A piece of steel has a length of 30 cm at 15∘C. At 90∘C its length increases by 0.027 cm. Its coefficient of linear expansion is

ΔT=90−15oC=75oC, Δl=0.027 cmΔl=L α ΔTΔlΔT × L= α ⇒ 0.02775×30 =α⇒ α = 12× 10−6/oC

A piece of steel has a length of 30 cm at 15∘C. At 90∘C its length increases by 0.027 cm. Its coefficient of linear expansion is

ΔT=90−15oC=75oC, Δl=0.027 cmΔl=L α ΔTΔlΔT × L= α ⇒ 0.02775×30 =α⇒ α = 12× 10−6/oC

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A piece of steel has a length of 30 cm at $15^{\circ} C . At 90^{\circ} C$ its length increases by 0.027 cm. Its coefficient of linear expansion is

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$12 \times 10^{- 6} /^{o} C$

$1.2 \times 10^{- 6} /^{o} C$

$12 \times 10^{6} /^{o} C$

$3.2 \times 10^{6} /^{o} C$

answer is A.

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Detailed Solution

$Δ T = {(90 - 15)}^{o} C = 75^{o} C, Δ l = 0.027 c m$

$Δ l = L α Δ T$

$\frac{Δ l}{Δ T \times L} = α \Rightarrow \frac{0.027}{75 \times 30} = α$

$\Rightarrow α = 12 \times 10^{- 6} /^{o} C$

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A piece of steel has a length of 30 cm at 15∘C. At 90∘C its length increases by 0.027 cm. Its coefficient of linear expansion is

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A piece of steel has a length of 30 cm at $15^{\circ} C . At 90^{\circ} C$ its length increases by 0.027 cm. Its coefficient of linear expansion is