SSC JE Mechanical Engineering 2014 Set 2 Paper

Question : 126 of 200

Marks: +1, -0

A jet of water issues from a nozzle with a velocity 20 m/s on a flat plate moving away from it at 10 m/s. The cross-sectional area of the jet is 0.01 m

^{2}

and the density of water = 1000 kg/m

^{3}

. The force developed on the plate in Newtons is :

2000
9810
5000
7000

Validate

Solution:

From the given value v = 20 m/s
v = jet velocity
plate velocity (u) = 10 m/s
area of jet = .01 m

^{2}

density of water = 1000 Kg/m

^{3}

force developed on plate

F = ρa(v – u)^2

= 1000 × 0.01 (20 -10)^2

= 1000 × .01 × 100 = 1000 N

second case blade are in series
force

F = ρav(v -u)

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