Question

A steel tape measures the length of a copper rod as $90\,cm$ when both are at ${10^0}C$ the calibration temperature for the tape. What would the tape read for the length of the rod when both are at ${30^0}C$ ? Given for steel, $\alpha = 1.2 \times {10^{ - 5}}{C^{ - 1}}$ and for copper $\alpha = 1.7 \times {10^{ - 5}}{C^{ - 1}}$
A. $90.01\,cm$
B. $89.90\,cm$
C. $90.22\,cm$
D. $89.80\,cm$

Answer 1

In order to solve this question, we will use the concept of linear expansion of metals on heating which states that, on heating a metal its linear dimension gets increased and its directly proportional to the temperature to which it is raised.

Formula used:
Mathematically linear expansion written as,
${L^1} = L(1 + \alpha \Delta T)$
where, ${L^1}$ is the new length when temperature is raised by $\Delta T$ with original length of $L$ . Here, $\alpha$ is known as the coefficient of linear expansion and its unit is $^0{C^{ - 1}}$ .

Complete step by step answer:
Since, we have given that the value of coefficients of linear expansion for steel ${\alpha _{steel}} = 1.2 \times {10^{ - 5}}{C^{ - 1}}$ and for copper its ${\alpha _{copper}} = 1.7 \times {10^{ - 5}}{C^{ - 1}}$ and we can see that ${\alpha _{copper}} > {\alpha _{steel}}$ which shows Copper will lengthen more as compared to steel as increased length is directly proportional to the coefficient of linear expansion in the formula ${L^1} = L(1 + \alpha \Delta T)$ .

Let us suppose that the final increased length of the copper and steel will be ${L_{copper}}$ and ${L_{steel}}$ . And it’s given that both together have an initial length of ${L_0} = 90cm$ and now, let us put the values of parameters in the formula ${L^1} = L(1 + \alpha \Delta T)$ for both steel and copper as:
${L_{copper}} = {L_0}(1 + {\alpha _{copper}}\Delta T) \to (i)$ And for steel its
$\Rightarrow {L_{steel}} = {L_0}(1 + {\alpha _{steel}}\Delta T) \to (ii)$
Subtract the equation $(ii)\,from\,(i)$ we get,
${L_{copper}} - {L_{steel}} = {L_0}({\alpha _{copper}} - {\alpha _{steel}})\Delta T$

Put the value of ${L_0} = 90cm$ and the increased temperature as given in question is $\Delta T = (30 - 10) = {20^0}C$ so put
$\Delta T = {20^0}C$ And ${L_0} = 90\,cm$ in equation ${L_{copper}} - {L_{steel}} = {L_0}({\alpha _{copper}} - {\alpha _{steel}})\Delta T$ we get,
${L_{copper}} - {L_{steel}} = 90 \times (1.7 - 1.2){10^{ - 5}} \times 20$
$\therefore {L_{copper}} - {L_{steel}} = 90.01cm$
So, the net increased length which will be read on tape is ${L_{copper}} - {L_{steel}} = 90.01\,cm$ .

Hence, the correct option is A.

Note: It should be remembered that, for every particular given material the value of coefficient of linear expansion $\alpha$ is different and larger the value of $\alpha$ larger the extension of length of the material on increasing the temperature. It should be noted that when material expands in area its expansion coefficient is known as coefficient of surface area expansion and denoted by $\beta$ and in case of volume expansion, it’s called coefficient of volume expansion and it’s denoted by $\gamma$.