Question

A pipe of uniform cross sectional of area a has a right angled bend, impure water of density δ 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.
• B.
• C.
• D. $\left( \frac { \mathrm { F } } { \sqrt { 2 } \delta \mathrm { a } } \right) ^ { 1 / 2 }$

Answer 1

Answer: (D)

$\left( \frac { \mathrm { F } } { \sqrt { 2 } \delta \mathrm { a } } \right) ^ { 1 / 2 }$

Answer 2

Force × Time = Change in momentum

or F × ∆t = ∆mv√2 or F = $\frac { \Delta \mathrm { m } } { \Delta \mathrm { t } } \mathrm { v } \sqrt { 2 }$

= $\mathrm { v } \sqrt { 2 } \times \rho \mathrm { a } \frac { \Delta \mathrm { l } } { \Delta \mathrm { t } }$

( ∆m = ρa∆L)

or F = = v² $\sqrt { 2 } \rho \mathrm { a }$

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