Question
Download Solution PDFA jet of liquid with a velocity of 25 m/s and its relative density 0.8 strikes normally a plate moving towards from it at 15 m/s. The jet has an area of 0.03 m2. The force exerted by plate on the jet is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Force exerted by a jet on the vertical moving plate in direction of the jet with velocity u is given by:
\(F = \rho a(v-u)^2\)
where ρ = Density of Liquid, v = velocity of Jet, u = velocity of plate, a = area of jet
Calculation:
Given:
v = 25 m/s, u = -15 m/s, a = 0.03 m2, Relative Density of Jet Liquid = 0.8
Density of Jet Liquid, ρ = 0.8 × 1000 = 800 Kg/m3
F = 800 × 0.03 (25 + 15)2
F = 38400 N
F = 38.4 KN
Last updated on Sep 23, 2024
-> JSSC JE Additional Result has been released for the Jharkhand Diploma Level Combined Competitive Examination-2023 (Regular and Backlog Recruitment) This is for the Advertisement No. 04/2023 and 05/2023.
-> The JSSC JE notification was released for 1562 (Regular+ Backlog) Junior Engineer vacancies.
-> Candidates applying for the said post will have to appear for the Jharkhand Diploma Level Combined Competitive Examination (JDLCCE).
-> Candidates with a diploma in the concerned engineering branch are eligible for this post. Prepare for the exam with JSSC JE Previous Year Papers.