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Question: An aluminium measuring rod, which is correct at \(5{}^\circ C\) measures the length of a line as \(8...

An aluminium measuring rod, which is correct at 5C5{}^\circ C measures the length of a line as 80cm80cm at 45C45{}^\circ C. If thermal coefficient of linear expansion of aluminium is 2.5×105C12.5\times {{10}^{-5}}{}^\circ {{C}^{-1}}, the correct length of the line is
A)80.08 B)79.92 C)81.12 D)79.62 \begin{aligned} & A)80.08 \\\ & B)79.92 \\\ & C)81.12 \\\ & D)79.62 \\\ \end{aligned}

Explanation

Solution

When an object undergoes change in temperature, the object tends to expand or contract. Coefficient of thermal expansion of an object refers to the ratio of strain of the object to the change in temperature experienced by the object. Strain of an object refers to the ratio of change in length of the object to the actual length of the object.

Formula used:
1)ε=ΔLL1)\varepsilon =\dfrac{\Delta L}{L}
2)α=ΔLLΔT2)\alpha =\dfrac{\Delta L}{L\Delta T}

Complete step-by-step answer:
When an object is heated or cooled, the object tends to deform in shape. Such an object is said to experience a strain. Strain of an object is defined as the ratio of change in length of the object to the actual length of the object. It is given by
ε=ΔLL\varepsilon =\dfrac{\Delta L}{L}
where
ε\varepsilon is the strain of an object
ΔL\Delta L is the change in length of the object
LL is the actual length of the object
Let this be equation 1.
Coefficient of thermal expansion of the material of an object is defined as the ratio of strain of the object to the change in temperature experienced by that object. It is given by
α=strainΔT=ΔLLΔT\alpha =\dfrac{strain}{\Delta T}=\dfrac{\Delta L}{L\Delta T}
where
α\alpha is the coefficient of thermal expansion of the material of an object
ΔL\Delta L is the change in length of the object
LL is the actual length of the object
ΔT\Delta T is the change in temperature experienced by the object
Let this be equation 2.
Coming to our question, we are provided with an aluminium rod, whose coefficient of thermal expansion is 2.5×105C12.5\times {{10}^{-5}}{}^\circ {{C}^{-1}}. This aluminium rod is used to measure the length of a line. It is said that this aluminium rod is correct in measuring the length of the given line, when it has a temperature of 5C5{}^\circ C. This means that the actual length of the line and the measured length of the line are equal, if measured using the given aluminium rod at 5C5{}^\circ C. It is also given that this rod measures the distance of the line as 80cm80cm at 45C45{}^\circ C. This suggests that the measured length of the line using the aluminium rod at 45C45{}^\circ C is equal to 80cm80cm. We are required to determine the actual length of the line.
From the information given above, we have
α=2.5×105C1 ΔT=45C5C=40C ΔLL=L80cmL=180cmL \begin{aligned} & \alpha =2.5\times {{10}^{-5}}{}^\circ {{C}^{-1}} \\\ & \Delta T=45{}^\circ C-5{}^\circ C=40{}^\circ C \\\ & \dfrac{\Delta L}{L}=\dfrac{L-80cm}{L}=1-\dfrac{80cm}{L} \\\ \end{aligned}
where
α\alpha is the coefficient of thermal expansion of aluminium
ΔL=(L80cm)\Delta L=(L-80cm), is the change in length of the line
LL is the actual length of the line
ΔT=40C\Delta T=40{}^\circ C, is the change in temperature experienced by the line
Substituting these values in equation 2, we have
α=ΔLLΔT2.5×105C1=180cmL40C180cmL=103L=80cm10.001=80cm0.999=80.08cm\alpha =\dfrac{\Delta L}{L\Delta T}\Rightarrow 2.5\times {{10}^{-5}}{}^\circ {{C}^{-1}}=\dfrac{1-\dfrac{80cm}{L}}{40{}^\circ C}\Rightarrow 1-\dfrac{80cm}{L}={{10}^{-3}}\Rightarrow L=\dfrac{80cm}{1-0.001}=\dfrac{80cm}{0.999}=80.08cm
Therefore, the actual length of the line is 80.08cm80.08cm. The correct option to be marked is AA.

So, the correct answer is “Option A”.

Note: For this question, students can also use the formula of linear expansion as follows:
Lt=L0(1+αΔT){{L}_{t}}={{L}_{0}}(1+\alpha \Delta T)
where
Lt{{L}_{t}} is the actual length of the line
L0{{L}_{0}} is the measured length of the line
α\alpha is the coefficient of thermal expansion
ΔT\Delta T is the change in temperature experienced by the line
This method would be a direct approach to the answer.