ISRO Mechanical & Aerospace 2011 - Question 59

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The number of degree of freedom of a five link plane mechanism with five revolute pairs is

Figure for question 59, ISRO Mechanical & Aerospace 2011

  • A. 3
  • B. 4
  • C. 1
  • D. 2

Answer: D. 2

Explanation: $F = 3(N-1) - 2J = 3(5-1) - 2(5) = 12 - 10 = 2$.

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