TGPSC HMWSSB Manager 2015 Paper 2 · Question 70
Question 70 of the TSPSC Manager (Engineering), HMWSSB, 2015: Paper 2 (Civil Engineering) (1 November 2015). Try it first, then open the answer and explanation.
A 50mm diameter jet having a velocity of 25m/s strikes a flat stationary plate, the normal of which is inclined at 60° to the axis of the jet. The normal force exerted on the plate in Newtons is
- 460N
- 360 N
- 640 N
- 630 N
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Answer: (1) 460N
With the plate normal at 60° to the jet, F_n = ρaV² cos 60° = 1000 × (π/4)(0.05)² × 25² × 0.5 ≈ 614 N (reading 60° as the angle of the plate itself gives ρaV² sin 60° ≈ 1063 N instead). The official key gives 460 N, which neither standard working reproduces, so treat this key with caution. (Official final key; some sources differ.)