A mercury thermometer reads ${{80}^{\circ }}C$ when the mercury is at 5.2 cm mark and ${{60}^{\circ }}C$ when the mercury is at 3.9 cm mark. The temperature (in$^{\circ }C$) when the mercury level is at the 2.6 cm mark is $4\times {{10}^{x}}$. Find the value of x.

Hint: Mercury is a material which linearly expands when there is a change in its temperature. Therefore, the length of mercury in the thermometer column will expand linearly with change in its temperature or we can also say that change in length of the mercury is directly proportional to the change in temperature.

Complete step by step answer:
There exist some material which linearly expand due to the change in their temperature. Mercury is one of those material.
Therefore, the length of mercury will expand linearly with change in its temperature or we can also say that change in length of the mercury is directly proportional to the change in temperature.
i.e. $\Delta l\propto \Delta T$.
Then,
$\Rightarrow \Delta l=k\Delta T$ ….. (i), where k is a proportionality constant.
It is given that the length of the mercury changes from 3.9 cm to 5.2 cm when the temperature is changed from ${{60}^{\circ }}C$ to ${{80}^{\circ }}C$.
This means that $\Delta l=5.2-3.9=1.3cm$ and $\Delta T={{80}^{\circ }}C-{{60}^{\circ }}C={{20}^{\circ }}C$.
By substituting these values in (i), we get that $1.3=20k$.
$\Rightarrow k=\dfrac{1.3}{20}=0.65\times {{10}^{-2}}cm{}^{\circ }{{C}^{-1}}$.
This means that $\Delta l=0.65\times {{10}^{-2}}\Delta T$ ….. (ii).
Let us say the mercury is expanded from 2.6 cm to 5.2 cm by changing the temperature form T to ${{80}^{\circ }}C$.
Then this means that $\Delta l=5.2-2.6=2.6cm$ and $\Delta T={{80}^{\circ }}C-T$.
Substitute these values in (ii).
$\Rightarrow 2.6=0.65\times {{10}^{-2}}\left( 80-T \right)$
$\Rightarrow \left( 80-T \right)=\dfrac{2.6}{0.65\times {{10}^{-2}}}=40$
$\Rightarrow T={{40}^{\circ }}C=4\times 10{}^{\circ }C$.
But it is given that $T=4\times {{10}^{x}}{}^{\circ }C$.

This means that $x=1$.

Note:We can also use the equation of thermal expansion which is given as $\dfrac{\Delta l}{l}=\alpha \Delta T$.
Here, $\dfrac{\Delta l}{l}$ is the fractional change in the length of the substance, $\alpha $ is called coefficient of linear thermal expansion and $\Delta T$ is the change in temperature.
For a given substance, the value of coefficient of linear expansion remains constant. Therefore, we can also use this equation to find the required value.