What is the percentage change in length of 1m iron rod if the temperature change by $100^{\circ}$ C ( $α = 2 \times 10^{- 5} /^{\circ}$ )C

$0.1 %$

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$0.2 %$

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$0.3 %$

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$0.4 %$

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Solution

The correct option is B
$0.2 %$

$l^{1} = l (1 + α Δ T) \Rightarrow l^{1} - l = l \times Δ T \Rightarrow \frac{Δ l}{l} = α Δ T = 2 \times 10^{- 5} \times 100 = 2 \times 10^{- 3}$
% change in length = $\frac{Δ l}{l} \times 100 = 2 \times 10^{- 3} \times 100 = 0.2 %$

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What is the percentage change in length of 1m iron rod if the temperature change by $100^{\circ}$ C ( $α = 2 \times 10^{- 5} /^{\circ}$ )C

Q. What is the percentage change in length of

1 m

iron rod if its temperature changes by

100^{\circ} C

? [

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for iron is

2 \times 10^{- 5} m /^{\circ} C

]

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. If a rod suffers a longitudinal strain of 2 x 10

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5 \times 10^{8}

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What is the percentage change in length of 1m iron rod if the temperature change by 100∘C (α=2×10−5/∘)C

The correct option is B 0.2% l1=l(1+αΔT)⇒l1−l=l×ΔT⇒Δll=αΔT=2×10−5×100=2×10−3 % change in length = Δll×100=2×10−3×100=0.2%

What is the percentage change in length of 1m iron rod if the temperature change by $100^{\circ}$ C ( $α = 2 \times 10^{- 5} /^{\circ}$ )C

The correct option is B
$0.2 %$

$l^{1} = l (1 + α Δ T) \Rightarrow l^{1} - l = l \times Δ T \Rightarrow \frac{Δ l}{l} = α Δ T = 2 \times 10^{- 5} \times 100 = 2 \times 10^{- 3}$
% change in length = $\frac{Δ l}{l} \times 100 = 2 \times 10^{- 3} \times 100 = 0.2 %$