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We would like to increase the length of a 15 cm long copper rod of cross-section $4 m m^{2} b y 1 m m .$ The energy absorbed by the rod if it is heated for this is $E_{1} .$ The energy absorbed by the rod if it is stretched slowly for this is $E_{2} .$ Then find $E_{1} / E_{2}$ is [Various parameters of copper are : Density $= 9 \times 10^{3} k g / m^{3},$ thermal coefficient of linear expansion $= 16 \times 10^{- 6} K^{- 1},$ Young’s modulus $= 135 \times 10^{9} P a$ and specific heat $= 400 J / k g - K]$

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answer is 500.

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Detailed Solution

Temp. is increased by $Δ θ$ then $Δ l = l α Δ θ \Rightarrow Δ θ = \frac{Δ l}{l α}$

$E_{1} = (ρ A l) S Δ θ = ρ A l S \frac{Δ l}{l α}, E_{2} = \frac{1}{2} (Y \frac{Δ l}{l}) (\frac{Δ l}{l 2}) \times A l = \frac{Y {(Δ l)}^{2} A}{2 l}$

$S o, \frac{E_{1}}{E_{2}} = \frac{ρ A l S Δ l \times 2 l}{l \times Y {(Δ l)}^{2} A} = \frac{2 ρ S l}{α (Δ l) Y} = 500$

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