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JEE Main 2026 (6 Apr, Shift 2) · Wire Combinations & Special Cases

Wire Combinations & Special Cases question from JEE Main 2026 (6 Apr, Shift 2)

Q3JEE Main 2026 · 6 APR · SHIFT 2Medium

Wire Combinations & Special Cases

Figure represents the extension (Δl)(\Delta l) of a wire of length 1 meter, suspended from the ceiling of the room at one end with a load WW connected to the other end. If the cross-sectional area of the wire is 10−5 m210^{-5}\text{ m}^2 then the Young's modulus of the wire is ‾ N/m2\underline{\qquad}\ \text{N/m}^2.

Figure

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Answer: (c)

  1. From the graph, Δl=1×10−4 m\Delta l = 1 \times 10^{-4}\text{ m} for W=10 NW = 10\text{ N}.
  2. Y=WLAΔl=10×110−5×10−4Y = \frac{WL}{A\Delta l} = \frac{10 \times 1}{10^{-5} \times 10^{-4}}
  3. Y=1.0×1010 N/m2Y = 1.0 \times 10^{10}\text{ N/m}^2, so the answer is (c).
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