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.

Answer 1

The percentage error if the scale is used

a)on a summer day is  $3.3 \times 10^{-2} \%$

b)on a winter day  is $1.1 \times 10^{-5}$

Explanation:

Given:

Initial temp,$t_1=16 ^{\circ}C$

Solution:

Let the correct length determined by a steel 16 ° C meter scale be L

A)Temperature on hot day of summer $t_2=46 ^{\circ}C$

So the temperature changes, $\Delta \theta =t_2$

$-t_1=30^{\circ}C$

Linear expansion ratio for steel

Given:

Let the correct length determined by a steel 16 ° C meter scale be L

Initial temp.  t1=16 oC

(a)Temperature on hot day of summer $t_2 = 46 ^{\circ}C$

So the temperature changes, ∆θ=t2

$-t_1 = 30^{\circ}C$

Linear expansion ratio for steel

$\alpha=1.1 \times 10^{-5} ^{\circ} C$

So change in length

$\Delta L=L \alpha \Delta \theta=L \times 1.1 \times 10^{-5} \times 30$

$\% \text { of error }=\left(\frac{\Delta L}{L} \times 100\right) \%$

= $\left(\frac{\ln \Delta \theta}{L} \times 100\right) \%$

$=\left(1.1 \times 10^{-5} \times 30 \times 100\right) \%$

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

(b) Winter-day temperature,$t_2 = 6 ^{\circ}C$

So the temperature changes, $\Delta \theta=t_2$

$- t_2=10^{\circ}C$

$\Delta L=L_{2}$

$-L=L \alpha \Delta \theta=L \times 1.1 \times 10^{-5} \times 30$

$\% \text { of error }=\left(\frac{\Delta L}{L} \times 100\right) \%$

$=\left(\frac{L \alpha \Delta \theta}{L} \times 100\right) \%$

$=\left(1.1 \times 10^{-5} \times 10 \times 100\right) \%$

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

Answer 2

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