Question

A stream of non-viscous liquid emerges from a very short outlet tube at the base of a large open tank, in which the depth of liquid is $h$ . The tube is at a fixed angle $\theta$ to the ground as shown in the figure. The maximum height of the stream $y$ is

• A. $hsi{n}^2\theta$
• B. $\text{h}\text{sin}2\theta$
• C. $\frac{1}{2}hsin\theta$
• D. $hta{n}^2\theta$

Answer 1

Answer: (A)

Let the speed of fluid at outlet is $v$ .
Applying Bernoulli's theorem,
$p_1+\rho gh+\frac{1}{2}\rho v^2=constant$
$p_0+\rho gh+0=p_0+0+\frac{1}{2}\rho v^2$
$\therefore v=\sqrt{2gh}$ ...(i)
At outlet,

$vsin\theta$ vertical velocity of the fluid at outlet,
$v^2=u^2+2g{h}_1$
For maximum height,
$0={\left(vsin\theta \right)}^2-2g{h}_1$
$\therefore h_1=\frac{v^{2}si{n}^2\theta }{2g}$
$=\frac{2ghsi{n}^2\theta }{2g}$ [from Equation (i)]
$=hsi{n}^2\theta$
$h_1=hsi{n}^2\theta$