Question

A tank, which is open at the top, contains a liquid up to a height $H$. A small hole is made in the side of a tank at a distance y below the liquid surface. The liquid emerging from the hole lands at a distance $x$ from the tank.

• A. If $y$ is increased from zero to $H$, $x$ will first increase and then decrease.
• B. x is maximum for $y=H/2$.
• C. The maximum value of $x$ is $H$.
• D. The maximum value of $x$ will depend on the density of the density of the liquid.

Answer 1

Answer: (A), (B), (C)

The correct options are
A If $y$ is increased from zero to $H$, $x$ will first increase and then decrease.
B x is maximum for $y=H/2$.
C The maximum value of $x$ is $H$.

The velocity of efflux $=v=\sqrt{2gy}$
The emerging liquid moves as a projectile and reaches the ground in time $t$, So we have
$H-y=\frac{1}{2}g{t}^2\Rightarrow t=\sqrt{\frac{2(H-y)}{g}}\Rightarrow x=vt=\sqrt{2gy}\times \sqrt{\frac{2(H-y)}{g}}=2\sqrt{(H-y)y}$
$\Rightarrow x=2\sqrt{(H-y)y}$
Differentiating $x$ w.r.t. to $y$, we have
$\frac{dx}{dy}=2\times \frac{1}{2\sqrt{(H-y)y}}\times (H-2y)\Rightarrow \frac{dx}{dy}=\frac{H-2y}{\sqrt{(H-y)y}}$
clearly
$\frac{dx}{dy}>0$ for $y<\frac{H}{2}$
and
$\frac{dx}{dy}<0$ for $y>\frac{H}{2}$
which implies that $x$ first increases and then decreases, as $y$ is increased from $0$ to $H$ $\Rightarrow$ option A is correct
for maxima or minima, we have
$\frac{dx}{dy}=0$
$\Rightarrow y=\frac{H}{2}$
Since the $x$ is first increasing which implies that $y=\frac{H}{2}$ is a point of maxima. $\Rightarrow$ option B is correct
$x_{max}=x(y=\frac{H}{2})$
$\Rightarrow x_{max}=2\sqrt{(H-\frac{H}{2})\frac{H}{2}}=H$ $\Rightarrow$ option C is correct.
It is clear from the above equation that $x_{max}$ is independent of density of the liquid.. $\Rightarrow$ option D is incorrect.