Mechanical Properties of Solids

JEE Main level, single correct, medium

9 of 118

Question 9 of 118

The two wires AA and BB of equal cross-section but of different materials are joined together. The ratio of Young's modulus of wire AA and wire BB is 20/1120/11. When the joined wire is kept under certain tension the elongations in the wires AA and BB are equal. If the length of wire AA is 2.22.2 m, then the length of wire BB is _______ m.

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Answer: (C) 1.211.21

Given that the wires have equal cross-sectional area AA and are under the same tension FF, the elongation ΔL\Delta L is given by:

ΔL=FLAY\Delta L = \dfrac{F L}{A Y}

Since the elongations in both wires are equal (ΔLA=ΔLB\Delta L_A = \Delta L_B):

FLAAYA=FLBAYB\dfrac{F L_A}{A Y_A} = \dfrac{F L_B}{A Y_B}

LAYA=LBYB\dfrac{L_A}{Y_A} = \dfrac{L_B}{Y_B}

LB=LA×YBYAL_B = L_A \times \dfrac{Y_B}{Y_A}

Substituting the given values LA=2.2L_A = 2.2 m and YAYB=2011\dfrac{Y_A}{Y_B} = \dfrac{20}{11}:

LB=2.2×1120L_B = 2.2 \times \dfrac{11}{20}

LB=24.220=1.21L_B = \dfrac{24.2}{20} = 1.21 m

Answer: 1.211.21

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