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JEE Main 2023 · Physics · Fluid Mechanics, Elasticity and Surface Tension · medium

A steel wire of length 2 m, cross-sectional area 1 mm^2 and Young's modulus 2×10^11 N m^-2 hangs from a rigid support. A 2 kg mass is attached to its…

Single correct option

A steel wire of length 2 m2\ \mathrm{m}, cross-sectional area 1 mm21\ \mathrm{mm^2} and Young's modulus 2×1011 N m−22\times10^{11}\ \mathrm{N\,m^{-2}} hangs from a rigid support. A 2 kg2\ \mathrm{kg} mass is attached to its lower end with the wire just taut and unstretched, and is then released suddenly. At the lowest point of its subsequent motion, the elastic energy stored in the wire is (g=10 m s−2g=10\ \mathrm{m\,s^{-2}})
  1. A. 4 mJ4\ \mathrm{mJ}
  2. B. 16 mJ16\ \mathrm{mJ}
  3. C. 8 mJ8\ \mathrm{mJ}
  4. D. 20 mJ20\ \mathrm{mJ}
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Answer: (C) 8 mJ8\ \mathrm{mJ}

Wire stiffness k=YAL=2×1011×10−62=105 N m−1k=\dfrac{YA}{L}=\dfrac{2\times10^{11}\times10^{-6}}{2}=10^5\ \mathrm{N\,m^{-1}}. At the lowest point the speed is zero, so the loss of gravitational energy equals the stored energy: mgx=12kx2mgx=\tfrac12kx^2, giving x=2mgk=4×10−4 mx=\dfrac{2mg}{k}=4\times10^{-4}\ \mathrm{m} (twice the static extension). Stored energy =12kx2=12×105×16×10−8=8×10−3 J=8 mJ=\tfrac12kx^2=\tfrac12\times10^5\times16\times10^{-8}=8\times10^{-3}\ \mathrm{J}=8\ \mathrm{mJ}. The static value 12kx02=2 mJ\tfrac12 kx_0^2=2\ \mathrm{mJ} is a common wrong pick, and mgx=8 mJmgx=8\ \mathrm{mJ} confirms the answer.
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