SSC JE Mechanical Engineering 2017 Set 2 Paper

Force exerted by a jet on a stationary inclined flat plate
Let a jet of water, coming out from the, strike an inclined flat plate as shown in figure.Let,
a = area of cross section of the jet $=\frac{\mathit{\pi }{\mathit{d}}^2}{4}$
v = velocity of jet in the direction of x
θ = angle between the jet and the plane
Then the mass of the water per sec stricking the plate = ρ × av
${\mathit{F}}_{\mathit{n}}$ = Mass of the jet striking per second × [initial velocity of the jet before striking in the direction of ${\mathit{F}}_{\mathit{n}}$ – final velocity of the jet after striking in the direction of ${\mathit{F}}_{\mathit{n}}$]
${\mathit{F}}_{\mathit{n}}$ = ρav [v sin θ – 0] ρav${}^2$ sinθ
If the force can be resolved into two components, one in the direction of the jet and the other perpendicular to the direction of the flow. Then we have,
${\mathit{F}}_{\mathit{x}}=\mathrm{\rho }\mathrm{a}{\mathrm{v}}^2{\sin }^2\mathit{\theta }$ (along the direction of the flow)
${\mathit{F}}_{\mathit{y}}=\mathrm{\rho }\mathrm{a}{\mathrm{v}}^2\sin \mathit{\theta }\cos \mathit{\theta }$ (perpendicular to flow)