The Young's modulus of steel is twice that of brass. Two wires of same length and of same area of cross-section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weight added to the steel and brass wires must be in the ratio of:
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Question

The Young's modulus of steel is twice that of brass. Two wires of same length and of same area of cross-section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weight added to the steel and brass wires must be in the ratio of:

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G1 Physics · Accepted Answer

Step 1: Extension of a wire is ΔL = FL/(AY). Step 2: Both wires have equal original length L and equal cross-sectional area A. Their lower ends are to remain at the same level, so the extensions must be equal. Step 3: FₛL/(AYₛ) = FᵦL/(AYᵦ) ⇒ Fₛ/Fᵦ = Yₛ/Yᵦ. Step 4: Given Yₛ = 2Yᵦ, so Fₛ/Fᵦ = 2/1. Final answer: (B) 2 : 1.