The extension of a circular bar tapering uniformly from diameter d₁ at one end to diameter d₂ at the other end and subjected to an axial pull of P is given by
δl = 4PE/ πl²
δl = 4πld²/PE
δl = 4Pl/πEd₁d₂
δl = 4PlE/ πd₁d₂
Correct Answer :
C. δl = 4Pl/πEd₁d₂
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