A Steel rod with Y=2.0×1011Nm−2 and α=10−5C -1 o of length 4m and area of cross-section 10cm2 is heated from 0°C to 400°C without being allowed to extend. The tension produced in the rod is x×105N where the value of X is –––––––

Original length=l0(1+αΔT)Compression =l0αΔT∴ stress =γ( strain )=γl0αΔTl0(1+αΔT)=γαΔT[∵α is very small ] So, F=stress × area =γαAΔT⇒F=10×10−42×101110-5(400)⇒F=8×105N=x×105N⇒x=8

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A Steel rod with $Y = 2 .0 \times 10^{11} {Nm}^{- 2} and α = 10^{- 5} {}^{o}C^{- 1}$ of length 4m and area of cross-section 10cm² is heated from 0°C to 400°C without being allowed to extend. The tension produced in the rod is $x \times 10^{5}$ N where the value of X is –––––––

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answer is 8.

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Detailed Solution

Original length $= l_{0} (1 + αΔT)$
Compression $= l_{0} αΔT$
$\begin{array}{l} ∴ stress = γ (strain) = \frac{{γl}_{0} αΔT}{l_{0} (1 + αΔT)} = γαΔT [∵ α is very small] \\ So, F=stress \times area = γαAΔT \\ \Rightarrow F = (10 \times 10^{- 4}) (2 \times 10^{11}) (10^{- 5}) (400) \\ \Rightarrow F = 8 \times 10^{5} N = x \times 10^{5} N \Rightarrow x = 8 \end{array}$