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JEE Main 2026 (4 Apr, Shift 2) · Wire Combinations & Special Cases

Wire Combinations & Special Cases question from JEE Main 2026 (4 Apr, Shift 2)

Q1JEE Main 2026 · 4 APR · SHIFT 2Medium

Wire Combinations & Special Cases

A metal string AA is suspended from a rigid support and its free end is attached to a block of mass MM. Second block having mass 2 M is suspended at the bottom of the first block using a string BB. The area of cross sections of strings AA and BB are same. The ratio of lengths of strings of AA to B is 2 and the ratio of their Young's moduli (YA/YB)(Y_A/Y_B) is 0.5. The ratio of elongations in AA to BB is ______.

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Answer: (d)

  1. Tension in A (holds both blocks) =3Mg= 3Mg; tension in B =2Mg= 2Mg.
  2. Same area: ΔlAΔlB=TATB⋅LALB⋅YBYA\frac{\Delta l_A}{\Delta l_B} = \frac{T_A}{T_B} \cdot \frac{L_A}{L_B} \cdot \frac{Y_B}{Y_A}
  3. =32×2×10.5=6= \frac{3}{2} \times 2 \times \frac{1}{0.5} = 6
  4. So the answer is (d) 6.
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