v=\sqrt{2gy}$ and $h=H-y$.
The range is
$R=v\sqrt{\frac{2h}{g}}$
$R=\sqrt{2gy}\sqrt{\frac{2(H-y)}{g}}$
$R^2=4(Hy-y^2)$
We know that $R_{max}=H.$
In the given problem,
$R=\frac{R_{max}}{2}$
$\Rightarrow R^2=\frac{R_{max}}{4}$
$\Rightarrow 4(Hy-y^2)=\frac{H^2}{4}$
$\Rightarrow y^2-Hy+\frac{H^2}{16}=0.$ ( is of the form $A{y}^2+By+C=0)$
The two roots $\left(y_1\&y_2\right)$ are the two depths, such that $y_2-y_1=x$
$\Rightarrow x=\frac{\sqrt{B^2-4AC}}{A}(A=1,B=-H,C=\frac{H^2}{16})$
$x=\frac{\sqrt{H^2-\frac{H^2}{4}}}{1}$
$x=\frac{\sqrt{3}H}{2}$