A meter scale made of steel reads accurately at $20^{\circ} C .$ In a sensitive experiment $,$ distances accurate up to $0.055 m m$ 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 $= 11 \times 10^{- 6}^{\circ} C .$

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Solution

$T_{1} = 20^{o} C, Δ L = 0.055 m m = 0.55 \times 10^{- 3} m$
$t_{2} = ?, α_{s t} = 11 \times 10^{- 6} /^{o} C$

We know, the change in length of scale will be:
$Δ L = L_{0} α Δ T$

According to the question:
$0.055 \times 10^{- 3} = 1 \times 11 \times 10^{- 6} \times (T_{1} + T_{2})$
$0.055 = (11 \times 10^{- 3} \times 20) + (11 \times 10^{- 3} \times T_{2})$
$T_{2} = 20 + 5 = 25^{o} C$
or $20 - 5 = 15^{o} C$
The expt. Can be performed from $15$ to $25^{o} C$

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A meter scale made of steel reads accurately at 20∘C. In a sensitive experiment, distances accurate up to 0.055mm in 1m 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.

T1=20oC,ΔL=0.055mm=0.55×10−3mt2=?,αst=11×10−6/oCWe know, the change in length of scale will be:ΔL=L0αΔTAccording to the question:0.055×10−3=1×11×10−6×(T1+T2)0.055=(11×10−3×20)+(11×10−3×T2)T2=20+5=25oCor 20−5=15oCThe expt. Can be performed from 15 to 25oC