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RRB Junior Engineer / DMS / CMA 2019 · Question 3524 of 3719

The horizontal component of tensile force in a wire that makes 60° with horizontal and is carrying a force of 20 kN is

  1. A10 kNCorrect
  2. B18 kN
  3. C30 kN
  4. D25 kN

Answer: A. 10 kN

Explanation

Calculation :- Given :- P = 20 kN, θ = 60° OX = 10 kN cosθ = ​ Hypotenuse base cos 60 = ∘ ​ OP OX ​ = 2 1 ​ 20 OX

Source: RRB JE 2019 (CBT 2) (ME) Previous Year Paper (31 Aug 2019) — prepp.in solved-paper PDF (answer key with explanations) · reliable-secondary

← Q3523View full paper (3719 questions)Q3525 →

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