Question

A gas of density $\rho$ flows with velocity $v$ along a pipe of cross - sectional area $S$ and bent to an angle of ${90}^{\circ }$ at point $A$. What force does the gas exert on the pipe at point $A$ ?

• A. $\sqrt{5}S{v}^2\rho$
• B. $\sqrt{2}S{v}^2\rho$
• C. $\sqrt{7}S{v}^2\rho$
• D. $\sqrt{3}S{v}^2\rho$

Answer 1

The correct option is B $\sqrt{2}S{v}^2\rho$


Initial velocity of gas,
$\overrightarrow{v_i}=v\hat{i}$
Final velocity of gas, after bending through ${90}^{\circ }$ is,
$\overrightarrow{v_f}=v\hat{j}$
So, change in velocity,
$\Delta \overrightarrow{v}=\overrightarrow{v_f}-\overrightarrow{v_i}=v\hat{j}-v\hat{i}$
Now, magnitude of change in velocity,
$|\Delta \overrightarrow{v}|=\sqrt{v^2+v^2}=\sqrt{2}v$
Magnitude of change in momentum of gas is,
$|\Delta \overrightarrow{p}|=m|\Delta \overrightarrow{v}|$
$\Rightarrow |\Delta \overrightarrow{p}|=\sqrt{2}mv$
Magnitude of force on pipe exerted by gas will be,
$F=\frac{\Delta p}{\Delta t}$
$\Rightarrow F=\frac{\sqrt{2}mv}{\Delta t}$ .............(1)
Mass $m$ of gas passing through cross section of pipe in time $\Delta t$,
$m=v\Delta tS\rho$ ............(2)
[ volume = length $\times$ Area ]
On putting (2) in (1)
$F=\frac{\sqrt{2}\times v\Delta tS\rho \times v}{\Delta t}$
$\Rightarrow F=\sqrt{2}S{v}^2\rho$