A wire of cross sectional area A, modulus of elasticity 2×1011Nm−2 and length 2 m is stretched between two vertical rigid supports. When a mass of 2 kg is suspended at the middle it sags lower from its original position making angle θ=1001 radian on the points of support. The value of A is ______ ×10−4m2 (consider x≪L ).
(given g=10m/s2 )
Answer: 1
Solution
Let the midpoint sag be x; each half-span (length L/2=1 m) makes angle θ=x/(L/2), so x=θ×1=0.01 m for θ=1/100 rad. Vertical equilibrium at the midpoint (mass 2 kg) gives 2Tsinθ=mg; with sinθ≈θ, T=2θmg=2×0.012×10=1000 N. The extension of each half-span is (L/2)2+x2−L/2≈Lx2, so the total extension is ΔL=L/2x2=1(0.01)2=1×10−4 m, giving strain ΔL/L=5×10−5. From Y=ΔL/LT/A:
