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JEE Main 2020 (9 Jan, Shift 2) · Elastic Potential Energy

Elastic Potential Energy question from JEE Main 2020 (9 Jan, Shift 2)

Q9JEE Main 2020 · 9 JAN · SHIFT 2Medium

Elastic Potential Energy

Two steel wires having same length are suspended from a ceiling under the same load. If the ratio of their energy stored per unit volume is 1: 4, the ratio of their diameters is:

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Answer: (d)

  1. u=S22Yu = \dfrac{S^2}{2Y} with S=Fπd2/4S = \dfrac{F}{\pi d^2/4}, so for the same load and material u∝1d4u \propto \dfrac{1}{d^4}.
  2. u1u2=(d2d1)4=14\dfrac{u_1}{u_2} = \left(\dfrac{d_2}{d_1}\right)^4 = \dfrac{1}{4}
  3. d1d2=41/4=2\dfrac{d_1}{d_2} = 4^{1/4} = \sqrt{2}
  4. So the answer is (d) 2:1\sqrt{2} : 1.
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