A thin metal wire of length l increases in length by $0.1$ on heating it from temperature $θ_{1}$ to $θ_{2}$ . If a plate of same material is heated from $θ_{1}$ to $θ_{2}$ , which has dimensions $2 l \times 3 l$ then percentage increase in its area is:

0.1

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0.05

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0.2

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0.6

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Solution

The correct option is C $0.2$
$Δ L = 0.1$

$Δ L = L \times Δ T$

$0.1 = L \times (θ_{2} - θ_{1})$

$L \times \frac{0.1}{θ_{2} - θ_{1}}$

For plate $2 L \times 3 l$

$Δ L = 2 L \times Δ T$

$Δ L = 2 \times L \frac{0.1}{(θ_{2} - θ_{1})} \times (θ_{2} - θ_{1})$

$= 0.2 L$

$Δ w = 0.3 L$

$\frac{Δ / A}{A} = \frac{Δ L}{L} + \frac{Δ w}{w} = \frac{0.2 L}{2 L} + \frac{0.3 L}{3 L}$

$= 0.1 + 0.1 = 0.2$

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A thin metal wire of length l increases in length by 0.1 on heating it from temperature θ1 to θ2. If a plate of same material is heated from θ1 to θ2 , which has dimensions 2l×3l then percentage increase in its area is:

The correct option is C 0.2ΔL=0.1 ΔL=L×ΔT 0.1=L×(θ2−θ1) L×0.1θ2−θ1 For plate 2L×3l ΔL=2L×ΔT ΔL=2×L0.1(θ2−θ1)×(θ2−θ1) =0.2L Δw=0.3L Δ/AA=ΔLL+Δww=0.2L2L+0.3L3L =0.1+0.1=0.2

A thin metal wire of length l increases in length by $0.1$ on heating it from temperature $θ_{1}$ to $θ_{2}$ . If a plate of same material is heated from $θ_{1}$ to $θ_{2}$ , which has dimensions $2 l \times 3 l$ then percentage increase in its area is: