Question

A horizontal jet of water is made to hit a vertical wall with a negligible rebound. If the speed of water from the jet is 'v', the diameter of the jet is 'd' and the density of water is 'Rho', then the force exerted on the wall by the jet of water is ??

Answer 1

We know that ,
Force exerted = Rate of change in linear momentum

= f = p/t = m•v/t = rho×v×l×A/t

= f = rho•A×v² [Since, l/t = v]

= f = rhoπ/4d²v² [ Since, A = π•d²/4]

Answer 2

Answer:

rho×pie×d²/4v²

Explanation:

Force exerted = change in momentum/ time

= Pf = final momentum

= Pi = initial momentum

= Pf- Pi/t

Pf = mv

= rho×pie(d/2)²×vt×v

Pi = 0

Force exerted = rho×pie×d²/4×v²t/t

= rho×pie×d²/4×v²

Hope it helps you...

Thank you....