A metre scale made of steel reads accurately at $20^{\circ} C$ . In a sensitive experiment, distances accurate up to 0.055 mm in 1 m are required. Find the range of temperature in which the experiment can be performed with this metre scale. Coefficient of linear expansion of steel $= 11 \times 10^{- 6}^{\circ} C^{- 1}$

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Solution

$T_{1} = 20^{\circ} C$

$Δ L$ = 0.055 mm

$= 0.055 \times 10^{- 3} m$

$T_{2} = ?$

$α_{s t} = 11 \times 10^{- 6} /^{\circ} C$

we know $Δ L = L_{0} α Δ T$

$\Rightarrow 0.055 \times 10^{- 3} = 1 \times 11 \times 10^{- 6} \times T_{1} \pm T_{2})$

$\Rightarrow 5 \times 10^{- 3} = (20 \pm T_{2}) \times 10^{- 3}$

$\Rightarrow 20 \pm T_{2} = 5$

$\Rightarrow T_{2} = 20 + 5 = 25^{\circ} C$

or 20 - 5 = $15^{\circ}$

Hence the experiment can be performed from $15^{\circ}$ to $25^{\circ}$ C

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A metre scale made of steel reads accurately at 20∘C. In a sensitive experiment, distances accurate up to 0.055 mm in 1 m are required. Find the range of temperature in which the experiment can be performed with this metre scale. Coefficient of linear expansion of steel =11×10−6 ∘C−1

T1=20∘ C ΔL = 0.055 mm =0.055×10−3m T2=? αst=11×10−6/∘C we know ΔL=L0 αΔT ⇒ 0.055×10−3=1×11×10−6×T1±T2) ⇒ 5×10−3=(20±T2)×10−3 ⇒ 20±T2=5 ⇒ T2=20+5=25∘C or 20 - 5 = 15∘ Hence the experiment can be performed from 15∘ to 25∘ C