A wire A has length 0.314 m and Young's modulus 2×10¹⁰ N·m⁻². It is connected in series with another wire B whose length and Young's modulus are both twice those of A. Both wires have radius 0.2 mm. A mass of 0.8 kg is suspended from the free end of the combination. Neglect the masses of the wires and take g = 10 m·s⁻². What is the total extension of the two-wire combination?
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Question

A wire A has length 0.314 m and Young's modulus 2×10¹⁰ N·m⁻². It is connected in series with another wire B whose length and Young's modulus are both twice those of A. Both wires have radius 0.2 mm. A mass of 0.8 kg is suspended from the free end of the combination. Neglect the masses of the wires and take g = 10 m·s⁻². What is the total extension of the two-wire combination?

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G1 Physics · Accepted Answer

The same tension acts in both wires: F = mg = 0.8×10 = 8 N. The cross-sectional area of each wire is A = πr² = 3.14×(2×10⁻⁴)². For wire A: ΔL(A) = FL ÷ (AY). For wire B: L(B) = 2L, Y(B) = 2Y. Hence, ΔL(B) = F(2L) ÷ [A(2Y)] = FL ÷ (AY) = ΔL(A). Therefore, ΔL(total) = 2FL ÷ (AY). Substitute: ΔL(total) = 2×8×0.314 ÷ [3.14×(2×10⁻⁴)²×2×10¹⁰] = 2×10⁻³ m = 2 mm. Correct option: B.