A metre scale made of steel is calibrated at $20^{o} C$ to give correct reading. Find the distance between $50 c m$ mark and $51 c m$ mark if the scale is used at $10^{o} C$ . Coefficient of linear expansion of steel is $1.1 \times 10^{- 5}^{o} C^{- 1}$

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Solution

$t_{1} = 20^{o} C, t_{2} = 10^{o} C, L_{1} = 1 c m = 0.01 m, L_{2} = ?$
$α_{s t e e l} = 1.1 \times 10^{- 5} /^{o} C$
$L_{2} = L_{1} (1 + α_{s t e e l} Δ T) - 0.01 (1 + 101 \times 10^{- 5} \times 10) = 0.01 + 0.01 \times 1.1 \times 10^{- 4}$
$= 10^{4} \times 10^{- 5} + 1.1 \times 10^{- 5} = 10^{- 5} (10000 + 1.1) = 10001.1$
$= 1.000111 \times 10^{- 2} m = 1.000111 c m$

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A metre scale made of steel is calibrated at 20oC to give correct reading. Find the distance between 50cm mark and 51cm mark if the scale is used at 10oC. Coefficient of linear expansion of steel is 1.1×10−5 oC−1

t1=20oC,t2=10oC,L1=1cm=0.01m,L2=?αsteel=1.1×10−5/oCL2=L1(1+αsteelΔT)−0.01(1+101×10−5×10)=0.01+0.01×1.1×10−4=104×10−5+1.1×10−5=10−5(10000+1.1)=10001.1=1.000111×10−2m=1.000111cm

A metre scale made of steel is calibrated at $20^{o} C$ to give correct reading. Find the distance between $50 c m$ mark and $51 c m$ mark if the scale is used at $10^{o} C$ . Coefficient of linear expansion of steel is $1.1 \times 10^{- 5}^{o} C^{- 1}$