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. $h \, sin^{2} \theta $
• B. $\text{h} \, \text{sin} 2 \theta $
• C. $\frac{1}{2 \, }hsin \theta \, $
• D. $h \, tan^{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 \, $
$p_{0}+\rho gh+0=p_{0}+0+\frac{1}{2}\rho v^{2}$
$\therefore v=\sqrt{2 g h}$ ...(i)
At outlet,

$vsin \theta $ vertical velocity of the fluid at outlet,
$v^{2}=u^{2}+2gh_{1}$
For maximum height,
$0=\left(v sin \theta \right)^{2}-2gh_{1}$
$\therefore h_{1}=\frac{v^{2} sin^{2} \theta }{2 g}$
$=\frac{2 g h sin^{2} \theta }{2 g}$ [from Equation (i)]
$=hsin^{2} \theta $
$h_{1}=hsin^{2} \theta $