Question

An L-shaped tube with a small orifice is held in a water stream as shown in the figure. The upper end of the tube is $10.6\text{cm}$above the surface of water. What will be the height of the jet of water coming from the orifice? (Velocity of water stream is $2.45{\text{m/s and take g=10 m/s}}^2$)

• A. zero
• B. $10.6\text{cm}$
• C. $19.41\text{cm}$
• D. $30.01\text{cm}$

Answer 1

The correct option is C $19.41\text{cm}$

Considering the entrance of the tube (Q) be the datum and $x$ be the height of the jet.

When the jet reaches maximum height i.e at point P velocity of the liquid becomes zero and pressure will be atmospheric$\left(P_o\right)$.
Applying the bernoulli equation between point P and Q in terms of head,
$\frac{P_P}{\rho g}+\frac{V_P^2}{2g}+Z_P=\frac{P_Q}{\rho g}+\frac{V_Q^2}{2g}+Z_Q$
$\frac{P_o}{\rho g}+\frac{(0{)}^2}{2g}+(x+10.6\times {10}^{-2}+y)=\frac{P_o+\rho gy}{\rho g}+\frac{V_Q^2}{2g}+0$
On rearranging the above equation,
$\frac{V_Q^2}{2g}=x+(10.6\times {10}^{-2})$
Substituting $V_Q=2.45m/s$,
$\frac{(2.45{)}^2}{2g}=x+(10.6\times {10}^{-2})$
$\Rightarrow x=0.1941m=19.41cm$