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JEE Main 2016 (9 Apr, Shift 1) · Wire Combinations & Special Cases

Wire Combinations & Special Cases question from JEE Main 2016 (9 Apr, Shift 1)

Q26JEE Main 2016 · 9 APR · SHIFT 1Difficult

Wire Combinations & Special Cases

A uniformly tapering conical wire is made from a material of Young’s modulus Y and has a normal, unextended length L. The radii, at the upper and lower ends of this conical wire, have values R and 3 R, respectively. The upper end of the wire is fixed to a rigid support and a mass M is suspended from its lower end. The equilibrium extended length, of this wire, would equal:

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Answer: (b)

  1. Radius at distance xx from the top: r=R(1+2xL)r = R\left(1 + \frac{2x}{L}\right); tension is MgMg throughout.
  2. ΔL=∫0LMg dxπr2Y=MgπYR2∫0Ldx(1+2x/L)2\Delta L = \int_0^L \frac{Mg\,dx}{\pi r^2 Y} = \frac{Mg}{\pi Y R^2}\int_0^L \frac{dx}{(1 + 2x/L)^2}
  3. ∫0Ldx(1+2x/L)2=L2(1−13)=L3\int_0^L \frac{dx}{(1 + 2x/L)^2} = \frac{L}{2}\left(1 - \frac{1}{3}\right) = \frac{L}{3} (same as ΔL=MgLπYR1R2\Delta L = \frac{MgL}{\pi Y R_1R_2} with R1R2=3R2R_1R_2 = 3R^2)
  4. Extended length =L(1+13MgπYR2)= L\left(1 + \frac{1}{3}\frac{Mg}{\pi Y R^2}\right), so the answer is (b).
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