SSC JE Mechanical Engineering 2018 Set 5 Paper

F= 3(n – 1) – 2l – h
F = 3(3 – 1) – 2 × 1 – 2
F = 2
F = total degrees of freedom in the mechanism
n = number of links (including the frame)
l = number of lower pairs (one degree of freedom)
h = number of higher pairs (two degrees of freedom)