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JEE Main 2018 · Physics · Fluid Mechanics, Elasticity and Surface Tension · medium

A steel rod of cross-sectional area 4 cm^2 is held between two rigid walls, just touching them, at 20^circC. The rod is heated uniformly to 70^circC.…

Single correct option

A steel rod of cross-sectional area 4 cm24\ \mathrm{cm^2} is held between two rigid walls, just touching them, at 20∘C20^\circ\mathrm{C}. The rod is heated uniformly to 70∘C70^\circ\mathrm{C}. For steel, Y=2×1011 N m−2Y=2\times10^{11}\ \mathrm{N\,m^{-2}} and α=1.2×10−5 ∘C−1\alpha=1.2\times10^{-5}\ {}^\circ\mathrm{C}^{-1}. The force exerted by the rod on each wall is
  1. A. 96 kN96\ \mathrm{kN}
  2. B. 48 kN48\ \mathrm{kN}
  3. C. 24 kN24\ \mathrm{kN}
  4. D. 4.8 kN4.8\ \mathrm{kN}
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Answer: (B) 48 kN48\ \mathrm{kN}

The walls prevent the free expansion ΔL=LαΔT\Delta L=L\alpha\Delta T, so the rod is compressed by the same amount. Strain =αΔT=1.2×10−5×50=6×10−4=\alpha\Delta T=1.2\times10^{-5}\times50=6\times10^{-4}. Stress =Y×strain=2×1011×6×10−4=1.2×108 N m−2=Y\times\text{strain}=2\times10^{11}\times6\times10^{-4}=1.2\times10^{8}\ \mathrm{N\,m^{-2}}. Force =stress×A=1.2×108×4×10−4=4.8×104 N=48 kN=\text{stress}\times A=1.2\times10^{8}\times4\times10^{-4}=4.8\times10^{4}\ \mathrm{N}=48\ \mathrm{kN}. The length of the rod cancels out, so it is not needed. The area must be converted from cm2\mathrm{cm^2} to m2\mathrm{m^2}, which is the usual source of a factor-of-1010 error.
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