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Standard XII

Physics

Thermal Expansion

A steel beam ...

Question

A steel beam is $5 m$ long at a temperature of $20^{o} C$ . On a hot day, the temperature rises to $40^{o} C$ . What is the change in the length of the beam due to thermal expansion?

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Solution

Here, $l_{1} = 5 m; Δ T = 40 - 20 = 20^{o} C;$
$α = 1.2 \times 10^{- 5}^{o} C^{- 1}$
If $l_{2}$ is length of the beam at $40^{o} C$ , the $l_{2} = l_{1} (1 + α Δ T)$
$l_{2} = l_{1} (1 + α Δ T)$
$= 5 \times (1 + 1.2 \times 10^{- 5} \times 20)$
$= 5 \times (1 + 2.4 + 10^{- 4}) = 5.0012 m$
Therefore, increase in length of the beam,
$Δ l = l_{2} - l_{1} = 5.0012 - 5$
$= 0.0012 m = 1.2 m m$

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A steel beam is 5 m long at a temperature of 20oC. On a hot day, the temperature rises to 40oC. What is the change in the length of the beam due to thermal expansion?

Here, l1=5m;ΔT=40−20=20oC;α=1.2×10−5oC−1If l2 is length of the beam at 40oC, the l2=l1(1+αΔT)l2=l1(1+αΔT)=5×(1+1.2×10−5×20)=5×(1+2.4+10−4)=5.0012mTherefore, increase in length of the beam,Δl=l2−l1=5.0012−5=0.0012m=1.2mm

A steel beam is $5 m$ long at a temperature of $20^{o} C$ . On a hot day, the temperature rises to $40^{o} C$ . What is the change in the length of the beam due to thermal expansion?