The correct answer is (a) Ef = 4fLv̅²/2D
For explanation I would say: Ef = 4fLv̅²/2D is the energy loss due to friction in a circular conduit. The frictional losses are present due to sudden contraction or enlargement of the flow cross-sections. Where, v̅ = Average velocity, D = Diameter of the conduit, L = Length of the conduit and f = friction factor.