Question

A pipe of uniform cross sectional area a has a right angled bend. Impure water of density $\rho$ flows through the horizontal portion of the pipe. The distance covered by water per second at the bend to keep the pipe in equilibrium against an applied force F at the bend is :

• A. $\frac{\rho aF}{\sqrt{2}}$
• B. $\sqrt{2}\frac{\rho }{F}$
• C. ${\left(\frac{\sqrt{2}F}{\rho a}\right)}^{\frac{1}{2}}$
• D. ${\left(\frac{F}{\sqrt{2}\rho a}\right)}^{\frac{1}{2}}$

Answer 1

Answer: (D)

${\left(\frac{F}{\sqrt{2}\rho a}\right)}^{\frac{1}{2}}$

Force $\times$ time$=$ Change in momentum.

$F\times \Delta t=\Delta mv\sqrt{2}$

$F=\frac{\Delta m}{\Delta t}v\sqrt{2}$

$F=\rho a\frac{\Delta l}{\Delta t}\times v\sqrt{2}$ as $(\Delta m=\rho a\Delta l)$

$F=\rho av\times v\sqrt{2}$ as $(\frac{\Delta l}{\Delta t}=v)$

$F=\rho a\sqrt{2}{v}^2$

$\therefore v={\left(\frac{F}{\sqrt{2}\rho a}\right)}^{1/2}$