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Question: A steel tape calibrated at \[20^\circ C\] . On a cold day when the temperature is \[ - 15\;^\circ C\...

A steel tape calibrated at 20∘C20^\circ C . On a cold day when the temperature is −15  ∘C - 15\;^\circ C , The percentage error in the tape is −x 1000 %- \dfrac{{x{\text{ }}}}{{1000{\text{ }}}}\% . Find xx .
Coefficient of linear thermal expansion for steel, α=12×10−6K−1\alpha = 12 \times {10^{ - 6}}{K^{ - 1}}

Explanation

Solution

If the temperature is raised the length of an object increases in a certain direction – this is the linear expansion of length of that object. Use the formula of the change in length of the steel tape given in the problem w.r.t the temperature change.
Find the formula of the percentage error for this expansion and put the value of length expansion.
Compare the percentage error you get in your calculation with the given value of that in the problem and find the required term.

Formula used:
If the length at temperature t1{t_1} is ll , the length at the temperature t2{t_2} is l′l' and the coefficient of linear expansion is α\alpha ,
The change in length, (l′−l)=αl(t2−t1)(l' - l) = \alpha l({t_2} - {t_1})
The percentage error = Δll×100\dfrac{{\Delta l}}{l} \times 100
Where, Δl=(l′−l)\Delta l = (l' - l)

Complete step by step answer:
For this problem, let the length at temperature t1{t_1} is ll , the length at the temperature t2{t_2} is l′l' and the coefficient of linear expansion is α\alpha ,
Then the change in length, (l′−l)=αl(t2−t1)(l' - l) = \alpha l({t_2} - {t_1})
α\alpha is a constant term. Its value is fixed for a material.

Given, t1=20∘C{t_1} = 20^\circ C
t2=−15∘C{t_2} = - 15^\circ C
α=12×10−6K−1\alpha = 12 \times {10^{ - 6}}{K^{ - 1}}
(l′−l)=12×10−6×l×(−15−20)(l' - l) = 12 \times {10^{ - 6}} \times l \times ( - 15 - 20)
⇒(l′−l)l=−420×10−6\Rightarrow \dfrac{{(l' - l)}}{l} = - 420 \times {10^{ - 6}}
The percentage error = Δll×100\dfrac{{\Delta l}}{l} \times 100
Where, Δl=(l′−l)\Delta l = (l' - l)

∴\therefore The percentage error =−420×10−6×100 = - 420 \times {10^{ - 6}} \times 100
⇒−421000%\Rightarrow - \dfrac{{42}}{{1000}}\%
In the problem the percentage error is given by, −x 1000 %- \dfrac{{x{\text{ }}}}{{1000{\text{ }}}}\%.
Hence, the value of x=42x = 42.

Note: If the temperature of a solid object is raised, the expansion of the object to a certain distance is called the linear expansion. It is shown that the expansion of the length due to the rise of the temperature is directly proportional to the initial length of the object at a certain temperature and the change in temperature.
The coefficient of linear expansion defined by the ratio of the expansion in length to the product of initial length and raised temperature.