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Question
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.
Solution
(a) Let the correct length measured by a metre scale made up of steel 16 °C be L.
Initial temperature, t1 = 16 °C
Temperature on a hot summer day, t2 = 46 °C
So, change in temperature, Δθ = t2
\[-\]t1 = 30 °C
Coefficient of linear expansion of steel,
\[\alpha\]= 1.1 × 10–5 °C-1
Therefore, change in length,
ΔL = L αΔθ = L × 1.1 × 10–5 × 30
% of error =`((ΔL)/L × 100) %`
= `((L∝Δθ)/L × 100)%`
=[1.1 × 10-5 × 30 ×100]%
=3.3 × 10-2 %
(b) Temperature on a winter day, t2 = 6 °C
So, change in temperature, Δθ = t1
\[-\]t2 = 10 °C
ΔL = L2
\[-\] L1 = L αΔθ = L × 1.1 × 10–5 × 10
`text/"% of error" = (ΔL/L × 100)%`
\[ = \left( \frac{L\alpha \Delta\theta}{L} \times 100 \right) % \]
\[ = 1 . 1 \times {10}^{- 5} \times 10 \times 100 % \]
\[ = 1 . 1 \times {10}^{- 2} \]
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