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Physics · Mechanical Properties of Matter

JEE Main 2026 — 6 April, Morning Shift — Question 7

The two wires A and B of equal cross-section but of different materials are joined together. The ratio of Young's modulus of wire A and wire B is 20/11. When the joined wire is kept under certain tension the elongations in the wires A and B are equal. If the length of wire A is 2.2 m, then the length of wire B is ______ m.

  1. Option A:

    1.1

  2. Option B:

    2.22

  3. Option C:

    1.21

    Correct
  4. Option D:

    4.44

Answer: C

Step-by-step solution

ΔL=FLAY\Delta L = \frac{FL}{AY}. For equal elongation, LA/YA=LB/YBL_A/Y_A = L_B/Y_B. So LB=LA(YB/YA)=2.2×(11/20)=1.21L_B = L_A (Y_B/Y_A) = 2.2 \times (11/20) = 1.21 m.

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
6 April, Morning Shift
Subject
Physics
Chapter
Mechanical Properties of Matter
Topic
Stress,Strain and Modulus of Elasticity

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