Two wires A and B have equal cross-sectional areas but are made of different materials. Their Young's moduli satisfy $\frac{Y_{A}}{Y_{B}}=\frac{20}{11}$ . The wires are joined and subjected to the same tension, and their elongations are found to be equal. If the length of wire A is 2.2 m, what is the length of wire B?
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Question

Two wires A and B have equal cross-sectional areas but are made of different materials. Their Young's moduli satisfy $\frac{Y_{A}}{Y_{B}}=\frac{20}{11}$ . The wires are joined and subjected to the same tension, and their elongations are found to be equal. If the length of wire A is 2.2 m, what is the length of wire B?

Options

G1 Physics · Accepted Answer

For a wire under tension F, ΔL = FL ÷ (YA). The two wires have the same tension F and the same cross-sectional area A. Their elongations are equal, so L(A) ÷ Y(A) = L(B) ÷ Y(B). Hence, L(B) = L(A)Y(B) ÷ Y(A). Given, Y(A) ÷ Y(B) = 20 ÷ 11, so Y(B) ÷ Y(A) = 11 ÷ 20. Therefore, L(B) = 2.2×11 ÷ 20 = 1.21 m. Correct option: C.