An intense stream of water of cross-sectional area A strikes a wall at an angle $θ$ with the normal to the wall and returns back elastically. If the density of water is $ρ$ and its velocity is $v$ , then the force exerted on the wall will be :

2 $A v ρ cos θ$
2 $A v^{2} ρ cos θ$
2 $A v^{2} ρ$
2 $A v ρ$

Question

An intense stream of water of cross-sectional area A strikes a wall at an angle $θ$ with the normal to the wall and returns back elastically. If the density of water is $ρ$ and its velocity is $v$ , then the force exerted on the wall will be :

2 $A v ρ cos θ$
2 $A v^{2} ρ cos θ$
2 $A v^{2} ρ$
2 $A v ρ$

Toppr Admin · Accepted Answer

Impulse on wall due to -quot-dm-quot- mass of the water is-I-x394-P-dmvcos-x3B8-x2212-x2212-dmvcos-x3B8-I-x394-P-2dmvcos-x3B8-Force -dIdt-dpdt-2vcos-x3B8-dmdtF-2vcos-x3B8-x3C1-Adxdt-2-x3C1-Av2cos-x3B8-

An intense stream of water of cross-sectional area A strikes a wall at an angle θ with the normal to the wall and returns back elastically. If the density of water is ρ and its velocity is v, then the force exerted on the wall will be :2Avρcosθ2Av2ρcosθ2 Av2ρ2Avρ

Impulse on wall due to "dm" mass of the water is:I=ΔP=dmvcosθ−(−dmvcosθ)I=ΔP=2dmvcosθForce =dIdt=dpdt=2vcosθdmdtF=2vcosθ(ρ.Adxdt)=2ρAv2cosθ.

Impulse on wall due to $" d m "$ mass of the water is:
$I = Δ P = d m v c o s θ - (- d m v c o s θ)$
$I = Δ P = 2 d m v c o s θ$
Force = $\frac{d I}{d t} = \frac{d p}{d t} = 2 v c o s θ \frac{d m}{d t}$
$F = 2 v c o s θ (ρ . A \frac{d x}{d t}) = 2 ρ A v^{2} c o s θ .$