A metre scale is made up of steel and measures correct length at $16^{\circ} C$ . What will be the percentage error if this scale is used (a) on a summer day when the temperature is $46^{\circ} C$ and (b) on a winter day when the temperature is $6^{\circ} C$ ? Coefficient of linear expansion of steel $= 11 \times 10^{- 6}^{\circ} C^{- 1}$

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Solution

(a) Length at $16^{\circ} C$ = L,

$t_{1} = 16^{\circ} C, t_{2} = 46^{\circ}$

$α = 1.1 \times 10^{- 5} /^{\circ} C$

$Δ L = L α Δ θ = L \times 1.1 \times 10^{- 5} \times 30$

% of error $= (\frac{Δ L}{L} \times 100) %$

$= (\frac{L α Δ θ}{L} \times 100) %$

$= [1.1 \times 10^{- 5} \times 30 \times 100] %$

$= 3.3 \times 10^{- 2} % = 0.033 %$

(b) $t_{2} = 6^{\circ} C$

% error $= (\frac{Δ L}{L} \times 100) %$

$= (\frac{L α Δ θ}{L} \times 100) %$

$= - 1.1 \times 10^{- 5} \times 10 \times 100 %$

$= - 1.1 \times 10^{- 2} = - 0.011 %$

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A metre scale is made up of steel and measures correct length at 16∘C. What will be the percentage error if this scale is used (a) on a summer day when the temperature is 46∘C and (b) on a winter day when the temperature is 6∘C ? Coefficient of linear expansion of steel =11×10−6 ∘C−1

(a) Length at 16∘C = L, t1=16∘ C, t2=46∘ α=1.1×10−5/∘C ΔL=LαΔθ=L×1.1×10−5×30 % of error =(ΔLL×100)% =(Lα ΔθL×100)% =[1.1×10−5×30×100]% =3.3×10−2%=0.033% (b) t2=6∘ C % error =(ΔLL×100)% =(LαΔθL×100)% =−1.1×10−5×10×100% =−1.1×10−2=−0.011%