Question

An iron scale is calibrated at ${0^ \circ }C$. The length of a zinc rod is measured to be 100 cm by the scale when the rod and the scale both are at ${0^ \circ }C$. What will be the length of the rod as measured by the scale when both are at ${100^ \circ }C$. Given: ${\alpha _{iron}} = 1 \cdot 2 \times {10^{ - 5}}^ \circ {C^{ - 1}}$ and ${\alpha _{zinc}} = 2 \cdot 6 \times {10^{ - 5}}^ \circ {C^{ - 1}}$

Answer 1

In this question, there is thermal expansion of the zinc due to the rise in temperature which can be calculated by applying the formula for thermal expansion directly. However, since we are using iron scale to calibrate, we should also take into consideration there will be thermal expansion even in the iron scale too. So, we have to account for that, too, along with the expansion in zinc rod.

Complete step by step answer:
Whenever there is an increase in temperature, there is a tendency in the materials to change its shape by increasing its length, area or even volume. Here, we classify these expansions into three broad categories of expansion: i) Linear expansion: Here we consider the increase in length per unit rise in temperature. ii) Superficial expansion: Here we consider the increase in area per unit rise in temperature. iii) Volumetric expansion: Here we consider the increase in volume per unit rise in temperature.
In this problem, since the length is given, we have to consider the linear expansion of the metal.
The formula for the linear expansion is given by :
$\Delta l = l\alpha \Delta T$
where
$\Delta l$ is the change in the length $l$ is the original length at ${0^ \circ }C$ $\alpha$ is called as the coefficient of linear expansion $\Delta T$ is the temperature difference
Since the iron scale is used to calibrate, we have to consider the expansion in it first, so as to use it to measure the expansion in the zinc rod.
Change in length in the iron rod,
$\implies$ $\Delta {l_{iron}} = {l_{iron}}{\alpha _{iron}}\Delta T$
Given, ${\alpha _{iron}} = 1 \cdot 2 \times {10^{ - 5}}^ \circ {C^{ - 1}}$ and $\Delta T = {100^ \circ }C$, substituting them,
$\implies$ $\dfrac{{\Delta {l_{iron}}}}{{{l_{iron}}}} = {\alpha _{iron}}\Delta T$
$\implies$ $\dfrac{{\Delta {l_{iron}}}}{{{l_{iron}}}} = 1 \cdot 2 \times {10^{ - 5}} \times 100$
$\implies$ $\dfrac{{\Delta {l_{iron}}}}{{{l_{iron}}}} = 1 \cdot 2 \times {10^{ - 3}}$
This is the error that we can get when measuring the length with the iron scale. So, if we were to calculate the expansion in the zinc rod, we have to subtract this value from the length.
Change in length in zinc,
$\implies$ $\dfrac{{\Delta {l_{zinc}}}}{{{l_{zinc}}}} = {\alpha _{zinc}}\Delta T - {\left( {\dfrac{{\Delta l}}{l}} \right)_{iron}}$
Length of zinc rod, ${l_{zinc}} = 100cm$
Coefficient of expansion of zinc, ${\alpha _{zinc}} = 2 \cdot 6 \times {10^{ - 4}}^ \circ {C^{ - 1}}$
Change in temperature, $\Delta T = {100^ \circ }C$
$\implies$ $\dfrac{{\Delta {l_{iron}}}}{{{l_{iron}}}} = 1 \cdot 2 \times {10^{ - 3}}$
Substituting,
$\implies$ $\dfrac{{\Delta {l_{zinc}}}}{{100}} = 2 \cdot 6 \times {10^{ - 5}} \times 100 - 1 \cdot 2 \times {10^{ - 3}}$
$\implies$ $\dfrac{{\Delta {l_{zinc}}}}{{100}} = 2 \cdot 6 \times {10^{ - 3}} - 1 \cdot 2 \times {10^{ - 3}} = 1 \cdot 4 \times {10^{ - 3}}$
$\therefore \Delta {l_{zinc}} = 100 \times 1 \cdot 4 \times {10^{ - 3}} = 0 \cdot 14cm$
Hence, the length of the zinc rod measured by the iron scale at ${100^ \circ }C$ is equal to:
$\Delta {l_{zinc}} = 0 \cdot 14cm$

Note: Just like we have the coefficient of linear expansion, we have two more coefficients of expansion for area and volume, represented by $\beta$ and $\gamma$ respectively.
The relationship among these coefficients of expansion is given by:
$\dfrac{\alpha }{1} = \dfrac{\beta }{2} = \dfrac{\gamma }{3}$