A water jet of cross-section area 0.01 m² impinges normally on a plate and falls down grazing the plate tangentially. If a force of 490 N is produced as a result of this impact, the velocity of the jet is-

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Question

This question was previously asked in

BDL MT Mechanical 16 April 2022 Official Paper

Answer 1

Concept:

Force exerted by a jet on the vertical plate having velocity of jet v is given by:

$F = ρ a (v - u)^{2}$

where ρ = Density of Liquid, v = velocity of Jet, a = area of jet

Calculation:

Given:

a = 0.01 m2, Density of Water = 1000 Kg/m³, F = 490 N

490 = 1000 × 0.01 × v2

490 = 10v²

v = 7 m/s