A metal beam is 6000 mm long at a temperature of 22-C- On a hot day- the temperature rises to 42-C- What is the change in the length of the beam due to thermal expansion-Take -metal - 1-6- 10-5 -C-1

We know that rise in length is Δ l=lαΔ T Now Δ T=42 20=20^oC #160;Length =6000mm=600cm Δ L=600× 1.6× 10^ 5× 20 Δ L=0.192 cm #160; ...

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Standard XII

Physics

Thermal Expansion

A metal beam ...

Question

A metal beam is $6000 m m$ long at a temperature of $22^{\circ} C$ . On a hot day, the temperature rises to $42^{\circ} C$ . What is the change in the length of the beam due to thermal expansion?
(Take $α_{m e t a l} = 1.6 \times 10^{- 5}^{\circ} C^{- 1}$ )

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Solution

We know that rise in length is $Δ l = l α Δ T$
Now $Δ T = 42 - 20 = 20^{o} C$
Length =6000mm=600cm
$Δ L = 600 \times 1.6 \times 10^{- 5} \times 20$
$Δ L = 0.192 c m$

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A metal beam is 6000 mm long at a temperature of 22∘C. On a hot day, the temperature rises to 42∘C. What is the change in the length of the beam due to thermal expansion?(Take αmetal=1.6×10−5 ∘C−1)

We know that rise in length is Δl=lαΔTNow ΔT=42−20=20oC Length =6000mm=600cmΔL=600×1.6×10−5×20ΔL=0.192cm

A metal beam is $6000 m m$ long at a temperature of $22^{\circ} C$ . On a hot day, the temperature rises to $42^{\circ} C$ . What is the change in the length of the beam due to thermal expansion?
(Take $α_{m e t a l} = 1.6 \times 10^{- 5}^{\circ} C^{- 1}$ )

We know that rise in length is $Δ l = l α Δ T$
Now $Δ T = 42 - 20 = 20^{o} C$
Length =6000mm=600cm
$Δ L = 600 \times 1.6 \times 10^{- 5} \times 20$
$Δ L = 0.192 c m$